Is it because all the molecular orbitals that are naturally occupied are all considered to be bonding orbitals? So for benzene, since the lowest three orbitals are all occupied (due to benzene having 6 pi electrons), they're all considered bonding, and any higher unoccupied are considered antibonding?

Is this the right way to think about this? If not, how?


1 Answer 1


No. It is because the lowest three MOs are lower in energy than the corresponding original AOs. In other words, the electrons would rather be there than belong to the separate atoms, as if they wanted the atoms to be together rather than apart. Actually, there is no "as if". These electrons are the force that brings the atoms together. That's exactly what "bonding" means.

The occupancy of orbitals is irrelevant. Sure, the lower-energy ones are occupied first, and they are also more likely to be bonding, but that's about it. Think of the $\ce{O2}$ molecule: it has two electrons sitting on the antibonding $\pi^*$ MO (which also happens to be doubly degenerate, hence the triplet and stuff). On the other hand, boranes are not unlikely to have empty bonding orbitals.

So it goes.

  • 1
    $\begingroup$ Boranes are actually so arranged that their bonding orbitals are all filled. The use of 3c-2e bonding leads to fewer and more strongly bonding orbitals than would be the case with just pairwise interactions. However, electropositive metal borides commonly have boron structures whose bonding orbitals are filled only by adding electrons, which leads to predominantly ionic metal-boron bnding. $\endgroup$ Commented Sep 24, 2021 at 9:56

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