## What hybridization is generally utilized by the central atom in a square planar molecule?

Answer:

Always think about the directionality of each orbital when considering the hybridization. Their directionalities should add up vectorially to give you the appropriate directions and dimensionalities.

I got \(s p^{2} d\) which is also shown below.

DETERMINING THE GEOMETRY

One thing to be careful about is what the \(d\)-electron count is, so that we can realize what geometry this is.

Since each \(\mathrm{Cl}^{-}\) contributes a \(-1\) charge, but each ammine ligand is neutral, platinum is a \(+2 \text { oxidation state. }\) Its electron configuration is then \([\ldots] 5 d^{8}\).

It is known that \(d^{8}\) heavy transition metals form square planar complexes. As a result, this cannot be \(s p^{3}\) hybridization, which is more typical of tetrahedral geometries and a possible error.

DETERMINING THE HYBRIDIZATION

The hybridization should be based on the geometry. We let the complex be on the \(xy\)-plane.

- The \(6 s\) orbital is involved by default. Treat it as if it were an “identity” element.
- Given that the \(6 p_{x}\) and \(6 p_{y}\) orbitals also lie along the coordinate axes, they would also be involved.
- Given that the \(5 d_{x^{2}-y^{2}}\) orbital lies along the coordinate axes (where the ligands lie as well), the \(d_{x^{2}-y^{2}}\) has to be involved.

Thus, the hybridization is \(s p^{2} d\). The \(p^{\prime}\)s and \(d\) provide the directionality and symmetry required to get four identical orbitals along the \(x\) and \(y\) axes.