1

CHAPTER 2

ATOMIC STRUCTURE

21 Bohrrsquos Atomic Model

22 Quantum Mechanical Model

23 Electronic Configuration

2

21 Bohrrsquos Atomic

Model

3

Bohrrsquos Atomic Model

At the end of this topic students should be able to-

4

In 1913 a young Dutch physicist

Niels Bohr proposed a theory of

atom that shook the scientific world

The atomic model he described

had electrons circling a central

nucleus that contains positively

charged protons

Bohr also proposed that these orbits can only

occur at specifically ldquopermittedrdquo levels only

according to the energy levels of the electron

and explain successfully the lines in the

hydrogen spectrum

BOHRrsquoS ATOMIC MODELS

5

1 Electron moves in circular orbits about the nucleus In

moving in the orbit the electron does not radiate any

energy and does not absorb any energy

H

Nucleus

(proton) H11

BOHRrsquoS ATOMIC

POSTULATES

6

2 The energy of an electron in a hydrogen atom is

quantised that is the electron has only a fixed set of

allowed orbits called stationary states

n=1

n=2

n=3

H Nucleus

(proton)

[ orbit = stationary state = energy level = shell ]

BOHRrsquoS ATOMIC

POSTULATES

7

3 At ordinary conditions the electron is at the ground state

(lowest level) If energy is supplied electron absorbed

the energy and is promoted from a lower energy level to

a higher ones (Electron is excited)

4 Electron at its excited states is unstable It will fall back

to lower energy level and released a specific amount of

energy in the form of light The energy of the photon

equals the energy difference between levels

BOHRrsquoS ATOMIC POSTULATES

8

1 Electron moves in circular orbits about the nucleus In

moving in the orbit the electron does not radiate any

energy and does not absorb any energy

1 The energy of an electron in a hydrogen atom is

quantised that is the electron has only a fixed set of

allowed orbits called stationary states

[ orbit = stationary state = energy level = shell ]

BOHRrsquoS ATOMIC MODELS

9

Ground state

the state in which the electrons have their lowest energy

Excited state

the state in which the electrons have shifted from a lower

energy level to a higher energy level

Energy level

energy associated with a specific orbit or state

Energy levels in an atom

10

The energy of an electron in its level is given by

RH (Rydberg constant) or A = 218times10-18J

n (principal quantum number) = 1 2 3 hellipinfin (integer)

Note

n identifies the orbit of electron

Energy is zero if electron is located infinitely far from nucleus

Energy associated with forces of attraction are taken to be

negative (thus negative sign)

THE ENERGY LEVEL

11

At the end of this topic students should be able to-

Level No Learning Outcomes

C2 21 d Describe the formation of line spectrum of hydrogen atom

C2 21 e Illustrate the formation of Lyman BalmerPaschen Brackett and

Pfund series

C3 21 f Calculate the energy change of an electron during transition

ΔE = RH (1n12 - 1n2

2)

where RH = 218 x 10-18 J

C3 amp C4 21 g Calculate the photon of energy emitted by an electron that produces a particular wavelength during transition

ΔE = hν where ν= cλ

12

Emission Spectra

Emission Spectra

Continuous

Spectra

Line

Spectra

13

Continuous Spectrum

A spectrum consists all wavelength components

(containing an unbroken sequence of frequencies) of the

visible portion of the electromagnetic spectrum are

present

It is produced by incandescent solids liquids and

compressed gases

14

Regions of the Electromagnetic Spectrum

15

When white light from

incandescent lamp is passed

through a slit then a prism it

separates into a spectrum

The white light spread out into

a rainbow of colours produces

a continuous spectrum

The spectrum is continuous in

that all wavelengths are

presents and each colour

merges into the next without a

break

FORMATION OF CONTINUOUS SPECTRUM

16

Line Spectrum (atomic spectrum)

A spectrum consists of discontinuous amp discrete lines produced by

excited atoms and ions as the electrons fall back to a lower energy

level The radiation emitted is only at a specific wavelength or

frequency It means each line corresponds to a specific wavelength

or frequency

Line spectrum are composed of only a few wavelengths giving a

series of discrete line separated by blank areas

17

prismfilm

The emitted light (photons) is then separated into its components by

a prism Each component is focused at a definite position according

to its wavelength and forms as an image on the photographic plate

The images are called spectral lines

FORMATION OF ATOMIC LINE SPECTRUM

18

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

En

erg

y

When an electrical discharge is passed through a sample of

hydrogen gas at low pressure hydrogen molecules decompose to

form hydrogen atoms

Radiant energy (a

quantum of energy)

absorbed by the atom (or

electron) causes the

electron to move from a

lower-energy state to a

higher-energy state

Hydrogen atom is said to

be at excited state (very

unstable)

19

FORMATION OF ATOMIC LINE SPECTRUM

Emission of photon

n = 2

n = 3

n = 4

n = 5

n = 6n = infin

En

erg

y

When the electrons fall

back to lower energy

levels radiant energies

(photons) are emitted in

the form of light

(electromagnetic radiation

of a particular frequency or

wavelength)

20

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum λE

Energy

21

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum

Balmer Series

λ E

Energy

22

Emission series of hydrogen atom

n = 1

n = 2

n = 3

n = 4

n = infin

Lyman series

Balmer series

Brackett series

Paschen series

Pfund series

23

Exercise Complete the following table

Series n1 n2

Spectrum

region

Lyman 1 234hellip Ultraviolet

Balmer 2 345hellip Visible light

Paschen 3 456hellip Infrared

Brackett 4 567hellip Infrared

Pfund 5 678hellip Infrared

24

The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series

Exercise

Line

spectrumE λ

Specify the increasing order of the radiant energy

frequency and wavelength of the emitted photon

Which of the line that corresponds to

i) the shortest wavelength

ii) the lowest frequency

A B C D E

ν

25

Describe the transitions of electrons that lead to

the lines W and Y respectively

Solution

Line

spectrum

W Y

Exercise

Balmer series

26

Homework

Calculate En for n = 1 2 3 and 4 Make a one-

dimensional graph showing energy at different

values of n increasing vertically On this graph

indicate by vertical arrows transitions that lead to

lines in

a) Lyman series

b) Paschen series

27

In Lyman series the frequency of the convergence of

spectral lines can be used to find the ionisation energy of

hydrogen atom

IE = hνinfin

The frequency of the first line of the Lyman series gt the

frequency of the first line of the Balmer series

Significance of Atomic Spectra

Lyman Series

Line

spectrum λ E

Balmer Series

νinfin

28

Line

spectrum

ABCDE

Exercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon

Describe the transition that gives rise to the line

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

3

Bohrrsquos Atomic Model

At the end of this topic students should be able to-

4

In 1913 a young Dutch physicist

Niels Bohr proposed a theory of

atom that shook the scientific world

The atomic model he described

had electrons circling a central

nucleus that contains positively

charged protons

Bohr also proposed that these orbits can only

occur at specifically ldquopermittedrdquo levels only

according to the energy levels of the electron

and explain successfully the lines in the

hydrogen spectrum

BOHRrsquoS ATOMIC MODELS

5

1 Electron moves in circular orbits about the nucleus In

moving in the orbit the electron does not radiate any

energy and does not absorb any energy

H

Nucleus

(proton) H11

BOHRrsquoS ATOMIC

POSTULATES

6

2 The energy of an electron in a hydrogen atom is

quantised that is the electron has only a fixed set of

allowed orbits called stationary states

n=1

n=2

n=3

H Nucleus

(proton)

[ orbit = stationary state = energy level = shell ]

BOHRrsquoS ATOMIC

POSTULATES

7

3 At ordinary conditions the electron is at the ground state

(lowest level) If energy is supplied electron absorbed

the energy and is promoted from a lower energy level to

a higher ones (Electron is excited)

4 Electron at its excited states is unstable It will fall back

to lower energy level and released a specific amount of

energy in the form of light The energy of the photon

equals the energy difference between levels

BOHRrsquoS ATOMIC POSTULATES

8

1 Electron moves in circular orbits about the nucleus In

moving in the orbit the electron does not radiate any

energy and does not absorb any energy

1 The energy of an electron in a hydrogen atom is

quantised that is the electron has only a fixed set of

allowed orbits called stationary states

[ orbit = stationary state = energy level = shell ]

BOHRrsquoS ATOMIC MODELS

9

Ground state

the state in which the electrons have their lowest energy

Excited state

the state in which the electrons have shifted from a lower

energy level to a higher energy level

Energy level

energy associated with a specific orbit or state

Energy levels in an atom

10

The energy of an electron in its level is given by

RH (Rydberg constant) or A = 218times10-18J

n (principal quantum number) = 1 2 3 hellipinfin (integer)

Note

n identifies the orbit of electron

Energy is zero if electron is located infinitely far from nucleus

Energy associated with forces of attraction are taken to be

negative (thus negative sign)

THE ENERGY LEVEL

11

At the end of this topic students should be able to-

Level No Learning Outcomes

C2 21 d Describe the formation of line spectrum of hydrogen atom

C2 21 e Illustrate the formation of Lyman BalmerPaschen Brackett and

Pfund series

C3 21 f Calculate the energy change of an electron during transition

ΔE = RH (1n12 - 1n2

2)

where RH = 218 x 10-18 J

C3 amp C4 21 g Calculate the photon of energy emitted by an electron that produces a particular wavelength during transition

ΔE = hν where ν= cλ

12

Emission Spectra

Emission Spectra

Continuous

Spectra

Line

Spectra

13

Continuous Spectrum

A spectrum consists all wavelength components

(containing an unbroken sequence of frequencies) of the

visible portion of the electromagnetic spectrum are

present

It is produced by incandescent solids liquids and

compressed gases

14

Regions of the Electromagnetic Spectrum

15

When white light from

incandescent lamp is passed

through a slit then a prism it

separates into a spectrum

The white light spread out into

a rainbow of colours produces

a continuous spectrum

The spectrum is continuous in

that all wavelengths are

presents and each colour

merges into the next without a

break

FORMATION OF CONTINUOUS SPECTRUM

16

Line Spectrum (atomic spectrum)

A spectrum consists of discontinuous amp discrete lines produced by

excited atoms and ions as the electrons fall back to a lower energy

level The radiation emitted is only at a specific wavelength or

frequency It means each line corresponds to a specific wavelength

or frequency

Line spectrum are composed of only a few wavelengths giving a

series of discrete line separated by blank areas

17

prismfilm

The emitted light (photons) is then separated into its components by

a prism Each component is focused at a definite position according

to its wavelength and forms as an image on the photographic plate

The images are called spectral lines

FORMATION OF ATOMIC LINE SPECTRUM

18

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

En

erg

y

When an electrical discharge is passed through a sample of

hydrogen gas at low pressure hydrogen molecules decompose to

form hydrogen atoms

Radiant energy (a

quantum of energy)

absorbed by the atom (or

electron) causes the

electron to move from a

lower-energy state to a

higher-energy state

Hydrogen atom is said to

be at excited state (very

unstable)

19

FORMATION OF ATOMIC LINE SPECTRUM

Emission of photon

n = 2

n = 3

n = 4

n = 5

n = 6n = infin

En

erg

y

When the electrons fall

back to lower energy

levels radiant energies

(photons) are emitted in

the form of light

(electromagnetic radiation

of a particular frequency or

wavelength)

20

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum λE

Energy

21

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum

Balmer Series

λ E

Energy

22

Emission series of hydrogen atom

n = 1

n = 2

n = 3

n = 4

n = infin

Lyman series

Balmer series

Brackett series

Paschen series

Pfund series

23

Exercise Complete the following table

Series n1 n2

Spectrum

region

Lyman 1 234hellip Ultraviolet

Balmer 2 345hellip Visible light

Paschen 3 456hellip Infrared

Brackett 4 567hellip Infrared

Pfund 5 678hellip Infrared

24

The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series

Exercise

Line

spectrumE λ

Specify the increasing order of the radiant energy

frequency and wavelength of the emitted photon

Which of the line that corresponds to

i) the shortest wavelength

ii) the lowest frequency

A B C D E

ν

25

Describe the transitions of electrons that lead to

the lines W and Y respectively

Solution

Line

spectrum

W Y

Exercise

Balmer series

26

Homework

Calculate En for n = 1 2 3 and 4 Make a one-

dimensional graph showing energy at different

values of n increasing vertically On this graph

indicate by vertical arrows transitions that lead to

lines in

a) Lyman series

b) Paschen series

27

In Lyman series the frequency of the convergence of

spectral lines can be used to find the ionisation energy of

hydrogen atom

IE = hνinfin

The frequency of the first line of the Lyman series gt the

frequency of the first line of the Balmer series

Significance of Atomic Spectra

Lyman Series

Line

spectrum λ E

Balmer Series

νinfin

28

Line

spectrum

ABCDE

Exercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon

Describe the transition that gives rise to the line

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

4

In 1913 a young Dutch physicist

Niels Bohr proposed a theory of

atom that shook the scientific world

The atomic model he described

had electrons circling a central

nucleus that contains positively

charged protons

Bohr also proposed that these orbits can only

occur at specifically ldquopermittedrdquo levels only

according to the energy levels of the electron

and explain successfully the lines in the

hydrogen spectrum

BOHRrsquoS ATOMIC MODELS

5

1 Electron moves in circular orbits about the nucleus In

moving in the orbit the electron does not radiate any

energy and does not absorb any energy

H

Nucleus

(proton) H11

BOHRrsquoS ATOMIC

POSTULATES

6

2 The energy of an electron in a hydrogen atom is

quantised that is the electron has only a fixed set of

allowed orbits called stationary states

n=1

n=2

n=3

H Nucleus

(proton)

[ orbit = stationary state = energy level = shell ]

BOHRrsquoS ATOMIC

POSTULATES

7

3 At ordinary conditions the electron is at the ground state

(lowest level) If energy is supplied electron absorbed

the energy and is promoted from a lower energy level to

a higher ones (Electron is excited)

4 Electron at its excited states is unstable It will fall back

to lower energy level and released a specific amount of

energy in the form of light The energy of the photon

equals the energy difference between levels

BOHRrsquoS ATOMIC POSTULATES

8

1 Electron moves in circular orbits about the nucleus In

moving in the orbit the electron does not radiate any

energy and does not absorb any energy

1 The energy of an electron in a hydrogen atom is

quantised that is the electron has only a fixed set of

allowed orbits called stationary states

[ orbit = stationary state = energy level = shell ]

BOHRrsquoS ATOMIC MODELS

9

Ground state

the state in which the electrons have their lowest energy

Excited state

the state in which the electrons have shifted from a lower

energy level to a higher energy level

Energy level

energy associated with a specific orbit or state

Energy levels in an atom

10

The energy of an electron in its level is given by

RH (Rydberg constant) or A = 218times10-18J

n (principal quantum number) = 1 2 3 hellipinfin (integer)

Note

n identifies the orbit of electron

Energy is zero if electron is located infinitely far from nucleus

Energy associated with forces of attraction are taken to be

negative (thus negative sign)

THE ENERGY LEVEL

11

At the end of this topic students should be able to-

Level No Learning Outcomes

C2 21 d Describe the formation of line spectrum of hydrogen atom

C2 21 e Illustrate the formation of Lyman BalmerPaschen Brackett and

Pfund series

C3 21 f Calculate the energy change of an electron during transition

ΔE = RH (1n12 - 1n2

2)

where RH = 218 x 10-18 J

C3 amp C4 21 g Calculate the photon of energy emitted by an electron that produces a particular wavelength during transition

ΔE = hν where ν= cλ

12

Emission Spectra

Emission Spectra

Continuous

Spectra

Line

Spectra

13

Continuous Spectrum

A spectrum consists all wavelength components

(containing an unbroken sequence of frequencies) of the

visible portion of the electromagnetic spectrum are

present

It is produced by incandescent solids liquids and

compressed gases

14

Regions of the Electromagnetic Spectrum

15

When white light from

incandescent lamp is passed

through a slit then a prism it

separates into a spectrum

The white light spread out into

a rainbow of colours produces

a continuous spectrum

The spectrum is continuous in

that all wavelengths are

presents and each colour

merges into the next without a

break

FORMATION OF CONTINUOUS SPECTRUM

16

Line Spectrum (atomic spectrum)

A spectrum consists of discontinuous amp discrete lines produced by

excited atoms and ions as the electrons fall back to a lower energy

level The radiation emitted is only at a specific wavelength or

frequency It means each line corresponds to a specific wavelength

or frequency

Line spectrum are composed of only a few wavelengths giving a

series of discrete line separated by blank areas

17

prismfilm

The emitted light (photons) is then separated into its components by

a prism Each component is focused at a definite position according

to its wavelength and forms as an image on the photographic plate

The images are called spectral lines

FORMATION OF ATOMIC LINE SPECTRUM

18

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

En

erg

y

When an electrical discharge is passed through a sample of

hydrogen gas at low pressure hydrogen molecules decompose to

form hydrogen atoms

Radiant energy (a

quantum of energy)

absorbed by the atom (or

electron) causes the

electron to move from a

lower-energy state to a

higher-energy state

Hydrogen atom is said to

be at excited state (very

unstable)

19

FORMATION OF ATOMIC LINE SPECTRUM

Emission of photon

n = 2

n = 3

n = 4

n = 5

n = 6n = infin

En

erg

y

When the electrons fall

back to lower energy

levels radiant energies

(photons) are emitted in

the form of light

(electromagnetic radiation

of a particular frequency or

wavelength)

20

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum λE

Energy

21

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum

Balmer Series

λ E

Energy

22

Emission series of hydrogen atom

n = 1

n = 2

n = 3

n = 4

n = infin

Lyman series

Balmer series

Brackett series

Paschen series

Pfund series

23

Exercise Complete the following table

Series n1 n2

Spectrum

region

Lyman 1 234hellip Ultraviolet

Balmer 2 345hellip Visible light

Paschen 3 456hellip Infrared

Brackett 4 567hellip Infrared

Pfund 5 678hellip Infrared

24

The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series

Exercise

Line

spectrumE λ

Specify the increasing order of the radiant energy

frequency and wavelength of the emitted photon

Which of the line that corresponds to

i) the shortest wavelength

ii) the lowest frequency

A B C D E

ν

25

Describe the transitions of electrons that lead to

the lines W and Y respectively

Solution

Line

spectrum

W Y

Exercise

Balmer series

26

Homework

Calculate En for n = 1 2 3 and 4 Make a one-

dimensional graph showing energy at different

values of n increasing vertically On this graph

indicate by vertical arrows transitions that lead to

lines in

a) Lyman series

b) Paschen series

27

In Lyman series the frequency of the convergence of

spectral lines can be used to find the ionisation energy of

hydrogen atom

IE = hνinfin

The frequency of the first line of the Lyman series gt the

frequency of the first line of the Balmer series

Significance of Atomic Spectra

Lyman Series

Line

spectrum λ E

Balmer Series

νinfin

28

Line

spectrum

ABCDE

Exercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon

Describe the transition that gives rise to the line

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

5

1 Electron moves in circular orbits about the nucleus In

moving in the orbit the electron does not radiate any

energy and does not absorb any energy

H

Nucleus

(proton) H11

BOHRrsquoS ATOMIC

POSTULATES

6

2 The energy of an electron in a hydrogen atom is

quantised that is the electron has only a fixed set of

allowed orbits called stationary states

n=1

n=2

n=3

H Nucleus

(proton)

[ orbit = stationary state = energy level = shell ]

BOHRrsquoS ATOMIC

POSTULATES

7

3 At ordinary conditions the electron is at the ground state

(lowest level) If energy is supplied electron absorbed

the energy and is promoted from a lower energy level to

a higher ones (Electron is excited)

4 Electron at its excited states is unstable It will fall back

to lower energy level and released a specific amount of

energy in the form of light The energy of the photon

equals the energy difference between levels

BOHRrsquoS ATOMIC POSTULATES

8

1 Electron moves in circular orbits about the nucleus In

moving in the orbit the electron does not radiate any

energy and does not absorb any energy

1 The energy of an electron in a hydrogen atom is

quantised that is the electron has only a fixed set of

allowed orbits called stationary states

[ orbit = stationary state = energy level = shell ]

BOHRrsquoS ATOMIC MODELS

9

Ground state

the state in which the electrons have their lowest energy

Excited state

the state in which the electrons have shifted from a lower

energy level to a higher energy level

Energy level

energy associated with a specific orbit or state

Energy levels in an atom

10

The energy of an electron in its level is given by

RH (Rydberg constant) or A = 218times10-18J

n (principal quantum number) = 1 2 3 hellipinfin (integer)

Note

n identifies the orbit of electron

Energy is zero if electron is located infinitely far from nucleus

Energy associated with forces of attraction are taken to be

negative (thus negative sign)

THE ENERGY LEVEL

11

At the end of this topic students should be able to-

Level No Learning Outcomes

C2 21 d Describe the formation of line spectrum of hydrogen atom

C2 21 e Illustrate the formation of Lyman BalmerPaschen Brackett and

Pfund series

C3 21 f Calculate the energy change of an electron during transition

ΔE = RH (1n12 - 1n2

2)

where RH = 218 x 10-18 J

C3 amp C4 21 g Calculate the photon of energy emitted by an electron that produces a particular wavelength during transition

ΔE = hν where ν= cλ

12

Emission Spectra

Emission Spectra

Continuous

Spectra

Line

Spectra

13

Continuous Spectrum

A spectrum consists all wavelength components

(containing an unbroken sequence of frequencies) of the

visible portion of the electromagnetic spectrum are

present

It is produced by incandescent solids liquids and

compressed gases

14

Regions of the Electromagnetic Spectrum

15

When white light from

incandescent lamp is passed

through a slit then a prism it

separates into a spectrum

The white light spread out into

a rainbow of colours produces

a continuous spectrum

The spectrum is continuous in

that all wavelengths are

presents and each colour

merges into the next without a

break

FORMATION OF CONTINUOUS SPECTRUM

16

Line Spectrum (atomic spectrum)

