Doesn't the fact that the lone pair on the nitrogen in amides occupy a p orbital contradict Hund's rules?

Question

Since the $\ce{sp^2}$ hybridized orbitals are lower in energy than the p orbital shouldn't the $\ce{sp^2}$ orbitals fill first? Why is this not the case - two electrons fill the p orbitals while all three $\ce{sp^2}$ orbitals are singly occupied? If you look at Carbene (CH2) it's $\ce{sp^2}$ hybridized with a lone pair in one of these orbitals. However, the higher energy p orbital is empty - as would be predicted is Hund's rules are followed. Is it something to do with the conjugation in amides?

Answer 1

two electrons fill the p orbitals while all three $\ce{sp^2}$ orbitals are singly occupied

Not really, hybridization occurs in response to bonding, so we can only talk about an atom's hybridization when it is in a molecule. In amides the nitrogen $\ce{sp^2}$ orbitals form molecular orbitals (MO's) with whatever else they are bound to. In the case of a simple amide with the nitrogen bound to hydrogen, the MO would involve a nitrogen $\ce{sp^2}$ orbital and a hydrogen 1s orbital and 2 electrons would occupy this orbital (one from the nitrogen $\ce{sp^2}$ orbital and one from the hydrogen 1s orbital) - so what was the $\ce{sp^2}$ orbital will be filled with 2 shared electrons when a bond is formed. Therefore the 3 $\ce{sp^2}$ orbitals, now MO's, and the p orbital on nitrogen will each be filled with 2 electrons in the molecule.

Carbene (CH2) it's $\ce{sp^2}$ hybridized with a lone pair in one of these orbitals

Carbenes can exist in either a singlet or triplet state, as shown below. There are 2 possible triplets depending upon how the carbene is hybridized, a bent $\ce{sp^2}$ hybridized carbene, or a linear $\ce{sp}$ hybridized carbene. In the case of methylene ($\ce{:CH2}$), and actually for most carbenes, the triplet states are lower in energy due to Hund's rule (don't pair electrons unless you must); and the triplet has one electron in the p orbital and one in the $\ce{sp^2}$ orbital.