chapter 7 powerpoint le m

Quantum Theory and the Electronic Structure of Atoms

Chapter 7

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

PowerPoint Lecture Presentation byJ. David RobertsonUniversity of Missouri

Properties of Waves

Wavelength () is the distance between identical points on successive waves.

Amplitude is the vertical distance from the midline of a wave to the peak or trough.

7.1

Properties of Waves

Frequency () is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s).

The speed (u) of the wave = x 7.1

Maxwell (1873), proposed that visible light consists of electromagnetic waves.

Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves.

Speed of light (c) in vacuum = 3.00 x 108 m/s

All electromagnetic radiation x c

7.1

x = c

= c/ = 3.00x108 m/s/6.0x104 Hz = 5.0 x 103 m

Radio wave

A photon has a frequency of 6.0 x 104 Hz. Convertthis frequency into wavelength (nm). Does this frequencyfall in the visible region? or Is this AM or FM frequency?

= 5.0 x 1012 nm

7.1

Mystery #1, “Black Body Problem”: Frequency dependence of radiation from a heated body Classical Physics couldn’t explain it : UV Catastrophe.

Solved by Planck witth a Quantum Theory (1900)

Energy (light) is emitted or absorbed in discrete units or a packet (quantum). “Energy of an atom is quantized.”

E = nh

h = 6.63 x 10-34 J•s Planck’s constant n: 1,2,3,4,5 etc.

7.1

What’s nature of light? According to classical physics, it is a wave because of the following properties.

• Reflection

• Refraction

• Diffraction: a result of interference

However, the wave nature of light couldn’t explain the photoelectric effect.

Light has both:1. wave nature 2. particle nature

h = KE + BE

Mystery #2, “Photoelectric Effect”Solved by Einstein in 1905

Photon is a “particle” of light

KE = h - BE

h

KE e-

7.2

Electrical Current generated by light:

Presence of the Threshold frequency

Absence of Time Lag

E = h

Alkali metals works the best.

E = h x

E = 6.63 x 10-34 (J•s) x 3.00 x 10 8 (m/s) / 0.154 x 10-9 (m)

E = 1.29 x 10 -15 J/photon = (1.29 x 10 -15 J/photon)(6.022 x 1023photon/mol) = 7.77x108 J/mol = 7.77x105 kJ/mol = 777 MJ/mol

Cf. Energy of a UV light with 200 nm: 598kJ/mol Energy of a red light with 671 nm (Li): 180kJ/mol Chemical Bond Energy: 500~1,000 kJ/mol

E = h x c /

7.2

When copper is bombarded with high-energy electrons, X rays are emitted. Calculate the energy (in joules) associated with the photons if the wavelength of the X rays is 0.154 nm.

Why medicines are stored in a brown glass bottles?

• Brown glass blocks high energy portion (green and blue) of the visible light by absorbing G & B. It also absorb red portion substantially to impart brown color. Many chemicals can be destroyed by EMW with higher energy (Blue ,Violet, UV and others).

• Brown (Dark Red) light has much less energy (<200kJ/mol), thus cannot harm many substances. Normally 200 kJ/mol or more of energy is needed to break many chemical bonds.

• Red Green Blue Lower Higher Energy

7.3

Mystery #3: Line Emission Spectrum of H Atoms

HOW are they generated?

Johann Balmer (1825-1989) solved the jigsaw puzzle

(396) 410 434 496 656 .

The Jigsaw puzzle of the H emission spectra was solved by Balmer (1885) without knowing its implacations

1/λ = 1.097x107 (1/22-1/n2) m-1

= 0.01097 (0.25-1/n2) nm-1

where n = 3, λ = 656 nm

4, λ = 486

5, λ = 434

6, λ = 410

1. e- can only have specific (quantized) energy values

2. light is emitted as e- moves from one energy level to a lower energy level

Bohr’s Model of the Atom (1913)

En = -RH ( )1n2

n (principal quantum number) = 1,2,3,…

RH (Rydberg constant) = 2.18 x 10-18J7.3

E = h

E = h

7.3

Ephoton = E = Ef - Ei

Ef = -RH ( )1n2

f

Ei = -RH ( )1n2

i

i fE = RH( )

1n2

1n2

nf = 1

ni = 2

nf = 1

ni = 3

nf = 2

ni = 3

7.3

Ephoton = 2.18 x 10-18 J x (1/25 - 1/9)

Ephoton = E = -1.55 x 10-19 J

= 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J

= 1280 nm Is this IR or UV?

Calculate the wavelength (in nm) of a photon emitted by a hydrogen atom when its electron drops from the n = 5 state to the n = 3 state.

Ephoton = h x c /

= h x c / Ephoton

i fE = RH( )

1n2

1n2

Ephoton =

7.3

The Dual Nature of Matter: Wave-Particle Duality: “Wavicle”

• Einstein: Light wave is Light Particle

(1905) (“Photon”)

E = hν = hc/ λ

• De Broglie: Electron (Particle) is Wave

(1923) λ = h/mu

De Broglie (1924) reasoned that e- is both particle and wave.

2r = n = h/mu

u = velocity of e-

m = mass of e-

Why is e- energy quantized?

7.4

= h/mu

= 6.63 x 10-34 / (2.5 x 10-3 x 15.6)

= 1.7 x 10-32 m = 1.7 x 10-23 nm

What is the de Broglie wavelength (in nm) associated with a 2.5 g Ping-Pong ball traveling at 15.6 m/s?

