What kinds of bonds does the \(d_{x^{2}-y^{2}}\) orbital form in transition metal complexes?
Answer:
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 y axis is the internuclear axis in the coordinates of the outer (non-central) atoms.