atomic structure note
TRANSCRIPT
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2.1 Bohr’s atomic model
2.2 Quantum mechanical model
2.3 Electronic configuration
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At the end of this topic students should be able to:-
a) Describe the Bohr’s atomic model.
b) Explain the existence of electron energy levels in an atom.
c) Calculate the energy of electron using:
d) Describe the formation of line spectrum of hydrogen atom.
e) Calculate the energy change of an electron during transition.
f) Calculate the photon of energy emitted by an electron that produces a particular wavelength during transition.
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2n
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g)perform calculations involving the rydberg
equation for lyman, balmer, paschen, brackett
and pfund series.
h)calculate the ionisation energy of hydrogen
atom from lyman series.
i) state the weaknesses of bohr's atomic model.
j) state the dual nature of electron using de
broglie's postulate and heisenberg's uncertainty
principle
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In 1913, a young Dutch physicist, Niels Böhr proposed a theory of atom that shock the scientific world.
The atomic model he described had electrons circling a central nucleus that contains positively charged protons.
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Böhr also proposed that these orbits can only
occur at specifically “permitted” levels only
according to the energy levels of the electron
and explain successfully the lines in the
hydrogen spectrum.
BOHR’S ATOMIC MODELS
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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.
Postulates
H
Nucleus
(proton) H 1
1
BOHR’S ATOMIC MODELS
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ii) 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. ( e can only exist on specific orbit,not between orbit)
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n=1
n=2
n=3
H Nucleus
(proton)
[ orbit = stationary state = energy level = shell ]
BOHR’S ATOMIC MODELS
Postulates
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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 to the energy difference between the levels.
BOHR’S ATOMIC MODELS
Postulates
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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
Points to Remember
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• The energy of an electron in its level is given by:
RH (Rydberg constant) or A = 2.18 10-18J.
n (principal quantum number) = 1, 2, 3 …. (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)
2Hnn
1RE
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THE BOHR ATOM
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• 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:
2
i
Hin
1RE
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THE BOHR ATOM
E = Ef - Ei
2
f
Hfn
1RE2
i
H2
f
Hn
1R
n
1RE
2
f
2
i
Hn
1
n
1RE
where
Thus, ni = initial orbit
nf = final orbit
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• 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.
where; h (Planck’s constant) =6.63 10-34 J s = frequency
c
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THE BOHR ATOM
E = h
Where; c (speed of light) = 3.00 108 ms-1
Thus, hcΔE
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n =1 n = 2 n = 3 n = 4
Electron is excited from lower to higher
energy level. A specific amount of energy
is absorbed
E = h = E1-E3 (+ve)
Electron falls from higher to lower energy level .
A photon of energy is released.
E = h = E3-E1 (-ve)
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Energy level diagram for the hydrogen atom
Pote
ntial en
erg
y
n = 1
n = 2
n = 3
n = 4
n =
Energy
released
Energy
absorbed
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Example
Calculate:
i) The E of an e- has when it occupies
when it was at n=3 & n=4.
ii) The E of the photon emitted when
one mole e- of drops from the fourth E
level to third E level.
iii) The frequency & wavelength of this
photon.
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Solution :
2Hnn
1RE
i) at n=3, E3 = -RH 1 = - 2.18 x 10-18 J
32 9
= -2.422 x 10-19 J
at n=4, E4 = -RH 1 = - 2.18 x 10-18 J
42 16
= -1.363 x 10-19J
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n = 4
n = 3
ΔE = Ef – Ei = E3 – E4
= -2.422x 10-19J- (-1.363 x 10-19J)
= -1.06 x 10-19J
-ve sign indicates that E is released
when e- falls.