What is the de Broglie wavelength (in nm) associated with an electron moving at a speed of 2x108 m/s?

m in kgh in J•s u in (m/s)

7.4 = 6.63 x 10-34 / (9.11x10-35 x 2x108) = 0.004 nm

Chemistry in Action: Electron Microscopy

STM image of iron atomson copper surface

e = 0.004 nm

Chemistry in Action: Element from the SunIn 1868, Pierre Janssen detected a new dark line in the solar emission spectrum that did not match known emission lines

In 1895, William Ramsey discovered helium in a mineral of uranium (from alpha decay).

Mystery element was named Helium

H: 656 486 434 410

He: 668 588 502 447

Chemistry in Action: Laser – The Splendid Light

Laser light is (1) intense, (2) monoenergetic, and (3) coherent

Schrodinger Wave EquationIn 1926 Schrodinger wrote an equation that described both the particle and wave nature of the e-

Wave function () describes:

1. energy of e- with a given

2. probability of finding e- in a volume of space

Schrodinger’s equation can only be solved exactly for the hydrogen atom. Must approximate its solution for multi-electron systems.

7.5

Schrodinger Wave Equation

fn(n, l, ml, ms)

principal quantum number n

n = 1, 2, 3, 4, ….

n=1 n=2 n=3

7.6

distance of e- from the nucleus

e- density (1s orbital) falls off rapidly as distance from nucleus increases

Where 90% of thee- density is foundfor the 1s orbital

7.6

= fn(n, l, ml, ms)

angular momentum quantum number l

for a given value of n, l = 0, 1, 2, 3, … n-1

n = 1, l = 0n = 2, l = 0 or 1

n = 3, l = 0, 1, or 2

Shape of the “volume” of space that the e- occupies

l = 0 s orbitall = 1 p orbitall = 2 d orbitall = 3 f orbital


7.6

l = 0 (s orbitals)

l = 1 (p orbitals)

7.6

l = 2 (d orbitals)

7.6

= fn(n, l, ml, ms)

magnetic quantum number ml

for a given value of lml = -l, …., 0, …. +l

orientation of the orbital in space

if l = 1 (p orbital), ml = -1, 0, or 1if l = 2 (d orbital), ml = -2, -1, 0, 1, or 2


7.6

ml = -1 ml = 0 ml = 1

ml = -2 ml = -1 ml = 0 ml = 1 ml = 27.6

= fn(n, l, ml, ms)

spin quantum number ms

ms = +½ or -½


ms = -½ms = +½

7.6

Existence (and energy) of electron in atom is described by its unique wave function .

Pauli exclusion principle - no two electrons in an atomcan have the same four quantum numbers.


= fn(n, l, ml, ms)

Each seat is uniquely identified (E, R12, S8)Each seat can hold only one individual at a time

7.6


= fn(n, l, ml, ms)

Shell – electrons with the same value of n

Subshell – electrons with the same values of n and l

Orbital – electrons with the same values of n, l, and ml

How many electrons can an orbital hold?

If n, l, and ml are fixed, then ms = ½ or - ½

= (n, l, ml, ½)or= (n, l, ml, -½)

An orbital can hold 2 electrons 7.6

How many 2p orbitals are there in an atom?

2p

n=2

l = 1

If l = 1, then ml = -1, 0, or +1

3 orbitals

How many electrons can be placed in the 3d subshell?

3d

n=3

l = 2

If l = 2, then ml = -2, -1, 0, +1, or +2

5 orbitals which can hold a total of 10 e-

7.6

Energy of orbitals in a single electron atomEnergy only depends on principal quantum number n

En = -RH ( )1n2

n=1

n=2

n=3

7.7

l = 0 1 2 3

Energy of orbitals in a multi-electron atom

Energy depends on n and l

n=1 l = 0

n=2 l = 0n=2 l = 1

n=3 l = 0n=3 l = 1

n=3 l = 2

7.7

“Fill up” electrons in lowest energy orbitals (Aufbau principle)

H 1 electron

H 1s1

He 2 electrons

He 1s2

Li 3 electrons

Li 1s22s1

Be 4 electrons

Be 1s22s2

B 5 electrons

B 1s22s22p1

C 6 electrons

? ?

7.7

C 6 electrons

The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins (Hund’s rule).

C 1s22s22p2

N 7 electrons

N 1s22s22p3

O 8 electrons

O 1s22s22p4

F 9 electrons

F 1s22s22p5

Ne 10 electrons

Ne 1s22s22p6

7.7

Order of orbitals (filling) in multi-electron atom

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s7.7

Electron configuration is how the electrons are distributed among the various atomic orbitals in an atom.

1s1

principal quantumnumber n

angular momentumquantum number l

number of electronsin the orbital or subshell

Orbital diagram

H

1s1

7.8

What is the electron configuration of Mg?

Mg 12 electrons

1s < 2s < 2p < 3s < 3p < 4s

1s22s22p63s2 2 + 2 + 6 + 2 = 12 electrons

7.8

Abbreviated as [Ne]3s2 [Ne] 1s22s22p6

What are the possible quantum numbers for the last (outermost) electron in Cl?

Cl 17 electrons 1s < 2s < 2p < 3s < 3p < 4s

1s22s22p63s23p5 2 + 2 + 6 + 2 + 5 = 17 electrons

Last electron added to 3p orbital

n = 3 l = 1 ml = -1, 0, or +1 ms = ½ or -½

Outermost subshell being filled with electrons

7.8

Paramagnetic

unpaired electrons

2p

Diamagnetic

all electrons paired

2p7.8

Ne, ArO, S

chapter 7 powerpoint le m

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