Recently, I have been reading up on the Jahn-Teller effect and I have read articles and posts on it, such as What is the mathematical basis behind the Jahn-Teller effect?. However, many of these rigorously discuss the quantum mechanical basis of the effect (which is fundamentally what it is about). But I seek to understand it using the simpler more intuitive "chemistry" explanation, which is through invoking an understanding of the nature of the d orbitals. Perhaps, the two explanations, both QM and chemistry, are equivalent and they very likely are... But I would just like to focus on the latter.

The reference I have cited was one of the literature I have come across which discussed the chemistry explanation I have mentioned above:

When looking for example at the configuration of $\ce {d^{0}_{x^{2}−y^{2}}}$ $\ce {d^1_{z^{2}}}$ ($\ce {d^4}$) and $\ce {d^{1}_{x^{2}−y^{2}}}$ $\ce {d^2_{z^{2}}}$ ($\ce {d^9}$), the ligand(s) along the z axis are much more screened from the charge of the central metal ion than the other four ligands along the x and y axes, because most of the electron density will be concentrated in the $\ce {d_{z^2}}$ orbital between the metal and the two ligands on the z axis. Consequently, there will be greater electrostatic repulsion between the electron(s) in the $\ce {d_{z^2}}$ orbital and the ligands along the z axis than between the electrons in the $\ce {t_{2g}}$ set of orbitals pointing between the axes and the ligands on the x and y axes itself. The ligands on the z axis will thus move further away from the central metal ion and thus lower the symmetry (z-out or elongation Jahn−Teller distortion). This causes the $\ce {d_{z^{2}}}$ orbital to be more stable (lower in energy) than the $\ce {d_{x^{2}−y^{2}}}$ orbital, as is shown in Figure 2.

As can be read, their explanation points to the elongation of the axial bonds. However, they do not explain how does the other scenario (i.e. compression) result in detail and they simply stated what happens in the other scenario. From what I read from the paper, the argument presented can only lead to one conclusion, and that is the axial bonds elongate. Could only help to fill in the other half of the story? In other words, I would like to know how do we explain the Jahn-Teller effect for the compression of axial bonds using a chemistry-orbitals approach.


Freitag, R., & Conradie, J. (2013). Understanding the Jahn−Teller Effect in Octahedral Transition-Metal Complexes: A Molecular Orbital View of the Mn(β-diketonato)3 Complex. Journal of Chemical Education, 90, 1692-1696. doi:dx.doi.org/10.1021/ed400370p

  • 3
    $\begingroup$ I cannot access the article right now because the ACS website is acting up, but I don't actually like the explanation as quoted: it seems to be circular logic. For the compression case, though, you simply need to replace $z^2$ with $x^2-y^2$ and $z$-axis with $xy$-plane. As an aside, the Jahn–Teller theorem, by itself, does not predict whether elongation or compression is observed. It only predicts that a distortion from $O_\mathrm h$ to $D_\mathrm{4h}$ symmetry will occur. Both elongation and compression fulfil this requirement. $\endgroup$ Commented May 25, 2018 at 23:33
  • $\begingroup$ @orthocresol Thanks for the insightful comment. I can send you the paper if you could give me some mailing address. Also, have you actually heard of this explanation using the nature of d orbitals before? $\endgroup$ Commented May 25, 2018 at 23:42
  • $\begingroup$ @orthocresol So is there any way to predict axial elongation/compression? I have read that electron density measurements performed on the complex ion can help us determine the orbital occupation in the eg set, allowing us to decide on whether elongation (when $\ce {d_{z^2}}$ is populated) or compression (when $\ce {d_{x^{2}-y^{2}}}$ is populated) occurs. $\endgroup$ Commented Mar 25, 2019 at 3:13
  • $\begingroup$ @orthocresol Actually, it isn't quite right to say that it is compression along the z axis but rather an elongation along the x and y axes. Is that right? $\endgroup$ Commented Mar 25, 2019 at 3:17
  • $\begingroup$ (1) looking at the geometry is the easiest way (2) it is both at the same time, but it gets really wordy $\endgroup$ Commented Mar 25, 2019 at 9:34


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.