1. before we did: p2p2 M L & M S Microstate Table States (S, P, D) Spin multiplicity Terms 3 P, 1 D, 1 S Ground state term 3 P

2. For metal complexes we need to consider d 1 -d 10 d2d2 3 F, 3 P, 1 G, 1 D, 1 S For 3 or more electrons, this is a long tedious process But luckily this has been tabulated before…

3. Free Ion terms for various d configurations. We seek to understand The “Same as “ relationships What will happen when we put these free ion into a complex, octahedral or tetrahedral. The spectroscopy involved.

4. Explanation of the mirror symmetry behavior of the free ion GS terms. d n has same terms as d 10-n Easy to understand using the hole formalism. Recall that we are ignoring “core” electrons. We only work with partially filled subshells. Why? Because the core, complete set of atomic orbitals within a subshell is wholly occupied. All the orbital and spin angular momentum components cancel each other out and add to 0. Hole formalism: Consider d 3. That configuration includes many different orbital occupancies. Here is one of those occupancies. It will participate on one or more of the terms. -2012 - 1/2xxx 1/2 Note that L z is -3 We claim d 3 to yield the same terms as d 7

5. Explanation of the mirror symmetry behavior of the free ion GS terms. Aim: d 3 has same terms as d 10-3 = d 7 -2012 - 1/2xx 1/2xxxxx Note that L z is -3 d 7 allows for many different orbital occupancies. Here is one of them. We claim this provides the same L z and S z as one of the d 3 orbital occupancy schemes. Consider the above to be formed from a completed d subshell and thee holes to “create” the missing electrons. The holes look like this -2012 - 1/2hole 1/2

6. Selection rules (determine intensities) Laporte rule g  g forbidden (that is, d-d forbidden) but g  u allowed (that is, d-p allowed) Spin rule Transitions between states of different multiplicities forbidden Transitions between states of same multiplicities allowed These rules are relaxed by molecular vibrations, and spin-orbit coupling

7. Group theory analysis of term splitting in octahedral. Notice the correspondence to the symmetry analysis of the atomic orbitals when they are split by an octahedral field.

8. High Spin Ground States dndn Free ion GSOct. complexTet complex d0d01S1St 2g 0 e g 0 e0t20e0t20 d1d12D2Dt 2g 1 e g 0 e1t20e1t20 d2d23F3Ft 2g 2 e g 0 e2t20e2t20 d3d34F4Ft 2g 3 e g 0 e2t21e2t21 d4d45D5Dt 2g 3 e g 1 e2t22e2t22 d5d56S6St 2g 3 e g 2 e2t23e2t23 d6d65D5Dt 2g 4 e g 2 e3t23e3t23 d7d74F4Ft 2g 5 e g 2 e4t23e4t23 d8d83F3Ft 2g 6 e g 2 e4t24e4t24 d9d92D2Dt 2g 6 e g 3 e4t25e4t25 d 101S1St 2g 6 e g 4 e4t26e4t26 Recall that the set of d atomic electrons are split into a t 2g set and an e g set by repulsion with the negative octahedral ligands, similarly for tetrahedral environment. So also are the free ion terms split by their environment. We seek to understand the mirror like GS terms. d n and d 10-n have the same GS terms for the free ion.. How to explain this via the ligand splitting that we know about (t 2g and e g ).

9. High Spin Ground States Hole Formalism: The d 10 electronic configurations is a spherical cloud of d electrons. We can move to fewer electrons, d 9, d 8, etc. by creating holes in the spherical shell. Ex: d 9 = d 10 + “d-hole” d 10-n electron configuration is equivalent to spherical d 10 + hole n It is the hole n which determines the orbital and spin angular momentum and the terms. Think positive electrons. The term can be obtained from either the actual electronic configuration or the hole configuration. Electronic configuration d 1 has the term 2 D. Ligand splitting occurs for 2 D..

