I was recently listening to a talk about the following work[1] when the author mentioned that they explained the following. In this $\ce{Au(II)}$-porphyrin complex you observe two sets of $\ce{Au-N}$ bond lenghts. Meaning that there is some sort of distortion in the square plane. They explained it with a second order Jahn–Teller effect that would lead to a rhombohedral distortion.

So I started to think about that situation a bit more. Second-order JT effects are of course favored in heavy metals where the relativistic effects cause the $\mathrm{6s}$ orbital and the $\mathrm{5d}$ orbitals to come closer together in energy. But I always assumed that second order Jahn–Teller required two paired electrons. Nevermind, I'm still thinking about how this would look like. I assume, as it is a regular $\ce{d^9}$-system in a square plane geometry it will start from the tetrahedron with a strong first-order JT distortion to raise the $\mathrm{B_{1g}}$ orbital high enough that it can interact with the $\mathrm{6s}$ orbital.

But when I look for square plane distortions what comes up is a paper on rectangular and rhombohedral distortions on molecules. They say the new point group was $D_\mathrm{2h}$ but I cannnot find any $\mathrm{d}_{x^2-y^2}$ in this character table. And furthermore, at least to what I see what will happen in these non transition-metal complexes is a splitting of the two-fold degenerate low-lying orbitals.

I'm not really good with group theory or orbital interactions but would mixing the $\mathrm{d}_{x^2-y^2}$ orbital with the higher s-orbital according to second order Jahn–Teller cause a rhombohedral distortion?

  1. Preiß, S.; Melomedov, J.; Wünsche von Leupoldt, A.; Heinze, K. Gold(III) tetraarylporphyrin amino acid derivatives: ligand or metal centred redox chemistry?. Chem. Sci. 2016, 7 (1), 596–610. DOI: 10.1039/C5SC03429A.
  • $\begingroup$ Isn't it more likely that it is Oh going to a D2h? d9 config in tetrahedral complexes should not have any gain since you put either eg electron in each its degenerate orbital for zero gain. I am seriously rusty looking at character tables and sorting stuff like this out, so as a comment - are you certain you can rule out Oh -> D2h ? $\endgroup$ – Stian Yttervik Sep 18 '18 at 12:10
  • $\begingroup$ In $D_{2h}$, $x^2-y^2$ should be totally symmetric and so have $A_g$ symmetry label. From a symmetry perspective interacting /mixing with this will not change anything, neither will an S orbital as it is similarly totally symmetric. symmetry. $\endgroup$ – porphyrin Sep 18 '18 at 12:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.