s-p orbital mixing
/ ess-pee OR-bih-tal MIX-ing /
If you build the simplest molecular orbital diagram for a second-row diatomic, you predict that the head-on sigma2p orbital should sit below the sideways pi2p orbitals. For oxygen and fluorine that prediction is right. But for boron, carbon, and nitrogen it is wrong — there the pi2p orbitals actually lie lower. The fix is a subtle effect called s-p mixing, and ignoring it makes you mispredict whether B2 and C2 are magnetic.
Here is what happens. The 2s-derived sigma orbital and the 2p-derived sigma orbital have the same symmetry along the bond axis, so the rules let them interact with each other, not only with their atomic partners. When they do, they repel in energy, like two people forced apart in a crowded lift: the lower sigma (mostly 2s) drops further, and the upper sigma (mostly 2p) is pushed up. If that push is strong enough, it shoves the sigma2p orbital above the pi2p pair, flipping their order. The pi orbitals, having a different symmetry, are untouched by this mixing.
Whether the flip happens depends on the 2s-2p energy gap. Early in the period (B, C, N) the 2s and 2p energies are close, mixing is strong, and the order flips so pi2p lies below sigma2p. Later (O, F) the larger nuclear charge pulls the 2s far below the 2p, the gap is wide, mixing is weak, and the simple order sigma2p-below-pi2p holds. s-p mixing is the reason a single diagram does not fit all the second-row diatomics, and it is essential for getting B2 and C2 paramagnetic versus diamagnetic right.
Boron's B2 is observed to be paramagnetic. That only makes sense if the pi2p orbitals lie below sigma2p, so its last two electrons go one each into two degenerate pi2p orbitals (unpaired, by Hund's rule). Without s-p mixing the diagram would wrongly pair them in sigma2p and predict B2 diamagnetic.
Strong s-p mixing in light diatomics raises sigma2p above pi2p, which is needed to explain B2's magnetism.
s-p mixing changes only the order of the sigma2p and pi2p levels, not the bond order; the total count of bonding and antibonding electrons is the same either way, so N2's bond order is three regardless.