# What is the point group of hexachloridotitanate(III)?

I found the answer to this question to be $D_\mathrm{3h}$ but I am not convinced. Can someone explain me with a diagram if possible and explain why it doesn't belong to the octahedral point group $O_\mathrm{h}$?

• Six Cl can be arranged octahedrically all right, and the charges do not need to be arranged at all, so there must be some other reason... Jahn-Teller distortion, by any chance? – Ivan Neretin May 31 '17 at 12:44
• A figure would help here but if its regular then point group $O_h$ and if two opposite ligands extend or contract as @Ivan Neretin suggests then $D_{4h}$. $D_{3h}$ does not seem to fit at all but would for $\ce{TiCl5}$ – porphyrin May 31 '17 at 12:58
• Yeah, that too. If you distort an octahedron so as to keep the threefold axis, you get $D_{3d}$ at best. – Ivan Neretin May 31 '17 at 13:09
• Trigonal prismatic geometry would be D3h, but that would imply a very large distortion. – orthocresol May 31 '17 at 13:36
• – orthocresol May 31 '17 at 19:06

Getting straight to the point, I believe whichever reference you have that claims $D_\mathrm{3h}$ is incorrect.

Titanium(III) complexes were the first example presented in my coordination chemistry class to examine the energy difference between the $\mathrm{t_{2g}}$ and $\mathrm{e_g}$ orbitals, explain why it exists and where the colour of coordination compounds derives from including carefully examining the UV spectrum of titanium(III). Our lecturer, Professor Klüfers of the LMU Munich, explained that the hexaaquacomplex is practically octahedral in shape. However, there is not just a clean absorption band but rather a band with a shoulder. This is because the excited state of $\ce{[Ti(H2O)6]^3+}$ is Jahn–Teller distorted, having an uneven $\mathrm{e_g}$ population, meaning that two slightly different excitation energies exist.

One might go a step further and argue that even in the ground state a certain distortion should exist. If one were to pull (not push) the ligands in $z$-direction slightly closer to the central metal, this would destabilise any orbitals with $z$ contribution meaning that $\mathrm d_{xy}$ suddenly becomes the single most stable d orbital. This slight distortion which one might call anti-Jahn–Teller (since it is opposite to the classic Jahn–Teller distortion) could explain a reduction of symmetry from $O_\mathrm{h}$ to $D_{\mathrm{4h}}$.

$D_\mathrm{3h}$ symmetry is not consistent with octahedraloid coordination spheres at all. It is very common for pentacoordinated centres such as $\ce{PF5}$. The only way to have six ligands — as in hexachloridotitanate(III) — around a central metal in this point group would be a trigonal prism. That would be a very uncommon coordination sphere and I am sure I would have heard of it in the context of titanium(III) in said course above should it exist for this species. Indeed, the source orthocresol found and linked in the comments makes no mention of any distorted geometry at all:

The low-temperature structure results from rotations of the $\ce{[TiCl6]^3-}$ octahedra around a 4-fold axis. (Emphasis and square brackets mine)

Source: Amels, R.; Kremer, S.; Reinen, D. Jahn-Teller effect of titanium(3+) in octahedral coordination: a spectroscopic study of hexachlorotitanate (TiCl63-) complexes. Inorg. Chem. 1985, 24, 2751–2754. DOI: 10.1021/ic00212a009.

Hexachloridotitanate(|||) has D4h point group because structure distorted by Jahn -Teller distortion, two axial bonds get elongated. Point group can be determined as following..

• Aside from the illustration, this doesn't seem to add anything to the existing answer, which actually explains why there is JT distortion in the first place and cites literature sources, whereas your answer looks like a postulate. Also, please note that | and I are different symbols which are not interchangeable. In the absence of the proper Unicode symbol for Roman numeral three Ⅲ, typically three English letters I are allowed to be used: III. – andselisk Dec 3 '19 at 13:47