[blaisem]

Using the example of Cr³⁺ as a d³ metal, we have the following:

Term symbol: ⁴F

Symmetry in Octahedral splitting:

A_2g
T_2g
T_1g (groundstate)



source

Main question [1]: Is there a rule of thumb for how the splitting terms, T_1g, T_2g, and A_2g are derived, or is this something laymen of group theory read out of a table?

My understanding is that the lowest energy state is labeled T_1g because, in this configuration (see bottom picture), the three electrons in the t_2g orbital can be rearranged in three different ways, hence yielding the T term (where E is used for doubly degenerate combinations and A for a single possibility). However, if this were the rule, I would expect T_2g to have 3x2 possible arrangements; 3 for the t_2g and two for the e_g.  For A_2g, I would expect three possible arrangements for the t_2g term.

Follow up questions [2]: Why is A_2g used over A_1g?  We don't skip the "1" with the T-symmetries.

[3]: I have found a list for the terms of transitions under octahedral splitting for all d-configurations, and they are all gerade.  Is an ungerade configuration possible?

[4]: Is a pairing arrangement, say ²T_g considered a possible transition (see picture below), albeit weak of course due to being spin forbidden?  I believe have seen some Tanabe-Sugano diagrams where ²E_g transitions still appear in optical spectroscopy despite a quintett groundstate.  I can only presume this indicates a transition where a t_2g orbital electron is paired rather than excited to the e_g symmetry orbitals.

I drew this as an example of my understanding of the electronic transitions of a d³ metal:



Thanks for any advice.  Further links are of course also appreciated.

[Corribus]

I am happy to help with this problem but I'm out of town until Tuesday morning with nothing but a smart phone. If you don't get a response from someone else before then, I'll come to your rescue.

[blaisem]

Glad to read this, Corribus The exam is still 3 weeks out, so there's no rush.  As I've continued turning my thoughts over on this, checking various sources, and attempting to reread my professor's barely legible, handwritten notes in German, I've become increasingly paranoid that my current understanding of the material as presented above may actually be completely wrong.  So if my diagram above happens to be confusing and wrong, that would be why.  Of course if the diagram is correct, it's actually a jealous colleague writing this post to discredit me.

[mjc123]

I'm pretty rusty on my term symbols, so I'll leave the detailed answering to Corribus, but in the meantime there are a couple of general points of terminology that suggest you might not be understanding things quite right. (Have you studied symmetry and group theory?)

1. Do you appreciate the distinction between a state and a transition? You appear to use "transition" several times when you are referring to electronic states, e.g. "²E_g transitions". A transition is a change from one state to another, and is referred to using both states, so e.g. ⁴T_2g  ⁴T_1g is a transition between the two states indicated, but ⁴T_2g is a state, not a transition.

2. Your question 3: The d orbitals all have gerade symmetry in O_h, so all d configurations are g. All d-d transitions are g g and therefore parity-forbidden, though they may be vibronically allowed by coupling with vibrational states of u symmetry. Thus e.g. the pink colour of Co(H₂O)₆³⁺ is much less intense than the blue colour of CoCl₄^2-, which has no centre of symmetry.

3. Your question 2: It's not a question of skipping. The states are not labelled 1, 2 etc. in order of energy. These are symmetry labels, you can find them in the character tables. The third level is of the A_2g symmetry type, that's why it has that label. (An A_1g state would be totally symmetrical, and an incomplete shell cannot be that.) The T_1g and T_2g states are so called because they have these respective symmetries - not because they are the "first" and "second" T_g states. Likewise your "²T_g" is wrong - there is no T_g in O_h, it must be either ²T_1g or ²T_2g, even if it is the only ²T state.

[blaisem]

Have you studied symmetry and group theory?

Formally, no.  Just what I have tried to pick up through various classes.  I've studied certain things individually, but it's highly likely I am missing the bigger picture.

1. Do you appreciate the distinction between a state and a transition? You appear to use "transition" several times when you are referring to electronic states, e.g. "²E_g transitions". A transition is a change from one state to another, and is referred to using both states, so e.g. ⁴T_2g  ⁴T_1g is a transition between the two states indicated, but ⁴T_2g is a state, not a transition.

Noted. That makes sense, thank you.

2. Your question 3: The d orbitals all have gerade symmetry in O_h, so all d configurations are g. All d-d transitions are g g and therefore parity-forbidden, though they may be vibronically allowed by coupling with vibrational states of u symmetry.

Is the Laporte rule a significant suppressor of optical transitions? I ask this because in homework assignments involving Tanabe-Sugano diagrams and octahedral complexes, I had the impression we were more concerned about whether a transition is spin forbidden rather than Laporte.  Is it maybe trivial to restore the intensity of a "Laporte-forbidden" transition appreciably, such as substituting one of the ligands in hexaaquacobalt III to induce Jahn-Teller distortion?

