I'm trying to draw the six π-orbitals (3 bonding, which are filled, and 3 anti-bonding) of a benzene ring when it complexes with metal d-orbitals and which metal orbitals have the correct symmetry ($\sigma$, π, $\delta$) to overlap with each of them. I know that metal orbitals will only form a bonding interaction with ligand orbitals if they have matching symmetry but I am not sure which ones will interact.

I think a π bond is formed between $e_{1g}$ of benzene and $d_{xy}$ and $d_{yz}$. Also a $\sigma$ bond is formed between $a_{2u}$ of the benzene and $d_{z^2}$ and $a_{2u}$. No δ bond is formed.


1 Answer 1


In theory benzene and a transition metal could form σ, π, δ, and even φ bonds as the symmetry allows this.

Here are the orbitals of benzene, calculated at the PBE0/def2-TZVPP level of theory in the D6h point group with Gamess.

pi orbitals of benzene, total density

The (valence) orbitals of chromium on the same level of theory in the same point group are as follows:

valence orbitals of chromium

The linear combination of these orbitals would yield orbitals that have the above mentioned bond symmetries. We will ignore for now that they will most likely not occupied in a complex like bis(benzene)chromium.

  • benzene a2u + Cr s → σ
  • benzene a2u + Cr d → σ
  • benzene e1g + Cr dxz → π
  • benzene e1g + Cr dyz → π
  • benzene e2u + Cr dxy → δ
  • benzene e2u + Cr dx²-y² → δ

If we look at f-orbitals (that is of course not very feasible for chromium), we could construct another bond type:

  • benzene a1g + Cr fx(x2−3y2) → φ

The necessary orbital looks like this:

f-type orbital of chromium

Of course which of these bonds will form or not depends on the metal in question, the occupation of the orbitals and more. After all, there are also p orbitals which can overlap to form σ and π bonds.

Colour code: Occupied orbitals are shown in blue and orange, while virtual orbitals are shown in yellow and red. Phases have been assigned randomly. The total density of the molecule uses a violet contour.


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