1. Quantum Theory and the Electronic Structure of Atoms
PowerPoint Lecture Presentation by J. David Robertson University of Missouri
Quantum Theory and the Electronic Structure of Atoms
Chapter 7
Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

2. Properties of Waves
Wavelength (l) 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

3. The speed (u) of the wave = l x n
Properties of Waves
Frequency (n) is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s).
The speed (u) of the wave = l x n
7.1

4. Speed of light (c) in vacuum = 3.00 x 108 m/s
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
l x n = c
7.1

5. 7.1

6. l x n = c l = c/n l = 3.00x108 m/s/6.0x104 Hz l = 5.0 x 103 m
A photon has a frequency of 6.0 x 104 Hz. Convert
this frequency into wavelength (nm). Does this frequency
fall in the visible region? or Is this AM or FM frequency? l n
l x n = c
l = c/n
l = 3.00x108 m/s/6.0x104 Hz
l = 5.0 x 103 m
l = 5.0 x 1012 nm
Radio wave
7.1

7. Energy (light) is emitted or absorbed
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 = nhn
h = 6.63 x J•s
Planck’s constant
n: 1,2,3,4,5 etc.
7.1

8. Diffraction: a result of interference
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.

9. Photon is a “particle” of light
Mystery #2, “Photoelectric Effect” Solved by Einstein in 1905 hn
Electrical Current generated by light:
Presence of the Threshold frequency
Absence of Time Lag
KE e-
Light has both:
wave nature
particle nature
E = hn
Photon is a “particle” of light
hn = KE + BE
KE = hn - BE
Alkali metals works the best.
7.2

10. E = 6.63 x 10-34 (J•s) x 3.00 x 10 8 (m/s) / 0.154 x 10-9 (m)
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 nm.
E = h x n
E = h x c / l
E = 6.63 x (J•s) x 3.00 x 10 8 (m/s) / x 10-9 (m)
E = 1.29 x J/photon
= (1.29 x 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: kJ/mol
Energy of a red light with 671 nm (Li): 180kJ/mol
Chemical Bond Energy: 500~1,000 kJ/mol
7.2

11. 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

12. Mystery #3: Line Emission Spectrum of H Atoms HOW are they generated?
(396)
Johann Balmer ( ) solved the jigsaw puzzle
7.3

13. 7.3

14. 1/λ = 1.097x107 (1/22-1/n2) m-1 = 0.01097 (0.25-1/n2) nm-1
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
= (0.25-1/n2) nm-1
where n = 3, λ = 656 nm
4, λ = 486
5, λ = 434
6, λ = 410

15. n (principal quantum number) = 1,2,3,…
Bohr’s Model of
the Atom (1913)
e- can only have specific (quantized) energy values
light is emitted as e- moves from one energy level to a lower energy level
En = -RH
( ) 1 n2
n (principal quantum number) = 1,2,3,…
RH (Rydberg constant) = 2.18 x 10-18J
7.3

16. E = hn
E = hn
7.3

17. ( ) ( ) ( ) Ephoton = DE = Ef - Ei 1 Ef = -RH n2 1 Ei = -RH n2 1
nf = 1
ni = 3
nf = 2
ni = 3
Ef = -RH
( ) 1 n2 f
nf = 1
ni = 2
Ei = -RH
( ) 1 n2 i i f
DE = RH
( ) 1 n2
7.3

18. 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. i f
DE = RH
( ) 1 n2
Ephoton =
Ephoton = 2.18 x J x (1/25 - 1/9)
Ephoton = DE = x J
Ephoton = h x c / l
l = h x c / Ephoton
l = 6.63 x (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J
l = 1280 nm Is this IR or UV?
7.3

19. 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

20. Why is e- energy quantized?
De Broglie (1924) reasoned that e- is both particle and wave.
2pr = nl l = h/mu
u = velocity of e-
m = mass of e-
7.4

21. What is the de Broglie wavelength (in nm) associated with a 2
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?
l = h/mu
h in J•s
m in kg
u in (m/s)
l = 6.63 x / (2.5 x 10-3 x 15.6)
l = 1.7 x m = 1.7 x nm
l = 6.63 x / (9.11x10-35 x 2x108) = nm
7.4

22. Chemistry in Action: Electron Microscopy
le = nm
STM image of iron atoms
on copper surface
Average WL of visible light: ~ 500nm
So ordinary light microscope can’t image a virus or smaller object.
Small virus: 100nm
Electron traveling at a speed of ~0.7C has a WL of nm (from =h/(mv)) can take care of it.

