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I've only recently started learning about this and I'm stating to get confused. (The orbital in the middle belongs to Be) So you have a + (+) + which gives a sigma bonding orbital and a + (-) + which gives a sigma anti-bonding orbital. Why isn't there a + (-) - which would result in a non-bonding orbital?

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  • $\begingroup$ If you look up the MOs of the allyl cation, it's exactly the same system here, except that you're dealing with s orbitals instead of p orbitals. The nonbonding orbital has no coefficient on beryllium because the antibonding linear combination of hydrogen orbitals ($A_{1u}$) does not have an appropriate symmetry match with the beryllium 2s ($A_{1g}$). Of course, this treatment is a bit simplistic because it neglects the beryllium 2p orbitals entirely; DSVA's answer covers those nicely. $\endgroup$ – orthocresol Nov 14 '16 at 20:12
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Well, you need two sigma bonds here, and two anti-bonding ones. And there are actually two non-bonding orbitals here.

First of all: how you write it down doesn't really work, since there are p-orbitals in Be, not only s-orbitals. You need to look at them too, not only the s-orbitals. And I'm not quite sure if you did, but don't confuse Valence Bond and MO theory.

So for Valence Bond Theory:

We know this molecule is linear. Therefore we need two orbitals on the Be which can form bonds with the hydrogens and are in an 180° angle. So we do sp-hybridization and with one sp-orbital from Be and the s-orbital from hydrogen we can form a $\sigma$ and $\sigma^*$ orbital on each side. Two sigma bonds and two empty p-orbitals on Be which are non-bonding.

For MO Theory:

We construct a $\ce{H2}$ first, in this case we could write + + and + -, so we got a bonding and an antibonding orbital. Now we put our Be in the middle. And now it get's quite complicate to write down, so here on page 10 you can find the MO of $\ce{BeH2}$. You see that the bonding "$\ce{H2}$" orbital was combined with the 2s orbital of Be in an bonding and antibonding way, and the antibonding + - orbital of "$\ce{H2}$" was combined with a p-orbital in a binding and antibinding way. The two other p-orbitals aren't used and are non-bonding, since they would have equal constructive and deconstructive interaction with both "$\ce{H2}$" orbitals. In the end same result as with the valence bond theory.

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  • $\begingroup$ I've been over what it actually looks like, but if I was to start a MO diagram is there no other way to do it other than from starting with H2. $\endgroup$ – Shubham R Nov 14 '16 at 18:31
  • $\begingroup$ Of course there are, you could also just "throw" all three atoms together, but it get's much harder to understand which orbitals will combine to form a new one. In case of molecules with one central atom and "ligands" around it the best way to do this is combine the ligands first $\endgroup$ – DSVA Nov 14 '16 at 19:01
  • $\begingroup$ So if I was to do a MO for H2O, I'd take the same approach. Start with H2 then see how those overlap with the orbitals on the O atom. $\endgroup$ – Shubham R Nov 14 '16 at 19:15
  • $\begingroup$ Yes exactly. That's also described stepwise in the linked pdf starting on page 6, LGOs are the ligand group orbitals, the "H2" molecule. $\endgroup$ – DSVA Nov 14 '16 at 19:34
  • $\begingroup$ @ShubhamR You'll find that this approach is pretty general. For example in one of my answers here you'll see that I constructed the "F6" ligand group orbitals (elsewhere you might also see the terminology symmetry-adapted linear combinations) first before deciding which sulfur AOs to overlap them with. $\endgroup$ – orthocresol Nov 14 '16 at 20:15

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