Why is it the case that for Jahn-Teller effects to occur in transition metals there must be degeneracy in either the $\mathrm{t_{2g}}$ or $\mathrm{e_g}$ orbitals?

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    $\begingroup$ Jahn-Teller effect is way broader than transition metals, or t2g, or eg. $\endgroup$ – Ivan Neretin Nov 15 '19 at 20:02
  • $\begingroup$ What do you mean? But for transition metals there must be degeneracy $\endgroup$ – ChemEng Nov 15 '19 at 20:05
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    $\begingroup$ I mean just what I said, Sure, there must be degeneracy in any case. But even when occurring in transition metals, Jahn-Teller effect is not restricted to t2g/eg thing. $\endgroup$ – Ivan Neretin Nov 15 '19 at 20:07
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    $\begingroup$ A nice non-inorganic case of JT distortion, let me mention that, is that of cyclobutadiene: in its equilibrium structure (singlet state), it "escapes" antiaromaticity (and degenerancy) with asimmetry. The reason of JT distortion is pretty much the same in inorganic chemistry: an asymmetric (so, non-degenerate) state is energetically favoured compared to the degenerate and symmetric alternative $\endgroup$ – The_Vinz Nov 15 '19 at 20:24

In the abstract to the original Jahn-Teller paper it is stated:[1]

We shall show that stability and degeneracy are not possible simultaneously unless the molecule is a linear one.

Thus the Jahn-Teller effect is specifically about the removal of a degeneracy, and that that will always lower the total energy of the system. Thus you can not have a Jahn-Teller effect without a degeneracy in the reference, fully symmetric state.

[1]: H. A. Jahn, E. Teller, Proc. Royal Soc. A 1937, 161, 220–235. DOI: 10.1098/rspa.1937.0142.

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  • $\begingroup$ That is correct. But the JT theorem concerns the degeneracy of a state, not the degeneracy of two (or more) orbitals, which is what OP is asking about. There is almost certainly a direct link, which would be insightful to flesh out (I am unwilling to think about it properly now, although I think it should be easy to show that an uneven population of degenerate orbitals leads to a degenerate state). $\endgroup$ – orthocresol Nov 16 '19 at 1:05
  • $\begingroup$ Good point - I mentally missed the word orbital $\endgroup$ – Ian Bush Nov 16 '19 at 11:35

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