When we're taught MO theory in basic chemistry, a rule-of-thumb hammered into our heads is that an orbital takes up to 2 electrons with opposite spins. Sigma or pi bonds are represented by stable bonding orbitals. Orbitals sometimes have just the one electron (radicals) but are much more stable with the pair.

While this justifies the occurrence of one-electron bonds, like that of the dihydrogen cation, it does not smoothly explain the occurrence of three-electron bonds like in nitric oxide and dioxygen. If we draw a straightforward MO diagram for nitric oxide with the "two electrons per orbital" rule, we end up with a lone electron in a antibonding pi orbital. "Antibonding" seems to imply the lone electron is sitting at its host atom, but that's not the case. Where does this lone electron go? Does it "merge" with one of the bonding sigma or pi orbitals and if so how do we know which one? Is this just a shortcoming of MO theory?

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    $\begingroup$ There's nothing inherently wrong with an electron sitting in an antibonding orbital. All "antibonding" means is that when you put an electron in it, the bond order decreases. It doesn't necessarily say anything about where the electron is - it is still delocalised over both atoms. And no, you can't merge two orbitals. $\endgroup$ Jan 10, 2019 at 1:08
  • $\begingroup$ Yes, I'm aware that electrons can sit in antibonding orbitals. We often see them "cancelling" out bonding orbitals (e.g. why dihelium doesn't form). But here's a case of an unpaired electron sitting in an antibonding orbital that is known to participate in bonding. $\endgroup$
    – BatWannaBe
    Jan 10, 2019 at 1:12
  • $\begingroup$ Mmhmm. All that means is that the "antibonding-ness", if you will, is less than the "bonding-ness". And so you have a weak(er) bond. $\endgroup$ Jan 10, 2019 at 1:12
  • $\begingroup$ Well, we can use NO as an example. mgh-images.s3.amazonaws.com/9781111779740/508967-10-99EEI1.png We can ignore the bottom bit, because it's not relevant. There's a σ orbital with 2 electrons, but the corresponding σ* orbital at the very top is unfilled, so that makes one σ bond between N and O. For the π orbitals, there are 4 bonding electrons and 1 antibonding electron, so we have a (4-1)/2 = 1.5 π bond. Altogether we have a 2.5-bond between N and O. $\endgroup$ Jan 10, 2019 at 1:18
  • $\begingroup$ So how can we know where the electron goes in the molecule? What makes NO have a three-electron bond instead of just normal two-electron bonds plus a lone electron in an antibonding orbital pulling the nuclei apart? Or does MO theory simply not account for such detail? $\endgroup$
    – BatWannaBe
    Jan 10, 2019 at 1:26


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