3. Your question 2: It's not a question of skipping.These are symmetry labels, you can find them in the character tables. The third level is of the A_2g symmetry type, that's why it has that label. Likewise your "²T_g" is wrong - there is no T_g in O_h, it must be either ²T_1g or ²T_2g, even if it is the only ²T state.

I think I'm missing the connection between d-electron arrangements and their corresponding symmetry representation in a character table. I can determine a state's groundstate spectroscopic term based on the arrangement of the d-electrons; I thought it would be similarly possible to derive its symmetry representation based on the arrangement of d-electrons. I figured this because of the following in my lecture notes:



Thank you for clarifying the question on the numbering of the subscript.  That makes sense.

[mjc123]

Have you studied symmetry and group theory?

Formally, no.  Just what I have tried to pick up through various classes.  I've studied certain things individually, but it's highly likely I am missing the bigger picture.

I'm afraid without a knowledge of group theory it's difficult to understand the symmetry terms, you just have to learn them and accept them. They are shorthand for the different ways a state (or set of degenerate states) can transform under the various symmetry operations of a molecule.

My understanding is that the lowest energy state is labeled T1g because, in this configuration (see bottom picture), the three electrons in the t2g orbital can be rearranged in three different ways, hence yielding the T term (where E is used for doubly degenerate combinations and A for a single possibility). However, if this were the rule, I would expect T2g to have 3x2 possible arrangements; 3 for the t2g and two for the eg.  For A2g, I would expect three possible arrangements for the t2g term.

I've just looked up the Tanabe-Sugano diagram for d³, and you appear to have the labels the wrong way round. (Shows how rusty I am that I didn't spot it sooner.) The ground state is ⁴A_2g, then ⁴T_2g, then ⁴T_1g. In the ground state there is only one way to arrange the 3 electrons between the 3 orbitals. And the lowest doublet state is ²E_g.

But anyway, it isn't as simple as "how many ways can you arrange n electrons?" Take the spherical term, ⁴F - degeneracy 7. How many ways can you arrange 3 electrons in 5 orbitals? The answer isn't 7! (It's 10, with all the spins parallel. Notice on the diagram there's also a ⁴D state - degeneracy 3. A total of 10 states, but not all degenerate as one might naively think.)

[Corribus]

@blaisem

It looks like mjc123 gave you some quality assistance. I won't insert myself into the thread at this point unless there's anything else you need specific help on.

[blaisem]

Sorry for the late response.  I wanted to try to play around with the idea more, but I didn't see much success.  This post is quite long, and as I mention at the end, there's no obligation to explain everything to me fully--or anything at all-- but just maybe there's something interesting here, so I thought I'd share it.  Also writing this down is good mental practice.

I found a pattern in the Mulliken Symbols for the spin-allowed transitions for the ground state spectroscopic term of a high-spin complex, such that the Mulliken symbols in the following picture can be explained for the most part.



The pattern I found is as follows:

1. Draw the d-electrons in their respective orbitals as dictated by the groundstate spectroscopic term.
2. Repeat this for each excited state, maintaining the same spin multiplicity, as we are dealing with spin-allowed transitions
3. Examine the t_2g and e_g symmetries separately. The number of possible ways the electrons can be arranged is the basis of its Mulliken Symbol.  If both t_2g and e_g symbols are occupied by electrons, the one with the higher symmetry prevails.

We can examine the d² through d⁴ spin-allowed transitions as an example, and you can test it with any other d-electron count--it works.



Example: d²

• Beginning with the ground state, we examine the e_g and t_2g symmetries separately.  The e_g is not occupied, so we refer to the t_2g orbitals, where there are three ways to distribute the electrons among the t_2g orbitals, yielding a triply degenerate symbol, T.
• The first excited state can be done similarly, yielding E for the e_g orbitals and T for the t_2g orbitals.  The higher degeneracy, T, takes precedence.
• The final excited state occupies only the e_g orbital; there is one way to arrange the electrons, so it is labeled with A.
  

Perhaps the reasoning seems conspicuously contrived, but it works.  It works for all spin-allowed transitions from the ground state term for d¹ through d⁹ complexes.

There are three problems with it that I ran into.

