1. Quantum Theory and the Electronic Structure of Atoms
Chapter 5

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 = 3.00 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? l n
l x n = c
l = c/n
l = 3.00 x 108 m/s / 6.0 x 104 Hz
l = 5.0 x 103 m
l = 5.0 x 1012 nm
Radio wave
7.1

7. Mystery #1, “Black Body Problem” Solved by Planck in 1900
Energy (light) is emitted or absorbed in discrete units (quantum).
E = h x n
Planck’s constant (h)
h = 6.63 x J•s
7.1

8. Photon is a “particle” of light
Mystery #2, “Photoelectric Effect” Solved by Einstein in 1905 hn
Light has both:
wave nature
particle nature
KE e-
Photon is a “particle” of light
hn = KE + BE
KE = hn - BE
7.2

9. 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
7.2

10. Line Emission Spectrum of Hydrogen Atoms
7.3

11. 7.3

12. 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).

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

14. 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

15. 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)

16. 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

17. 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

18. 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

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

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

21. 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

22. l = 2 (d orbitals)
7.6

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

24. 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

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

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

27. 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

28. 7.6

29. 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

30. 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

31. 1 3 5 7 2 6 10 14 How many electrons can be in a sublevel?
Remember: A maximum of two electrons can be placed in an orbital.
s orbitals
p orbitals
d orbitals
f orbitals
Number of
orbitals 1 3 5 7
Number of electrons 2 6 10 14

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

33. Outermost subshell being filled with electrons
7.8

34. 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!

35. 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

36. Shorthand Notation
Step 1: Find the closest noble gas to the atom (or ion), WITHOUT GOING OVER the number of electrons in the atom (or ion). Write the noble gas in brackets [ ].
Step 2: Find where to resume by finding the next energy level.
Step 3: Resume the configuration until it’s finished.

37. 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

38. 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

39. “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

40. 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

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

42. 7.8

43. 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.

44. 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.

45. 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

46. 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 outermost energy level (not the highest orbital in energy!)

47. 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!