ii)
E released by 1 mol of e-,
ΔE = -1.06 x 10-19J x 6.023 x 1023 mol-1
= - 63 843.8 Jmol-1 = -63.8438 kJmol-1
E
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iii) Frequency, v = ΔE = 1.06 x 10-19J
h 6.63 x 10-34 Js
= 1.599 x 1014 s-1
= 1.599 x 1014Hz
# -ve sign of ΔE is ignored because frequency is
always +ve
Wavelength, λ = c = 3.0 x 108 ms-1
v 1.599 x 1014 s-1
= 1.876 x 10-6 m
= 1876 nm
# 1 m = 109 nm
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Exercises: 1) Calculate the energy of an electron in the second
energy level of a hydrogen atom. (-5.448 x 10-19 J) 2) Calculate the energy of an electron in the energy
level n = 6 of a 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: 1.55 x 10-19J) 4) Calculate the frequency and wavelength (nm) of
the radiation emitted in question 3.
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Emission Spectra
Continuous
Spectra
Line
Spectra
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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.
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Regions of the Electromagnetic Spectrum
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• 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 present and each colour merges into the next without a break.
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FORMATION OF CONTINUOUS SPECTRUM
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Line Spectrum (atomic spectrum)
A spectrum consists of discontinuous & 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
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prism film
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
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Line Emission Spectrum of Hydrogen Atoms
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FORMATION OF ATOMIC / LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5 n =
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 in an excited state
(very unstable).
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FORMATION OF ATOMIC / LINE SPECTRUM
Emission of photon
n = 2
n = 3
n = 4
n = 5
n = 6 n =
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)
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FORMATION OF ATOMIC / LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5 n =
Lyman Series
Emission of photon
Line
spectrum E
Energy
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FORMATION OF ATOMIC / LINE SPECTRUM
n = 1
n = 2
n = 3
n = 4
n = 5 n =
Lyman Series
Emission of photon
Line
spectrum
Balmer Series
E
Energy
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Emission series of hydrogen atom
n = 1
n = 2
n = 3
n = 4
n =
222
AE
23
AE3 24
4
AE
211
AE
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Lyman series
Balmer series
Brackett series
Paschen series
Pfund series
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Exercise: Complete the following table
Series n1 n2 Spectru
m region
Lyman 2,3,4,…
2 3,4,5,…
Paschen 4,5,6,… Infrared
4 5,6,7,… Infrared
5 6,7,8,… Infrared 32
ultraviolet
Visible Balmer
Brackett
Pfund
1
3
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The following diagram shows the line spectrum of hydrogen atom. Line A is the first line of the Lyman series.
Example 1
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?
Draw the energy level diagram for corresponding line
spectrum above.
Line
spectrum E
A B C D E
Line E
Line A
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A B C D E n = 1
n = 2
n = 3
n = 4
n = 5
n = 6
E
iii)
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Describe the transitions of electrons that
lead to the lines W, and Y, respectively.
Example 2
Solution
Line
spectrum
W Y
Balmer series
For W: transition of electron is from n=4 to n=2
For Y: electron shifts from n=7 to n=2
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Exercise The following diagram depicts the line spectrum of hydrogen atom. Line A is the first line of the Lyman series.
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Line
spectrum E
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
Line E
Line A
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Homework
Calculate En for n = 1, 2, 3, and 4. Draw a
diagram showing how the energy, at
different values of n, increases vertically
and indicate by vertical arrows the electron
transitions that lead to lines in:
a) Lyman series
b) Paschen series
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Significance of Atomic Spectra
• In Lyman series, the frequency of the
convergence of spectral lines can be used to
find the ionisation energy of hydrogen atom:
IE = h
• The frequency of the first line of the Lyman
series > the frequency of the first line of the
Balmer series.
38 Lyman Series
Line
spectrum E
Balmer Series
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Line
spectrum
A B C D E
Exercise
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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.
Line A.
The electron moves from n=4 to n=3.
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EXERCISES
Calculate En for n = 1, 2, 3, and 4.
