inorganic chemistry - What does the molecular orbital scheme of beryllium chloride and hydride look like?

Beryllium is a somewhat fascinating element since it is the only member of the second group that behaves somewhat non-metalish and, e.g. forms a somewhat covalent chloride and a covalent hydride.

In an answer to a different and unrelated question, I tried guessing what the bonding picture of $\ce{BeCl2}$ would be. Given the monomer’s linear shape, I was inclined to assume something that introduces a four-electron-three-centre bond, which has the nice side-advantage of allowing beryllium to contribute to bonding orbitals using its s orbital only. However, I would guess that the more traditional picture would include two $\mathrm{sp^2}$ type bonds to the neighbouring chlorines and then a nonzero amount of π backbonding making the MO similar to that of $\ce{CO2}$.

What does the actual, calculated MO scheme of $\ce{BeCl2}$ look like; what does $\ce{BeH2}$’s look like, how do they compare and what can we learn from them concerning the bonding properties of beryllium?

Answer

I calculated molecular orbitals for $\ce{BeCl2}$ at MP2/jun-cc-pVDZ level of theory; given below are the doubly occupied MOs at this level of theory along with their symmetry designations and energy in a.u. For some reason my Avogadro kept crashing when I tried to select an iso value I thought appropriate for the last two MOs; anyway these are completely localised $\ce{Be}$ atoms. If we look at the 8 valence orbitals, there is indeed delocalised $\pi$ bonding as you suggest, and do bear a striking resemblance to MOs for $\ce{CO2}$.

For $\ce{BeH2}$ I again ran into trouble generating nice visuals using the cube file in Avogadro. I will update this post with visuals as soon as possible (It would be great if someone can help out).

I got the following doubly occupied MOs at MP2/jun-cc-pVDZ level of theory: -4.644660 ($\ce{a_g}$) ; -0.504366 ($\ce{a_g}$) ; -0.469857 ($\ce{b_{3u}}$). However, it is evident from the symmetry designations that $\pi$-type orbitals aren't involved in bonding.

P.S: If you are interested, I can provide the input, output, and cube files on request.