Delocalization of π-electrons in nitrate ion

Question

Benzene and nitrate ion are given in my textbook as examples for the delocalization of π-electrons. Benzene, due to symmetry of its resonating structures, is simple enough. We assume that σ-electrons are localized and π-electrons are delocalized in the ring.

Each carbon atom promotes one electron from its $\mathrm{s}$ orbital to the empty $\mathrm{2p}$ orbital. It hybridizes two $\mathrm{p}$ orbitals with the $\mathrm{s}$ to form three $\mathrm{sp^2}$ orbitals which it then uses to form three σ-bonds, two with other carbons and one with a hydrogen.

The remaining unhybridized $\mathrm{p}$ orbital then sticks out above and below the plane of the ring. The electrons from all the six unhybridized $\mathrm{p}$ orbitals of the six carbons are then delocalized above and below the plane of the ring.

However, when I try to apply a similar reasoning to the nitrate anion, problems arise. The resonance Lewis structures of nitrate ion are:

Now, assuming again that only the π-electrons are delocalized, we would expect that only two electrons are delocalized (since there is only one double bond). But my textbook claims that each atom is $\mathrm{sp^2}$-hybridized, and as in benzene, one unhybridized $\mathrm{p}$ orbital per atom sticks out above and below the plane of the molecule.

Assuming this to be true, I presumed that two electrons would be localized and tried to arrange the remaining 22 electrons into a Lewis structure with $\mathrm{sp^2}$ hybridized atoms. I couldn't find any viable structures with only three lone pairs around each atom.

What is the exact mechanism for delocalization of electrons in nitrate? Is there a general scheme to this mechanism that applies to all such similar molecules such as $\ce{CO3^2-}?$

Answer 1

Unfortunately the key to understanding the delocalisation of electrons lies in understanding rudimentary molecular orbital theory. It is also necessary to understand that hybridisation is a mathematical concept which can be used to describe bonding. It is no necessity to have bonding.

Your explanation of benzene is the most common description, but it is somewhat incomplete. First, you are right, one can consider the carbon atoms having three (roughly) sp² hybrid orbitals, giving rise to two σ bonds to the neighbouring carbons and one σ bond to hydrogen. The remaining 6 perpendicular p orbitals form the delocalised π system. These orbitals form 6 new orbitals and due to symmetry constraints, i.e. D_6h, three of them are bonding and three of them are antibonding. In the π system, there are six electrons, so only the bonding orbitals will be occupied. This makes the extraordinary stability of these compounds. The following picture schematises this on the left side, while on the right side these are the orbitals obtained by a quantum chemical calculation. (At the DF-BP86/def2-SVP level of theory.)

The nitrate ion $\ce{NO3-}$ is what I like to call a Y-aromat since it has some similarity in the electronic structure. The bonding situation is very comparable to benzene.
The resonance structures, however, look much more complicated.

In quite good approximation, you can regard the nitrogen as having three sp² hybrid orbitals (and one p orbital) and the oxygen as (roughly) having two sp (and 2 p) orbitals. You follow the same scheme, therefore you obtain four equivalent nitrogen oxygen σ bonds. Each oxygen has one lone pair in the in-plane p orbital and one lone pair in the sp hybrid. The remaining out-of-plane p orbitals of nitrogen and oxygen build up the delocalised π system. This time there are only four orbitals. Because of symmetry constraints, i.e. D_3h, one is bonding, two are (close enough to) non-bonding, and one is antibonding.* With the remaining six electrons you only fill the bonding and non-bonding orbitals. The concept is similar. Below you find the schemes like the ones above for benzene.

Since the carbonate anion $\ce{CO3^{2-}}$ is isoelectronic to nitrate, approximately the same orbital picture will result.

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* Strictly speaking, there are no non-bonding orbitals. The classification can only be with respect to a bonding axis, and there are only two options: 1. there is no nodal plane (or surface) perpendicular to that bonding axis; it is then called bonding. 2. There is a nodal plane (or surface) perpendicular to that bonding axis; it is the called antibonding. A molecular orbital can be bonding respectively to one bonding axis, and antibonding with respect to another.
For example, the E" orbitals of nitrate have this feature, since the nodal surface curves through the nitrogen atom. The above depicted orbital on the left is bonding with respect to one of the $\ce{N-O}$ bonds and antibonding towards the other. (It is also bonding with respect to a $\ce{O\bond{~}O}$ interaction.) The orbital on the right is exactly opposite (orthogonal) to that.