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.

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.

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.

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.

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) 5 + 9 + 1 = 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.

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:

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.

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 9 + 21 + 5 + 9 + 1 = 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 …

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.

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

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.

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.

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.

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.

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”)

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

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.

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