# How do I determine the crystal field splitting for an arbitrary point group?

How can I figure out the degeneracy of the d orbitals for a site that has a given point group? Specifically I'm interested in $D_{3d}$ and $D_{3h}$, but it would be good to know how to do it in the general case.

If you already know the symmetry of your site then it is quite easy. In a lot of books (e.g. this one) and on this web site you can find the character tables of the point groups supplemented with two additional columns which show the transformation properties of the basis vectors (e.g. $\ce{p}$ orbitals), their rotations and their quadratic combinations (e.g. $\ce{d}$ orbitals). In the following picture I highlighted the important parts in the character table of the $\ce{D_{3\mathrm{h}}}$ group. So in the $\ce{D_{3\mathrm{h}}}$ group the $\ce{d_{z^2}}$ orbital transforms as the irreducible representation $A_{1}^{'}$ and the $\ce{p_{x}}$ and $\ce{p_{y}}$ orbitals transform as the irreducible representation $\ce{E^{'}}$.
After you have identified the irreducible representations of the $\ce{d}$ orbitals you can tell their degeneracy by identifying which $\ce{d}$ orbitals belong to the same irreduzible representation. If two or three $\ce{d}$ orbitals belong to the same irreducible representation they are degenerate. In the case of the $\ce{D_{3\mathrm{h}}}$ group shown below this means that $\ce{d_{z^2}}$ forms an energy level of its own, while $\ce{d_{x^2-y^2}}$ and $\ce{d_{xy}}$ (both belonging to $\ce{E^{'}}$) are degenerate and $\ce{d_{xz}}$ and $\ce{d_{yz}}$ (both belonging to $\ce{E^{''}}$) are degenerate. So, your $\ce{d}$ orbitals split into three energy levels, two of which are doubly degenerate, when your site has $\ce{D_{3\mathrm{h}}}$ symmetry.
• @bobthechemist As far as I know it is not possible to determine an energetic ordering of the levels by symmetry alone. What is possible is to reason that, the more directly the orbital lobes of a certain $\ce{d}$ orbital point towards the ligands the higher is the respective $\ce{d}$ orbital's energy (though this method is not very exact). Do you mean that? – Philipp Jul 21 '13 at 17:05