A spectrum consists of discontinuous amp discrete lines produced by

excited atoms and ions as the electrons fall back to a lower energy

level The radiation emitted is only at a specific wavelength or

frequency It means each line corresponds to a specific wavelength

or frequency

Line spectrum are composed of only a few wavelengths giving a

series of discrete line separated by blank areas

17

prismfilm

The emitted light (photons) is then separated into its components by

a prism Each component is focused at a definite position according

to its wavelength and forms as an image on the photographic plate

The images are called spectral lines

FORMATION OF ATOMIC LINE SPECTRUM

18

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

En

erg

y

When an electrical discharge is passed through a sample of

hydrogen gas at low pressure hydrogen molecules decompose to

form hydrogen atoms

Radiant energy (a

quantum of energy)

absorbed by the atom (or

electron) causes the

electron to move from a

lower-energy state to a

higher-energy state

Hydrogen atom is said to

be at excited state (very

unstable)

19

FORMATION OF ATOMIC LINE SPECTRUM

Emission of photon

n = 2

n = 3

n = 4

n = 5

n = 6n = infin

En

erg

y

When the electrons fall

back to lower energy

levels radiant energies

(photons) are emitted in

the form of light

(electromagnetic radiation

of a particular frequency or

wavelength)

20

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum λE

Energy

21

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum

Balmer Series

λ E

Energy

22

Emission series of hydrogen atom

n = 1

n = 2

n = 3

n = 4

n = infin

Lyman series

Balmer series

Brackett series

Paschen series

Pfund series

23

Exercise Complete the following table

Series n1 n2

Spectrum

region

Lyman 1 234hellip Ultraviolet

Balmer 2 345hellip Visible light

Paschen 3 456hellip Infrared

Brackett 4 567hellip Infrared

Pfund 5 678hellip Infrared

24

The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series

Exercise

Line

spectrumE λ

Specify the increasing order of the radiant energy

frequency and wavelength of the emitted photon

Which of the line that corresponds to

i) the shortest wavelength

ii) the lowest frequency

A B C D E

ν

25

Describe the transitions of electrons that lead to

the lines W and Y respectively

Solution

Line

spectrum

W Y

Exercise

Balmer series

26

Homework

Calculate En for n = 1 2 3 and 4 Make a one-

dimensional graph showing energy at different

values of n increasing vertically On this graph

indicate by vertical arrows transitions that lead to

lines in

a) Lyman series

b) Paschen series

27

In Lyman series the frequency of the convergence of

spectral lines can be used to find the ionisation energy of

hydrogen atom

IE = hνinfin

The frequency of the first line of the Lyman series gt the

frequency of the first line of the Balmer series

Significance of Atomic Spectra

Lyman Series

Line

spectrum λ E

Balmer Series

νinfin

28

Line

spectrum

ABCDE

Exercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon

Describe the transition that gives rise to the line

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

6

2 The energy of an electron in a hydrogen atom is

quantised that is the electron has only a fixed set of

allowed orbits called stationary states

n=1

n=2

n=3

H Nucleus

(proton)

[ orbit = stationary state = energy level = shell ]

BOHRrsquoS ATOMIC

POSTULATES

7

3 At ordinary conditions the electron is at the ground state

(lowest level) If energy is supplied electron absorbed

the energy and is promoted from a lower energy level to

a higher ones (Electron is excited)

4 Electron at its excited states is unstable It will fall back

to lower energy level and released a specific amount of

energy in the form of light The energy of the photon

equals the energy difference between levels

BOHRrsquoS ATOMIC POSTULATES

8

1 Electron moves in circular orbits about the nucleus In

moving in the orbit the electron does not radiate any

energy and does not absorb any energy

1 The energy of an electron in a hydrogen atom is

quantised that is the electron has only a fixed set of

allowed orbits called stationary states

[ orbit = stationary state = energy level = shell ]

BOHRrsquoS ATOMIC MODELS

9

Ground state

the state in which the electrons have their lowest energy

Excited state

the state in which the electrons have shifted from a lower

energy level to a higher energy level

Energy level

energy associated with a specific orbit or state

Energy levels in an atom

10

The energy of an electron in its level is given by

RH (Rydberg constant) or A = 218times10-18J

n (principal quantum number) = 1 2 3 hellipinfin (integer)

Note

n identifies the orbit of electron

Energy is zero if electron is located infinitely far from nucleus

Energy associated with forces of attraction are taken to be

negative (thus negative sign)

THE ENERGY LEVEL

11

At the end of this topic students should be able to-

Level No Learning Outcomes

C2 21 d Describe the formation of line spectrum of hydrogen atom

C2 21 e Illustrate the formation of Lyman BalmerPaschen Brackett and

Pfund series

C3 21 f Calculate the energy change of an electron during transition

ΔE = RH (1n12 - 1n2

2)

where RH = 218 x 10-18 J

C3 amp C4 21 g Calculate the photon of energy emitted by an electron that produces a particular wavelength during transition

ΔE = hν where ν= cλ

12

Emission Spectra

Emission Spectra

Continuous

Spectra

Line

Spectra

13

Continuous Spectrum

A spectrum consists all wavelength components

(containing an unbroken sequence of frequencies) of the

visible portion of the electromagnetic spectrum are

present

It is produced by incandescent solids liquids and

compressed gases

14

Regions of the Electromagnetic Spectrum

15

When white light from

incandescent lamp is passed

through a slit then a prism it

separates into a spectrum

The white light spread out into

a rainbow of colours produces

a continuous spectrum

The spectrum is continuous in

that all wavelengths are

presents and each colour

merges into the next without a

break

FORMATION OF CONTINUOUS SPECTRUM

16

Line Spectrum (atomic spectrum)

A spectrum consists of discontinuous amp discrete lines produced by

excited atoms and ions as the electrons fall back to a lower energy

level The radiation emitted is only at a specific wavelength or

frequency It means each line corresponds to a specific wavelength

or frequency

Line spectrum are composed of only a few wavelengths giving a

series of discrete line separated by blank areas

17

prismfilm

The emitted light (photons) is then separated into its components by

a prism Each component is focused at a definite position according

to its wavelength and forms as an image on the photographic plate

The images are called spectral lines

FORMATION OF ATOMIC LINE SPECTRUM

18

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

En

erg

y

When an electrical discharge is passed through a sample of

hydrogen gas at low pressure hydrogen molecules decompose to

form hydrogen atoms

Radiant energy (a

quantum of energy)

absorbed by the atom (or

electron) causes the

electron to move from a

lower-energy state to a

higher-energy state

Hydrogen atom is said to

be at excited state (very

unstable)

19

FORMATION OF ATOMIC LINE SPECTRUM

Emission of photon

n = 2

n = 3

n = 4

n = 5

n = 6n = infin

En

erg

y

When the electrons fall

back to lower energy

levels radiant energies

(photons) are emitted in

the form of light

(electromagnetic radiation

of a particular frequency or

wavelength)

20

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum λE

Energy

21

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum

Balmer Series

λ E

Energy

22

Emission series of hydrogen atom

n = 1

n = 2

n = 3

n = 4

n = infin

Lyman series

Balmer series

Brackett series

Paschen series

Pfund series

23

Exercise Complete the following table

Series n1 n2

Spectrum

region

Lyman 1 234hellip Ultraviolet

Balmer 2 345hellip Visible light

Paschen 3 456hellip Infrared

Brackett 4 567hellip Infrared

Pfund 5 678hellip Infrared

24

The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series

Exercise

Line

spectrumE λ

Specify the increasing order of the radiant energy

frequency and wavelength of the emitted photon

Which of the line that corresponds to

i) the shortest wavelength

ii) the lowest frequency

A B C D E

ν

25

Describe the transitions of electrons that lead to

the lines W and Y respectively

Solution

Line

spectrum

W Y

Exercise

Balmer series

26

Homework

Calculate En for n = 1 2 3 and 4 Make a one-

dimensional graph showing energy at different

values of n increasing vertically On this graph

indicate by vertical arrows transitions that lead to

lines in

a) Lyman series

b) Paschen series

27

In Lyman series the frequency of the convergence of

spectral lines can be used to find the ionisation energy of

hydrogen atom

IE = hνinfin

The frequency of the first line of the Lyman series gt the

frequency of the first line of the Balmer series

Significance of Atomic Spectra

Lyman Series

Line

spectrum λ E

Balmer Series

νinfin

28

Line

spectrum

ABCDE

Exercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon

Describe the transition that gives rise to the line

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

7

3 At ordinary conditions the electron is at the ground state

(lowest level) If energy is supplied electron absorbed

the energy and is promoted from a lower energy level to

a higher ones (Electron is excited)

4 Electron at its excited states is unstable It will fall back

to lower energy level and released a specific amount of

energy in the form of light The energy of the photon

equals the energy difference between levels

BOHRrsquoS ATOMIC POSTULATES

8

1 Electron moves in circular orbits about the nucleus In

moving in the orbit the electron does not radiate any

energy and does not absorb any energy

1 The energy of an electron in a hydrogen atom is

quantised that is the electron has only a fixed set of

allowed orbits called stationary states

[ orbit = stationary state = energy level = shell ]

BOHRrsquoS ATOMIC MODELS

9

Ground state

the state in which the electrons have their lowest energy

Excited state

the state in which the electrons have shifted from a lower

energy level to a higher energy level

Energy level

energy associated with a specific orbit or state

Energy levels in an atom

10

The energy of an electron in its level is given by

RH (Rydberg constant) or A = 218times10-18J

n (principal quantum number) = 1 2 3 hellipinfin (integer)

Note

n identifies the orbit of electron

Energy is zero if electron is located infinitely far from nucleus

Energy associated with forces of attraction are taken to be

negative (thus negative sign)

THE ENERGY LEVEL

11

At the end of this topic students should be able to-

Level No Learning Outcomes

C2 21 d Describe the formation of line spectrum of hydrogen atom

C2 21 e Illustrate the formation of Lyman BalmerPaschen Brackett and

Pfund series

C3 21 f Calculate the energy change of an electron during transition

ΔE = RH (1n12 - 1n2

2)

where RH = 218 x 10-18 J

C3 amp C4 21 g Calculate the photon of energy emitted by an electron that produces a particular wavelength during transition

ΔE = hν where ν= cλ

12

Emission Spectra

Emission Spectra

Continuous

Spectra

Line

Spectra

13

Continuous Spectrum

A spectrum consists all wavelength components

(containing an unbroken sequence of frequencies) of the

visible portion of the electromagnetic spectrum are

present

It is produced by incandescent solids liquids and

compressed gases

14

Regions of the Electromagnetic Spectrum

15

When white light from

incandescent lamp is passed

through a slit then a prism it

separates into a spectrum

The white light spread out into

a rainbow of colours produces

a continuous spectrum

The spectrum is continuous in

that all wavelengths are

presents and each colour

merges into the next without a

break

FORMATION OF CONTINUOUS SPECTRUM

16

Line Spectrum (atomic spectrum)

A spectrum consists of discontinuous amp discrete lines produced by

excited atoms and ions as the electrons fall back to a lower energy

level The radiation emitted is only at a specific wavelength or

frequency It means each line corresponds to a specific wavelength

or frequency

Line spectrum are composed of only a few wavelengths giving a

series of discrete line separated by blank areas

17

prismfilm

The emitted light (photons) is then separated into its components by

a prism Each component is focused at a definite position according

to its wavelength and forms as an image on the photographic plate

The images are called spectral lines

FORMATION OF ATOMIC LINE SPECTRUM

18

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

En

erg

y

When an electrical discharge is passed through a sample of

hydrogen gas at low pressure hydrogen molecules decompose to

form hydrogen atoms

Radiant energy (a

quantum of energy)

absorbed by the atom (or

electron) causes the

electron to move from a

lower-energy state to a

higher-energy state

Hydrogen atom is said to

be at excited state (very

unstable)

19

FORMATION OF ATOMIC LINE SPECTRUM

Emission of photon

n = 2

n = 3

n = 4

n = 5

n = 6n = infin

En

erg

y

When the electrons fall

back to lower energy

levels radiant energies

(photons) are emitted in

the form of light

(electromagnetic radiation

of a particular frequency or

wavelength)

20

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum λE

Energy

21

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum

Balmer Series

λ E

Energy

22

Emission series of hydrogen atom

n = 1

n = 2

n = 3

n = 4

n = infin

Lyman series

Balmer series

Brackett series

Paschen series

Pfund series

23

Exercise Complete the following table

Series n1 n2

Spectrum

region

Lyman 1 234hellip Ultraviolet

Balmer 2 345hellip Visible light

Paschen 3 456hellip Infrared

Brackett 4 567hellip Infrared

Pfund 5 678hellip Infrared

24

The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series

Exercise

Line

spectrumE λ

Specify the increasing order of the radiant energy

frequency and wavelength of the emitted photon

Which of the line that corresponds to

i) the shortest wavelength

ii) the lowest frequency

A B C D E

ν

25

Describe the transitions of electrons that lead to

the lines W and Y respectively

Solution

Line

spectrum

W Y

Exercise

Balmer series

26

Homework

Calculate En for n = 1 2 3 and 4 Make a one-

dimensional graph showing energy at different

values of n increasing vertically On this graph

indicate by vertical arrows transitions that lead to

lines in

a) Lyman series

b) Paschen series

27

In Lyman series the frequency of the convergence of

spectral lines can be used to find the ionisation energy of

hydrogen atom

IE = hνinfin

The frequency of the first line of the Lyman series gt the

frequency of the first line of the Balmer series

Significance of Atomic Spectra

Lyman Series

Line

spectrum λ E

Balmer Series

νinfin

28

Line

spectrum

ABCDE

Exercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon

Describe the transition that gives rise to the line

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

8

1 Electron moves in circular orbits about the nucleus In

moving in the orbit the electron does not radiate any

energy and does not absorb any energy

1 The energy of an electron in a hydrogen atom is

quantised that is the electron has only a fixed set of

allowed orbits called stationary states

[ orbit = stationary state = energy level = shell ]

BOHRrsquoS ATOMIC MODELS

9

Ground state

the state in which the electrons have their lowest energy

Excited state

the state in which the electrons have shifted from a lower

energy level to a higher energy level

Energy level

energy associated with a specific orbit or state

Energy levels in an atom

10

The energy of an electron in its level is given by

RH (Rydberg constant) or A = 218times10-18J

n (principal quantum number) = 1 2 3 hellipinfin (integer)

Note

n identifies the orbit of electron

Energy is zero if electron is located infinitely far from nucleus

Energy associated with forces of attraction are taken to be

negative (thus negative sign)

THE ENERGY LEVEL

11

At the end of this topic students should be able to-

Level No Learning Outcomes

C2 21 d Describe the formation of line spectrum of hydrogen atom

C2 21 e Illustrate the formation of Lyman BalmerPaschen Brackett and

Pfund series

C3 21 f Calculate the energy change of an electron during transition

ΔE = RH (1n12 - 1n2

2)

where RH = 218 x 10-18 J

C3 amp C4 21 g Calculate the photon of energy emitted by an electron that produces a particular wavelength during transition

ΔE = hν where ν= cλ

12

Emission Spectra

Emission Spectra

Continuous

Spectra

Line

Spectra

13

Continuous Spectrum

A spectrum consists all wavelength components

(containing an unbroken sequence of frequencies) of the

visible portion of the electromagnetic spectrum are

present

It is produced by incandescent solids liquids and

compressed gases

14

Regions of the Electromagnetic Spectrum

15

When white light from

incandescent lamp is passed

through a slit then a prism it

separates into a spectrum

The white light spread out into

a rainbow of colours produces

a continuous spectrum

The spectrum is continuous in

that all wavelengths are

presents and each colour

merges into the next without a

break

FORMATION OF CONTINUOUS SPECTRUM

16

Line Spectrum (atomic spectrum)

A spectrum consists of discontinuous amp discrete lines produced by

excited atoms and ions as the electrons fall back to a lower energy

level The radiation emitted is only at a specific wavelength or

frequency It means each line corresponds to a specific wavelength

or frequency

Line spectrum are composed of only a few wavelengths giving a

series of discrete line separated by blank areas

17

prismfilm

The emitted light (photons) is then separated into its components by

a prism Each component is focused at a definite position according

to its wavelength and forms as an image on the photographic plate

The images are called spectral lines

FORMATION OF ATOMIC LINE SPECTRUM

18

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

En

erg

y

When an electrical discharge is passed through a sample of

hydrogen gas at low pressure hydrogen molecules decompose to

form hydrogen atoms

Radiant energy (a

quantum of energy)

absorbed by the atom (or

electron) causes the

electron to move from a

lower-energy state to a

higher-energy state

Hydrogen atom is said to

be at excited state (very

unstable)

19

FORMATION OF ATOMIC LINE SPECTRUM

Emission of photon

n = 2

n = 3

n = 4

n = 5

n = 6n = infin

En

erg

y

When the electrons fall

back to lower energy

levels radiant energies

(photons) are emitted in

the form of light

(electromagnetic radiation

of a particular frequency or

wavelength)

20

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum λE

Energy

21

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum

Balmer Series

λ E

Energy

22

Emission series of hydrogen atom

n = 1

n = 2

n = 3

n = 4

n = infin

Lyman series

Balmer series

Brackett series

Paschen series

Pfund series

23

Exercise Complete the following table

Series n1 n2

Spectrum

region

Lyman 1 234hellip Ultraviolet

Balmer 2 345hellip Visible light

Paschen 3 456hellip Infrared

Brackett 4 567hellip Infrared

Pfund 5 678hellip Infrared

24

The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series

Exercise

Line

spectrumE λ

Specify the increasing order of the radiant energy

frequency and wavelength of the emitted photon

Which of the line that corresponds to

i) the shortest wavelength

ii) the lowest frequency

A B C D E

ν

25

Describe the transitions of electrons that lead to

the lines W and Y respectively

Solution

Line

spectrum

W Y

Exercise

Balmer series

26

Homework

Calculate En for n = 1 2 3 and 4 Make a one-

dimensional graph showing energy at different

values of n increasing vertically On this graph

indicate by vertical arrows transitions that lead to

lines in

a) Lyman series

b) Paschen series

27

In Lyman series the frequency of the convergence of

spectral lines can be used to find the ionisation energy of

hydrogen atom

IE = hνinfin

The frequency of the first line of the Lyman series gt the

frequency of the first line of the Balmer series

Significance of Atomic Spectra

Lyman Series

Line

spectrum λ E

Balmer Series

νinfin

28

Line

spectrum

ABCDE

Exercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon

Describe the transition that gives rise to the line

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

9

Ground state

the state in which the electrons have their lowest energy

Excited state

the state in which the electrons have shifted from a lower

energy level to a higher energy level

Energy level

energy associated with a specific orbit or state

Energy levels in an atom

10

The energy of an electron in its level is given by

RH (Rydberg constant) or A = 218times10-18J

n (principal quantum number) = 1 2 3 hellipinfin (integer)

Note

n identifies the orbit of electron

Energy is zero if electron is located infinitely far from nucleus

Energy associated with forces of attraction are taken to be

negative (thus negative sign)

THE ENERGY LEVEL

11

At the end of this topic students should be able to-

Level No Learning Outcomes

C2 21 d Describe the formation of line spectrum of hydrogen atom

C2 21 e Illustrate the formation of Lyman BalmerPaschen Brackett and

Pfund series

C3 21 f Calculate the energy change of an electron during transition

ΔE = RH (1n12 - 1n2

2)

where RH = 218 x 10-18 J

C3 amp C4 21 g Calculate the photon of energy emitted by an electron that produces a particular wavelength during transition

ΔE = hν where ν= cλ

12

Emission Spectra

Emission Spectra

Continuous

Spectra

Line

Spectra

13

Continuous Spectrum

A spectrum consists all wavelength components

(containing an unbroken sequence of frequencies) of the

visible portion of the electromagnetic spectrum are

present

It is produced by incandescent solids liquids and

compressed gases

14

Regions of the Electromagnetic Spectrum

15

When white light from

incandescent lamp is passed

through a slit then a prism it

separates into a spectrum

The white light spread out into

a rainbow of colours produces

a continuous spectrum

The spectrum is continuous in

that all wavelengths are

presents and each colour

merges into the next without a

break

FORMATION OF CONTINUOUS SPECTRUM

16

Line Spectrum (atomic spectrum)

A spectrum consists of discontinuous amp discrete lines produced by

excited atoms and ions as the electrons fall back to a lower energy

level The radiation emitted is only at a specific wavelength or

frequency It means each line corresponds to a specific wavelength

or frequency

Line spectrum are composed of only a few wavelengths giving a

series of discrete line separated by blank areas

17

prismfilm

The emitted light (photons) is then separated into its components by

a prism Each component is focused at a definite position according

to its wavelength and forms as an image on the photographic plate

The images are called spectral lines

FORMATION OF ATOMIC LINE SPECTRUM

18

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

En

erg

y

When an electrical discharge is passed through a sample of

hydrogen gas at low pressure hydrogen molecules decompose to

form hydrogen atoms

Radiant energy (a

quantum of energy)

absorbed by the atom (or

electron) causes the

electron to move from a

lower-energy state to a

higher-energy state

Hydrogen atom is said to

be at excited state (very

unstable)

19

FORMATION OF ATOMIC LINE SPECTRUM

Emission of photon

n = 2

n = 3

n = 4

n = 5

n = 6n = infin

En

erg

y

When the electrons fall

back to lower energy

levels radiant energies

(photons) are emitted in

the form of light

(electromagnetic radiation

of a particular frequency or

wavelength)

