Coordination geometry: why is of Cu(acac)2 square planar?

Question

Why is $\ce{Cu(acac)2}$ square planar, whereas $\ce{Al(acac)3}$ and $\ce{Fe(acac)3}$ form octahedral complexes? I see how a similar argument to $\ce{Pt}$ complexes would apply, i.e. with $\mathrm{d^9}$, there is only one electron that goes in the very destabilised $2\mathrm{b_{1g}}$ (corresponding to $\mathrm{d}_{x^2-y^2}$) in the square-planar orbital splitting, whereas the rest is preferentially stabilised. However, there are many examples of Cu(II) forming octahedral compounds, in which acac would still bridge only $90^\circ$. Is this something to do with the acac ligands being (pseudo-)aromatic?

Thanks!

Answer 1

With bidentate ligands of the form $\ce{LL'}$, it is difficult to force triple coordination to a copper(II) centre. I don't have data for acac offhand, but I have some data for en binding to Cu. (Source: some past exam question somewhere.)

$$\begin{array}{c|ccc} n & 1 & 2 & 3 \\ \hline \log K_n & 10.55 & 9.05 & -1.0 \end{array}$$

The exact numbers don't matter; only the trend does. (Which is why I'm lazy to find a better source.) The binding of the third ligand, which would necessitate coordination at the axial position, is very weak. To get to a $\ce{[Cu(acac)3]}$ complex, you would have to either 1) adopt a perfect octahedral environment and sacrifice the electronic stabilisation from the Jahn-Teller distortion; or 2) stretch two of the acac ligands to a longer length than they would like to be.

As for why axial elongation is favoured over compression, it is perhaps due to electronic reasons: under elongation, the $\mathrm{d}_{z^2}$ orbital is doubly occupied and the $\mathrm{d}_{x^2-y^2}$ orbital is singly occupied. On the other hand, under compression, $\mathrm{d}_{z^2}$ orbital is singly occupied.

Now, there is a second-order mixing which allows the $\mathrm{4s}$ orbital to mix with the $\mathrm{d}_{z^2}$, since they possess the same symmetry ($A_\mathrm{1g}$) under $D_\mathrm{4h}$ symmetry; in the process the $\mathrm{d}_{z^2}$ orbital is stabilised, and the $\mathrm{4s}$ destabilised. The destabilisation of the $\mathrm{4s}$ orbital doesn't matter, since it's empty. However, the stabilisation of the $\mathrm{d}_{z^2}$ orbital is, of course, more profound when there are more electrons in the $\mathrm{d}_{z^2}$ orbital, which corresponds to elongation.

Here's a diagram from my notes.