Why is the crystal field splitting energy larger for square planar than octahedral complexes?

Question

I recently came across a fact that for the same combination of metal and ligand, the crystal field splitting energy for square planar complexes is larger than that of the corresponding octahedral complex. In fact, I came to know that

Δ_sp is greater than Δ_o by a factor of 1.3 to 1.7.

Is it possible to explain why it is greater theoretically or was it just observed through experiments? And can that factor of 1.3 to 1.7 be justified?

I think in octahedral field placement of ligands both $\mathrm d_{x^2-y^2}$ and $\mathrm d_{z^2}$ orbitals of the metals face direct contact with the ligand (when compared to only $\mathrm d_{x^2-y^2}$ orbital in the corresponding square planar complex) and hence they must experience more repulsion. As the result, there must be greater enhancement of the energy levels and the splitting energy of the octahedral complex must be higher than the corresponding square planar complex. What is wrong in my reasoning?

Answer 1

Think of square planar as if two ligands from octahedral have been removed (the ligands above and below the plane). Now, the energy of the $\mathrm d_{z^2}$ orbital decreases as there are no ligands at the axial position. So, the order of energy changes too:

$$\mathrm d_{x^2-y^2} > \mathrm d_{xy} > \mathrm d_{z^2} > \mathrm d_{xz} = \mathrm d_{yz}.$$

The $\mathrm d_{z^2}$ which was once with $\mathrm d_{x^2-y^2}$ and $\mathrm d_{xy}$ which was once with $\mathrm d_{xz}$ and $\mathrm d_{yz}$ in octahedral splitting, has now split further which has increased the gap between the orbitals $(\mathrm d_{x^2-y^2})$ and $(\mathrm d_{xz}$ and $\mathrm d_{yz}),$ which means the splitting has increased.