20

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum λE

Energy

21

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum

Balmer Series

λ E

Energy

22

Emission series of hydrogen atom

n = 1

n = 2

n = 3

n = 4

n = infin

Lyman series

Balmer series

Brackett series

Paschen series

Pfund series

23

Exercise Complete the following table

Series n1 n2

Spectrum

region

Lyman 1 234hellip Ultraviolet

Balmer 2 345hellip Visible light

Paschen 3 456hellip Infrared

Brackett 4 567hellip Infrared

Pfund 5 678hellip Infrared

24

The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series

Exercise

Line

spectrumE λ

Specify the increasing order of the radiant energy

frequency and wavelength of the emitted photon

Which of the line that corresponds to

i) the shortest wavelength

ii) the lowest frequency

A B C D E

ν

25

Describe the transitions of electrons that lead to

the lines W and Y respectively

Solution

Line

spectrum

W Y

Exercise

Balmer series

26

Homework

Calculate En for n = 1 2 3 and 4 Make a one-

dimensional graph showing energy at different

values of n increasing vertically On this graph

indicate by vertical arrows transitions that lead to

lines in

a) Lyman series

b) Paschen series

27

In Lyman series the frequency of the convergence of

spectral lines can be used to find the ionisation energy of

hydrogen atom

IE = hνinfin

The frequency of the first line of the Lyman series gt the

frequency of the first line of the Balmer series

Significance of Atomic Spectra

Lyman Series

Line

spectrum λ E

Balmer Series

νinfin

28

Line

spectrum

ABCDE

Exercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon

Describe the transition that gives rise to the line

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

10

The energy of an electron in its level is given by

RH (Rydberg constant) or A = 218times10-18J

n (principal quantum number) = 1 2 3 hellipinfin (integer)

Note

n identifies the orbit of electron

Energy is zero if electron is located infinitely far from nucleus

Energy associated with forces of attraction are taken to be

negative (thus negative sign)

THE ENERGY LEVEL

11

At the end of this topic students should be able to-

Level No Learning Outcomes

C2 21 d Describe the formation of line spectrum of hydrogen atom

C2 21 e Illustrate the formation of Lyman BalmerPaschen Brackett and

Pfund series

C3 21 f Calculate the energy change of an electron during transition

ΔE = RH (1n12 - 1n2

2)

where RH = 218 x 10-18 J

C3 amp C4 21 g Calculate the photon of energy emitted by an electron that produces a particular wavelength during transition

ΔE = hν where ν= cλ

12

Emission Spectra

Emission Spectra

Continuous

Spectra

Line

Spectra

13

Continuous Spectrum

A spectrum consists all wavelength components

(containing an unbroken sequence of frequencies) of the

visible portion of the electromagnetic spectrum are

present

It is produced by incandescent solids liquids and

compressed gases

14

Regions of the Electromagnetic Spectrum

15

When white light from

incandescent lamp is passed

through a slit then a prism it

separates into a spectrum

The white light spread out into

a rainbow of colours produces

a continuous spectrum

The spectrum is continuous in

that all wavelengths are

presents and each colour

merges into the next without a

break

FORMATION OF CONTINUOUS SPECTRUM

16

Line Spectrum (atomic spectrum)

A spectrum consists of discontinuous amp discrete lines produced by

excited atoms and ions as the electrons fall back to a lower energy

level The radiation emitted is only at a specific wavelength or

frequency It means each line corresponds to a specific wavelength

or frequency

Line spectrum are composed of only a few wavelengths giving a

series of discrete line separated by blank areas

17

prismfilm

The emitted light (photons) is then separated into its components by

a prism Each component is focused at a definite position according

to its wavelength and forms as an image on the photographic plate

The images are called spectral lines

FORMATION OF ATOMIC LINE SPECTRUM

18

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

En

erg

y

When an electrical discharge is passed through a sample of

hydrogen gas at low pressure hydrogen molecules decompose to

form hydrogen atoms

Radiant energy (a

quantum of energy)

absorbed by the atom (or

electron) causes the

electron to move from a

lower-energy state to a

higher-energy state

Hydrogen atom is said to

be at excited state (very

unstable)

19

FORMATION OF ATOMIC LINE SPECTRUM

Emission of photon

n = 2

n = 3

n = 4

n = 5

n = 6n = infin

En

erg

y

When the electrons fall

back to lower energy

levels radiant energies

(photons) are emitted in

the form of light

(electromagnetic radiation

of a particular frequency or

wavelength)

20

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum λE

Energy

21

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum

Balmer Series

λ E

Energy

22

Emission series of hydrogen atom

n = 1

n = 2

n = 3

n = 4

n = infin

Lyman series

Balmer series

Brackett series

Paschen series

Pfund series

23

Exercise Complete the following table

Series n1 n2

Spectrum

region

Lyman 1 234hellip Ultraviolet

Balmer 2 345hellip Visible light

Paschen 3 456hellip Infrared

Brackett 4 567hellip Infrared

Pfund 5 678hellip Infrared

24

The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series

Exercise

Line

spectrumE λ

Specify the increasing order of the radiant energy

frequency and wavelength of the emitted photon

Which of the line that corresponds to

i) the shortest wavelength

ii) the lowest frequency

A B C D E

ν

25

Describe the transitions of electrons that lead to

the lines W and Y respectively

Solution

Line

spectrum

W Y

Exercise

Balmer series

26

Homework

Calculate En for n = 1 2 3 and 4 Make a one-

dimensional graph showing energy at different

values of n increasing vertically On this graph

indicate by vertical arrows transitions that lead to

lines in

a) Lyman series

b) Paschen series

27

In Lyman series the frequency of the convergence of

spectral lines can be used to find the ionisation energy of

hydrogen atom

IE = hνinfin

The frequency of the first line of the Lyman series gt the

frequency of the first line of the Balmer series

Significance of Atomic Spectra

Lyman Series

Line

spectrum λ E

Balmer Series

νinfin

28

Line

spectrum

ABCDE

Exercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon

Describe the transition that gives rise to the line

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

11

At the end of this topic students should be able to-

Level No Learning Outcomes

C2 21 d Describe the formation of line spectrum of hydrogen atom

C2 21 e Illustrate the formation of Lyman BalmerPaschen Brackett and

Pfund series

C3 21 f Calculate the energy change of an electron during transition

ΔE = RH (1n12 - 1n2

2)

where RH = 218 x 10-18 J

C3 amp C4 21 g Calculate the photon of energy emitted by an electron that produces a particular wavelength during transition

ΔE = hν where ν= cλ

12

Emission Spectra

Emission Spectra

Continuous

Spectra

Line

Spectra

13

Continuous Spectrum

A spectrum consists all wavelength components

(containing an unbroken sequence of frequencies) of the

visible portion of the electromagnetic spectrum are

present

It is produced by incandescent solids liquids and

compressed gases

14

Regions of the Electromagnetic Spectrum

15

When white light from

incandescent lamp is passed

through a slit then a prism it

separates into a spectrum

The white light spread out into

a rainbow of colours produces

a continuous spectrum

The spectrum is continuous in

that all wavelengths are

presents and each colour

merges into the next without a

break

FORMATION OF CONTINUOUS SPECTRUM

16

Line Spectrum (atomic spectrum)

A spectrum consists of discontinuous amp discrete lines produced by

excited atoms and ions as the electrons fall back to a lower energy

level The radiation emitted is only at a specific wavelength or

frequency It means each line corresponds to a specific wavelength

or frequency

Line spectrum are composed of only a few wavelengths giving a

series of discrete line separated by blank areas

17

prismfilm

The emitted light (photons) is then separated into its components by

a prism Each component is focused at a definite position according

to its wavelength and forms as an image on the photographic plate

The images are called spectral lines

FORMATION OF ATOMIC LINE SPECTRUM

18

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

En

erg

y

When an electrical discharge is passed through a sample of

hydrogen gas at low pressure hydrogen molecules decompose to

form hydrogen atoms

Radiant energy (a

quantum of energy)

absorbed by the atom (or

electron) causes the

electron to move from a

lower-energy state to a

higher-energy state

Hydrogen atom is said to

be at excited state (very

unstable)

19

FORMATION OF ATOMIC LINE SPECTRUM

Emission of photon

n = 2

n = 3

n = 4

n = 5

n = 6n = infin

En

erg

y

When the electrons fall

back to lower energy

levels radiant energies

(photons) are emitted in

the form of light

(electromagnetic radiation

of a particular frequency or

wavelength)

20

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum λE

Energy

21

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum

Balmer Series

λ E

Energy

22

Emission series of hydrogen atom

n = 1

n = 2

n = 3

n = 4

n = infin

Lyman series

Balmer series

Brackett series

Paschen series

Pfund series

23

Exercise Complete the following table

Series n1 n2

Spectrum

region

Lyman 1 234hellip Ultraviolet

Balmer 2 345hellip Visible light

Paschen 3 456hellip Infrared

Brackett 4 567hellip Infrared

Pfund 5 678hellip Infrared

24

The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series

Exercise

Line

spectrumE λ

Specify the increasing order of the radiant energy

frequency and wavelength of the emitted photon

Which of the line that corresponds to

i) the shortest wavelength

ii) the lowest frequency

A B C D E

ν

25

Describe the transitions of electrons that lead to

the lines W and Y respectively

Solution

Line

spectrum

W Y

Exercise

Balmer series

26

Homework

Calculate En for n = 1 2 3 and 4 Make a one-

dimensional graph showing energy at different

values of n increasing vertically On this graph

indicate by vertical arrows transitions that lead to

lines in

a) Lyman series

b) Paschen series

27

In Lyman series the frequency of the convergence of

spectral lines can be used to find the ionisation energy of

hydrogen atom

IE = hνinfin

The frequency of the first line of the Lyman series gt the

frequency of the first line of the Balmer series

Significance of Atomic Spectra

Lyman Series

Line

spectrum λ E

Balmer Series

νinfin

28

Line

spectrum

ABCDE

Exercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon

Describe the transition that gives rise to the line

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

12

Emission Spectra

Emission Spectra

Continuous

Spectra

Line

Spectra

13

Continuous Spectrum

A spectrum consists all wavelength components

(containing an unbroken sequence of frequencies) of the

visible portion of the electromagnetic spectrum are

present

It is produced by incandescent solids liquids and

compressed gases

14

Regions of the Electromagnetic Spectrum

15

When white light from

incandescent lamp is passed

through a slit then a prism it

separates into a spectrum

The white light spread out into

a rainbow of colours produces

a continuous spectrum

The spectrum is continuous in

that all wavelengths are

presents and each colour

merges into the next without a

break

FORMATION OF CONTINUOUS SPECTRUM

16

Line Spectrum (atomic spectrum)

A spectrum consists of discontinuous amp discrete lines produced by

excited atoms and ions as the electrons fall back to a lower energy

level The radiation emitted is only at a specific wavelength or

frequency It means each line corresponds to a specific wavelength

or frequency

Line spectrum are composed of only a few wavelengths giving a

series of discrete line separated by blank areas

17

prismfilm

The emitted light (photons) is then separated into its components by

a prism Each component is focused at a definite position according

to its wavelength and forms as an image on the photographic plate

The images are called spectral lines

FORMATION OF ATOMIC LINE SPECTRUM

18

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

En

erg

y

When an electrical discharge is passed through a sample of

hydrogen gas at low pressure hydrogen molecules decompose to

form hydrogen atoms

Radiant energy (a

quantum of energy)

absorbed by the atom (or

electron) causes the

electron to move from a

lower-energy state to a

higher-energy state

Hydrogen atom is said to

be at excited state (very

unstable)

19

FORMATION OF ATOMIC LINE SPECTRUM

Emission of photon

n = 2

n = 3

n = 4

n = 5

n = 6n = infin

En

erg

y

When the electrons fall

back to lower energy

levels radiant energies

(photons) are emitted in

the form of light

(electromagnetic radiation

of a particular frequency or

wavelength)

20

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum λE

Energy

21

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum

Balmer Series

λ E

Energy

22

Emission series of hydrogen atom

n = 1

n = 2

n = 3

n = 4

n = infin

Lyman series

Balmer series

Brackett series

Paschen series

Pfund series

23

Exercise Complete the following table

Series n1 n2

Spectrum

region

Lyman 1 234hellip Ultraviolet

Balmer 2 345hellip Visible light

Paschen 3 456hellip Infrared

Brackett 4 567hellip Infrared

Pfund 5 678hellip Infrared

24

The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series

Exercise

Line

spectrumE λ

Specify the increasing order of the radiant energy

frequency and wavelength of the emitted photon

Which of the line that corresponds to

i) the shortest wavelength

ii) the lowest frequency

A B C D E

ν

25

Describe the transitions of electrons that lead to

the lines W and Y respectively

Solution

Line

spectrum

W Y

Exercise

Balmer series

26

Homework

Calculate En for n = 1 2 3 and 4 Make a one-

dimensional graph showing energy at different

values of n increasing vertically On this graph

indicate by vertical arrows transitions that lead to

lines in

a) Lyman series

b) Paschen series

27

In Lyman series the frequency of the convergence of

spectral lines can be used to find the ionisation energy of

hydrogen atom

IE = hνinfin

The frequency of the first line of the Lyman series gt the

frequency of the first line of the Balmer series

Significance of Atomic Spectra

Lyman Series

Line

spectrum λ E

Balmer Series

νinfin

28

Line

spectrum

ABCDE

Exercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon

Describe the transition that gives rise to the line

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

13

Continuous Spectrum

A spectrum consists all wavelength components

(containing an unbroken sequence of frequencies) of the

visible portion of the electromagnetic spectrum are

present

It is produced by incandescent solids liquids and

compressed gases

14

Regions of the Electromagnetic Spectrum

15

When white light from

incandescent lamp is passed

through a slit then a prism it

separates into a spectrum

The white light spread out into

a rainbow of colours produces

a continuous spectrum

The spectrum is continuous in

that all wavelengths are

presents and each colour

merges into the next without a

break

FORMATION OF CONTINUOUS SPECTRUM

16

Line Spectrum (atomic spectrum)

A spectrum consists of discontinuous amp discrete lines produced by

excited atoms and ions as the electrons fall back to a lower energy

level The radiation emitted is only at a specific wavelength or

frequency It means each line corresponds to a specific wavelength

or frequency

Line spectrum are composed of only a few wavelengths giving a

series of discrete line separated by blank areas

17

prismfilm

The emitted light (photons) is then separated into its components by

a prism Each component is focused at a definite position according

to its wavelength and forms as an image on the photographic plate

The images are called spectral lines

FORMATION OF ATOMIC LINE SPECTRUM

18

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

En

erg

y

When an electrical discharge is passed through a sample of

hydrogen gas at low pressure hydrogen molecules decompose to

form hydrogen atoms

Radiant energy (a

quantum of energy)

absorbed by the atom (or

electron) causes the

electron to move from a

lower-energy state to a

higher-energy state

Hydrogen atom is said to

be at excited state (very

unstable)

19

FORMATION OF ATOMIC LINE SPECTRUM

Emission of photon

n = 2

n = 3

n = 4

n = 5

n = 6n = infin

En

erg

y

When the electrons fall

back to lower energy

levels radiant energies

(photons) are emitted in

the form of light

(electromagnetic radiation

of a particular frequency or

wavelength)

20

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum λE

Energy

21

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum

Balmer Series

λ E

Energy

22

Emission series of hydrogen atom

n = 1

n = 2

n = 3

n = 4

n = infin

Lyman series

Balmer series

Brackett series

Paschen series

Pfund series

23

Exercise Complete the following table

Series n1 n2

Spectrum

region

Lyman 1 234hellip Ultraviolet

Balmer 2 345hellip Visible light

Paschen 3 456hellip Infrared

Brackett 4 567hellip Infrared

Pfund 5 678hellip Infrared

24

The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series

Exercise

Line

spectrumE λ

Specify the increasing order of the radiant energy

frequency and wavelength of the emitted photon

Which of the line that corresponds to

i) the shortest wavelength

ii) the lowest frequency

A B C D E

ν

25

Describe the transitions of electrons that lead to

the lines W and Y respectively

Solution

Line

spectrum

W Y

Exercise

Balmer series

26

Homework

Calculate En for n = 1 2 3 and 4 Make a one-

dimensional graph showing energy at different

values of n increasing vertically On this graph

indicate by vertical arrows transitions that lead to

lines in

a) Lyman series

b) Paschen series

27

In Lyman series the frequency of the convergence of

spectral lines can be used to find the ionisation energy of

hydrogen atom

IE = hνinfin

The frequency of the first line of the Lyman series gt the

frequency of the first line of the Balmer series

Significance of Atomic Spectra

Lyman Series

Line

spectrum λ E

Balmer Series

νinfin

28

Line

spectrum

ABCDE

Exercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon

Describe the transition that gives rise to the line

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

14

Regions of the Electromagnetic Spectrum

15

When white light from

incandescent lamp is passed

through a slit then a prism it

separates into a spectrum

The white light spread out into

a rainbow of colours produces

a continuous spectrum

The spectrum is continuous in

that all wavelengths are

presents and each colour

merges into the next without a

break

FORMATION OF CONTINUOUS SPECTRUM

16

Line Spectrum (atomic spectrum)

A spectrum consists of discontinuous amp discrete lines produced by

excited atoms and ions as the electrons fall back to a lower energy

level The radiation emitted is only at a specific wavelength or

frequency It means each line corresponds to a specific wavelength

or frequency

Line spectrum are composed of only a few wavelengths giving a

series of discrete line separated by blank areas

17

prismfilm

The emitted light (photons) is then separated into its components by

a prism Each component is focused at a definite position according

to its wavelength and forms as an image on the photographic plate

The images are called spectral lines

FORMATION OF ATOMIC LINE SPECTRUM

18

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

En

erg

y

When an electrical discharge is passed through a sample of

hydrogen gas at low pressure hydrogen molecules decompose to

form hydrogen atoms

Radiant energy (a

quantum of energy)

absorbed by the atom (or

electron) causes the

electron to move from a

lower-energy state to a

higher-energy state

Hydrogen atom is said to

be at excited state (very

unstable)

19

FORMATION OF ATOMIC LINE SPECTRUM

Emission of photon

n = 2

n = 3

n = 4

n = 5

n = 6n = infin

En

erg

y

When the electrons fall

back to lower energy

levels radiant energies

(photons) are emitted in

the form of light

(electromagnetic radiation

of a particular frequency or

wavelength)

20

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum λE

Energy

21

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum

Balmer Series

λ E

Energy

22

Emission series of hydrogen atom

n = 1

n = 2

n = 3

n = 4

n = infin

Lyman series

Balmer series

Brackett series

Paschen series

Pfund series

23

Exercise Complete the following table

Series n1 n2

Spectrum

region

Lyman 1 234hellip Ultraviolet

Balmer 2 345hellip Visible light

Paschen 3 456hellip Infrared

Brackett 4 567hellip Infrared

Pfund 5 678hellip Infrared

24

The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series

Exercise

Line

spectrumE λ

Specify the increasing order of the radiant energy

frequency and wavelength of the emitted photon

Which of the line that corresponds to

i) the shortest wavelength

ii) the lowest frequency

A B C D E

ν

25

Describe the transitions of electrons that lead to

the lines W and Y respectively

Solution

Line

spectrum

W Y

Exercise

Balmer series

26

Homework

Calculate En for n = 1 2 3 and 4 Make a one-

dimensional graph showing energy at different

values of n increasing vertically On this graph

indicate by vertical arrows transitions that lead to

lines in

a) Lyman series

b) Paschen series

27

In Lyman series the frequency of the convergence of

spectral lines can be used to find the ionisation energy of

hydrogen atom

IE = hνinfin

The frequency of the first line of the Lyman series gt the

frequency of the first line of the Balmer series

Significance of Atomic Spectra

Lyman Series

Line

spectrum λ E

Balmer Series

νinfin

28

Line

spectrum

ABCDE

Exercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon

Describe the transition that gives rise to the line

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

15

When white light from

incandescent lamp is passed

through a slit then a prism it

separates into a spectrum

The white light spread out into

a rainbow of colours produces

a continuous spectrum

The spectrum is continuous in

that all wavelengths are

presents and each colour

merges into the next without a

break

FORMATION OF CONTINUOUS SPECTRUM

16

Line Spectrum (atomic spectrum)

A spectrum consists of discontinuous amp discrete lines produced by

excited atoms and ions as the electrons fall back to a lower energy

level The radiation emitted is only at a specific wavelength or

frequency It means each line corresponds to a specific wavelength

or frequency

Line spectrum are composed of only a few wavelengths giving a

series of discrete line separated by blank areas

17

prismfilm

The emitted light (photons) is then separated into its components by

a prism Each component is focused at a definite position according

to its wavelength and forms as an image on the photographic plate

The images are called spectral lines

FORMATION OF ATOMIC LINE SPECTRUM

18

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

En

erg

y

When an electrical discharge is passed through a sample of

hydrogen gas at low pressure hydrogen molecules decompose to

form hydrogen atoms

Radiant energy (a

quantum of energy)

absorbed by the atom (or

electron) causes the

electron to move from a

lower-energy state to a

higher-energy state

Hydrogen atom is said to

be at excited state (very

unstable)

19

FORMATION OF ATOMIC LINE SPECTRUM

Emission of photon

n = 2

n = 3

n = 4

n = 5

n = 6n = infin

En

erg

y

When the electrons fall

back to lower energy

levels radiant energies

(photons) are emitted in

the form of light

(electromagnetic radiation

of a particular frequency or

wavelength)

