# What are the molecular orbitals shaped like in diberyllium (mono)cation?

I think one way of describing the result is that the bond in $\ce{Be^+_2}$? cation is a resonance hybrid of two possible pi bonds. Each pi bond has a bond order of 1 so the resonance hybrid structure has two partial pi bonds. But then can it further be in a resonance hybrid of different rotations and so be an even cylinder?

I guess if you wanted to be silly you could say that that the electron configuration for each beryllium ion is like:

$$\underset{2s}{[\uparrow \downarrow]} \underset{(2p)^2}{[\uparrow \vert \;]} \underset{2p}{[\;]}$$

• The resonance hybrid of the two pi orientations does produce an even cylinder. You don't need to include other rotations. – f'' Jul 23 '16 at 23:31
• @f" Suppose I had two resonant pi orientations such as in carbon dioxide. Can I freely rotate the parts around each other then? – Steven Stewart-Gallus Jul 24 '16 at 6:15
• What do you mean by "freely rotate the parts around each other"? The electron density of the molecule has cylindrical symmetry, if that's what you're asking. – f'' Jul 24 '16 at 6:24

## 1 Answer

Short answer: the molecular orbitals will be $\sigma$ and $\sigma^*$ and thus cylindrically symmetric.

First off, your electron configuration is wrong. $\ce{Be2+}$ will have 3 electrons, so the $\ce{2p}$ orbitals need not be considered.

Here's a quick MO diagram:

So there aren't an $\pi$ bonds at all. There's a $\sigma$ and a $\sigma^*$ orbital, for a bond order of 0.5.

If I generate orbitals (using Avogadro) then I see something like this for the $\sigma$

And this is the $\sigma^*$