# What happens if an orbital has no entry in a point group table?

A while ago I cited a few passages from a book on $$f$$-orbital contributions in metal complexes. And although usually the $$4f$$-orbitals are too isolated to really do chemistry and the elctron density is mostly on the f-orbitals there should still be at least some interactions with the ligands to a point where the degenerecy is destroyed. What I'm currently facing is a mixed-anionic Ce(III) complex as capped square antiprism. I know this would be way better dealt with by using band structure calculations. So this not going to be a question about the use of this method but rather a problem I faced when I tried to treat the complex like a $$d$$-orbital complex.

So for this let us consider the folling conditions:

• The mixed-anionic situation would automatically create a non-symmetric polyhedron so we simply reduce it to 9 equal ligands $$L$$.
• For Ce(III) we have a small gap between $$4f$$- and $$5d$$-orbitals. That is why for example Ce(III) or Eu(II) doped luminescent materials are so dependend on their host materials.
• Therefore we expect quite some correlation between the $$f$$- and $$d$$-orbitals. And according to what I read effects like first and second order Jahn-Teller exist for Ce(III) complexes as well, but it becomes quite difficult to calculate them.

So we are talking about a qualitative picture here:

When I looked for examples for ligand field splitting in Ce(III) I found the other $$CN$$ = 9 isomer as tricapped trigonal prism: That means that we are looking for a $${D_{3h}}$$ point group:

$$\begin{array}{c|cccccc|cc} \hline D_\mathrm{3h} & E & 2C_3 & 3C_2 & \sigma_\mathrm{h} & 2S_3 & 3\sigma_\mathrm{v} & & \\ \hline \mathrm{A_1'} & 1 & 1 & 1 & 1 & 1 & 1 & & x^2+y^2,z^2 & x(x^2-3y^2) \\ \mathrm{A_2'} & 1 & 1 & -1 & 1 & 1 & -1 & R_z & & y(3x^2-y^2)\\ \mathrm{E'} & 2 & -1 & 0 & 2 & -1 & 0 & (x,y) & (x^2-y^2,xy)& (xz^2, yz^2), x(x^2+y^2), y(x^2+y^2) \\ \mathrm{A_1''} & 1 & 1 & 1 & -1 & -1 & -1 & & \\ \mathrm{A_2''} & 1 & 1 & -1 & -1 & -1 & 1 & z & & z^3, z(x^2+y^2) \\ \mathrm{E''} & 2 & -1 & 0 & -2 & 1 & 0 & (R_x,R_y) & (xz,yz) & xyz, z(x^2-y^2) \\ \hline \end{array}$$

That means for the $$d$$-orbitals we expect a splitting into $$A'_1$$ + $$E'$$ + $$E''$$ and for the $$f$$-orbitals into $$A'_1$$ + $$A'_2$$ + $$E'$$+ $$A''_2$$ + $$E''$$ And when we look into the group's computational results that matches the point group:

Source: M. Kotzian, N. Rösch, J. Phys. Chem. 1992, 96, 7288-7293.

So what about my case? I am looking for the point group $${C_{4v}}$$.

$$\begin{array}{c|ccccc|cc} \hline C_\mathrm{4v} & E & 2C_4 & C_2 & 2\sigma_\mathrm{v} & 2\sigma_\mathrm{d} & & \\ \hline \mathrm{A_1} & 1 & 1 & 1 & 1 & 1 & z & x^2+y^2, z^2 & z^3, z(x^2+y^2) \\ \mathrm{A_2} & 1 & 1 & 1 & -1 & -1 & R_z & \\ \mathrm{B_1} & 1 & -1 & 1 & 1 & -1 & & x^2-y^2 & z(x^2-y^2) \\ \mathrm{B_2} & 1 & -1 & 1 & -1 & 1 & & xy & xyz \\ \mathrm{E} & 2 & 0 & -2 & 0 & 0 & (x,y),(R_x,R_y) & (xz,yz) & (xz^2, yz^2), (xy^2, x^2y),(x^3, y^3) \\ \hline \end{array}$$

The $$f$$-orbitals can be described with two different sets, the general set and the cubic set leading to:

General Set: $$f_{z^3}$$, $$f_{xz^2}$$, $$f_{yz^2}$$, $$f_{y(3x^2-y^2)}$$, $$f_{x(x^2-3y^2)}$$, $$f_{xyz}$$, $$f_{z(x^2-y^2)}$$

Cubic Set: $$f_{x^3}$$, $$f_{y^3}$$, $$f_{z^3}$$, $$f_{x(z^2-y^2)}$$, $$f_{y(z^2-x^2)}$$, $$f_{z(x^2-y^2)}$$ and $$f_{xyz}$$

Now if I try to match them to the point group I won't be able to include the $$f_{y(3x^2-y^2)}$$ and the $$f_{x(x^2-3y^2)}$$ for the general set and for the cubic set.

What does that mean now? The first similar example I came up with is the $$t_{2g}$$ problem for σ-only $$d$$-metal complexes. As long as we don't have a π-bonded ligand that will give us the $$t_{2g}$$ symmetry three d-metal orbitals will be unpaired and simply won't change in energy. But in that example they are still there. They have a representation in the point group but just no matching ligand. Here we are talking about the opposite case, they would be there in general but they have no representation. So are they just left at the $$f$$-orbital energy level then? And how would I give them a Mulliken symbol in the end?

Sidenote: I also tried a simple computational calculation using ORCA but I guess that I cannot simply print the orbitals for systems that involve $$f$$-orbitals and further more strong correlations. I tried to increase the base to a point where it also considered the $$g$$-orbitals to make sure that it would certainly include the $$f$$-orbitals but it also went up to the point where it calculated the $$f$$-orbitals for the oxide-ligand (I chose oxide as example). And in the end I ended up with an MO-diagram that had 300+ energy levels. So perhaps should be discussed much more qualitatively at this point.

• The $x((x^2-3y^2)$ and $y(3x^2-y^2)$ both belong to E. The direct product of $B_1 \otimes E = E$ in $C_{4v}$ point group. – porphyrin May 21 '19 at 7:38
• Could you please elaborate how you knew that both were E? I guess since they are somehow degenerate this will give you E but why did you then multiply it with B1 here? – Justanotherchemist May 21 '19 at 10:20
• because the squared terms are $B_1$ in $C_{4V}$. – porphyrin May 21 '19 at 13:26