[science123]

My professor told that a molecule will try to be in the lowest energy and most stable state. So if we say that all the atoms in ring are sp2 due to resonance, then we have to count the lone pairs of both the nitrogen (the lone pairs have to be in the p orbital since the 3 sp2 orbitals are used to form bonds). So if we count both the lone pairs, and the two double bonds we would have total 8 pi electrons making it anti-aromatic. So the question is can we count the lone pair of only one nitrogen to have 6 pi electrons making the ring aromatic and more stable or we consider both the nitrogens as sp3 and make the molecule non-aromatic (which is at least more stable than being anti-aromatic) ?

[Smrt guy]

In order to have continuous overlap of the orbitals, both N lone pairs would have to be incorporated, making the molecule anti-aromatic.  The molecule will bend out of planarity and/or orient the lone pairs such that continuous overlap is not possible, thus resulting in a nonaromatic structure.

[science123]

In order to have continuous overlap of the orbitals, both N lone pairs would have to be incorporated, making the molecule anti-aromatic.  The molecule will bend out of planarity and/or orient the lone pairs such that continuous overlap is not possible, thus resulting in a nonaromatic structure.

Do u mean to say that it will try to have the lone pair in an sp3 hybridized orbital in order to escape anti-aromaticity?

[Arctic-Nation]

Exactly. The nitrogen atoms will be sp3 hybridized, though I think inversion will still be possible.