23. 7.6

24. Chemistry in Action: Element from the Sun
In 1868, Pierre Janssen detected a new dark line in the solar emission spectrum that did not match known emission lines
Mystery element was named Helium
In 1895, William Ramsey discovered helium in a mineral of uranium (from alpha decay). H:
Pierre Janssen( ), French Astronomer,
Solar Eclipse in1868
Sir William Ramsay, KCB FRSE (2 October 1852 – 23 July 1916) was a Scottish chemist
Noble gases
NP 1904
He:

25. Chemistry in Action: Laser – The Splendid Light
Laser light is (1) intense, (2) monoenergetic, and (3) coherent

26. Schrodinger Wave Equation
In 1926 Schrodinger wrote an equation that described both the particle and wave nature of the e-
Wave function (Y) describes:
. energy of e- with a given Y
. 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

27. Schrodinger Wave Equation
Y = fn(n, l, ml, ms)
principal quantum number n
n = 1, 2, 3, 4, ….
distance of e- from the nucleus
n=1
n=2
n=3
7.6

28. e- density (1s orbital) falls off rapidly
Where 90% of the
e- density is found
for the 1s orbital
e- density (1s orbital) falls off rapidly
as distance from nucleus increases
7.6

29. Schrodinger Wave Equation
Y = fn(n, l, ml, ms)
angular momentum quantum number l
for a given value of n, l = 0, 1, 2, 3, … n-1
l = s orbital
l = p orbital
l = d orbital
l = f orbital
n = 1, l = 0
n = 2, l = 0 or 1
n = 3, l = 0, 1, or 2
Shape of the “volume” of space that the e- occupies
7.6

30. l = 0 (s orbitals)
l = 1 (p orbitals)
7.6

31. l = 2 (d orbitals)
7.6

32. Schrodinger Wave Equation
Y = fn(n, l, ml, ms)
magnetic quantum number ml
for a given value of l
ml = -l, …., 0, …. +l
if l = 1 (p orbital), ml = -1, 0, or 1
if l = 2 (d orbital), ml = -2, -1, 0, 1, or 2
orientation of the orbital in space
7.6

33. ml = -1
ml = 0
ml = 1
ml = -2
ml = -1
ml = 0
ml = 1
ml = 2
7.6

34. Schrodinger Wave Equation
Y = fn(n, l, ml, ms)
spin quantum number ms
ms = +½ or -½
ms = +½
ms = -½
7.6

35. Schrodinger Wave Equation
Y = fn(n, l, ml, ms)
Existence (and energy) of electron in atom is described
by its unique wave function Y.
Pauli exclusion principle - no two electrons in an atom
can have the same four quantum numbers.
Each seat is uniquely identified (E, R12, S8)
Each seat can hold only one individual at a time
7.6

36. 7.6

37. Schrodinger Wave Equation
Y = 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
Y = (n, l, ml, ½)
or Y = (n, l, ml, -½)
An orbital can hold 2 electrons
7.6

38. How many 2p orbitals are there in an atom?
If l = 1, then ml = -1, 0, or +1 2p
3 orbitals
l = 1
How many electrons can be placed in the 3d subshell?
n=3
If l = 2, then ml = -2, -1, 0, +1, or +2 3d
5 orbitals which can hold a total of 10 e-
l = 2
7.6

39. ( ) Energy of orbitals in a single electron atom 1 En = -RH n2
Energy only depends on principal quantum number n
l =
n=3
n=2
En = -RH
( ) 1 n2
n=1
7.7

40. Energy of orbitals in a multi-electron atom
Energy depends on n and l
n=3 l = 2
n=3 l = 1
n=3 l = 0
n=2 l = 1
n=2 l = 0
n=1 l = 0
7.7

41. “Fill up” electrons in lowest energy orbitals (Aufbau principle)? ?
Li 3 electrons
Be 4 electrons
C 6 electrons
B 5 electrons
B 1s22s22p1
Be 1s22s2
Li 1s22s1
H 1 electron
He 2 electrons
He 1s2
H 1s1
7.7

42. The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins (Hund’s rule).
F 9 electrons
Ne 10 electrons
C 6 electrons
O 8 electrons
N 7 electrons
Ne 1s22s22p6
C 1s22s22p2
N 1s22s22p3
O 1s22s22p4
F 1s22s22p5
7.7

43. Order of orbitals (filling) in multi-electron atom
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s
7.7

44. in the orbital or subshell
Electron configuration is how the electrons are distributed among the various atomic orbitals in an atom.
number of electrons
in the orbital or subshell
1s1
principal quantum
number n
angular momentum
quantum number l
Orbital diagram
1s1 H
7.8

45. What is the electron configuration of Mg?
Mg 12 electrons
1s < 2s < 2p < 3s < 3p < 4s
1s22s22p63s2
= 12 electrons
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
= 17 electrons
Last electron added to 3p orbital
n = 3
l = 1
ml = -1, 0, or +1
ms = ½ or -½
7.8

46. Outermost subshell being filled with electrons
7.8

47. 7.8

48. Paramagnetic Diamagnetic unpaired electrons all electrons paired 2p 2p
Ne, Ar
O, S
7.8