1. Where Are We Going…? Week 6: Orbitals and Terms
Russell-Saunders coupling of orbital and spin angular momenta
Free-ion terms for p2
Week 7: Terms and ionization energies
Free-ion terms for d2
Ionization energies for 2p and 3d elements
Week 8: Terms and levels
Spin-orbit coupling
Total angular momentum
Week 9: Levels and ionization energies
j-j coupling
Ionization energies for 6p elements
Spend time talking about upcoming topics.
Explain about 4 main topics (content driven), then 3 “interegnums”.
The interegnums are designed to look at some interesting chemistry, relevant to everyday life, or important real world processes, or recent research. The interegnums also tie topics together and usually rely on combining knowledge about several topics.
Both content and interegnums are assessable.

2. MS p2 1 -1 2 ML -2 Pauli forbidden Indistinguishable-1 2 ML -2
Pauli forbidden
Indistinguishable
Demos: charles law (2.3), CO2 density (2.1), star wars (4.3)
Brief mention of 3 states of matter.

3. Working Out Allowed Terms
Pick highest available ML: there is a term with L equal to this ML
Highest ML = 2  L = 2  D term
For this ML: pick highest MS: this term has S equal to this M2
Highest MS = 0  S = 0  2S+1 = 1: 1D term
Term must be complete:
For L = 2, ML = 2, 1, 0, -1, -2
For S = 0, Ms = 0
Repeat 1-3 until all microstates are used up
Highest ML = 1  L = 1  P term
Highest MS = 1  S = 1  2S+1 = 3: 3P term
Strike out 9 microstates (MS = 1, 0, -1 for each ML = 1, 0, -1)
Left with ML = 0  L = 0  S term
This has MS = 0  S = 0  2S+1 = 1: 1S term }
for each value of Ms, strike out microstates with these ML values
Demos: charles law (2.3), CO2 density (2.1), star wars (4.3)
Brief mention of 3 states of matter.

4. MS p2 1 -1 2 ML -2 1D 3P 1S
Demos: charles law (2.3), CO2 density (2.1), star wars (4.3)
Brief mention of 3 states of matter.

5. Check The configuration p2 gives rise to 15 microstates
These give belong to three terms:
1D is composed of 5 states (MS = 0 for each of ML = 2, 1, 0, -1, -2
3P is composed of 9 states (MS = 1, 0, -1 for each of ML = 1, 0, -1
1S is composed of 1 state (MS = 0, ML = 0)
= 15
The three terms differ in energy:
Lowest energy term is 3P as it has highest S (unpaired electrons)
Demos: charles law (2.3), CO2 density (2.1), star wars (4.3)
Brief mention of 3 states of matter.

6. d2 and f2
For p2
6 ways of placing 1st electron, 5 ways of placing 2nd electron (Pauli)
Divide by two because of indistinguishabillty:
Number of combinations for 2 identical objects with 6 choices: nC2
For d2
10 ways of placing 1st electron, 9 ways of placing 2nd electron (Pauli)
Divide by two because of indistinguishabillty:
Demos: charles law (2.3), CO2 density (2.1), star wars (4.3)
Brief mention of 3 states of matter.
For f2
14 ways of placing 1st electron, 13 ways of placing 2nd electron (Pauli)
Divide by two because of indistinguishabillty:

7. MS d2 1 -1 4 3 2 ML -2 -3 -4
Demos: charles law (2.3), CO2 density (2.1), star wars (4.3)
Brief mention of 3 states of matter.

8. d2 and f2 Terms The configuration d2 gives rise to 45 microstates
These give belong to five terms:
1G (L = 4: 2L+1 = 9, S = 0: 2S+1 = 1  9 × 1 = 9 states
3F (L = 3: 2L+1 = 7, S = 1: 2S+1 = 3  7 × 3 = 21 states
1D (L = 2: 2L+1 = 5, S = 0: 2S+1 = 1  5 × 1 = 5 states
3P (L = 1: 2L + 1 = 3, S = 1: 2S+1 = 3  3 × 3 = 9 states
1S (L = 0: 2L +1 = 1, S = 0: 2S+1 = 1  1 × 1 = 1 states
= 45
Demos: charles law (2.3), CO2 density (2.1), star wars (4.3)
Brief mention of 3 states of matter.
The configuration f2 gives rise to 91 microstates
These give belong to seven terms:
1I, 3H, 1G, 3F, 1D, 3P, 1S
L: 0, 1, 2, 3, 4, 5, 6 …
code: S, P, D, F, G, H, I …

9. Hund’s First and Second Rules
1st Rule: the term with the highest value of S lies lowest
lower repulsion between electrons with parallel spins
2nd Rule: if the first rule is ambiguous, the term with the maximum L is lowest
lower repulsion between electrons with like-rotation
p2: 1D 3P 1S
d2: 3F 1D 3P 1S
f2: 3H 1G 3F 1D 3P 1S
Demos: charles law (2.3), CO2 density (2.1), star wars (4.3)
Brief mention of 3 states of matter.