1. The numeric subscript "1" and "2" don't seem to follow a strict pattern, failing to account for certain exceptions.

• Comparing the d² and d³ states above, the T_1g and T_2g seem to follow different patterns.  This occurs in multiple other examples outside of the high-spin groundstate spectroscopic term, and I can't seem to find a consistent pattern for them.
• My problem is I don't fully understand how the symmetry is applied in general, not to mention when we are talking about a partially filled shell.
• For the issue of applying the symmetry generally, the rules are stated to be more complicated for double and triply degenerate representations -- unfortunately, I didn't find the exact rules for the doubly and triply degenerate case.  In my mind, I consider the t_2g orbitals in an xyz-coordinate system, and I am performing a C₂ rotation along the X-axis to test for a "1" or "2" subscript for each orbital individually.  For the d_xy and d_xz orbitals, this elicits a -1 transformation, corresponding to the appropriate "2" subscript; however, d_yz remains 1 in this instance, so the determination of the overall sign for these degenerate orbitals must follow some other rule.  Maybe both x- and y-axes need to be tested and combined somehow.
• Taking the problem further to include electrons occupying the orbitals, when we compare the d² groundstate and the d³ first spin-allowed excited state, both possess the same Mulliken symbol, only with their numeric subscripts changed. In the d² case in particular, I don't know why it has a "1" subscript when the symmetries for this set of orbitals is t_2g. I assume there is something about how the electrons are distributed that gives orbitals different priorities when testing for their subscript.  At this point, I tried tracking the orbitals via their m_L values, so I could figure out which orbital is occupied in which state and see if testing those orbitals for "1" or "2" got me anywhere. I couldn't find a way to keep the ground state along with all excited states consistent with the original spectroscopic term for both spin and angular moment quantum numbers, which is a puzzling problem in its own right.

2. The pattern falls apart for low spin states

• Taking low-spin d⁴ as an example, we have a ³H₂ spectroscopic term.  The corresponding spin-allowed transitions result in ³E and ³T₂ symmetry states.  This doesn't follow my pattern above at all, which would predict both excited states to be labeled as triply degenerate.

3. Similar inconsistencies as seen in low-spin complexes also occur in higher term symbols or tetrahedral complexes.

When I first approached this topic, I didn't have any handle on the concepts involved.  Talking with mjc prompted more reading, and along with his advice, I do feel more comfortable now, even if I don't understand things fully.  I have the feeling I've gone pretty far on a tangent, beyond what's required and possibly to a depth and detail where explaining it to me would require an inordinate amount of time from any advisers here. In case this doesn't go any further, I wanted to write I already feel something was accomplished and am grateful for that.

[mjc123]

First of all, let me say that for someone who hasn't studied group theory you have done very well to get this far, and establish for yourself some empirical rules of thumb for the electronic states. I will just make some comments, from a group theoretical perspective, that would not be intuitively obvious, and may help supplement your rules of thumb.

1. For degenerate cases, all members of the set must be considered together in determining the symmetry. For example

In my mind, I consider the t2g orbitals in an xyz-coordinate system, and I am performing a C2 rotation along the X-axis to test for a "1" or "2" subscript for each orbital individually.  For the dxy and dxz orbitals, this elicits a -1 transformation, corresponding to the appropriate "2" subscript; however, dyz remains 1 in this instance, so the determination of the overall sign for these degenerate orbitals must follow some other rule.

In this case, the overall character under C2x is the sum of those for the individual orbitals, i.e. -1 -1 +1 = -1, corresponding to T_2g. The set {dxy, dxz, dyz} has T₂g symmetry.

2. There are mathematical rules in group theory for combining states of given symmetry, to obtain the symmetry of the resultant state. So just because you have electrons in t_2g orbitals doesn't mean the state will be T_2g. So for d² you have two t_2g electrons, but the state is T_1g - that's just how it works.
In fact the full answer is T_2g x T_2g = A_1g + E_g + T_1g + T_2g (total of nine because 3 x 3 = 9). The A_1g, E_g and T_2g states are singlet states, while ³T_1g is the ground state.
However, there are some simple rules of thumb you can add to those you already have to help out. (I'm not certain they're valid in every case, I haven't checked, but they seem to be for the high spin cases you have examined.)
(i) A single electron in the t_2g level gives T_2g. Two t_2g electrons give T_1g.
(ii) If you have two electrons in e_g, with A_2g symmetry, the combination with A_2g converts T_1g to T_2g and vice versa (e.g. the third state of d³ and the second state of d⁴).

3. As mentioned before, simple consideration of distributing electrons between orbitals doesn't always give you the answer. Consider the case above for t_2g² singlet states. We could naively conceive two triply degenerate states, one with two antiparallel electrons in the same orbital, and one with the electrons in different orbitals. In fact we get three states, with degeneracies of 1, 2 and 3, and it is hard - or I find it hard - to visualise them in terms of occupancy of orbitals. But that's what comes out of the group theory (and, presumably, the spectroscopy).

[blaisem]

Ok, that all makes sense.  Thanks especially for that rule for triply degenerate states.

In fact the full answer is T_2g x T_2g = A_1g + E_g + T_1g + T_2g (total of nine because 3 x 3 = 9). The A_1g, E_g and T_2g states are singlet states, while ³T_1g is the ground state.

One question: Do you have a link where you found these symmetry states?

[mjc123]

http://www.webqc.org/symmetrypointgroup-oh.html
Which are singlet and which triplet is harder to work out - I just got them from the T-S diagram.