1. Quantum Theory and the Electronic Structure of Atoms
Chapter 7

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. 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. Relationship Between Frequency and Wavelength
Frequency and Wavelength are inversely related
C = ln
Where c = speed of light, l is the wavelength, and n is the frequency

7. Example 1 C = ln l = 3.00 x 108 m/s / 6.0 x 104 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?
C = ln
l = 3.00 x 108 m/s / 6.0 x 104 Hz
l = 5.0 x 103 m
l = 5.0 x 1012 nm
No, this is a radio wave
7.1

8. Example 2 3.0 x 108 m/s = (7.0 x 10-7 m)(n) n=4.29 x 1014 Hz
Red light has a wavelength that ranges from 7.0 x 10-7 m to 6.5 x 10-7 m. What is the frequency of light found in the red region?
3.0 x 108 m/s = (7.0 x 10-7 m)(n)
n=4.29 x 1014 Hz
3.0 x 108 m/s = (6.5 x 10-7 m)(n)
n=4.62 x 1014 Hz
nrange=4.29 x 1014 Hz x 1014 Hz

9. Example 3
A gamma ray has a frequency of 3.0 x 1021 Hz. What is the wavelength, in nm, of the ray? 3.0 x 108 = l(3.0 x 1021 s-1) l= 1.0 x m l= 1.0 x 10-4 nm

10. Mystery #1, “Black Body Problem” Solved by Planck in 1900
Energy had been thought to be emitted continuously at all wavelengths. However, energy (light) is emitted or absorbed in discrete units called quanta.
E = h x n
Planck’s constant (h)
h = x J•s
7.1

11. Mystery #2, “Photoelectric Effect” Solved by Einstein in 1905hn
KE e-
Discovered that light has both:
wave nature
particle nature
Photon is a “particle” of light
7.2

12. Relationship Between Energy and Frequency
E = hn h=Planck’s Constant h=6.626 x J s This constant was determined experimentally

13. Example 1
What is the energy of a wave that has a frequency of 1.05 x 1016 Hz?
E = hn
E = (6.626 x J s)(1.05 x 1016 s-1)
E = 6.96 x J

14. Example 2
What is the wavelength of a photon of light that has 4.79 x J of energy? What color light is represented?
E = hn
4.79 x J = (6.626 x J s) n
n=7.23 x 1014 Hz
c = ln
3.0 x 108 m/s = l (7.23 x 1014 s-1)
l = 4.1 x 10-7 m; violet

15. Line Emission Spectrum of Hydrogen Atoms
7.3

16. 7.3

17. 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

18. E = hn
E = hn
7.3

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

20. ( ) DE = RH 1 n2 Ephoton = Ephoton = 2.18 x 10-18 J x (1/25 - 1/9)
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
7.3

21. Why is e- energy quantized?
De Broglie (1924) reasoned that the e- is both particle and wave (he called this the wave/particle duality).
= h/mn
h=Planck’s Constant
m=mass
7.4

22. 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?
l = h/mn
h in J•s
m in kg
n in (m/s)
l = 6.63 x J s/ (2.5 x 10-3 kg x 15.6 m/s)
l = 1.7 x m = 1.7 x nm
7.4

23. 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).

24. Chemistry in Action: Electron Microscopy
le = nm
STM image of iron atoms
on copper surface

25. Heisenberg’s Uncertainty Principle
Heisenberg stated that it is impossible to know precisely both the velocity and the position of a particle
*If you measure the position of a particle it will be disturbed therefore altering the velocity

26. Schrodinger Wave Equation
In 1926 Schrodinger wrote an equation that described both the particle and wave nature of the e-
EY = HY
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. Probability of Finding an Electron
Max Born: later found it was more useful to determine the probability of finding an electron in a certain location Y2

28. QUANTUM NUMBERS
The shape, size, and energy of each orbital is a function of 3 quantum numbers which describe the location of an electron within an atom or ion n (principal) ---> energy level l (orbital) ---> shape of orbital ml (magnetic) ---> designates a particular suborbital The fourth quantum number is not derived from the wave function s (spin) ---> spin of the electron (clockwise or counterclockwise: ½ or – ½)

29. 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

30. 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

31. 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

32. Types of Orbitals (l)
s orbital
p orbital
d orbital

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

34. p Orbitals 3py orbital this is a p sublevel with 3 orbitals
These are called x, y, and z
There is a PLANAR NODE thru the nucleus, which is an area of zero probability of finding an electron
3py orbital

35. p Orbitals
The three p orbitals lie 90o apart in space

36. l = 2 (d orbitals)
7.6

37. f Orbitals
For l = 3, f sublevel with 7 orbitals

38. 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

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

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

41. 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

42. 7.6

43. 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

44. 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

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

46. “Fill up” electrons in lowest energy orbitals (Aufbau principle)
7.7

47. The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins (Hund’s rule).
7.7

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

49. Why are d and f orbitals always in lower energy levels?
d and f orbitals require LARGE amounts of energy
It’s better (lower in energy) to skip a sublevel that requires a large amount of energy (d and f orbtials) for one in a higher level but lower energy
This is the reason for the diagonal rule! BE SURE TO FOLLOW THE ARROWS IN ORDER!

50. 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

51. Outermost subshell being filled with electrons
7.8

52. Paramagnetic Diamagnetic unpaired electrons all electrons paired 2p 2p
7.8

53. 7.8

54. Exceptions to the Aufbau Principle
Remember d and f orbitals require LARGE amounts of energy
If we can’t fill these sublevels, then the next best thing is to be HALF full (one electron in each orbital in the sublevel)
There are many exceptions, but the most common ones are
d4 and d9
For the purposes of this class, we are going to assume that ALL atoms (or ions) that end in d4 or d9 are exceptions to the rule. This may or may not be true, it just depends on the atom.

55. Exceptions to the Aufbau Principle
d4 is one electron short of being HALF full In order to become more stable (require less energy), one of the closest s electrons will actually go into the d, making it d5 instead of d4. For example: Cr would be [Ar] 4s2 3d4, but since this ends exactly with a d4 it is an exception to the rule. Thus, Cr should be [Ar] 4s1 3d5. Procedure: Find the closest s orbital. Steal one electron from it, and add it to the d. This process is called hydridization

56. Write the shorthand notation for: Cu W Au
Try These!
Write the shorthand notation for: Cu W Au
[Ar] 4s1 3d10
[Xe] 6s1 4f14 5d5
[Xe] 6s1 4f14 5d10

57. Exceptions to the Aufbau Principle
The next most common are f1 and f8
The electron goes into the next d orbital
Example:
La [Xe]6s2 5d1
Gd [Xe]6s2 4f7 5d1

58. Keep an Eye On Those Ions!
Electrons are lost or gained like they always are with ions… negative ions have gained electrons, positive ions have lost electrons
The electrons that are lost or gained should be added/removed from the highest energy level (not the highest orbital in energy!)

59. Keep an Eye On Those Ions!
Tin
Atom: [Kr] 5s2 4d10 5p2
Sn+4 ion: [Kr] 4d10
Sn+2 ion: [Kr] 5s2 4d10
Note that the electrons came out of the highest energy level, not the highest energy orbital!