20

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum λE

Energy

21

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum

Balmer Series

λ E

Energy

22

Emission series of hydrogen atom

n = 1

n = 2

n = 3

n = 4

n = infin

Lyman series

Balmer series

Brackett series

Paschen series

Pfund series

23

Exercise Complete the following table

Series n1 n2

Spectrum

region

Lyman 1 234hellip Ultraviolet

Balmer 2 345hellip Visible light

Paschen 3 456hellip Infrared

Brackett 4 567hellip Infrared

Pfund 5 678hellip Infrared

24

The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series

Exercise

Line

spectrumE λ

Specify the increasing order of the radiant energy

frequency and wavelength of the emitted photon

Which of the line that corresponds to

i) the shortest wavelength

ii) the lowest frequency

A B C D E

ν

25

Describe the transitions of electrons that lead to

the lines W and Y respectively

Solution

Line

spectrum

W Y

Exercise

Balmer series

26

Homework

Calculate En for n = 1 2 3 and 4 Make a one-

dimensional graph showing energy at different

values of n increasing vertically On this graph

indicate by vertical arrows transitions that lead to

lines in

a) Lyman series

b) Paschen series

27

In Lyman series the frequency of the convergence of

spectral lines can be used to find the ionisation energy of

hydrogen atom

IE = hνinfin

The frequency of the first line of the Lyman series gt the

frequency of the first line of the Balmer series

Significance of Atomic Spectra

Lyman Series

Line

spectrum λ E

Balmer Series

νinfin

28

Line

spectrum

ABCDE

Exercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon

Describe the transition that gives rise to the line

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

16

Line Spectrum (atomic spectrum)

A spectrum consists of discontinuous amp discrete lines produced by

excited atoms and ions as the electrons fall back to a lower energy

level The radiation emitted is only at a specific wavelength or

frequency It means each line corresponds to a specific wavelength

or frequency

Line spectrum are composed of only a few wavelengths giving a

series of discrete line separated by blank areas

17

prismfilm

The emitted light (photons) is then separated into its components by

a prism Each component is focused at a definite position according

to its wavelength and forms as an image on the photographic plate

The images are called spectral lines

FORMATION OF ATOMIC LINE SPECTRUM

18

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

En

erg

y

When an electrical discharge is passed through a sample of

hydrogen gas at low pressure hydrogen molecules decompose to

form hydrogen atoms

Radiant energy (a

quantum of energy)

absorbed by the atom (or

electron) causes the

electron to move from a

lower-energy state to a

higher-energy state

Hydrogen atom is said to

be at excited state (very

unstable)

19

FORMATION OF ATOMIC LINE SPECTRUM

Emission of photon

n = 2

n = 3

n = 4

n = 5

n = 6n = infin

En

erg

y

When the electrons fall

back to lower energy

levels radiant energies

(photons) are emitted in

the form of light

(electromagnetic radiation

of a particular frequency or

wavelength)

20

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum λE

Energy

21

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum

Balmer Series

λ E

Energy

22

Emission series of hydrogen atom

n = 1

n = 2

n = 3

n = 4

n = infin

Lyman series

Balmer series

Brackett series

Paschen series

Pfund series

23

Exercise Complete the following table

Series n1 n2

Spectrum

region

Lyman 1 234hellip Ultraviolet

Balmer 2 345hellip Visible light

Paschen 3 456hellip Infrared

Brackett 4 567hellip Infrared

Pfund 5 678hellip Infrared

24

The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series

Exercise

Line

spectrumE λ

Specify the increasing order of the radiant energy

frequency and wavelength of the emitted photon

Which of the line that corresponds to

i) the shortest wavelength

ii) the lowest frequency

A B C D E

ν

25

Describe the transitions of electrons that lead to

the lines W and Y respectively

Solution

Line

spectrum

W Y

Exercise

Balmer series

26

Homework

Calculate En for n = 1 2 3 and 4 Make a one-

dimensional graph showing energy at different

values of n increasing vertically On this graph

indicate by vertical arrows transitions that lead to

lines in

a) Lyman series

b) Paschen series

27

In Lyman series the frequency of the convergence of

spectral lines can be used to find the ionisation energy of

hydrogen atom

IE = hνinfin

The frequency of the first line of the Lyman series gt the

frequency of the first line of the Balmer series

Significance of Atomic Spectra

Lyman Series

Line

spectrum λ E

Balmer Series

νinfin

28

Line

spectrum

ABCDE

Exercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon

Describe the transition that gives rise to the line

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

17

prismfilm

The emitted light (photons) is then separated into its components by

a prism Each component is focused at a definite position according

to its wavelength and forms as an image on the photographic plate

The images are called spectral lines

FORMATION OF ATOMIC LINE SPECTRUM

18

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

En

erg

y

When an electrical discharge is passed through a sample of

hydrogen gas at low pressure hydrogen molecules decompose to

form hydrogen atoms

Radiant energy (a

quantum of energy)

absorbed by the atom (or

electron) causes the

electron to move from a

lower-energy state to a

higher-energy state

Hydrogen atom is said to

be at excited state (very

unstable)

19

FORMATION OF ATOMIC LINE SPECTRUM

Emission of photon

n = 2

n = 3

n = 4

n = 5

n = 6n = infin

En

erg

y

When the electrons fall

back to lower energy

levels radiant energies

(photons) are emitted in

the form of light

(electromagnetic radiation

of a particular frequency or

wavelength)

20

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum λE

Energy

21

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum

Balmer Series

λ E

Energy

22

Emission series of hydrogen atom

n = 1

n = 2

n = 3

n = 4

n = infin

Lyman series

Balmer series

Brackett series

Paschen series

Pfund series

23

Exercise Complete the following table

Series n1 n2

Spectrum

region

Lyman 1 234hellip Ultraviolet

Balmer 2 345hellip Visible light

Paschen 3 456hellip Infrared

Brackett 4 567hellip Infrared

Pfund 5 678hellip Infrared

24

The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series

Exercise

Line

spectrumE λ

Specify the increasing order of the radiant energy

frequency and wavelength of the emitted photon

Which of the line that corresponds to

i) the shortest wavelength

ii) the lowest frequency

A B C D E

ν

25

Describe the transitions of electrons that lead to

the lines W and Y respectively

Solution

Line

spectrum

W Y

Exercise

Balmer series

26

Homework

Calculate En for n = 1 2 3 and 4 Make a one-

dimensional graph showing energy at different

values of n increasing vertically On this graph

indicate by vertical arrows transitions that lead to

lines in

a) Lyman series

b) Paschen series

27

In Lyman series the frequency of the convergence of

spectral lines can be used to find the ionisation energy of

hydrogen atom

IE = hνinfin

The frequency of the first line of the Lyman series gt the

frequency of the first line of the Balmer series

Significance of Atomic Spectra

Lyman Series

Line

spectrum λ E

Balmer Series

νinfin

28

Line

spectrum

ABCDE

Exercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon

Describe the transition that gives rise to the line

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

18

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

En

erg

y

When an electrical discharge is passed through a sample of

hydrogen gas at low pressure hydrogen molecules decompose to

form hydrogen atoms

Radiant energy (a

quantum of energy)

absorbed by the atom (or

electron) causes the

electron to move from a

lower-energy state to a

higher-energy state

Hydrogen atom is said to

be at excited state (very

unstable)

19

FORMATION OF ATOMIC LINE SPECTRUM

Emission of photon

n = 2

n = 3

n = 4

n = 5

n = 6n = infin

En

erg

y

When the electrons fall

back to lower energy

levels radiant energies

(photons) are emitted in

the form of light

(electromagnetic radiation

of a particular frequency or

wavelength)

20

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum λE

Energy

21

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum

Balmer Series

λ E

Energy

22

Emission series of hydrogen atom

n = 1

n = 2

n = 3

n = 4

n = infin

Lyman series

Balmer series

Brackett series

Paschen series

Pfund series

23

Exercise Complete the following table

Series n1 n2

Spectrum

region

Lyman 1 234hellip Ultraviolet

Balmer 2 345hellip Visible light

Paschen 3 456hellip Infrared

Brackett 4 567hellip Infrared

Pfund 5 678hellip Infrared

24

The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series

Exercise

Line

spectrumE λ

Specify the increasing order of the radiant energy

frequency and wavelength of the emitted photon

Which of the line that corresponds to

i) the shortest wavelength

ii) the lowest frequency

A B C D E

ν

25

Describe the transitions of electrons that lead to

the lines W and Y respectively

Solution

Line

spectrum

W Y

Exercise

Balmer series

26

Homework

Calculate En for n = 1 2 3 and 4 Make a one-

dimensional graph showing energy at different

values of n increasing vertically On this graph

indicate by vertical arrows transitions that lead to

lines in

a) Lyman series

b) Paschen series

27

In Lyman series the frequency of the convergence of

spectral lines can be used to find the ionisation energy of

hydrogen atom

IE = hνinfin

The frequency of the first line of the Lyman series gt the

frequency of the first line of the Balmer series

Significance of Atomic Spectra

Lyman Series

Line

spectrum λ E

Balmer Series

νinfin

28

Line

spectrum

ABCDE

Exercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon

Describe the transition that gives rise to the line

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

19

FORMATION OF ATOMIC LINE SPECTRUM

Emission of photon

n = 2

n = 3

n = 4

n = 5

n = 6n = infin

En

erg

y

When the electrons fall

back to lower energy

levels radiant energies

(photons) are emitted in

the form of light

(electromagnetic radiation

of a particular frequency or

wavelength)

20

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum λE

Energy

21

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum

Balmer Series

λ E

Energy

22

Emission series of hydrogen atom

n = 1

n = 2

n = 3

n = 4

n = infin

Lyman series

Balmer series

Brackett series

Paschen series

Pfund series

23

Exercise Complete the following table

Series n1 n2

Spectrum

region

Lyman 1 234hellip Ultraviolet

Balmer 2 345hellip Visible light

Paschen 3 456hellip Infrared

Brackett 4 567hellip Infrared

Pfund 5 678hellip Infrared

24

The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series

Exercise

Line

spectrumE λ

Specify the increasing order of the radiant energy

frequency and wavelength of the emitted photon

Which of the line that corresponds to

i) the shortest wavelength

ii) the lowest frequency

A B C D E

ν

25

Describe the transitions of electrons that lead to

the lines W and Y respectively

Solution

Line

spectrum

W Y

Exercise

Balmer series

26

Homework

Calculate En for n = 1 2 3 and 4 Make a one-

dimensional graph showing energy at different

values of n increasing vertically On this graph

indicate by vertical arrows transitions that lead to

lines in

a) Lyman series

b) Paschen series

27

In Lyman series the frequency of the convergence of

spectral lines can be used to find the ionisation energy of

hydrogen atom

IE = hνinfin

The frequency of the first line of the Lyman series gt the

frequency of the first line of the Balmer series

Significance of Atomic Spectra

Lyman Series

Line

spectrum λ E

Balmer Series

νinfin

28

Line

spectrum

ABCDE

Exercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon

Describe the transition that gives rise to the line

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

20

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum λE

Energy

21

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum

Balmer Series

λ E

Energy

22

Emission series of hydrogen atom

n = 1

n = 2

n = 3

n = 4

n = infin

Lyman series

Balmer series

Brackett series

Paschen series

Pfund series

23

Exercise Complete the following table

Series n1 n2

Spectrum

region

Lyman 1 234hellip Ultraviolet

Balmer 2 345hellip Visible light

Paschen 3 456hellip Infrared

Brackett 4 567hellip Infrared

Pfund 5 678hellip Infrared

24

The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series

Exercise

Line

spectrumE λ

Specify the increasing order of the radiant energy

frequency and wavelength of the emitted photon

Which of the line that corresponds to

i) the shortest wavelength

ii) the lowest frequency

A B C D E

ν

25

Describe the transitions of electrons that lead to

the lines W and Y respectively

Solution

Line

spectrum

W Y

Exercise

Balmer series

26

Homework

Calculate En for n = 1 2 3 and 4 Make a one-

dimensional graph showing energy at different

values of n increasing vertically On this graph

indicate by vertical arrows transitions that lead to

lines in

a) Lyman series

b) Paschen series

27

In Lyman series the frequency of the convergence of

spectral lines can be used to find the ionisation energy of

hydrogen atom

IE = hνinfin

The frequency of the first line of the Lyman series gt the

frequency of the first line of the Balmer series

Significance of Atomic Spectra

Lyman Series

Line

spectrum λ E

Balmer Series

νinfin

28

Line

spectrum

ABCDE

Exercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon

Describe the transition that gives rise to the line

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

21

FORMATION OF ATOMIC LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5n = infin

Lyman Series

Emission of photon

Line

spectrum

Balmer Series

λ E

Energy

22

Emission series of hydrogen atom

n = 1

n = 2

n = 3

n = 4

n = infin

Lyman series

Balmer series

Brackett series

Paschen series

Pfund series

23

Exercise Complete the following table

Series n1 n2

Spectrum

region

Lyman 1 234hellip Ultraviolet

Balmer 2 345hellip Visible light

Paschen 3 456hellip Infrared

Brackett 4 567hellip Infrared

Pfund 5 678hellip Infrared

24

The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series

Exercise

Line

spectrumE λ

Specify the increasing order of the radiant energy

frequency and wavelength of the emitted photon

Which of the line that corresponds to

i) the shortest wavelength

ii) the lowest frequency

A B C D E

ν

25

Describe the transitions of electrons that lead to

the lines W and Y respectively

Solution

Line

spectrum

W Y

Exercise

Balmer series

26

Homework

Calculate En for n = 1 2 3 and 4 Make a one-

dimensional graph showing energy at different

values of n increasing vertically On this graph

indicate by vertical arrows transitions that lead to

lines in

a) Lyman series

b) Paschen series

27

In Lyman series the frequency of the convergence of

spectral lines can be used to find the ionisation energy of

hydrogen atom

IE = hνinfin

The frequency of the first line of the Lyman series gt the

frequency of the first line of the Balmer series

Significance of Atomic Spectra

Lyman Series

Line

spectrum λ E

Balmer Series

νinfin

28

Line

spectrum

ABCDE

Exercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon

Describe the transition that gives rise to the line

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

22

Emission series of hydrogen atom

n = 1

n = 2

n = 3

n = 4

n = infin

Lyman series

Balmer series

Brackett series

Paschen series

Pfund series

23

Exercise Complete the following table

Series n1 n2

Spectrum

region

Lyman 1 234hellip Ultraviolet

Balmer 2 345hellip Visible light

Paschen 3 456hellip Infrared

Brackett 4 567hellip Infrared

Pfund 5 678hellip Infrared

24

The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series

Exercise

Line

spectrumE λ

Specify the increasing order of the radiant energy

frequency and wavelength of the emitted photon

Which of the line that corresponds to

i) the shortest wavelength

ii) the lowest frequency

A B C D E

ν

25

Describe the transitions of electrons that lead to

the lines W and Y respectively

Solution

Line

spectrum

W Y

Exercise

Balmer series

26

Homework

Calculate En for n = 1 2 3 and 4 Make a one-

dimensional graph showing energy at different

values of n increasing vertically On this graph

indicate by vertical arrows transitions that lead to

lines in

a) Lyman series

b) Paschen series

27

In Lyman series the frequency of the convergence of

spectral lines can be used to find the ionisation energy of

hydrogen atom

IE = hνinfin

The frequency of the first line of the Lyman series gt the

frequency of the first line of the Balmer series

Significance of Atomic Spectra

Lyman Series

Line

spectrum λ E

Balmer Series

νinfin

28

Line

spectrum

ABCDE

Exercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon

Describe the transition that gives rise to the line

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

23

Exercise Complete the following table

Series n1 n2

Spectrum

region

Lyman 1 234hellip Ultraviolet

Balmer 2 345hellip Visible light

Paschen 3 456hellip Infrared

Brackett 4 567hellip Infrared

Pfund 5 678hellip Infrared

24

The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series

Exercise

Line

spectrumE λ

Specify the increasing order of the radiant energy

frequency and wavelength of the emitted photon

Which of the line that corresponds to

i) the shortest wavelength

ii) the lowest frequency

A B C D E

ν

25

Describe the transitions of electrons that lead to

the lines W and Y respectively

Solution

Line

spectrum

W Y

Exercise

Balmer series

26

Homework

Calculate En for n = 1 2 3 and 4 Make a one-

dimensional graph showing energy at different

values of n increasing vertically On this graph

indicate by vertical arrows transitions that lead to

lines in

a) Lyman series

b) Paschen series

27

In Lyman series the frequency of the convergence of

spectral lines can be used to find the ionisation energy of

hydrogen atom

IE = hνinfin

The frequency of the first line of the Lyman series gt the

frequency of the first line of the Balmer series

Significance of Atomic Spectra

Lyman Series

Line

spectrum λ E

Balmer Series

νinfin

28

Line

spectrum

ABCDE

Exercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon

Describe the transition that gives rise to the line

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

24

The following diagram depicts the line spectrum of hydrogen atom Line A is the first line of the Lyman series

Exercise

Line

spectrumE λ

Specify the increasing order of the radiant energy

frequency and wavelength of the emitted photon

Which of the line that corresponds to

i) the shortest wavelength

ii) the lowest frequency

A B C D E

ν

25

Describe the transitions of electrons that lead to

the lines W and Y respectively

Solution

Line

spectrum

W Y

Exercise

Balmer series

26

Homework

Calculate En for n = 1 2 3 and 4 Make a one-

dimensional graph showing energy at different

values of n increasing vertically On this graph

indicate by vertical arrows transitions that lead to

lines in

a) Lyman series

b) Paschen series

27

In Lyman series the frequency of the convergence of

spectral lines can be used to find the ionisation energy of

hydrogen atom

IE = hνinfin

The frequency of the first line of the Lyman series gt the

frequency of the first line of the Balmer series

Significance of Atomic Spectra

Lyman Series

Line

spectrum λ E

Balmer Series

νinfin

28

Line

spectrum

ABCDE

Exercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon

Describe the transition that gives rise to the line

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

25

Describe the transitions of electrons that lead to

the lines W and Y respectively

Solution

Line

spectrum

W Y

Exercise

Balmer series

26

Homework

Calculate En for n = 1 2 3 and 4 Make a one-

dimensional graph showing energy at different

values of n increasing vertically On this graph

indicate by vertical arrows transitions that lead to

lines in

a) Lyman series

b) Paschen series

27

In Lyman series the frequency of the convergence of

spectral lines can be used to find the ionisation energy of

hydrogen atom

IE = hνinfin

The frequency of the first line of the Lyman series gt the

frequency of the first line of the Balmer series

Significance of Atomic Spectra

Lyman Series

Line

spectrum λ E

Balmer Series

νinfin

28

Line

spectrum

ABCDE

Exercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon

Describe the transition that gives rise to the line

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

26

Homework

Calculate En for n = 1 2 3 and 4 Make a one-

dimensional graph showing energy at different

values of n increasing vertically On this graph

indicate by vertical arrows transitions that lead to

lines in

a) Lyman series

b) Paschen series

27

In Lyman series the frequency of the convergence of

spectral lines can be used to find the ionisation energy of

hydrogen atom

IE = hνinfin

The frequency of the first line of the Lyman series gt the

frequency of the first line of the Balmer series

Significance of Atomic Spectra

Lyman Series

Line

spectrum λ E

Balmer Series

νinfin

28

Line

spectrum

ABCDE

Exercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon

Describe the transition that gives rise to the line

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

27

In Lyman series the frequency of the convergence of

spectral lines can be used to find the ionisation energy of

hydrogen atom

IE = hνinfin

The frequency of the first line of the Lyman series gt the

frequency of the first line of the Balmer series

Significance of Atomic Spectra

Lyman Series

Line

spectrum λ E

Balmer Series

νinfin

28

Line

spectrum

ABCDE

Exercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon

Describe the transition that gives rise to the line

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

28

Line

spectrum

ABCDE

Exercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon

Describe the transition that gives rise to the line

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

29

Radiant energy emitted when the electron moves from

higher-energy state to lower-energy state is given by the

difference in energy between energy levels

Energy calculation

ΔE = Ef - Ei

where

Thus

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

30

The amount of energy released by the electron is called a photon of energy

A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength

whereh (Planckrsquos constant) =663 times 10-34 J sν = frequency

Energy calculation

ΔE = hν

Wherec (speed of light) = 300times108 ms-1

Thus

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

31

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher

energy level A specific amount of energy

is absorbed

ΔE = hν = E1-E3 (+ve)

Electron falls from higher to lower energy level

A photon of energy is released

ΔE = hν = E3-E1 (-ve)

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

32

Energy level diagram for the hydrogen atom

Pote

ntial energ

y

n = 1

n = 2

n = 3

n = 4

n = infin

Energy

released

Energy

absorbed

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

33

Exercises

1) Calculate the energy of an electron in the second energy level of a hydrogen atom (-5448 x 10-19 J)

1) Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom

3) Calculate the energy change (J) that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom

(answer 155 x 10-19J)

4) Calculate the frequency and wavelength (nm) of the radiation emitted in question 3

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

34

At the end of this topic students should be able to-

Level No Learning Outcomes

C3 amp C4 21 h Perform calculations involving the Rydberg equation for Lyman Balmer Paschen Brackett and Pfund series

1 λ = RH (1n12 - 1n2

2)

where RH = 1097 x 107 m-1 and n1ltn2

C3 21 i Calculate the ionisation energy of hydrogen atom from

Lyman series

C2 21 j State the limitation of Bohrrsquos atomic model

C2 21 k State the dual nature of electron using the Brogliersquos

postulate and Hesseinbergrsquos uncertainty

principle

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

35

Wavelength emitted by the transition of electron

between two energy levels is calculated using

Rydberg equation

Rydberg Equation

RH = 1097 times 107 m-1

λ = wavelength

Since λ should have a positive value thus n1 lt n2

where

1λ = RH (1ni2 ndash 1nf

2)

