# What kinds of bonds does the d_(x^2-y^2) orbital form in transition metal complexes?

Aug 5, 2017

As it is aligned along coordinate axes, the ${d}_{{x}^{2} - {y}^{2}}$ usually forms $\sigma$ bonds in preference to $\pi$ bonds, seeing as how ligands bond along the coordinate axes as the internuclear axes.

An example is in square planar transition metal complexes... ...or in octahedral transition metal complexes: One "exception" (among others) is trigonal bipyramidal transition metal complexes, where the ${d}_{{x}^{2} - {y}^{2}}$ does not quite provide a direct $\sigma$ interaction. In that case, it only is a partially direct overlap and not an ideal $\sigma$ bond. The ${d}_{{x}^{2} - {y}^{2}}$ orbitals would lie along the coordinate axes defined by ligand "$2$" and the wedge bond, whereas the ligands would lie on the corners of the dashed triangle.

See this answer for further detail on the ${d}_{{x}^{2} - {y}^{2}}$ orbital.

On the other hand, the ${d}_{x y}$, ${d}_{x z}$, and ${d}_{y z}$ tend to form $\pi$ bonds, and rarely $\delta$ bonds. One note here is that usually the internuclear axis is the $z$ axis, but with some exceptions.

The above diagram for octahedral complexes uses the convention that in polyatomic compounds, the $\boldsymbol{y}$ axis is the internuclear axis in the coordinates of the outer (non-central) atoms.
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