10. High Spin Ground States Again, d 10-n electron configuration is equivalent to spherical d 10 + hole n It is the hole n which determines the orbital and spin angular momentum and the terms. The term can be obtained from either the actual electronic configuration or the hole configuration. But note that ligand splitting of holes will yield inverted splitting relative to electrons. Principle: Electrons are repelled by ligands, but holes attracted to ligands.. Ex: d 3 has t 2g 3 e g 0 electronic configuration (electrons repel ligands). Ex: high spin d 7 is either electronic configuration of t 2g 5 e g 2 (ligands destabilize electrons, favoring the t 2g orbitals) or hole 3 with t 2g 1 e g 2 hole configuration (ligands stabilize positive holes, favoring the e g )

11. High Spin Ground States dndn Free ion GSOct. complexTet complex d0d01S1St 2g 0 e g 0 e0t20e0t20 d1d12D2Dt 2g 1 e g 0 e1t20e1t20 d2d23F3Ft 2g 2 e g 0 e2t20e2t20 d3d34F4Ft 2g 3 e g 0 e2t21e2t21 d4d45D5Dt 2g 3 e g 1 e2t22e2t22 d5d56S6St 2g 3 e g 2 e2t23e2t23 d6d65D5Dt 2g 4 e g 2 e3t23e3t23 d7d74F4Ft 2g 5 e g 2 e4t23e4t23 d8d83F3Ft 2g 6 e g 2 e4t24e4t24 d9d92D2Dt 2g 6 e g 3 e4t25e4t25 d 101S1St 2g 6 e g 4 e4t26e4t26 GS has t 2g negative electron, repelling the ligands. Symmetry is T 2g for d 1 octahedral complex GS has e g positive d-hole, attracted to ligands. Symmetry designation of d 9 octahedral complex is E g = t 2g = e g

12. High Spin Ground States, d 1 and d 9 tetrahedral complexes dndn Free ion GSOct. complexTet complex d0d01S1St 2g 0 e g 0 e0t20e0t20 d1d12D2Dt 2g 1 e g 0 e1t20e1t20 d2d23F3Ft 2g 2 e g 0 e2t20e2t20 d3d34F4Ft 2g 3 e g 0 e2t21e2t21 d4d45D5Dt 2g 3 e g 1 e2t22e2t22 d5d56S6St 2g 3 e g 2 e2t23e2t23 d6d65D5Dt 2g 4 e g 2 e3t23e3t23 d7d74F4Ft 2g 5 e g 2 e4t23e4t23 d8d83F3Ft 2g 6 e g 2 e4t24e4t24 d9d92D2Dt 2g 6 e g 3 e4t25e4t25 d 101S1St 2g 6 e g 4 e4t26e4t26 We have obtained the GS sym for octahedral d 1 and d 9. Now consider tetrahedral. Recall the reversed tetrahedral orbital splitting. For tetrahedral ligand field e electron destabilized less by ligands, t 2 more so. Opposite for holes. For d 1 the electron is e GS. Symmetry is thus E for the d 1 tet complex. For d 9 or d-hole 1 the d-hole symmetry is t 2. The GS complex symmetry is T 2. = e = t 2

13. High Spin Ground States: d 4 and d 6 octahedral and tetrahedral dndn Free ion GSOct. complexTet complex d0d01S1St 2g 0 e g 0 e0t20e0t20 d1d12D2Dt 2g 1 e g 0 e1t20e1t20 d2d23F3Ft 2g 2 e g 0 e2t20e2t20 d3d34F4Ft 2g 3 e g 0 e2t21e2t21 d4d45D5Dt 2g 3 e g 1 e2t22e2t22 d5d56S6St 2g 3 e g 2 e2t23e2t23 d6d65D5Dt 2g 4 e g 2 e3t23e3t23 d7d74F4Ft 2g 5 e g 2 e4t23e4t23 d8d83F3Ft 2g 6 e g 2 e4t24e4t24 d9d92D2Dt 2g 6 e g 3 e4t25e4t25 d 101S1St 2g 6 e g 4 e4t26e4t26 Further hole discussion. A d 4 can be considered as spherical d 5 with a positive hole embedded in it. In octahedral complex the d-hole is e g symmetry. In tetrahedral the d- hole is t 2 symmetry. The GS of the octahedral d 4 complex will have symmetry of the e g hole, E g. Spin is taken from the free ion yielding 5 E g. Reversed splitting in tetrahedral leads to t 2 for hole in d 4 tetrahedral complexes. Taking spin from the free ion GS yields 5 T 2. The d 6 is regarded as a single electron (t 2g ) on top of the spherical d 5 half shell (t 2g 3 e g 2 ). For tetrahedral we have one electron (e) on top of half shell. But the spherical shell has spin which must be allowed for. = t 2g = e = e g = t 2 Using the high spin multiplicities from the free ion this means 5 T 2g for octahedral d 6 and, similarly, 5 E for the tetrahedral d 6.