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

36

Calculate the wavelength in nanometers of the spectrum

of hydrogen corresponding to ni = 2 and nf = 4 in the

Rydberg equation

Example

Solution

Rydberg equation

1λ = RH (1ni2 ndash 1nf

2)

ni = 2 nf = 4

RH = 1097 x 10m7

1λ = RH (122 ndash 142)

= RH(14-116)

λ = 486m x 102 m

= 486nm

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

37

Use the Rydberg equation to calculate the wavelength of the

spectral line of hydrogen atom that would result when an

electron drops from the fourth orbit to the second orbit then

identified the series the line would be found

Example

Solution

1λ = RH (1n12 ndash 1n2

2)

n1 = 2 n2 = 4

1λ = 1097 x 107 (122 ndash 142)

λ = 486 x 10-7 m

= 486 nm

e dropped to the second orbit (n=2)

gtgtgt Balmer series

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

38

EXAMPLE 3

Calculate the wavelengths of the fourth line in the

Balmer series of hydrogen

n1 = 2 n2 = 6

RH = 1097 x 107m-1

λ = 410 x 10-7 m

RH 22 62

1 11=

λ

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

39

Different values of RH and its usage

1 RH = 1097 times 107 m-1

RH n21

n22

1 11=

λ

RH = 218 x 10-18 J

n1 lt n2

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

40

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom

ΔE = 458 x 10-19 J

ΔE = (663 times 10-34Js)X(300times108 ms-1)

RH n21

n22

1 11=

λ

521097 x 107

22

1 11

=λ

1

λ= 02303 X 107 m-1

X (02303 X 107 m-1)

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

41

Calculate what is

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum

in Lyman series

Wave number = 1wavelength

EXERCISE

For Lyman series n1 = 1

amp n2 = infin

Ans

i 9116 x10-8m

ii 329 x1015 s-1

iii 10970 X 107 m-1

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

42

Definition Ionization energy is the minimum energy

required to remove one mole of electron from one mole

of gaseous atomion

M (g) rarr M+ (g) + e ΔH = +ve

The hydrogen atom is said to be ionised when electron

is removed from its ground state (n = 1) to n = infin

At n = infin the potential energy of electron is zero here

the nucleus attractive force has no effect on the electron

(electron is free from nucleus)

Ionization Energy

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

43

n1 = 1 n2 = infin

∆E = RH (1n12 ndash 1n2

2)

= 218 X 10 -18 (112 ndash 1 infin 2)

= 218 X 10 -18 (1 ndash 0)

= 218 X 10 -18 J

Ionisation energy

= 218 X 10 -18x 602 X 1023J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Example

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

44

λ1

1 st lineConvergent limit

Finding ionisation energy experimentally

λinfin

Ionisation energy is determined by detecting

the wavelength of the convergence point

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

45

1097 1066 1052 1027 974 822

wave number (x106 m-1)

The Lyman series of the spectrum of hydrogen is shown

above Calculate the ionisation energy of hydrogen from

the spectrum

Example

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

46

ΔE = hcλ

=h x c λ = h x c x wave no

= 6626 x 10-34 J s x 3 x 108 m s-1 x 1097x 106 m-1

= 21806x 10-20 J

= 218 x 10-18J

Ionisation energy

= 218 X 10 -18x 602 X 1023 J mol-1

=1312 x 106 J mol-1

= 1312 kJ mol-1

Solution

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

47

Compute the ionisation energy of hydrogen atom in kJ molminus1

Exercise

Solution

J

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

48

The weakness of Bohrrsquos Theory

1 His theory could not be extended to predict the energy

levels and spectra of atoms and ions with more than

one electron It only can explain the hydrogen spectrum

or ions contain one electron eg He+ Li2+

1 Electrons are restricted to orbit the nucleus at certain

fixed distances

1 It cannot explain for the dual nature of electron

1 It cannot explain for the extra lines formed in the

hydrogen spectrum

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

49

Davisson amp Germer observed the diffraction of

electrons when a beam of electrons was directed at a

nickel crystal Diffraction patterns produced by

scattering electrons from crystals are very similar to

those produced by scattering X-rays from crystals This

experiment demonstrated that electrons do indeed

possess wavelike properties

Thus can the lsquopositionrsquo of a wave be specified

Point to Ponder

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

50

de Brogliersquos Postulate

In 1924 Louis de Broglie proposed that not only light but all

matter has a dual nature and possesses both wave and

corpuscular properties De Broglie deduced that the particle

and wave properties are related by the expression

h = Planck constant (J s)

m = particle mass (kg)

μ = velocity (ms)

λ = wavelength of a matter wave

λ

=

h

m

μ

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

51

Heisenbergrsquos Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain

Stated mathematically

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum

= Δmv

h = Planck constant

h

4

π

Δx Δp ge

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

52

22 QUANTUM

MECHANICAL MODEL

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

53

At the end of this topic students should be able to-

Level No Learning Outcomes

C1 22 a Define the term orbital

C2 22 b Explain all four quantum numbers of an electron in an orbital

i) principal quantum number n

ii) angular momentum quantum number ℓ

iii) magnetic quantum number m

iv) electron spin quantum number s

C2 22 c Sketch the 3-D shapes of sp and d orbitals

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

54

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron

Definition

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

55

Each of the electrons in an atom is described and

characterised by a set of four quantum numbers namely

a) principal quantum number n

b) angular momentum quantum number ℓ

c) magnetic quantum number m

d) electron spin quantum number s

Quantum Numbers

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

56

The value of n determines the energy of an orbital and thereby

the energy of the electron in that particular orbital

The principal quantum number may have only integral values n

=1 2 3 hellip infin

Principal Quantum Number n

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

57

Angular Momentum Quantum Number ℓ

- Alternative name Subsidiary Azimuthal Orbital

Quantum Number

- The value of ℓ indicates the shape of the atomic orbital (AO) the

types of orbitals and the angular momentum of the electron

- The allowed values of ℓ are 0 1 2hellip (nminus1)

Letters are assigned to different numerical values of ℓ

Numerical value of ℓ Symbol

0

1

2

3

Orbital shape

s

p

d

f

spherical

dumbbell

cloverleaf

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

58

Angular Momentum Quantum Number ℓ

- ℓ is dependant on n (ie 0 le ℓ lt n)

If n = 1 ℓ can only be 0 (s-orbital)

If n = 2 ℓ can be 0 or 1 giving rise to two subshells

(s and p-orbitals) of slightly different energy

If n = 3 ℓ can be 0 1 or 2 (there are three subshells

(s p and d-orbitals)

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

59

Magnetic Quantum Number m

The direction or orientation of the magnetic field is determined

by the value of m

Possible values of m depend on the value of For a given

m can be minusℓ hellip 0 hellip + ℓ

(minus ℓ le m le + ℓ)

If ℓ = 0 m can only be 0 rArr one orbital in s-subshell

If ℓ = 1 m can be minus1 0 +1 rArr three orbitals in p-subshell

If ℓ = 2 m can be minus2 minus1 0 +1 +2 rArr five orbitals in d-subshell

60

The number of m values indicates the number of orbitals in

a subshell with a particular value

The values of n = 2 and = 1 indicate that we have a 2p-

subshell and in this subshell we have three 2p-orbitals

(because there are three values of m given by -1 0 and

+1)

Magnetic Quantum Number m

61

Electron Spin Quantum Number s

The value of s determines the direction of spinning motions of an

electron (either clockwise or counter clockwise) which is spinning

on its own axes as Earth does

The electron spin quantum number has a value of

+1

2-

1

2or

62

Atomic orbitals with the same energy (ie the same value

of n and ℓ) are said to be degenerated Therefore there

are (2 ℓ +1) degenerate orbitals for each value of ℓ

The maximum number of electrons in a particular energy

level n is given by the expression as follows

max no of eminus = 2n2

Points to Remember

63

Shell nℓ

(ℓltn)

Orbital

notation

m

(-ℓ le m le +ℓ)

No of

degenerated

orbitals

K

L

M

2

1

3

Exercise Complete the following table

64

Exercise

State whether or not each of the following symbols is an

acceptable designation for an atomic orbital Explain what

is wrong with the unacceptable symbols

b) 6g

a) 2d

c) 7s

d) 5i

65

Shape of Atomic Orbitals

a) s orbitals

Spherical shape with the nucleus at the centre

The probability of finding electrons at the distance r from the nucleus is the same from all direction

When ℓ = 0

As n increases s orbital

gets larger

Shape of s orbital

with different n

66

b) p orbitals

When ℓ = 1

dumbbell shaped

three p-orbitals px py and pz

correspond m of -1 0 and +1

As n increases the p-orbitals get larger

All p-orbitals have a node at the nucleus

Shape of Atomic Orbitals

67

Shape of Atomic Orbitals

shape four d orbitals have four lobes (perpendicular)

one d orbital has two major lobes along z axis

and a donut-shaped girdles the centre

When ℓ = 2

m = -2 -1012

the orbitals are dyz dxz dxy dx2-y2 dz2

68

69

Shape of Atomic Orbitals

70

Electronic Configuration

At the end of this topic students should

be able to-

Level No Learning Outcomes

C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos

Exclusion Principle

C3 23 b Predict the electronic configuration of atoms and

monotaomic ions using spdf notation

C3 23 c Justify the anomalous electronic configurations of

chromium and copper

71

Representing Electronic Configuration

Method 1 Orbital diagram

O8

1s 2s 2p

Method 2 spdf notation

O8 1s 2s 2p2 2 4

box

platform

Concentric circle

72

Rules for Assigning Electrons to Orbitals

i) Aufbau Principle

Electrons fill the lowest energy orbitals first and other

orbitals in order of ascending energy

The order of filling orbitals is

1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s

1s

2s

3s

4s

5s

2p

3p

4p

5p

3d

4d

5d

4f

5f

1s 2s 2p

73

Relative Energy Level of Atomic Orbitals

en

erg

y

n=1

n=2

n=3

n=4

1s

2s 2p

3s

4s

3p

4p

3d

4d

en

erg

y

n=1

n=2

n=3

n=4

1s

2s2p

3s

4s

3p

4p

3d

4d5s

Orbital energy levels

in the H atom

Orbital energy levels

in a many-electron atom

74

ii) Pauli Exclusion Principle

Rules for Assigning Electrons to Orbitals

No two electrons in an atom can have the same four

quantum numbers (n m s)

1s

a b c

e(a)

e(b)

e(c)

n ℓ m s

1 0

01

0

01

0

0

12

12

12

( )

)(

75

iii) Hundrsquos Rule

Rules for Assigning Electrons to Orbitals

Only when all the degenerate orbitals (a group of

orbitals of identical energy eg three p-orbitals and five d-

orbitals) contain an electron do the electrons begin to

occupy these orbitals in pairs The electrons in half-filled

orbitals have the same spins that is parallel spins

2p

76

Indicate which of the following orbital diagrams are

acceptable or unacceptable for an atom in ground state

Explain what mistakes have been made in each and draw

the correct orbital diagram

Exercise

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

77

Draw lsquoelectrons-in-boxesrsquo diagram of the electronic

configuration of titanium Ti (Z = 22) Also write the ground-

state electronic configurations for Ti and Ti2+ ion

Exercise

IMPORTANT

In an empty atom the 4s orbital has a lower

energy compared to that of the 3d orbital That is

why electrons fill the 4s orbital first before filling

the 3d orbital

However once electrons isare added to the 3d

orbital the 4s electrons are repelled to a higher

energy level The 3d orbitals now have lower

energy than 4s

78

79

Points to remember

The electronic configuration of atom or monatomic ion at

ground state

rArr Distribution of electrons obeys Aufbau principle Pauli

exclusion principle and Hundrsquos rule

Each atomic orbital can only accommodate a maximum of 2

electrons

Atomic orbital is a 3-D region in space around the nucleus

where there is a high probability of finding an electron

Assigning electrons to subshells

s-orbital rArr a max of 2 electrons (ns2)

p-orbitals rArr a max of 6 electrons (np6)

d-orbitals rArr a max of 10 electrons (nd10)

80

The Anomalous Electronic Configurations of

Cr and Cu

Cr and Cu have electron configurations which are

inconsistent with the Aufbau principle The anomalous

are explained on the basis that a filled or half-filled orbital

is more stable

Element Expected Observedactual

Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1

Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1

81

24Cr 18[Ar]

The actual orbital notation

24Cr 18[Ar]

Half filled orbital is more stable

(possesses an extra added stability)

3d 4s

3d 4s

Chromium predicted orbital notation

82

Copper predicted orbital notation

Cu [Ar]

The actual orbital notation

Cu [Ar]

4s3d

3d 4s

Full filled orbital is more stable

(possesses an extra added stability)

83

z = 21

z = 30

84

Write the ground-state electronic configuration and

explain the anomalous case for Cr (Z=24) and Cu ( Z=29)

Exercise

Writing Electronic Configuration for Negative Ion

Add electron according to Aufbau Principle

Example

i Cl-

ii O2-

Writing Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example

i Mg2+

i K+

i Fe2+

61

Electron Spin Quantum Number s

The value of s determines the direction of spinning motions of an

electron (either clockwise or counter clockwise) which is spinning

on its own axes as Earth does

The electron spin quantum number has a value of

+1

2-

1

2or

62

Atomic orbitals with the same energy (ie the same value

of n and ℓ) are said to be degenerated Therefore there

are (2 ℓ +1) degenerate orbitals for each value of ℓ

The maximum number of electrons in a particular energy

level n is given by the expression as follows

max no of eminus = 2n2

Points to Remember

63

Shell nℓ

(ℓltn)

Orbital

notation

m

(-ℓ le m le +ℓ)

No of

degenerated

orbitals

K

L

M

2

1

3

Exercise Complete the following table

64

Exercise

State whether or not each of the following symbols is an

acceptable designation for an atomic orbital Explain what

is wrong with the unacceptable symbols

b) 6g

a) 2d

c) 7s

d) 5i

65

Shape of Atomic Orbitals

a) s orbitals

Spherical shape with the nucleus at the centre

The probability of finding electrons at the distance r from the nucleus is the same from all direction

When ℓ = 0

As n increases s orbital

gets larger

Shape of s orbital

with different n

66

b) p orbitals

When ℓ = 1

dumbbell shaped

three p-orbitals px py and pz

correspond m of -1 0 and +1

As n increases the p-orbitals get larger

All p-orbitals have a node at the nucleus

Shape of Atomic Orbitals

67

Shape of Atomic Orbitals

shape four d orbitals have four lobes (perpendicular)

one d orbital has two major lobes along z axis

and a donut-shaped girdles the centre

When ℓ = 2

m = -2 -1012

the orbitals are dyz dxz dxy dx2-y2 dz2

68

69

Shape of Atomic Orbitals

70

Electronic Configuration

At the end of this topic students should

be able to-

Level No Learning Outcomes

C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos

Exclusion Principle

C3 23 b Predict the electronic configuration of atoms and

monotaomic ions using spdf notation

C3 23 c Justify the anomalous electronic configurations of

chromium and copper

71

Representing Electronic Configuration

Method 1 Orbital diagram

O8

1s 2s 2p

Method 2 spdf notation

O8 1s 2s 2p2 2 4

box

platform

Concentric circle

72

Rules for Assigning Electrons to Orbitals

i) Aufbau Principle

Electrons fill the lowest energy orbitals first and other

orbitals in order of ascending energy

The order of filling orbitals is

1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s

1s

2s

3s

4s

5s

2p

3p

4p

5p

3d

4d

5d

4f

5f

1s 2s 2p

73

Relative Energy Level of Atomic Orbitals

en

erg

y

n=1

n=2

n=3

n=4

1s

2s 2p

3s

4s

3p

4p

3d

4d

en

erg

y

n=1

n=2

n=3

n=4

1s

2s2p

3s

4s

3p

4p

3d

4d5s

Orbital energy levels

in the H atom

Orbital energy levels

in a many-electron atom

74

ii) Pauli Exclusion Principle

Rules for Assigning Electrons to Orbitals

No two electrons in an atom can have the same four

quantum numbers (n m s)

1s

a b c

e(a)

e(b)

e(c)

n ℓ m s

1 0

01

0

01

0

0

12

12

12

( )

)(

75

iii) Hundrsquos Rule

Rules for Assigning Electrons to Orbitals

Only when all the degenerate orbitals (a group of

orbitals of identical energy eg three p-orbitals and five d-

orbitals) contain an electron do the electrons begin to

occupy these orbitals in pairs The electrons in half-filled

orbitals have the same spins that is parallel spins

2p

76

Indicate which of the following orbital diagrams are

acceptable or unacceptable for an atom in ground state

Explain what mistakes have been made in each and draw

the correct orbital diagram

Exercise

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

77

Draw lsquoelectrons-in-boxesrsquo diagram of the electronic

configuration of titanium Ti (Z = 22) Also write the ground-

state electronic configurations for Ti and Ti2+ ion

Exercise

IMPORTANT

In an empty atom the 4s orbital has a lower

energy compared to that of the 3d orbital That is

why electrons fill the 4s orbital first before filling

the 3d orbital

However once electrons isare added to the 3d

orbital the 4s electrons are repelled to a higher

energy level The 3d orbitals now have lower

energy than 4s

78

79

Points to remember

The electronic configuration of atom or monatomic ion at

ground state

rArr Distribution of electrons obeys Aufbau principle Pauli

exclusion principle and Hundrsquos rule

Each atomic orbital can only accommodate a maximum of 2

electrons

Atomic orbital is a 3-D region in space around the nucleus

where there is a high probability of finding an electron

Assigning electrons to subshells

s-orbital rArr a max of 2 electrons (ns2)

p-orbitals rArr a max of 6 electrons (np6)

d-orbitals rArr a max of 10 electrons (nd10)

80

The Anomalous Electronic Configurations of

Cr and Cu

Cr and Cu have electron configurations which are

inconsistent with the Aufbau principle The anomalous

are explained on the basis that a filled or half-filled orbital

is more stable

Element Expected Observedactual

Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1

Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1

81

24Cr 18[Ar]

The actual orbital notation

24Cr 18[Ar]

Half filled orbital is more stable

(possesses an extra added stability)

3d 4s

3d 4s

Chromium predicted orbital notation

82

Copper predicted orbital notation

Cu [Ar]

The actual orbital notation

Cu [Ar]

4s3d

3d 4s

Full filled orbital is more stable

(possesses an extra added stability)

83

z = 21

z = 30

84

Write the ground-state electronic configuration and

explain the anomalous case for Cr (Z=24) and Cu ( Z=29)

Exercise

Writing Electronic Configuration for Negative Ion

Add electron according to Aufbau Principle

Example

i Cl-

ii O2-

Writing Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example

i Mg2+

i K+

i Fe2+

62

Atomic orbitals with the same energy (ie the same value

of n and ℓ) are said to be degenerated Therefore there

are (2 ℓ +1) degenerate orbitals for each value of ℓ

The maximum number of electrons in a particular energy

level n is given by the expression as follows

max no of eminus = 2n2

Points to Remember

63

Shell nℓ

(ℓltn)

Orbital

notation

m

(-ℓ le m le +ℓ)

No of

degenerated

orbitals

K

L

M

2

1

3

Exercise Complete the following table

64

Exercise

State whether or not each of the following symbols is an

acceptable designation for an atomic orbital Explain what

is wrong with the unacceptable symbols

b) 6g

a) 2d

c) 7s

d) 5i

65

Shape of Atomic Orbitals

a) s orbitals

Spherical shape with the nucleus at the centre

The probability of finding electrons at the distance r from the nucleus is the same from all direction

When ℓ = 0

As n increases s orbital

gets larger

Shape of s orbital

with different n

66

b) p orbitals

When ℓ = 1

dumbbell shaped

three p-orbitals px py and pz

correspond m of -1 0 and +1

As n increases the p-orbitals get larger

All p-orbitals have a node at the nucleus

Shape of Atomic Orbitals

67

Shape of Atomic Orbitals

shape four d orbitals have four lobes (perpendicular)

one d orbital has two major lobes along z axis

and a donut-shaped girdles the centre

When ℓ = 2

m = -2 -1012

the orbitals are dyz dxz dxy dx2-y2 dz2

68

69

Shape of Atomic Orbitals

70

Electronic Configuration

At the end of this topic students should

be able to-

Level No Learning Outcomes

C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos

Exclusion Principle

C3 23 b Predict the electronic configuration of atoms and

monotaomic ions using spdf notation

C3 23 c Justify the anomalous electronic configurations of

chromium and copper

71

Representing Electronic Configuration

Method 1 Orbital diagram

O8

1s 2s 2p

Method 2 spdf notation

O8 1s 2s 2p2 2 4

box

platform

Concentric circle

72

Rules for Assigning Electrons to Orbitals

i) Aufbau Principle

Electrons fill the lowest energy orbitals first and other

orbitals in order of ascending energy

The order of filling orbitals is

1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s

1s

2s

3s

4s

5s

2p

3p

4p

5p

3d

4d

5d

4f

5f

1s 2s 2p

73

Relative Energy Level of Atomic Orbitals

en

erg

y

n=1

n=2

n=3

n=4

1s

2s 2p

3s

4s

3p

4p

3d

4d

en

erg

y

n=1

n=2

n=3

n=4

1s

2s2p

3s

4s

3p

4p

3d

4d5s

Orbital energy levels

in the H atom

Orbital energy levels

in a many-electron atom

74

ii) Pauli Exclusion Principle

Rules for Assigning Electrons to Orbitals

No two electrons in an atom can have the same four

quantum numbers (n m s)

1s

a b c

e(a)

e(b)

e(c)

n ℓ m s

1 0

01

0

01

0

0

12

12

12

( )