Draw an energy level diagram and line
spectrum for the transition of electron that
lead to the formation of 4 lines in :
a) Pfund series
b) Paschen series
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• Wavelength emitted by the transition of
electron between two energy levels is
calculated using Rydberg equation:
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Rydberg Equation
RH = 1.097 107 m-1
= wavelength
Since should have a positive value thus n1 < n2
where
2
2
2
1
H n
1
n
1 R
λ
1
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Exercise
Calculate
a) the wavelength in nm
b) the frequency
c) the energy
that associated with the second line in the Balmer series of the hydrogen spectrum.
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Solution (a)
Second line of
Balmer series:
the transition of
electron is from
n2=4 to n1=2
= RH
1 1
n12
n22
1
= (1.097x107 m 1)
1 1
22
42
1
= x 1
109 m
nm
= 486 nm
4.86x10 7 m
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Frequency, v = c
= 3.0 x 10 8 ms-1
4.86 x 10-7 m
= 6.173 x 1014 s-1
Energy, ΔE = hv
= 6.63 x 10 -34Js x 6.173 x 1014s-1
= 4.093 x 10 -19J
Solution (c)
Solution (b)
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Example
Calculate the wavelength, in nanometers of
the spectrum of hydrogen corresponding to
ni = 2 and nf = 4 in the Rydberg equation.
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Solution:
Rydberg equation:
1/λ = RH (1/ni2 – 1/nf2)
ni = 2 nf = 4
RH = 1.097 x 107m
1/λ = RH (1/22 – 1/42)
= RH(1/4-1/16)
λ = 4.86m x 102 m
= 486nm
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Example
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. Name the series of
line formed.
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Solution:
1/λ = RH (1/n1 2 – 1/n2 2)
n1 = 2 n2 = 4
1/λ = 1.097 x 107 (1/22 – 1/42)
λ = 4.86 x 10-7 m
= 486 nm
*e dropped to the second orbit (n=2),
>>> Balmer series
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EXAMPLE 3
Calculate the wavelengths of the fourth line in
the Balmer series of hydrogen.
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n1 = 2 n2 = 6
RH = 1.097 x 107m-1
λ = 4.10 x 10-7 m
RH 22 62
1 1 1 =
λ
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Different values of RH and its usage
1. RH = 1.097 107 m-1
2
f
2
i
Hn
1
n
1RE
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RH n2
1 n2
2
1 1 1 =
λ
RH = 2.18 x 10-18 J
n1 < n2
ni – initial orbit
nf - final orbit
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EXAMPLE 4
Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom.
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RH n2
1 n2
2
1 1 1 =
λ
52 1.097 x 107
22
1 1 1 =
λ
1
λ = 0.2303 X 107 m-1
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ΔE = 4.58 x 10-19 J
hcΔE
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ΔE = (6.63 10-34Js)X(3.00 108 ms-1) X (0.2303 X 107 m-1)
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For Lyman series n1=1 and n2 = ∞ calculate;
i ) Wavelength
ii ) Frequency
iii ) Wave number of the last line of hydrogen spectrum in Lyman series
Wave number = 1/wavelength
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EXERCISE 5
Ans: i) 9.116 x 10-8 m ii) 3.29 x 1015 s-1 iii) 1.0970 x 107 m-1
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The weakness of Bohr’s Theory
• Bohr was successful in introducing the idea of
quantum energy and in explaining the lines of
hydrogen spectrum.
• His theory could not be extended to predict the
energy levels and spectra of atoms and ions
with more than one electron.
• His theory can only explain the hydrogen
spectrum or ions contain one electron eg He+,
Li2+.
• Modern quantum mechanics retain Bohr’s
concept of discrete energy states and energy
involved during transition of electrons but totally
reject the circular orbits he introduced. 51
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At the end of this topic students
should be able to:-
1) Define the term orbital.
2) State all the four quantum numbers of
an electron in an orbital.
3) Sketch the shape of s, p and d orbitals
with the correct orientations.
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2n
1
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Atomic Orbital
An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron.
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Definition
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ATOMIC ORBITAL
•Orbital - region in space around the
nucleus where there is a high
probability of finding an electron.
