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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:

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

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    $\begingroup$ The resonance hybrid of the two pi orientations does produce an even cylinder. You don't need to include other rotations. $\endgroup$ – f'' Jul 23 '16 at 23:31
  • $\begingroup$ @f" Suppose I had two resonant pi orientations such as in carbon dioxide. Can I freely rotate the parts around each other then? $\endgroup$ – Steven Stewart-Gallus Jul 24 '16 at 6:15
  • $\begingroup$ 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. $\endgroup$ – f'' Jul 24 '16 at 6:24
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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:

MO picture for Be2+

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 $\sigma$

enter image description here

And this is the $\sigma^*$

enter image description here

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