Question

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

Answer 1

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 `sp^2d`, which is also shown below.

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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 `"Cl"^-` contributes a `-1` charge, but each ammine ligand is neutral, platinum is a `bb(+2)` oxidation state. Its electron configuration is then `[...]5d^8`.

It is known that `d^8` heavy transition metals form square planar complexes. As a result, this cannot be `sp^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 `6s` orbital is involved by default. Treat it as if it were an "identity" element.
• Given that the `6p_x` and `6p_y` orbitals also lie along the coordinate axes, they would also be involved.
• Given that the `5d_(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 `color(blue)(sp^2d)`. The `p`'s and `d` provide the directionality and symmetry required to get four identical orbitals along the `x` and `y` axes.