)(

75

iii) Hundrsquos Rule

Rules for Assigning Electrons to Orbitals

Only when all the degenerate orbitals (a group of

orbitals of identical energy eg three p-orbitals and five d-

orbitals) contain an electron do the electrons begin to

occupy these orbitals in pairs The electrons in half-filled

orbitals have the same spins that is parallel spins

2p

76

Indicate which of the following orbital diagrams are

acceptable or unacceptable for an atom in ground state

Explain what mistakes have been made in each and draw

the correct orbital diagram

Exercise

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

77

Draw lsquoelectrons-in-boxesrsquo diagram of the electronic

configuration of titanium Ti (Z = 22) Also write the ground-

state electronic configurations for Ti and Ti2+ ion

Exercise

IMPORTANT

In an empty atom the 4s orbital has a lower

energy compared to that of the 3d orbital That is

why electrons fill the 4s orbital first before filling

the 3d orbital

However once electrons isare added to the 3d

orbital the 4s electrons are repelled to a higher

energy level The 3d orbitals now have lower

energy than 4s

78

79

Points to remember

The electronic configuration of atom or monatomic ion at

ground state

rArr Distribution of electrons obeys Aufbau principle Pauli

exclusion principle and Hundrsquos rule

Each atomic orbital can only accommodate a maximum of 2

electrons

Atomic orbital is a 3-D region in space around the nucleus

where there is a high probability of finding an electron

Assigning electrons to subshells

s-orbital rArr a max of 2 electrons (ns2)

p-orbitals rArr a max of 6 electrons (np6)

d-orbitals rArr a max of 10 electrons (nd10)

80

The Anomalous Electronic Configurations of

Cr and Cu

Cr and Cu have electron configurations which are

inconsistent with the Aufbau principle The anomalous

are explained on the basis that a filled or half-filled orbital

is more stable

Element Expected Observedactual

Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1

Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1

81

24Cr 18[Ar]

The actual orbital notation

24Cr 18[Ar]

Half filled orbital is more stable

(possesses an extra added stability)

3d 4s

3d 4s

Chromium predicted orbital notation

82

Copper predicted orbital notation

Cu [Ar]

The actual orbital notation

Cu [Ar]

4s3d

3d 4s

Full filled orbital is more stable

(possesses an extra added stability)

83

z = 21

z = 30

84

Write the ground-state electronic configuration and

explain the anomalous case for Cr (Z=24) and Cu ( Z=29)

Exercise

Writing Electronic Configuration for Negative Ion

Add electron according to Aufbau Principle

Example

i Cl-

ii O2-

Writing Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example

i Mg2+

i K+

i Fe2+

63

Shell nℓ

(ℓltn)

Orbital

notation

m

(-ℓ le m le +ℓ)

No of

degenerated

orbitals

K

L

M

2

1

3

Exercise Complete the following table

64

Exercise

State whether or not each of the following symbols is an

acceptable designation for an atomic orbital Explain what

is wrong with the unacceptable symbols

b) 6g

a) 2d

c) 7s

d) 5i

65

Shape of Atomic Orbitals

a) s orbitals

Spherical shape with the nucleus at the centre

The probability of finding electrons at the distance r from the nucleus is the same from all direction

When ℓ = 0

As n increases s orbital

gets larger

Shape of s orbital

with different n

66

b) p orbitals

When ℓ = 1

dumbbell shaped

three p-orbitals px py and pz

correspond m of -1 0 and +1

As n increases the p-orbitals get larger

All p-orbitals have a node at the nucleus

Shape of Atomic Orbitals

67

Shape of Atomic Orbitals

shape four d orbitals have four lobes (perpendicular)

one d orbital has two major lobes along z axis

and a donut-shaped girdles the centre

When ℓ = 2

m = -2 -1012

the orbitals are dyz dxz dxy dx2-y2 dz2

68

69

Shape of Atomic Orbitals

70

Electronic Configuration

At the end of this topic students should

be able to-

Level No Learning Outcomes

C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos

Exclusion Principle

C3 23 b Predict the electronic configuration of atoms and

monotaomic ions using spdf notation

C3 23 c Justify the anomalous electronic configurations of

chromium and copper

71

Representing Electronic Configuration

Method 1 Orbital diagram

O8

1s 2s 2p

Method 2 spdf notation

O8 1s 2s 2p2 2 4

box

platform

Concentric circle

72

Rules for Assigning Electrons to Orbitals

i) Aufbau Principle

Electrons fill the lowest energy orbitals first and other

orbitals in order of ascending energy

The order of filling orbitals is

1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s

1s

2s

3s

4s

5s

2p

3p

4p

5p

3d

4d

5d

4f

5f

1s 2s 2p

73

Relative Energy Level of Atomic Orbitals

en

erg

y

n=1

n=2

n=3

n=4

1s

2s 2p

3s

4s

3p

4p

3d

4d

en

erg

y

n=1

n=2

n=3

n=4

1s

2s2p

3s

4s

3p

4p

3d

4d5s

Orbital energy levels

in the H atom

Orbital energy levels

in a many-electron atom

74

ii) Pauli Exclusion Principle

Rules for Assigning Electrons to Orbitals

No two electrons in an atom can have the same four

quantum numbers (n m s)

1s

a b c

e(a)

e(b)

e(c)

n ℓ m s

1 0

01

0

01

0

0

12

12

12

( )

)(

75

iii) Hundrsquos Rule

Rules for Assigning Electrons to Orbitals

Only when all the degenerate orbitals (a group of

orbitals of identical energy eg three p-orbitals and five d-

orbitals) contain an electron do the electrons begin to

occupy these orbitals in pairs The electrons in half-filled

orbitals have the same spins that is parallel spins

2p

76

Indicate which of the following orbital diagrams are

acceptable or unacceptable for an atom in ground state

Explain what mistakes have been made in each and draw

the correct orbital diagram

Exercise

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

77

Draw lsquoelectrons-in-boxesrsquo diagram of the electronic

configuration of titanium Ti (Z = 22) Also write the ground-

state electronic configurations for Ti and Ti2+ ion

Exercise

IMPORTANT

In an empty atom the 4s orbital has a lower

energy compared to that of the 3d orbital That is

why electrons fill the 4s orbital first before filling

the 3d orbital

However once electrons isare added to the 3d

orbital the 4s electrons are repelled to a higher

energy level The 3d orbitals now have lower

energy than 4s

78

79

Points to remember

The electronic configuration of atom or monatomic ion at

ground state

rArr Distribution of electrons obeys Aufbau principle Pauli

exclusion principle and Hundrsquos rule

Each atomic orbital can only accommodate a maximum of 2

electrons

Atomic orbital is a 3-D region in space around the nucleus

where there is a high probability of finding an electron

Assigning electrons to subshells

s-orbital rArr a max of 2 electrons (ns2)

p-orbitals rArr a max of 6 electrons (np6)

d-orbitals rArr a max of 10 electrons (nd10)

80

The Anomalous Electronic Configurations of

Cr and Cu

Cr and Cu have electron configurations which are

inconsistent with the Aufbau principle The anomalous

are explained on the basis that a filled or half-filled orbital

is more stable

Element Expected Observedactual

Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1

Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1

81

24Cr 18[Ar]

The actual orbital notation

24Cr 18[Ar]

Half filled orbital is more stable

(possesses an extra added stability)

3d 4s

3d 4s

Chromium predicted orbital notation

82

Copper predicted orbital notation

Cu [Ar]

The actual orbital notation

Cu [Ar]

4s3d

3d 4s

Full filled orbital is more stable

(possesses an extra added stability)

83

z = 21

z = 30

84

Write the ground-state electronic configuration and

explain the anomalous case for Cr (Z=24) and Cu ( Z=29)

Exercise

Writing Electronic Configuration for Negative Ion

Add electron according to Aufbau Principle

Example

i Cl-

ii O2-

Writing Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example

i Mg2+

i K+

i Fe2+

64

Exercise

State whether or not each of the following symbols is an

acceptable designation for an atomic orbital Explain what

is wrong with the unacceptable symbols

b) 6g

a) 2d

c) 7s

d) 5i

65

Shape of Atomic Orbitals

a) s orbitals

Spherical shape with the nucleus at the centre

The probability of finding electrons at the distance r from the nucleus is the same from all direction

When ℓ = 0

As n increases s orbital

gets larger

Shape of s orbital

with different n

66

b) p orbitals

When ℓ = 1

dumbbell shaped

three p-orbitals px py and pz

correspond m of -1 0 and +1

As n increases the p-orbitals get larger

All p-orbitals have a node at the nucleus

Shape of Atomic Orbitals

67

Shape of Atomic Orbitals

shape four d orbitals have four lobes (perpendicular)

one d orbital has two major lobes along z axis

and a donut-shaped girdles the centre

When ℓ = 2

m = -2 -1012

the orbitals are dyz dxz dxy dx2-y2 dz2

68

69

Shape of Atomic Orbitals

70

Electronic Configuration

At the end of this topic students should

be able to-

Level No Learning Outcomes

C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos

Exclusion Principle

C3 23 b Predict the electronic configuration of atoms and

monotaomic ions using spdf notation

C3 23 c Justify the anomalous electronic configurations of

chromium and copper

71

Representing Electronic Configuration

Method 1 Orbital diagram

O8

1s 2s 2p

Method 2 spdf notation

O8 1s 2s 2p2 2 4

box

platform

Concentric circle

72

Rules for Assigning Electrons to Orbitals

i) Aufbau Principle

Electrons fill the lowest energy orbitals first and other

orbitals in order of ascending energy

The order of filling orbitals is

1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s

1s

2s

3s

4s

5s

2p

3p

4p

5p

3d

4d

5d

4f

5f

1s 2s 2p

73

Relative Energy Level of Atomic Orbitals

en

erg

y

n=1

n=2

n=3

n=4

1s

2s 2p

3s

4s

3p

4p

3d

4d

en

erg

y

n=1

n=2

n=3

n=4

1s

2s2p

3s

4s

3p

4p

3d

4d5s

Orbital energy levels

in the H atom

Orbital energy levels

in a many-electron atom

74

ii) Pauli Exclusion Principle

Rules for Assigning Electrons to Orbitals

No two electrons in an atom can have the same four

quantum numbers (n m s)

1s

a b c

e(a)

e(b)

e(c)

n ℓ m s

1 0

01

0

01

0

0

12

12

12

( )

)(

75

iii) Hundrsquos Rule

Rules for Assigning Electrons to Orbitals

Only when all the degenerate orbitals (a group of

orbitals of identical energy eg three p-orbitals and five d-

orbitals) contain an electron do the electrons begin to

occupy these orbitals in pairs The electrons in half-filled

orbitals have the same spins that is parallel spins

2p

76

Indicate which of the following orbital diagrams are

acceptable or unacceptable for an atom in ground state

Explain what mistakes have been made in each and draw

the correct orbital diagram

Exercise

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

77

Draw lsquoelectrons-in-boxesrsquo diagram of the electronic

configuration of titanium Ti (Z = 22) Also write the ground-

state electronic configurations for Ti and Ti2+ ion

Exercise

IMPORTANT

In an empty atom the 4s orbital has a lower

energy compared to that of the 3d orbital That is

why electrons fill the 4s orbital first before filling

the 3d orbital

However once electrons isare added to the 3d

orbital the 4s electrons are repelled to a higher

energy level The 3d orbitals now have lower

energy than 4s

78

79

Points to remember

The electronic configuration of atom or monatomic ion at

ground state

rArr Distribution of electrons obeys Aufbau principle Pauli

exclusion principle and Hundrsquos rule

Each atomic orbital can only accommodate a maximum of 2

electrons

Atomic orbital is a 3-D region in space around the nucleus

where there is a high probability of finding an electron

Assigning electrons to subshells

s-orbital rArr a max of 2 electrons (ns2)

p-orbitals rArr a max of 6 electrons (np6)

d-orbitals rArr a max of 10 electrons (nd10)

80

The Anomalous Electronic Configurations of

Cr and Cu

Cr and Cu have electron configurations which are

inconsistent with the Aufbau principle The anomalous

are explained on the basis that a filled or half-filled orbital

is more stable

Element Expected Observedactual

Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1

Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1

81

24Cr 18[Ar]

The actual orbital notation

24Cr 18[Ar]

Half filled orbital is more stable

(possesses an extra added stability)

3d 4s

3d 4s

Chromium predicted orbital notation

82

Copper predicted orbital notation

Cu [Ar]

The actual orbital notation

Cu [Ar]

4s3d

3d 4s

Full filled orbital is more stable

(possesses an extra added stability)

83

z = 21

z = 30

84

Write the ground-state electronic configuration and

explain the anomalous case for Cr (Z=24) and Cu ( Z=29)

Exercise

Writing Electronic Configuration for Negative Ion

Add electron according to Aufbau Principle

Example

i Cl-

ii O2-

Writing Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example

i Mg2+

i K+

i Fe2+

65

Shape of Atomic Orbitals

a) s orbitals

Spherical shape with the nucleus at the centre

The probability of finding electrons at the distance r from the nucleus is the same from all direction

When ℓ = 0

As n increases s orbital

gets larger

Shape of s orbital

with different n

66

b) p orbitals

When ℓ = 1

dumbbell shaped

three p-orbitals px py and pz

correspond m of -1 0 and +1

As n increases the p-orbitals get larger

All p-orbitals have a node at the nucleus

Shape of Atomic Orbitals

67

Shape of Atomic Orbitals

shape four d orbitals have four lobes (perpendicular)

one d orbital has two major lobes along z axis

and a donut-shaped girdles the centre

When ℓ = 2

m = -2 -1012

the orbitals are dyz dxz dxy dx2-y2 dz2

68

69

Shape of Atomic Orbitals

70

Electronic Configuration

At the end of this topic students should

be able to-

Level No Learning Outcomes

C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos

Exclusion Principle

C3 23 b Predict the electronic configuration of atoms and

monotaomic ions using spdf notation

C3 23 c Justify the anomalous electronic configurations of

chromium and copper

71

Representing Electronic Configuration

Method 1 Orbital diagram

O8

1s 2s 2p

Method 2 spdf notation

O8 1s 2s 2p2 2 4

box

platform

Concentric circle

72

Rules for Assigning Electrons to Orbitals

i) Aufbau Principle

Electrons fill the lowest energy orbitals first and other

orbitals in order of ascending energy

The order of filling orbitals is

1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s

1s

2s

3s

4s

5s

2p

3p

4p

5p

3d

4d

5d

4f

5f

1s 2s 2p

73

Relative Energy Level of Atomic Orbitals

en

erg

y

n=1

n=2

n=3

n=4

1s

2s 2p

3s

4s

3p

4p

3d

4d

en

erg

y

n=1

n=2

n=3

n=4

1s

2s2p

3s

4s

3p

4p

3d

4d5s

Orbital energy levels

in the H atom

Orbital energy levels

in a many-electron atom

74

ii) Pauli Exclusion Principle

Rules for Assigning Electrons to Orbitals

No two electrons in an atom can have the same four

quantum numbers (n m s)

1s

a b c

e(a)

e(b)

e(c)

n ℓ m s

1 0

01

0

01

0

0

12

12

12

( )

)(

75

iii) Hundrsquos Rule

Rules for Assigning Electrons to Orbitals

Only when all the degenerate orbitals (a group of

orbitals of identical energy eg three p-orbitals and five d-

orbitals) contain an electron do the electrons begin to

occupy these orbitals in pairs The electrons in half-filled

orbitals have the same spins that is parallel spins

2p

76

Indicate which of the following orbital diagrams are

acceptable or unacceptable for an atom in ground state

Explain what mistakes have been made in each and draw

the correct orbital diagram

Exercise

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

77

Draw lsquoelectrons-in-boxesrsquo diagram of the electronic

configuration of titanium Ti (Z = 22) Also write the ground-

state electronic configurations for Ti and Ti2+ ion

Exercise

IMPORTANT

In an empty atom the 4s orbital has a lower

energy compared to that of the 3d orbital That is

why electrons fill the 4s orbital first before filling

the 3d orbital

However once electrons isare added to the 3d

orbital the 4s electrons are repelled to a higher

energy level The 3d orbitals now have lower

energy than 4s

78

79

Points to remember

The electronic configuration of atom or monatomic ion at

ground state

rArr Distribution of electrons obeys Aufbau principle Pauli

exclusion principle and Hundrsquos rule

Each atomic orbital can only accommodate a maximum of 2

electrons

Atomic orbital is a 3-D region in space around the nucleus

where there is a high probability of finding an electron

Assigning electrons to subshells

s-orbital rArr a max of 2 electrons (ns2)

p-orbitals rArr a max of 6 electrons (np6)

d-orbitals rArr a max of 10 electrons (nd10)

80

The Anomalous Electronic Configurations of

Cr and Cu

Cr and Cu have electron configurations which are

inconsistent with the Aufbau principle The anomalous

are explained on the basis that a filled or half-filled orbital

is more stable

Element Expected Observedactual

Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1

Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1

81

24Cr 18[Ar]

The actual orbital notation

24Cr 18[Ar]

Half filled orbital is more stable

(possesses an extra added stability)

3d 4s

3d 4s

Chromium predicted orbital notation

82

Copper predicted orbital notation

Cu [Ar]

The actual orbital notation

Cu [Ar]

4s3d

3d 4s

Full filled orbital is more stable

(possesses an extra added stability)

83

z = 21

z = 30

84

Write the ground-state electronic configuration and

explain the anomalous case for Cr (Z=24) and Cu ( Z=29)

Exercise

Writing Electronic Configuration for Negative Ion

Add electron according to Aufbau Principle

Example

i Cl-

ii O2-

Writing Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example

i Mg2+

i K+

i Fe2+

66

b) p orbitals

When ℓ = 1

dumbbell shaped

three p-orbitals px py and pz

correspond m of -1 0 and +1

As n increases the p-orbitals get larger

All p-orbitals have a node at the nucleus

Shape of Atomic Orbitals

67

Shape of Atomic Orbitals

shape four d orbitals have four lobes (perpendicular)

one d orbital has two major lobes along z axis

and a donut-shaped girdles the centre

When ℓ = 2

m = -2 -1012

the orbitals are dyz dxz dxy dx2-y2 dz2

68

69

Shape of Atomic Orbitals

70

Electronic Configuration

At the end of this topic students should

be able to-

Level No Learning Outcomes

C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos

Exclusion Principle

C3 23 b Predict the electronic configuration of atoms and

monotaomic ions using spdf notation

C3 23 c Justify the anomalous electronic configurations of

chromium and copper

71

Representing Electronic Configuration

Method 1 Orbital diagram

O8

1s 2s 2p

Method 2 spdf notation

O8 1s 2s 2p2 2 4

box

platform

Concentric circle

72

Rules for Assigning Electrons to Orbitals

i) Aufbau Principle

Electrons fill the lowest energy orbitals first and other

orbitals in order of ascending energy

The order of filling orbitals is

1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s

1s

2s

3s

4s

5s

2p

3p

4p

5p

3d

4d

5d

4f

5f

1s 2s 2p

73

Relative Energy Level of Atomic Orbitals

en

erg

y

n=1

n=2

n=3

n=4

1s

2s 2p

3s

4s

3p

4p

3d

4d

en

erg

y

n=1

n=2

n=3

n=4

1s

2s2p

3s

4s

3p

4p

3d

4d5s

Orbital energy levels

in the H atom

Orbital energy levels

in a many-electron atom

74

ii) Pauli Exclusion Principle

Rules for Assigning Electrons to Orbitals

No two electrons in an atom can have the same four

quantum numbers (n m s)

1s

a b c

e(a)

e(b)

e(c)

n ℓ m s

1 0

01

0

01

0

0

12

12

12

( )

)(

75

iii) Hundrsquos Rule

Rules for Assigning Electrons to Orbitals

Only when all the degenerate orbitals (a group of

orbitals of identical energy eg three p-orbitals and five d-

orbitals) contain an electron do the electrons begin to

occupy these orbitals in pairs The electrons in half-filled

orbitals have the same spins that is parallel spins

2p

76

Indicate which of the following orbital diagrams are

acceptable or unacceptable for an atom in ground state

Explain what mistakes have been made in each and draw

the correct orbital diagram

Exercise

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

77

Draw lsquoelectrons-in-boxesrsquo diagram of the electronic

configuration of titanium Ti (Z = 22) Also write the ground-

state electronic configurations for Ti and Ti2+ ion

Exercise

IMPORTANT

In an empty atom the 4s orbital has a lower

energy compared to that of the 3d orbital That is

why electrons fill the 4s orbital first before filling

the 3d orbital

However once electrons isare added to the 3d

orbital the 4s electrons are repelled to a higher

energy level The 3d orbitals now have lower

energy than 4s

78

79

Points to remember

The electronic configuration of atom or monatomic ion at

ground state

rArr Distribution of electrons obeys Aufbau principle Pauli

exclusion principle and Hundrsquos rule

Each atomic orbital can only accommodate a maximum of 2

electrons

Atomic orbital is a 3-D region in space around the nucleus

where there is a high probability of finding an electron

Assigning electrons to subshells

s-orbital rArr a max of 2 electrons (ns2)

p-orbitals rArr a max of 6 electrons (np6)

d-orbitals rArr a max of 10 electrons (nd10)

80

The Anomalous Electronic Configurations of

Cr and Cu

Cr and Cu have electron configurations which are

inconsistent with the Aufbau principle The anomalous

are explained on the basis that a filled or half-filled orbital

is more stable

Element Expected Observedactual

Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1

Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1

81

24Cr 18[Ar]

The actual orbital notation

24Cr 18[Ar]

Half filled orbital is more stable

(possesses an extra added stability)

3d 4s

3d 4s

Chromium predicted orbital notation

82

Copper predicted orbital notation

Cu [Ar]

The actual orbital notation

Cu [Ar]

4s3d

3d 4s

Full filled orbital is more stable

(possesses an extra added stability)

