# Why do tetrahedral complexes have approximately 4/9 the field split of octahedral complexes?

I am trying to calculate the relationship between the octahedral field splitting parameter ($\Delta_\mathrm{o}$) and the square planar field splitting parameter ($\Delta_\mathrm{sp}$) and thought a good way to start would be to understand why $\Delta_\mathrm{t}=4/9\Delta_\mathrm{o}$ for tetrahedral fields but couldn't find any information on it.

How does one arrive at this particular ratio mathematically?

• – Loong Oct 4 '16 at 18:32
• I decided to edit and vote for reopening. The linked duplicate asks why one is greater than the other, this one asks about the numerical difference and how it is derived. – Jan Dec 23 '16 at 1:24

The ratio is derived in The angular overlap model. How to use it and why J. Chem. Educ., vol. 51, page 633-640.

First, relative expressions (Shaffer Angular Overlap Factors) are derived for overlap integrals between metal and ligand orbitals as a function of angles.

For ligand sigma orbitals, the expressions are a functions of 2 angles, specifying the direction of the ligands in the coordinate system of the metal.

For pi-x and pi-y ligand orbits, there is a third angle for orientation about the ligand-metal axis.

Then, assuming the ligand-metal distances are the same in both the tetrahedral and octahedral cases, and considering the coordinate geometry of the ligands in the two cases, the relationship:

$$\Delta_{T} = -4/9 \Delta_{O}$$

is obtained.

For another discussion of this ratio in the framework of the angular overlap model, as well as extension to other geometries such as square planar, see A New Look at Structure and Bonding in Transition Metal Complexes, Advances in Inorganic Chemistry and Radiochemistry, Volume 21, pages 113-146.