•Where electron can be expected to be
found.
•Orbit – the path of an e- as it travels
round the nucleus of an atom.
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QUANTUM NUMBERS
• Quanta - discrete amounts of E that an e-
absorbs as it moves up an E level or releases
when it moves down to lower E level.
• The positions and orbits of e- referred as E state
and described by 4 quantum numbers :
i. Principal quantum number (n)
ii. Angular momentum/Azimuthal quantum
number (l)
iii. Magnetic quantum number (m)
iv. Spin quantum number (s)
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PRINCIPAL QUANTUM NUMBER, n
- Determines the energy & size of an orbital.
- n is large – size of atom is larger & distance e- from nucleus is greater.
- The principal quantum number may have only integral values: n =1, 2, 3, …,
.
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n 1 2 3 4
shell K L M N
Orbital size
Energy increases
PRINCIPAL QUANTUM NUMBER, n
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ANGULAR MOMENTUM QUANTUM NUMBER, l
58
Alternative name: Subsidiary / Azimuthal / Orbital
Quantum Number
- Indicates the shape of the atomic orbital, the
types of orbitals, and the angular momentum
of the electron.
- The allowed values of l are 0, 1, 2,…, (n 1).
- Depends on value of n.
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ANGULAR MOMENTUM QUANTUM NUMBER, l
59
Symbol Orbital shape
0 s spherical
1 p dumbbell
2 d cloverleaf
3 f ~
Numerical
value of l
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Example
60
Shell, n Sub-shell, l Called as
2
4
0
1
2s
2p
0
1
2
3
4s
4p
4d
4f
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MAGNETIC QUANTUM NUMBER, m
•Determines the orientation of orbital in
space.
•Permitted value for m depend on the value of l.
•It has integer value ranging from –l to +l .
•Indicates the maximum number of orbitals for a particular value of l.
•Example :
l = 1, m = -1, 0, 1
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ELECTRON SPIN QUANTUM NUMBER, s
•Determines the
direction of spinning
motions of an
electron on its own
axis.
•Clockwise or counter
clockwise.
•The electron spin
quantum number
has a value of : 62
+ 1
2 -
1
2 or
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The relationship between the values of n, , and m
n
(<n)
Orbital
notati
on
m
(- m +)
No. of
degenerat
ed orbitals
1 0 1s 0 1
2 0 2s 0 1
1 2p 1,0,-1 3
3
0 3s 0 1
1 3p 1,0,-1 3
2 3d 2,1,0,-1,-2 5 63
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Example
64
n l shap
e
m s No. of
orbital
No. of
e-
1
2
3
4
1s 0
0
1
0
1
2
0
1
2
3
0
2s
3s
2p
3p
3d
4s
4p
4d
4f
0
0
0
-1, 0, 1
-1, 0, 1
-1, 0, 1
-2, -1, 0, 1, 2
-2, -1, 0, 1, 2
-3, -2, -1, 0, 1,
2, 3
1/2
1/2
1/2
1/2
1/2
1/2
1/2
1/2
1/2
1/2
1 2
1
3 8
18
32
1
3
3
1
5
7
5
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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.
65
b) 6g
a) 2d
c) 7s
d) 5i
n=2 l=2
(l < n)
n=6
n=7
n=5
l=4
l=0
l=6
unacceptable
acceptable
unacceptable
acceptable
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SHAPE OF ATOMIC ORBITALS
a) s orbitals
When l = 0
Spherical shape.
As n increases, s orbital gets larger
66
Shape of s orbital with different n
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SHAPE OF ATOMIC ORBITALS
b) p orbitals
When l = 1
Dumbbell shaped
Consists of 3 p orbital, each with the
same size, shape and energy; they differ
from one another only in orientation - px,
py, and pz.
Correspond m of -1, 0, and +1.