14. Energy ligand field strength d 1  d 6 d 4  d 9 Orgel diagram for d 1, d 4, d 6, d 9 in both octahederal and tetrahedral complexes.   D d 4, d 9 tetrahedral or T 2 or E T 2g or E g or d 4, d 9 octahedral T2T2 E d 1, d 6 tetrahedral EgEg T 2g d 1, d 6 octahedral Here is what we have: d 1, d 4, d 6, d 9 all have a d shaped object – either a d electron or d-hole – and thus are designated as a D term for the free ion high spin GS. Ligand splitting causes t 2(g) electron to be lower than e (g), reversed in tetrahedral, reversed also for d-holes. Now look at Orgel diagram for these configurations. Look at right part of diagram first An octahederal field splits d 1 into T 2g (GS) and E g d 6 into T 2g (GS) and E g A tetrahedral splits d 4 into T 2 (GS) and E d 9 into T 2 (GS) and E An octahederal field splits d 1 into T 2 and E (GS) d 6 into T 2 and E (GS) A tetrahedral splits d 4 into T 2g and E g (GS) d 9 into T 2g and E g (GS) 0 Both of these represent an electron on top of a spherical shell. Both of these represent a d-hole (reversal) in a tetrahedral field (a second reversal). d-hole reversal Tetrahedral reversal.

15. High Spin Ground States: d 2, d 3, d 6, and d 7 We have taken care of the d 0, d 1, d 4, d 5, d 6, d 9, and d 10 configurations. Now have to do d 2, d 3, d 6, and d 7 configurations. It turns out that all we have to do is solve d 2. We saw earlier that F in octahedral environment splits to A 2g + T 1g + T 2g ; in tetrahedral we would get A 2 + T 1 + T 2. Our problem is the energy ordering. Which is GS? Thus the 3 F GS for d 2 splits into 3 A 2g + 3 T 1g + 3 T 2g. The 4 F GS for d 3 splits into 4 A 2g + 4 T 1g + 4 T 2g. Where did the spin multiplicities come from?? But it is not so easy. Here is our approach: We know the symmetry of the GS of the free d 2 ion. How? We can get the terms for d 2 using the methods applied earlier to p 2, etc. They are 3 F, 1 D, 3 P, 1 G, 1 S. We identify the GS as 3 F. How? But how do we decide on what becomes the GS after the splitting due to the ligands? We use a correlation diagram. It shows the affect of increasing the ligand field strength from zero (free ion) to very high where energy ordering is determined solely by the occupancy of the t 2g and the e g orbitals. d 2 and d 7 both are electrons on top of a spherical shell yielding a splitting pattern: 1, 2, 3 d 3 and d 8 are both two d-holes in a spherical shell, yielding reversed splitting: 3, 2, 1

16. d2d2 Real complexes We have two electrons in the e g orbitals. It can be shown that these give rise to 1 A 1g, 1 E g, and 3 A 2g which have same energy in strong ligand field. Connect the terms of the same symmetry without crossing. Splitting of free ion terms. Similarly, splitting occurs for these occupancies. We have included the 3 T 1g originating from the 3 P. We will need it immediately. Same symmetry as lower energy 3 T 1g from the 3 F. Free Ion termsFree Ion terms Configurations based on splitting of d electrons. Dominant in very strong fields.