83

z = 21

z = 30

84

Write the ground-state electronic configuration and

explain the anomalous case for Cr (Z=24) and Cu ( Z=29)

Exercise

Writing Electronic Configuration for Negative Ion

Add electron according to Aufbau Principle

Example

i Cl-

ii O2-

Writing Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example

i Mg2+

i K+

i Fe2+

67

Shape of Atomic Orbitals

shape four d orbitals have four lobes (perpendicular)

one d orbital has two major lobes along z axis

and a donut-shaped girdles the centre

When ℓ = 2

m = -2 -1012

the orbitals are dyz dxz dxy dx2-y2 dz2

68

69

Shape of Atomic Orbitals

70

Electronic Configuration

At the end of this topic students should

be able to-

Level No Learning Outcomes

C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos

Exclusion Principle

C3 23 b Predict the electronic configuration of atoms and

monotaomic ions using spdf notation

C3 23 c Justify the anomalous electronic configurations of

chromium and copper

71

Representing Electronic Configuration

Method 1 Orbital diagram

O8

1s 2s 2p

Method 2 spdf notation

O8 1s 2s 2p2 2 4

box

platform

Concentric circle

72

Rules for Assigning Electrons to Orbitals

i) Aufbau Principle

Electrons fill the lowest energy orbitals first and other

orbitals in order of ascending energy

The order of filling orbitals is

1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s

1s

2s

3s

4s

5s

2p

3p

4p

5p

3d

4d

5d

4f

5f

1s 2s 2p

73

Relative Energy Level of Atomic Orbitals

en

erg

y

n=1

n=2

n=3

n=4

1s

2s 2p

3s

4s

3p

4p

3d

4d

en

erg

y

n=1

n=2

n=3

n=4

1s

2s2p

3s

4s

3p

4p

3d

4d5s

Orbital energy levels

in the H atom

Orbital energy levels

in a many-electron atom

74

ii) Pauli Exclusion Principle

Rules for Assigning Electrons to Orbitals

No two electrons in an atom can have the same four

quantum numbers (n m s)

1s

a b c

e(a)

e(b)

e(c)

n ℓ m s

1 0

01

0

01

0

0

12

12

12

( )

)(

75

iii) Hundrsquos Rule

Rules for Assigning Electrons to Orbitals

Only when all the degenerate orbitals (a group of

orbitals of identical energy eg three p-orbitals and five d-

orbitals) contain an electron do the electrons begin to

occupy these orbitals in pairs The electrons in half-filled

orbitals have the same spins that is parallel spins

2p

76

Indicate which of the following orbital diagrams are

acceptable or unacceptable for an atom in ground state

Explain what mistakes have been made in each and draw

the correct orbital diagram

Exercise

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

77

Draw lsquoelectrons-in-boxesrsquo diagram of the electronic

configuration of titanium Ti (Z = 22) Also write the ground-

state electronic configurations for Ti and Ti2+ ion

Exercise

IMPORTANT

In an empty atom the 4s orbital has a lower

energy compared to that of the 3d orbital That is

why electrons fill the 4s orbital first before filling

the 3d orbital

However once electrons isare added to the 3d

orbital the 4s electrons are repelled to a higher

energy level The 3d orbitals now have lower

energy than 4s

78

79

Points to remember

The electronic configuration of atom or monatomic ion at

ground state

rArr Distribution of electrons obeys Aufbau principle Pauli

exclusion principle and Hundrsquos rule

Each atomic orbital can only accommodate a maximum of 2

electrons

Atomic orbital is a 3-D region in space around the nucleus

where there is a high probability of finding an electron

Assigning electrons to subshells

s-orbital rArr a max of 2 electrons (ns2)

p-orbitals rArr a max of 6 electrons (np6)

d-orbitals rArr a max of 10 electrons (nd10)

80

The Anomalous Electronic Configurations of

Cr and Cu

Cr and Cu have electron configurations which are

inconsistent with the Aufbau principle The anomalous

are explained on the basis that a filled or half-filled orbital

is more stable

Element Expected Observedactual

Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1

Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1

81

24Cr 18[Ar]

The actual orbital notation

24Cr 18[Ar]

Half filled orbital is more stable

(possesses an extra added stability)

3d 4s

3d 4s

Chromium predicted orbital notation

82

Copper predicted orbital notation

Cu [Ar]

The actual orbital notation

Cu [Ar]

4s3d

3d 4s

Full filled orbital is more stable

(possesses an extra added stability)

83

z = 21

z = 30

84

Write the ground-state electronic configuration and

explain the anomalous case for Cr (Z=24) and Cu ( Z=29)

Exercise

Writing Electronic Configuration for Negative Ion

Add electron according to Aufbau Principle

Example

i Cl-

ii O2-

Writing Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example

i Mg2+

i K+

i Fe2+

68

69

Shape of Atomic Orbitals

70

Electronic Configuration

At the end of this topic students should

be able to-

Level No Learning Outcomes

C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos

Exclusion Principle

C3 23 b Predict the electronic configuration of atoms and

monotaomic ions using spdf notation

C3 23 c Justify the anomalous electronic configurations of

chromium and copper

71

Representing Electronic Configuration

Method 1 Orbital diagram

O8

1s 2s 2p

Method 2 spdf notation

O8 1s 2s 2p2 2 4

box

platform

Concentric circle

72

Rules for Assigning Electrons to Orbitals

i) Aufbau Principle

Electrons fill the lowest energy orbitals first and other

orbitals in order of ascending energy

The order of filling orbitals is

1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s

1s

2s

3s

4s

5s

2p

3p

4p

5p

3d

4d

5d

4f

5f

1s 2s 2p

73

Relative Energy Level of Atomic Orbitals

en

erg

y

n=1

n=2

n=3

n=4

1s

2s 2p

3s

4s

3p

4p

3d

4d

en

erg

y

n=1

n=2

n=3

n=4

1s

2s2p

3s

4s

3p

4p

3d

4d5s

Orbital energy levels

in the H atom

Orbital energy levels

in a many-electron atom

74

ii) Pauli Exclusion Principle

Rules for Assigning Electrons to Orbitals

No two electrons in an atom can have the same four

quantum numbers (n m s)

1s

a b c

e(a)

e(b)

e(c)

n ℓ m s

1 0

01

0

01

0

0

12

12

12

( )

)(

75

iii) Hundrsquos Rule

Rules for Assigning Electrons to Orbitals

Only when all the degenerate orbitals (a group of

orbitals of identical energy eg three p-orbitals and five d-

orbitals) contain an electron do the electrons begin to

occupy these orbitals in pairs The electrons in half-filled

orbitals have the same spins that is parallel spins

2p

76

Indicate which of the following orbital diagrams are

acceptable or unacceptable for an atom in ground state

Explain what mistakes have been made in each and draw

the correct orbital diagram

Exercise

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

77

Draw lsquoelectrons-in-boxesrsquo diagram of the electronic

configuration of titanium Ti (Z = 22) Also write the ground-

state electronic configurations for Ti and Ti2+ ion

Exercise

IMPORTANT

In an empty atom the 4s orbital has a lower

energy compared to that of the 3d orbital That is

why electrons fill the 4s orbital first before filling

the 3d orbital

However once electrons isare added to the 3d

orbital the 4s electrons are repelled to a higher

energy level The 3d orbitals now have lower

energy than 4s

78

79

Points to remember

The electronic configuration of atom or monatomic ion at

ground state

rArr Distribution of electrons obeys Aufbau principle Pauli

exclusion principle and Hundrsquos rule

Each atomic orbital can only accommodate a maximum of 2

electrons

Atomic orbital is a 3-D region in space around the nucleus

where there is a high probability of finding an electron

Assigning electrons to subshells

s-orbital rArr a max of 2 electrons (ns2)

p-orbitals rArr a max of 6 electrons (np6)

d-orbitals rArr a max of 10 electrons (nd10)

80

The Anomalous Electronic Configurations of

Cr and Cu

Cr and Cu have electron configurations which are

inconsistent with the Aufbau principle The anomalous

are explained on the basis that a filled or half-filled orbital

is more stable

Element Expected Observedactual

Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1

Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1

81

24Cr 18[Ar]

The actual orbital notation

24Cr 18[Ar]

Half filled orbital is more stable

(possesses an extra added stability)

3d 4s

3d 4s

Chromium predicted orbital notation

82

Copper predicted orbital notation

Cu [Ar]

The actual orbital notation

Cu [Ar]

4s3d

3d 4s

Full filled orbital is more stable

(possesses an extra added stability)

83

z = 21

z = 30

84

Write the ground-state electronic configuration and

explain the anomalous case for Cr (Z=24) and Cu ( Z=29)

Exercise

Writing Electronic Configuration for Negative Ion

Add electron according to Aufbau Principle

Example

i Cl-

ii O2-

Writing Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example

i Mg2+

i K+

i Fe2+

69

Shape of Atomic Orbitals

70

Electronic Configuration

At the end of this topic students should

be able to-

Level No Learning Outcomes

C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos

Exclusion Principle

C3 23 b Predict the electronic configuration of atoms and

monotaomic ions using spdf notation

C3 23 c Justify the anomalous electronic configurations of

chromium and copper

71

Representing Electronic Configuration

Method 1 Orbital diagram

O8

1s 2s 2p

Method 2 spdf notation

O8 1s 2s 2p2 2 4

box

platform

Concentric circle

72

Rules for Assigning Electrons to Orbitals

i) Aufbau Principle

Electrons fill the lowest energy orbitals first and other

orbitals in order of ascending energy

The order of filling orbitals is

1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s

1s

2s

3s

4s

5s

2p

3p

4p

5p

3d

4d

5d

4f

5f

1s 2s 2p

73

Relative Energy Level of Atomic Orbitals

en

erg

y

n=1

n=2

n=3

n=4

1s

2s 2p

3s

4s

3p

4p

3d

4d

en

erg

y

n=1

n=2

n=3

n=4

1s

2s2p

3s

4s

3p

4p

3d

4d5s

Orbital energy levels

in the H atom

Orbital energy levels

in a many-electron atom

74

ii) Pauli Exclusion Principle

Rules for Assigning Electrons to Orbitals

No two electrons in an atom can have the same four

quantum numbers (n m s)

1s

a b c

e(a)

e(b)

e(c)

n ℓ m s

1 0

01

0

01

0

0

12

12

12

( )

)(

75

iii) Hundrsquos Rule

Rules for Assigning Electrons to Orbitals

Only when all the degenerate orbitals (a group of

orbitals of identical energy eg three p-orbitals and five d-

orbitals) contain an electron do the electrons begin to

occupy these orbitals in pairs The electrons in half-filled

orbitals have the same spins that is parallel spins

2p

76

Indicate which of the following orbital diagrams are

acceptable or unacceptable for an atom in ground state

Explain what mistakes have been made in each and draw

the correct orbital diagram

Exercise

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

77

Draw lsquoelectrons-in-boxesrsquo diagram of the electronic

configuration of titanium Ti (Z = 22) Also write the ground-

state electronic configurations for Ti and Ti2+ ion

Exercise

IMPORTANT

In an empty atom the 4s orbital has a lower

energy compared to that of the 3d orbital That is

why electrons fill the 4s orbital first before filling

the 3d orbital

However once electrons isare added to the 3d

orbital the 4s electrons are repelled to a higher

energy level The 3d orbitals now have lower

energy than 4s

78

79

Points to remember

The electronic configuration of atom or monatomic ion at

ground state

rArr Distribution of electrons obeys Aufbau principle Pauli

exclusion principle and Hundrsquos rule

Each atomic orbital can only accommodate a maximum of 2

electrons

Atomic orbital is a 3-D region in space around the nucleus

where there is a high probability of finding an electron

Assigning electrons to subshells

s-orbital rArr a max of 2 electrons (ns2)

p-orbitals rArr a max of 6 electrons (np6)

d-orbitals rArr a max of 10 electrons (nd10)

80

The Anomalous Electronic Configurations of

Cr and Cu

Cr and Cu have electron configurations which are

inconsistent with the Aufbau principle The anomalous

are explained on the basis that a filled or half-filled orbital

is more stable

Element Expected Observedactual

Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1

Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1

81

24Cr 18[Ar]

The actual orbital notation

24Cr 18[Ar]

Half filled orbital is more stable

(possesses an extra added stability)

3d 4s

3d 4s

Chromium predicted orbital notation

82

Copper predicted orbital notation

Cu [Ar]

The actual orbital notation

Cu [Ar]

4s3d

3d 4s

Full filled orbital is more stable

(possesses an extra added stability)

83

z = 21

z = 30

84

Write the ground-state electronic configuration and

explain the anomalous case for Cr (Z=24) and Cu ( Z=29)

Exercise

Writing Electronic Configuration for Negative Ion

Add electron according to Aufbau Principle

Example

i Cl-

ii O2-

Writing Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example

i Mg2+

i K+

i Fe2+

70

Electronic Configuration

At the end of this topic students should

be able to-

Level No Learning Outcomes

C3 23 a Explain Aufbau principle Hundrsquos rule and Paulirsquos

Exclusion Principle

C3 23 b Predict the electronic configuration of atoms and

monotaomic ions using spdf notation

C3 23 c Justify the anomalous electronic configurations of

chromium and copper

71

Representing Electronic Configuration

Method 1 Orbital diagram

O8

1s 2s 2p

Method 2 spdf notation

O8 1s 2s 2p2 2 4

box

platform

Concentric circle

72

Rules for Assigning Electrons to Orbitals

i) Aufbau Principle

Electrons fill the lowest energy orbitals first and other

orbitals in order of ascending energy

The order of filling orbitals is

1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s

1s

2s

3s

4s

5s

2p

3p

4p

5p

3d

4d

5d

4f

5f

1s 2s 2p

73

Relative Energy Level of Atomic Orbitals

en

erg

y

n=1

n=2

n=3

n=4

1s

2s 2p

3s

4s

3p

4p

3d

4d

en

erg

y

n=1

n=2

n=3

n=4

1s

2s2p

3s

4s

3p

4p

3d

4d5s

Orbital energy levels

in the H atom

Orbital energy levels

in a many-electron atom

74

ii) Pauli Exclusion Principle

Rules for Assigning Electrons to Orbitals

No two electrons in an atom can have the same four

quantum numbers (n m s)

1s

a b c

e(a)

e(b)

e(c)

n ℓ m s

1 0

01

0

01

0

0

12

12

12

( )

)(

75

iii) Hundrsquos Rule

Rules for Assigning Electrons to Orbitals

Only when all the degenerate orbitals (a group of

orbitals of identical energy eg three p-orbitals and five d-

orbitals) contain an electron do the electrons begin to

occupy these orbitals in pairs The electrons in half-filled

orbitals have the same spins that is parallel spins

2p

76

Indicate which of the following orbital diagrams are

acceptable or unacceptable for an atom in ground state

Explain what mistakes have been made in each and draw

the correct orbital diagram

Exercise

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

77

Draw lsquoelectrons-in-boxesrsquo diagram of the electronic

configuration of titanium Ti (Z = 22) Also write the ground-

state electronic configurations for Ti and Ti2+ ion

Exercise

IMPORTANT

In an empty atom the 4s orbital has a lower

energy compared to that of the 3d orbital That is

why electrons fill the 4s orbital first before filling

the 3d orbital

However once electrons isare added to the 3d

orbital the 4s electrons are repelled to a higher

energy level The 3d orbitals now have lower

energy than 4s

78

79

Points to remember

The electronic configuration of atom or monatomic ion at

ground state

rArr Distribution of electrons obeys Aufbau principle Pauli

exclusion principle and Hundrsquos rule

Each atomic orbital can only accommodate a maximum of 2

electrons

Atomic orbital is a 3-D region in space around the nucleus

where there is a high probability of finding an electron

Assigning electrons to subshells

s-orbital rArr a max of 2 electrons (ns2)

p-orbitals rArr a max of 6 electrons (np6)

d-orbitals rArr a max of 10 electrons (nd10)

80

The Anomalous Electronic Configurations of

Cr and Cu

Cr and Cu have electron configurations which are

inconsistent with the Aufbau principle The anomalous

are explained on the basis that a filled or half-filled orbital

is more stable

Element Expected Observedactual

Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1

Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1

81

24Cr 18[Ar]

The actual orbital notation

24Cr 18[Ar]

Half filled orbital is more stable

(possesses an extra added stability)

3d 4s

3d 4s

Chromium predicted orbital notation

82

Copper predicted orbital notation

Cu [Ar]

The actual orbital notation

Cu [Ar]

4s3d

3d 4s

Full filled orbital is more stable

(possesses an extra added stability)

83

z = 21

z = 30

84

Write the ground-state electronic configuration and

explain the anomalous case for Cr (Z=24) and Cu ( Z=29)

Exercise

Writing Electronic Configuration for Negative Ion

Add electron according to Aufbau Principle

Example

i Cl-

ii O2-

Writing Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example

i Mg2+

i K+

i Fe2+

71

Representing Electronic Configuration

Method 1 Orbital diagram

O8

1s 2s 2p

Method 2 spdf notation

O8 1s 2s 2p2 2 4

box

platform

Concentric circle

72

Rules for Assigning Electrons to Orbitals

i) Aufbau Principle

Electrons fill the lowest energy orbitals first and other

orbitals in order of ascending energy

The order of filling orbitals is

1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s

1s

2s

3s

4s

5s

2p

3p

4p

5p

3d

4d

5d

4f

5f

1s 2s 2p

73

Relative Energy Level of Atomic Orbitals

en

erg

y

n=1

n=2

n=3

n=4

1s

2s 2p

3s

4s

3p

4p

3d

4d

en

erg

y

n=1

n=2

n=3

n=4

1s

2s2p

3s

4s

3p

4p

3d

4d5s

Orbital energy levels

in the H atom

Orbital energy levels

in a many-electron atom

74

ii) Pauli Exclusion Principle

Rules for Assigning Electrons to Orbitals

No two electrons in an atom can have the same four

quantum numbers (n m s)

1s

a b c

e(a)

e(b)

e(c)

n ℓ m s

1 0

01

0

01

0

0

12

12

12

( )

)(

75

iii) Hundrsquos Rule

Rules for Assigning Electrons to Orbitals

Only when all the degenerate orbitals (a group of

orbitals of identical energy eg three p-orbitals and five d-

orbitals) contain an electron do the electrons begin to

occupy these orbitals in pairs The electrons in half-filled

orbitals have the same spins that is parallel spins

2p

76

Indicate which of the following orbital diagrams are

acceptable or unacceptable for an atom in ground state

Explain what mistakes have been made in each and draw

the correct orbital diagram

Exercise

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

77

Draw lsquoelectrons-in-boxesrsquo diagram of the electronic

configuration of titanium Ti (Z = 22) Also write the ground-

state electronic configurations for Ti and Ti2+ ion

Exercise

IMPORTANT

In an empty atom the 4s orbital has a lower

energy compared to that of the 3d orbital That is

why electrons fill the 4s orbital first before filling

the 3d orbital

However once electrons isare added to the 3d

orbital the 4s electrons are repelled to a higher

energy level The 3d orbitals now have lower

energy than 4s

78

79

Points to remember

The electronic configuration of atom or monatomic ion at

ground state

rArr Distribution of electrons obeys Aufbau principle Pauli

exclusion principle and Hundrsquos rule

Each atomic orbital can only accommodate a maximum of 2

electrons

Atomic orbital is a 3-D region in space around the nucleus

where there is a high probability of finding an electron

Assigning electrons to subshells

s-orbital rArr a max of 2 electrons (ns2)

p-orbitals rArr a max of 6 electrons (np6)

d-orbitals rArr a max of 10 electrons (nd10)

80

The Anomalous Electronic Configurations of

Cr and Cu

Cr and Cu have electron configurations which are

inconsistent with the Aufbau principle The anomalous

are explained on the basis that a filled or half-filled orbital

is more stable

Element Expected Observedactual

Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1

Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1

81

24Cr 18[Ar]

The actual orbital notation

24Cr 18[Ar]

Half filled orbital is more stable

(possesses an extra added stability)

3d 4s

3d 4s

Chromium predicted orbital notation

82

Copper predicted orbital notation

Cu [Ar]

The actual orbital notation

Cu [Ar]

4s3d

3d 4s

Full filled orbital is more stable

(possesses an extra added stability)

83

z = 21

z = 30

84

Write the ground-state electronic configuration and

explain the anomalous case for Cr (Z=24) and Cu ( Z=29)

Exercise

Writing Electronic Configuration for Negative Ion

Add electron according to Aufbau Principle

Example

i Cl-

ii O2-

Writing Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example

i Mg2+

i K+

i Fe2+

72

Rules for Assigning Electrons to Orbitals

i) Aufbau Principle

Electrons fill the lowest energy orbitals first and other

orbitals in order of ascending energy

The order of filling orbitals is

1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s

1s

2s

3s

4s

5s

2p

3p

4p

5p

3d

4d

5d

4f

5f

1s 2s 2p

73

Relative Energy Level of Atomic Orbitals

en

erg

y

n=1

n=2

n=3

n=4

1s

2s 2p

3s

4s

3p

4p

3d

4d

en

erg

y

n=1

n=2

n=3

n=4

1s

2s2p

3s

4s

3p

4p

3d

4d5s

Orbital energy levels

in the H atom

Orbital energy levels

in a many-electron atom

74

ii) Pauli Exclusion Principle

Rules for Assigning Electrons to Orbitals

No two electrons in an atom can have the same four

quantum numbers (n m s)

1s

a b c

e(a)

e(b)

e(c)

n ℓ m s

1 0

01

0

01

0

0

12

12

12

( )