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SHAPE OF ATOMIC ORBITALS
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c) d - orbitals
When l = 2, m = -2, -1, 0, 1, 2
The orbitals are: dyz, dxz, dxy, dx2-y2, dz2
SHAPE OF ATOMIC ORBITALS
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2.3 Electronic Configuration
At the end of this topic students should
be able to:-
a. State Aufbou principle, Hund’s rule and Pauli ‘s exclusion principle.
b. Apply the rules in (a) to fill electrons into atomic orbital.
c. Write the electronic configuration of atoms and monoatomic ions using spdf notation.
d) Explain the anomalous electronic configurations of chromium and copper.
71
2n
1
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Representing Electronic Configuration
Method 1: Orbital diagram
72
O: 8
1s 2s 2p
Method 2: spdf notation
O: 8 1s 2s 2p 2 2 4
platform
Concentric circle
box
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Rules for Assigning Electrons to Orbitals
73
ii) Aufbau Principle
Electrons fill the lowest energy orbitals first and other
orbitals in order of ascending energy.
The order of filling orbitals is:
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s
1s 2s 2p 1s
2s
3s
4s
5s
2p
3p
4p
5p
3d
4d
5d
4f
5f
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Relative Energy Level of Atomic Orbitals
74
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
2s 2p
3s
4s
3p
4p
3d
4d 5s
Orbital energy levels
in the H atom
Orbital energy levels
in a many-electron atom
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ii) Pauli Exclusion Principle
75
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 m s
1 0
0 1
0
0 1
0
0
1 2
1 2
1 2
( ) , , ,
) ( , , ,
+
+
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iii) Hund’s Rule
76
Rules for Assigning Electrons to Orbitals
Only when all the degenerate orbitals (a group of
orbitals of identical energy e.g. 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
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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:
77
Exercise
1s 2s 2p
1s 2s 2p
1s 2s 2p
?
?
1s 2s 2p
?
1s 2s 2p
?
1s 2s 2p
?
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Draw ‘electrons-in-boxes’ diagram of the electronic
configuration of titanium, Ti (Z = 22). Also, write the ground-
state electronic configurations for Ti and Ti2+ ion.
78
Exercise
1s 2s 2p
Ti:
3s 3p 4s
3d
1s 2 2s 2 2p 6 Ti: 3s 2 3p 6 3d 2 4s 2
1s 2 2s 2 2p 6 Ti2+: 3s 2 3p 6 3d 2
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Points to remember
• The electronic configuration of atom or monatomic ion at
ground state
Distribution of electrons obeys Aufbau principle, Pauli
exclusion principle and Hund’s 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 a max of 2 electrons (ns2)
p-orbitals a max of 6 electrons (np6)
d-orbitals a max of 10 electrons (nd10) 79
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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 completely filled or half-
filled orbital is more stable.
Element Expected Observed/actual
Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1
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For Chromium : 24 e-
Expected :
24Cr = 1s2 2s2 2p6 3s2 3p6 4s2 3d4
Orbital diagram:
24Cr = [ Ar ]
Observed :
24Cr = 1s2 2s2 2p6 3s2 3p6 4s1 3d5
Orbital diagram:
24Cr = [ Ar ]
81
3d 4s
3d 4s
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For Copper : 29 e-
Expected :
29Cu = 1s2 2s2 2p6 3s2 3p6 4s2 3d9
Orbital diagram:
29Cu = [ Ar ]
Observed :
29Cu = 1s2 2s2 2p6 3s2 3p6 4s1 3d10
Orbital diagram:
29Cu = [ Ar ]
82
3d 4s
3d 4s
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• Orbitals that are half–filled (s1, p3 or d5)
or completely filled (s2, p6 or d10) have
extra stability due to the equal
symmetrical distribution of charge
around an atom.
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84
z = 21
z = 30
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1. How many 2p orbitals are there in an atom?
2. How many electrons can be placed in the 3d sub-shell?
Thinking question
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3.What is the electron configuration of Mg?
4.What are the possible quantum numbers for the last
(outermost) electron in Cl?
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THE END
87