10. Ground Terms Quickly L: 0, 1, 2, 3, 4, 5, 6 …
code: S, P, D, F, G, H, I …
Ground Terms Quickly
Draw out 2l+1 boxes and label with ml values
Place electrons in boxes to maximize S (1st Rule)
Occupy from left to right to maximize L (2nd Rule)
Add ms to get S and ml to get L 1 -1
p1:
L = 1, S = ½  2S+1 = 2
 2P
p2:
L = 1, S = ½ + ½ = 1 2S+1 = 3
 3P
Demos: charles law (2.3), CO2 density (2.1), star wars (4.3)
Brief mention of 3 states of matter.
p3:
L = 0, S = 3 × ½ = 3/2 2S+1 = 4
 4S
p4:
L = 1, S = ½ + ½ = 1 2S+1 = 3
 3P

11. pn Ground Terms L: 0, 1, 2, 3, 4, 5, 6 … code: S, P, D, F, G, H, I … 1-1 L S
ground term p0 1S p1
1/2 2P p2 3P p3
3/2 4S p4 p5 p6
Demos: charles law (2.3), CO2 density (2.1), star wars (4.3)
Brief mention of 3 states of matter.

12. dn Ground Terms L: 0, 1, 2, 3, 4, 5, 6 … code: S, P, D, F, G, H, I … 2-1 -2 L S
ground term d0 1S d1
1/2 2D d2 3 3F d3
3/2 4F d4 5D d5
5/2 6S d6 d7 d8 d9
d10
Demos: charles law (2.3), CO2 density (2.1), star wars (4.3)
Brief mention of 3 states of matter.

13. Ionization Energies: (i) Hund’s 1st Rule
p-block ionization energies: M  M+
Overall increase across each period
Increase in e- / nuclear attraction acts to increase IE and
Increase in e- / e- repulsion acts to lower IE
Increase in nuclear charge is generally more important
Decrease between p3 and p4 (“half filled shell stability”)
Demos: charles law (2.3), CO2 density (2.1), star wars (4.3)
Brief mention of 3 states of matter.

14. pn  pn-1 Effect of ionization-1 L S ΔS p1
1/2 p2 p3
3/2 p4 p5 p6
-1/2
-1/2
-1/2
+1/2
+1/2
+1/2
Overall increase across each period
Increase in e- / nuclear attraction acts to increase IE and
Increase in e- / e- repulsion acts to lower IE
Increase in nuclear charge is in general more important
BUT ionization of p4 increases S
 large reduction in repulsion and so ionization is easier
Demos: charles law (2.3), CO2 density (2.1), star wars (4.3)
Brief mention of 3 states of matter.

15. Ionization Energies: (ii) Hund’s 2nd Rule
3d-block ionization energies: M2+  M3+
Demos: charles law (2.3), CO2 density (2.1), star wars (4.3)
Brief mention of 3 states of matter.
Overall increase with lower IE for d6 than for d5 (1st Rule)
d1-d5 and d6-d10 are not linear (“1/4 and 3/4 shell effects”)

16. dn Ground Terms L: 0, 1, 2, 3, 4, 5, 6 … code: S, P, D, F, G, H, I … 2-1 -2 L S ΔS ΔL d0 d1
1/2
-1/2 d2 3 d3
3/2 d4 d5
5/2 d6
+1/2 d7 d8 d9
d10 -1
harder +1
easier
easier +2 -2
harder -1
harder
Demos: charles law (2.3), CO2 density (2.1), star wars (4.3)
Brief mention of 3 states of matter. +1
easier +2
easier

17. Ionization Energies: fn Lanthanide ionization energies: M2+  M3+
“3/4 shell”
“1/4 shell”
“half filled shell”
Demos: charles law (2.3), CO2 density (2.1), star wars (4.3)
Brief mention of 3 states of matter.

18. Summary Ground terms Occupy orbital ml values to maximize S and L
Ionization energy and Hund’s 1st Rule
“Half filled shell stability” is really reflection of higher e-/e- repulsion when electrons have to pair
Ionization energy and Hund’s 2nd Rule
“1/4” and “3/4 shell stability” is really a reflection of higher e-/e- repulsion when electrons do not orbit in the same direction
Task!
Work out ground terms and explain IE for fn elements
Very important equation - one that you need to remember
surely this should depend on the gas ? … No, as we’ll see soon