(ii) The stepwise stability constants of the following complexes in aqueous solution at $\pu{25 °C}$ are given below:

$$ \begin{array}{cll} \hline \ce{M} & \ce{[M(en)2(H2O)2]^2+} & \ce{[M(en)3]^2+} \\ \hline \ce{Cu} & \pu{1E9} & 0.1~\text{(estimated)} \\ \ce{Ni} & \pu{1E6} & \pu{1E4} \\ \hline \end{array} $$

I got this data from an assignment problem. In the case of copper it makes complete sense for the stability constant of formation of $\ce{[Cu(en)_3]^{2+}}$ to be that low given that we have $\mathrm{d^9}$ configuration and strong Jahn–Teller (JT) distortion leading to strain in the rings formed in the complex.

But I can't figure out if there'll be any distortion in the $\ce{[Ni(en)_3]}$ ($\mathrm{d^8}$ configuration). Initially, I assumed there'll be no distortion and the complex would be satisfied with the chelation alone (compromising on the stability provided by JT distortion). But then some of my friends pointed out that $K$ reduces from $10^6$ to $10^4$ so there must be some kind of distortion. After some googling, the only article with some data I could find is by Sohail et al. [1].

In the linked article, there are no comments on JT distortion. Instead, packing effects have been held responsible for the natural distortion of the complex (the last line under Comment section). But I am not entirely convinced; so my questions are:

  1. Would the $\mathrm{e_g}$ orbitals be symmetrically or asymmetrically filled?

  2. If asymmetrically, then why is the JT distortion so profound in the copper(II) complex but weaker in the nickel(II) complex? And if it's symmetrically filled, then what is responsible for the decrease in the stepwise stability constant?


  1. Sohail, M.; Molloy, K. C.; Mazhar, M.; Kociok-Köhn, G.; Khosa, M. K. Tris(ethylenediamine)nickel(II) tetraiodocadmate(II). Acta Cryst E Struct Rep Online 2006, 62 (2), m394–m396. https://doi.org/10/b82dng. (Open Access)
  • $\begingroup$ In the $d^9$ complex the $e_g$ orbitals are equally split in energy (one up , one down by same amount) by the Jahn Teller effect causing an overall lowering in energy, i..e. 2 electrons lowered and one increased in energy. In $d^8$ one electron is in each $e_g$ orbital so if split there is would be no change in overall energy so this will not happen. (The $t_{2g}$ are also split but as they are full there is no overall energy change). $\endgroup$
    – porphyrin
    Nov 1, 2019 at 8:58
  • $\begingroup$ @porphyrin what do you mean by "if split there is would be...." ? $\endgroup$
    – Archer
    Nov 1, 2019 at 9:02
  • $\begingroup$ The $eg$ is doubly degenerate and splits by a very small amount (far, far less than $e_g \to t_{2g}$ split) into two levels (one up and one down by same amount) with a perturbation such as Jahn Teller. (Splitting is far less than spin pairing energy). One electron in each level has the same energy as if no perturbation is present. $\endgroup$
    – porphyrin
    Nov 1, 2019 at 12:14

1 Answer 1


There is a key difference between the $\mathrm d^8$ state and the $\mathrm d^9$ state which lies in exactly that extra electron. The result of the extra electron is that instead of nickel(II)’s $\mathrm{(t_{2g})^6\ (e_g^*)^2}$ state copper(II) has a $\mathrm{(t_{2g})^6\ (e_g^*)^3}$ state. There are two $\mathrm{e_g^*}$ orbitals, thus occupying each with a single electron with parallel spins leads to no degeneracy problems. An additional electron will mean that one of the two $\mathrm{e_g^*}$ orbitals is doubly occupied while the other is singly occupied—but this is a degenerate state as the energy is the same whether it is $\mathrm d_{z^2}$ or $\mathrm d_{x^2-y^2}$ that is doubly occupied.

This degeneracy of possible states in copper(II)—or any other $\mathrm d^9$ system—is the underlying reason for Jahn-Teller distortions. By lowering symmetry from $O_\mathrm h$ to $D_\mathrm{4h}$, these two orbitals are no longer degenerate and they can (and must) diverge in energy. Thus, you have a clear ‘winner’—the lower-energy orbital which receives an electron pair—and a ‘loser’—the higher-energy orbital which receives the single electron. Looking only at $\mathrm d_{z^2}$ and $\mathrm d_{x^2-y^2}$ it doesn’t matter whether the symmetry reduction is performed by moving the $z$-ligands further away or closer but it does make a difference for the other three d orbitals ($\mathrm d_{xy}, \mathrm d_{xz}, \mathrm d_{yz}$) for which lengthening the $\ce{M-L}$ bonds along the $z$ axis proves favourable overall.

A $\mathrm d^8$ system at first glance has nothing to gain by this distortion. Rather than having a stabilisation that benefits two electrons at the expense of one, it can only stabilise one electron at the expense of one. This is energetically not favourable as there is nothing to be gained (but probably something to be lost) so a ‘simple’ Jahn-Teller distortion does not occur; nickel(II) prefers to remain octahedral in $\ce{[ML6]}$ systems.

Note however that there is one possibility when such a stabilisation is beneficial: if the ‘distortion’ is of such a large magnitude that the energy difference between $\mathrm d_{x^2-y^2}$ and the next-lowest orbital is larger than the spin-pairing energy. In this case, we could say that nickel(II) drops from a high-spin state to a low-spin state, although these terms are not usually used in this context. Geometrically, this ‘distortion’ is so significant that we term the resulting complex square planar rather than distorted octahedron; in general there is no longer any evidence of coordination bonding along the $z$ axis. A number of nickel(II) complexes with strong mid-field or strong-field ligands are indeed square planar for this reason such as $\ce{[Ni(dmg)2]}$, the dimethylglyoxime (or diacetyldioxime) complex.


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