)(

75

iii) Hundrsquos Rule

Rules for Assigning Electrons to Orbitals

Only when all the degenerate orbitals (a group of

orbitals of identical energy eg three p-orbitals and five d-

orbitals) contain an electron do the electrons begin to

occupy these orbitals in pairs The electrons in half-filled

orbitals have the same spins that is parallel spins

2p

76

Indicate which of the following orbital diagrams are

acceptable or unacceptable for an atom in ground state

Explain what mistakes have been made in each and draw

the correct orbital diagram

Exercise

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

77

Draw lsquoelectrons-in-boxesrsquo diagram of the electronic

configuration of titanium Ti (Z = 22) Also write the ground-

state electronic configurations for Ti and Ti2+ ion

Exercise

IMPORTANT

In an empty atom the 4s orbital has a lower

energy compared to that of the 3d orbital That is

why electrons fill the 4s orbital first before filling

the 3d orbital

However once electrons isare added to the 3d

orbital the 4s electrons are repelled to a higher

energy level The 3d orbitals now have lower

energy than 4s

78

79

Points to remember

The electronic configuration of atom or monatomic ion at

ground state

rArr Distribution of electrons obeys Aufbau principle Pauli

exclusion principle and Hundrsquos rule

Each atomic orbital can only accommodate a maximum of 2

electrons

Atomic orbital is a 3-D region in space around the nucleus

where there is a high probability of finding an electron

Assigning electrons to subshells

s-orbital rArr a max of 2 electrons (ns2)

p-orbitals rArr a max of 6 electrons (np6)

d-orbitals rArr a max of 10 electrons (nd10)

80

The Anomalous Electronic Configurations of

Cr and Cu

Cr and Cu have electron configurations which are

inconsistent with the Aufbau principle The anomalous

are explained on the basis that a filled or half-filled orbital

is more stable

Element Expected Observedactual

Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1

Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1

81

24Cr 18[Ar]

The actual orbital notation

24Cr 18[Ar]

Half filled orbital is more stable

(possesses an extra added stability)

3d 4s

3d 4s

Chromium predicted orbital notation

82

Copper predicted orbital notation

Cu [Ar]

The actual orbital notation

Cu [Ar]

4s3d

3d 4s

Full filled orbital is more stable

(possesses an extra added stability)

83

z = 21

z = 30

84

Write the ground-state electronic configuration and

explain the anomalous case for Cr (Z=24) and Cu ( Z=29)

Exercise

Writing Electronic Configuration for Negative Ion

Add electron according to Aufbau Principle

Example

i Cl-

ii O2-

Writing Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example

i Mg2+

i K+

i Fe2+

73

Relative Energy Level of Atomic Orbitals

en

erg

y

n=1

n=2

n=3

n=4

1s

2s 2p

3s

4s

3p

4p

3d

4d

en

erg

y

n=1

n=2

n=3

n=4

1s

2s2p

3s

4s

3p

4p

3d

4d5s

Orbital energy levels

in the H atom

Orbital energy levels

in a many-electron atom

74

ii) Pauli Exclusion Principle

Rules for Assigning Electrons to Orbitals

No two electrons in an atom can have the same four

quantum numbers (n m s)

1s

a b c

e(a)

e(b)

e(c)

n ℓ m s

1 0

01

0

01

0

0

12

12

12

( )

)(

75

iii) Hundrsquos Rule

Rules for Assigning Electrons to Orbitals

Only when all the degenerate orbitals (a group of

orbitals of identical energy eg three p-orbitals and five d-

orbitals) contain an electron do the electrons begin to

occupy these orbitals in pairs The electrons in half-filled

orbitals have the same spins that is parallel spins

2p

76

Indicate which of the following orbital diagrams are

acceptable or unacceptable for an atom in ground state

Explain what mistakes have been made in each and draw

the correct orbital diagram

Exercise

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

77

Draw lsquoelectrons-in-boxesrsquo diagram of the electronic

configuration of titanium Ti (Z = 22) Also write the ground-

state electronic configurations for Ti and Ti2+ ion

Exercise

IMPORTANT

In an empty atom the 4s orbital has a lower

energy compared to that of the 3d orbital That is

why electrons fill the 4s orbital first before filling

the 3d orbital

However once electrons isare added to the 3d

orbital the 4s electrons are repelled to a higher

energy level The 3d orbitals now have lower

energy than 4s

78

79

Points to remember

The electronic configuration of atom or monatomic ion at

ground state

rArr Distribution of electrons obeys Aufbau principle Pauli

exclusion principle and Hundrsquos rule

Each atomic orbital can only accommodate a maximum of 2

electrons

Atomic orbital is a 3-D region in space around the nucleus

where there is a high probability of finding an electron

Assigning electrons to subshells

s-orbital rArr a max of 2 electrons (ns2)

p-orbitals rArr a max of 6 electrons (np6)

d-orbitals rArr a max of 10 electrons (nd10)

80

The Anomalous Electronic Configurations of

Cr and Cu

Cr and Cu have electron configurations which are

inconsistent with the Aufbau principle The anomalous

are explained on the basis that a filled or half-filled orbital

is more stable

Element Expected Observedactual

Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1

Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1

81

24Cr 18[Ar]

The actual orbital notation

24Cr 18[Ar]

Half filled orbital is more stable

(possesses an extra added stability)

3d 4s

3d 4s

Chromium predicted orbital notation

82

Copper predicted orbital notation

Cu [Ar]

The actual orbital notation

Cu [Ar]

4s3d

3d 4s

Full filled orbital is more stable

(possesses an extra added stability)

83

z = 21

z = 30

84

Write the ground-state electronic configuration and

explain the anomalous case for Cr (Z=24) and Cu ( Z=29)

Exercise

Writing Electronic Configuration for Negative Ion

Add electron according to Aufbau Principle

Example

i Cl-

ii O2-

Writing Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example

i Mg2+

i K+

i Fe2+

74

ii) Pauli Exclusion Principle

Rules for Assigning Electrons to Orbitals

No two electrons in an atom can have the same four

quantum numbers (n m s)

1s

a b c

e(a)

e(b)

e(c)

n ℓ m s

1 0

01

0

01

0

0

12

12

12

( )

)(

75

iii) Hundrsquos Rule

Rules for Assigning Electrons to Orbitals

Only when all the degenerate orbitals (a group of

orbitals of identical energy eg three p-orbitals and five d-

orbitals) contain an electron do the electrons begin to

occupy these orbitals in pairs The electrons in half-filled

orbitals have the same spins that is parallel spins

2p

76

Indicate which of the following orbital diagrams are

acceptable or unacceptable for an atom in ground state

Explain what mistakes have been made in each and draw

the correct orbital diagram

Exercise

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

77

Draw lsquoelectrons-in-boxesrsquo diagram of the electronic

configuration of titanium Ti (Z = 22) Also write the ground-

state electronic configurations for Ti and Ti2+ ion

Exercise

IMPORTANT

In an empty atom the 4s orbital has a lower

energy compared to that of the 3d orbital That is

why electrons fill the 4s orbital first before filling

the 3d orbital

However once electrons isare added to the 3d

orbital the 4s electrons are repelled to a higher

energy level The 3d orbitals now have lower

energy than 4s

78

79

Points to remember

The electronic configuration of atom or monatomic ion at

ground state

rArr Distribution of electrons obeys Aufbau principle Pauli

exclusion principle and Hundrsquos rule

Each atomic orbital can only accommodate a maximum of 2

electrons

Atomic orbital is a 3-D region in space around the nucleus

where there is a high probability of finding an electron

Assigning electrons to subshells

s-orbital rArr a max of 2 electrons (ns2)

p-orbitals rArr a max of 6 electrons (np6)

d-orbitals rArr a max of 10 electrons (nd10)

80

The Anomalous Electronic Configurations of

Cr and Cu

Cr and Cu have electron configurations which are

inconsistent with the Aufbau principle The anomalous

are explained on the basis that a filled or half-filled orbital

is more stable

Element Expected Observedactual

Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1

Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1

81

24Cr 18[Ar]

The actual orbital notation

24Cr 18[Ar]

Half filled orbital is more stable

(possesses an extra added stability)

3d 4s

3d 4s

Chromium predicted orbital notation

82

Copper predicted orbital notation

Cu [Ar]

The actual orbital notation

Cu [Ar]

4s3d

3d 4s

Full filled orbital is more stable

(possesses an extra added stability)

83

z = 21

z = 30

84

Write the ground-state electronic configuration and

explain the anomalous case for Cr (Z=24) and Cu ( Z=29)

Exercise

Writing Electronic Configuration for Negative Ion

Add electron according to Aufbau Principle

Example

i Cl-

ii O2-

Writing Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example

i Mg2+

i K+

i Fe2+

75

iii) Hundrsquos Rule

Rules for Assigning Electrons to Orbitals

Only when all the degenerate orbitals (a group of

orbitals of identical energy eg three p-orbitals and five d-

orbitals) contain an electron do the electrons begin to

occupy these orbitals in pairs The electrons in half-filled

orbitals have the same spins that is parallel spins

2p

76

Indicate which of the following orbital diagrams are

acceptable or unacceptable for an atom in ground state

Explain what mistakes have been made in each and draw

the correct orbital diagram

Exercise

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

77

Draw lsquoelectrons-in-boxesrsquo diagram of the electronic

configuration of titanium Ti (Z = 22) Also write the ground-

state electronic configurations for Ti and Ti2+ ion

Exercise

IMPORTANT

In an empty atom the 4s orbital has a lower

energy compared to that of the 3d orbital That is

why electrons fill the 4s orbital first before filling

the 3d orbital

However once electrons isare added to the 3d

orbital the 4s electrons are repelled to a higher

energy level The 3d orbitals now have lower

energy than 4s

78

79

Points to remember

The electronic configuration of atom or monatomic ion at

ground state

rArr Distribution of electrons obeys Aufbau principle Pauli

exclusion principle and Hundrsquos rule

Each atomic orbital can only accommodate a maximum of 2

electrons

Atomic orbital is a 3-D region in space around the nucleus

where there is a high probability of finding an electron

Assigning electrons to subshells

s-orbital rArr a max of 2 electrons (ns2)

p-orbitals rArr a max of 6 electrons (np6)

d-orbitals rArr a max of 10 electrons (nd10)

80

The Anomalous Electronic Configurations of

Cr and Cu

Cr and Cu have electron configurations which are

inconsistent with the Aufbau principle The anomalous

are explained on the basis that a filled or half-filled orbital

is more stable

Element Expected Observedactual

Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1

Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1

81

24Cr 18[Ar]

The actual orbital notation

24Cr 18[Ar]

Half filled orbital is more stable

(possesses an extra added stability)

3d 4s

3d 4s

Chromium predicted orbital notation

82

Copper predicted orbital notation

Cu [Ar]

The actual orbital notation

Cu [Ar]

4s3d

3d 4s

Full filled orbital is more stable

(possesses an extra added stability)

83

z = 21

z = 30

84

Write the ground-state electronic configuration and

explain the anomalous case for Cr (Z=24) and Cu ( Z=29)

Exercise

Writing Electronic Configuration for Negative Ion

Add electron according to Aufbau Principle

Example

i Cl-

ii O2-

Writing Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example

i Mg2+

i K+

i Fe2+

76

Indicate which of the following orbital diagrams are

acceptable or unacceptable for an atom in ground state

Explain what mistakes have been made in each and draw

the correct orbital diagram

Exercise

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

77

Draw lsquoelectrons-in-boxesrsquo diagram of the electronic

configuration of titanium Ti (Z = 22) Also write the ground-

state electronic configurations for Ti and Ti2+ ion

Exercise

IMPORTANT

In an empty atom the 4s orbital has a lower

energy compared to that of the 3d orbital That is

why electrons fill the 4s orbital first before filling

the 3d orbital

However once electrons isare added to the 3d

orbital the 4s electrons are repelled to a higher

energy level The 3d orbitals now have lower

energy than 4s

78

79

Points to remember

The electronic configuration of atom or monatomic ion at

ground state

rArr Distribution of electrons obeys Aufbau principle Pauli

exclusion principle and Hundrsquos rule

Each atomic orbital can only accommodate a maximum of 2

electrons

Atomic orbital is a 3-D region in space around the nucleus

where there is a high probability of finding an electron

Assigning electrons to subshells

s-orbital rArr a max of 2 electrons (ns2)

p-orbitals rArr a max of 6 electrons (np6)

d-orbitals rArr a max of 10 electrons (nd10)

80

The Anomalous Electronic Configurations of

Cr and Cu

Cr and Cu have electron configurations which are

inconsistent with the Aufbau principle The anomalous

are explained on the basis that a filled or half-filled orbital

is more stable

Element Expected Observedactual

Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1

Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1

81

24Cr 18[Ar]

The actual orbital notation

24Cr 18[Ar]

Half filled orbital is more stable

(possesses an extra added stability)

3d 4s

3d 4s

Chromium predicted orbital notation

82

Copper predicted orbital notation

Cu [Ar]

The actual orbital notation

Cu [Ar]

4s3d

3d 4s

Full filled orbital is more stable

(possesses an extra added stability)

83

z = 21

z = 30

84

Write the ground-state electronic configuration and

explain the anomalous case for Cr (Z=24) and Cu ( Z=29)

Exercise

Writing Electronic Configuration for Negative Ion

Add electron according to Aufbau Principle

Example

i Cl-

ii O2-

Writing Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example

i Mg2+

i K+

i Fe2+

77

Draw lsquoelectrons-in-boxesrsquo diagram of the electronic

configuration of titanium Ti (Z = 22) Also write the ground-

state electronic configurations for Ti and Ti2+ ion

Exercise

IMPORTANT

In an empty atom the 4s orbital has a lower

energy compared to that of the 3d orbital That is

why electrons fill the 4s orbital first before filling

the 3d orbital

However once electrons isare added to the 3d

orbital the 4s electrons are repelled to a higher

energy level The 3d orbitals now have lower

energy than 4s

78

79

Points to remember

The electronic configuration of atom or monatomic ion at

ground state

rArr Distribution of electrons obeys Aufbau principle Pauli

exclusion principle and Hundrsquos rule

Each atomic orbital can only accommodate a maximum of 2

electrons

Atomic orbital is a 3-D region in space around the nucleus

where there is a high probability of finding an electron

Assigning electrons to subshells

s-orbital rArr a max of 2 electrons (ns2)

p-orbitals rArr a max of 6 electrons (np6)

d-orbitals rArr a max of 10 electrons (nd10)

80

The Anomalous Electronic Configurations of

Cr and Cu

Cr and Cu have electron configurations which are

inconsistent with the Aufbau principle The anomalous

are explained on the basis that a filled or half-filled orbital

is more stable

Element Expected Observedactual

Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1

Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1

81

24Cr 18[Ar]

The actual orbital notation

24Cr 18[Ar]

Half filled orbital is more stable

(possesses an extra added stability)

3d 4s

3d 4s

Chromium predicted orbital notation

82

Copper predicted orbital notation

Cu [Ar]

The actual orbital notation

Cu [Ar]

4s3d

3d 4s

Full filled orbital is more stable

(possesses an extra added stability)

83

z = 21

z = 30

84

Write the ground-state electronic configuration and

explain the anomalous case for Cr (Z=24) and Cu ( Z=29)

Exercise

Writing Electronic Configuration for Negative Ion

Add electron according to Aufbau Principle

Example

i Cl-

ii O2-

Writing Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example

i Mg2+

i K+

i Fe2+

IMPORTANT

In an empty atom the 4s orbital has a lower

energy compared to that of the 3d orbital That is

why electrons fill the 4s orbital first before filling

the 3d orbital

However once electrons isare added to the 3d

orbital the 4s electrons are repelled to a higher

energy level The 3d orbitals now have lower

energy than 4s

78

79

Points to remember

The electronic configuration of atom or monatomic ion at

ground state

rArr Distribution of electrons obeys Aufbau principle Pauli

exclusion principle and Hundrsquos rule

Each atomic orbital can only accommodate a maximum of 2

electrons

Atomic orbital is a 3-D region in space around the nucleus

where there is a high probability of finding an electron

Assigning electrons to subshells

s-orbital rArr a max of 2 electrons (ns2)

p-orbitals rArr a max of 6 electrons (np6)

d-orbitals rArr a max of 10 electrons (nd10)

80

The Anomalous Electronic Configurations of

Cr and Cu

Cr and Cu have electron configurations which are

inconsistent with the Aufbau principle The anomalous

are explained on the basis that a filled or half-filled orbital

is more stable

Element Expected Observedactual

Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1

Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1

81

24Cr 18[Ar]

The actual orbital notation

24Cr 18[Ar]

Half filled orbital is more stable

(possesses an extra added stability)

3d 4s

3d 4s

Chromium predicted orbital notation

82

Copper predicted orbital notation

Cu [Ar]

The actual orbital notation

Cu [Ar]

4s3d

3d 4s

Full filled orbital is more stable

(possesses an extra added stability)

83

z = 21

z = 30

84

Write the ground-state electronic configuration and

explain the anomalous case for Cr (Z=24) and Cu ( Z=29)

Exercise

Writing Electronic Configuration for Negative Ion

Add electron according to Aufbau Principle

Example

i Cl-

ii O2-

Writing Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example

i Mg2+

i K+

i Fe2+

79

Points to remember

The electronic configuration of atom or monatomic ion at

ground state

rArr Distribution of electrons obeys Aufbau principle Pauli

exclusion principle and Hundrsquos rule

Each atomic orbital can only accommodate a maximum of 2

electrons

Atomic orbital is a 3-D region in space around the nucleus

where there is a high probability of finding an electron

Assigning electrons to subshells

s-orbital rArr a max of 2 electrons (ns2)

p-orbitals rArr a max of 6 electrons (np6)

d-orbitals rArr a max of 10 electrons (nd10)

80

The Anomalous Electronic Configurations of

Cr and Cu

Cr and Cu have electron configurations which are

inconsistent with the Aufbau principle The anomalous

are explained on the basis that a filled or half-filled orbital

is more stable

Element Expected Observedactual

Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1

Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1

81

24Cr 18[Ar]

The actual orbital notation

24Cr 18[Ar]

Half filled orbital is more stable

(possesses an extra added stability)

3d 4s

3d 4s

Chromium predicted orbital notation

82

Copper predicted orbital notation

Cu [Ar]

The actual orbital notation

Cu [Ar]

4s3d

3d 4s

Full filled orbital is more stable

(possesses an extra added stability)

83

z = 21

z = 30

84

Write the ground-state electronic configuration and

explain the anomalous case for Cr (Z=24) and Cu ( Z=29)

Exercise

Writing Electronic Configuration for Negative Ion

Add electron according to Aufbau Principle

Example

i Cl-

ii O2-

Writing Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example

i Mg2+

i K+

i Fe2+

80

The Anomalous Electronic Configurations of

Cr and Cu

Cr and Cu have electron configurations which are

inconsistent with the Aufbau principle The anomalous

are explained on the basis that a filled or half-filled orbital

is more stable

Element Expected Observedactual

Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1

Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1

81

24Cr 18[Ar]

The actual orbital notation

24Cr 18[Ar]

Half filled orbital is more stable

(possesses an extra added stability)

3d 4s

3d 4s

Chromium predicted orbital notation

82

Copper predicted orbital notation

Cu [Ar]

The actual orbital notation

Cu [Ar]

4s3d

3d 4s

Full filled orbital is more stable

(possesses an extra added stability)

83

z = 21

z = 30

84

Write the ground-state electronic configuration and

explain the anomalous case for Cr (Z=24) and Cu ( Z=29)

Exercise

Writing Electronic Configuration for Negative Ion

Add electron according to Aufbau Principle

Example

i Cl-

ii O2-

Writing Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example

i Mg2+

i K+

i Fe2+

81

24Cr 18[Ar]

The actual orbital notation

24Cr 18[Ar]

Half filled orbital is more stable

(possesses an extra added stability)

3d 4s

3d 4s

Chromium predicted orbital notation

82

Copper predicted orbital notation

Cu [Ar]

The actual orbital notation

Cu [Ar]

4s3d

3d 4s

Full filled orbital is more stable

(possesses an extra added stability)

83

z = 21

z = 30

84

Write the ground-state electronic configuration and

explain the anomalous case for Cr (Z=24) and Cu ( Z=29)

Exercise

Writing Electronic Configuration for Negative Ion

Add electron according to Aufbau Principle

Example

i Cl-

ii O2-

Writing Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example

i Mg2+

i K+

i Fe2+

82

Copper predicted orbital notation

Cu [Ar]

The actual orbital notation

Cu [Ar]

4s3d

3d 4s

Full filled orbital is more stable

(possesses an extra added stability)

83

z = 21

z = 30

84

Write the ground-state electronic configuration and

explain the anomalous case for Cr (Z=24) and Cu ( Z=29)

Exercise

Writing Electronic Configuration for Negative Ion

Add electron according to Aufbau Principle

Example

i Cl-

ii O2-

Writing Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example

i Mg2+

i K+

i Fe2+

83

z = 21

z = 30

84

Write the ground-state electronic configuration and

explain the anomalous case for Cr (Z=24) and Cu ( Z=29)

Exercise

Writing Electronic Configuration for Negative Ion

Add electron according to Aufbau Principle

Example

i Cl-

ii O2-

Writing Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example

i Mg2+

i K+

i Fe2+

84

Write the ground-state electronic configuration and

explain the anomalous case for Cr (Z=24) and Cu ( Z=29)

Exercise

Writing Electronic Configuration for Negative Ion

Add electron according to Aufbau Principle

Example

i Cl-

ii O2-

Writing Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example

i Mg2+

i K+

i Fe2+

Writing Electronic Configuration for Negative Ion

Add electron according to Aufbau Principle

Example

i Cl-

ii O2-

Writing Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example

i Mg2+

i K+

i Fe2+

Writing Electronic Configuration for Positive Ions

Remove electron from the outermost orbital (largest value of n)

Example

i Mg2+

i K+

i Fe2+