17. F P Ligand field strength (D q ) Energy Orgel diagram for d 2, d 3, d 7, d 8 ions d 2, d 7 tetrahedral d 2, d 7 octahedral d 3, d 8 octahedral d 3, d 8 tetrahedral 0 A 2 or A 2g T 1 or T 1g T 2 or T 2g A 2 or A 2g T 2 or T 2g T 1 or T 1g And now d 2 and d 7 in tetrahedral (reversed due to tetrahedral field) and d 3 and d 8 in octahedral (reversed due to d-holes). Note the reversed ordering of the splitting coming from F (T 1 /T 2 /A 2 ). The lower T 1(g) now aims up and should cross the upper T 1(g) but does not due to interaction with the upper T 1(g). Now have strong curvature to avoid crossing. Same symmetry; crossing forbidden T 1 or T 1g First look at d 2 and d 7 in octahedral (2 elecs on a spherical cloud) and d 3 and d 8 in tetrahedral (double reversal: d-holes and tetrahedral) All states shown are of the same spin. Transitions occur between them but weakly. Note the weak interaction of the two T 1, the curvature. This curvature will complicate interpretation of spectra.

18. Move to Tanabe-Sugano diagrams. d 1 – d 3 and d 8 – d 9 which have only high spin GS are easier. Here is d 2. Correlation diagram for d 2. Convert to Tanabe-Sugano. Tanabe-Sugano

19. Electronic transitions and spectra d 2 Tanabe-Sugano diagram V(H 2 O) 6 3+, a d 2 complex

20. Configurations having only high spin GS d1d1 d9d9 d3d3 d2d2 d8d8 Note the two lines curving away from each other.

21. Configurations having either high or low spin GS The limit between high spin and low spin

22. Determining  o from spectra d1d1 d9d9 One transition allowed of energy  o Exciting electron from t 2g to e g Exciting d-hole from e g to t 2g Exciting electron from t 2g to e g

23. Lowest energy transition =  o mixing Determining  o from spectra Here the mixing is not a problem since the “mixed” state is not involved in the excitation.

24. Ground state and excited state mixing which we saw earlier. E (T 1g  A 2g ) - E (T 1g  T 2g ) =  o For d 2 and d 7 (=d 5 +d 2 ) which involves mixing of the two T 1g states, unavoidable problem. Make sure you can identify the transitions!! But note that the difference in energies of two excitations is  o.

25. Can use T-S to calculate Ligand Field Splitting. Ex: d 2, V(H 2 O) 6 3+ Again, the root, basic problem is that the two T 1 s have affected each other via mixing. The energy gap depends to some extent on the mixing! E/B  O /B Technique: Fit the observed energies to the diagram. We must find a value of the splitting parameter,  o /B, which provides two excitations with the ratio of 25,700/17,800 = 1.44 Observed spectrum    17,800 cm -1  2 : 25,700 cm -1 First, clearly  1 should correspond to 3 T 1  3 T 2 But note that the  2 could correspond to either 3 T 1  3 A 2 or 3 T 1  3 T 1. The ratio of  2 /  1 = 1.44 is obtained at  o / B= 31 Now can use excitation energies For  1 : E/B = 17,800 cm -1 /B = 29 yielding B = 610 cm -1 By using 31 =  o /B =  o /610 obtain  o = 19,000 cm -1

26. The d 5 case All possible transitions forbidden Very weak signals, faint color

27. Jahn-Teller Effect found if there is an asymmetrically occupied e set. Can produce two transitions. This picture is in terms of the orbitals. Now for one derived from the terms.

28. Continue with d 9 GS will have d-hole in either of the two e g orbitals. ES puts d- hole in either of the three t 2g orbitals. For example, the GS will have the d-hole in the x 2 -y 2 orbital which is closer to the ligands.

29. Some examples of spectra

30. Charge transfer spectra LMCT MLCT Ligand character Metal character Ligand character Much more intense bands