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

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

Δsp is greater than Δo 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?

EDIT:

My attempt:

In Octahedral field placement of ligands,both dx2-y2 and dz2 orbitals of the metals face direct contact with the ligand (when compared to only dx2-y2 orbital in the corresponding square planar complex) and hence they must experience more repulsion.Due to this,there must be greater enhancement of the energy levels and hence the splitting energy of the octahedral complex must be higher than the corresponding square planar complex.What is wrong in my reasoning?

• I try to guess the answer with this reasoning. When four ligands L are at the corners of a horizontal square and around the metal M, the bond length ML takes what I call a reference value. If I add a 5th and a 6th ligand above and under the metal, they may interact with the four first ligands, and repel them a tiny bit, so that the bond ML is also increased. And the intensity of the $\Delta$ effect is slightly decreased – Maurice May 16 '20 at 11:45