Forming the resonance structures for the nitrite anion Yesterday I learned about resonance structures and came across this problem.

The resonance structures are formed when there is a delocalized π electron cloud. I understand that for benzene, there are two resonance structures. For making π bonds I know that there are three main reasons:


The two orbitals should overlap each other.
There should only be one electron for each orbital.
The two orbitals should be in one level.


In benzene there are six $\mathrm{p}_z$ unhybridized orbitals, therefore they can share their electrons with each other (delocalize the π bond electrons). So benzene has 2 resonance structures.

But I don't understand how this works for the nitrite anion, $\ce{NO2-}$.



In $\ce{NO2-}$ there is also π electron delocalization, so there are 2 resonance structures. Now I am going to say how I think the delocalization is happening.

When the first structure changes into the second structure, nitrogen has to make a π bond with the right side oxygen with each of the $\mathrm{p}_z$ orbitals before, but now that π bond's two electrons take to the right side oxygen, and then there are 2 electrons in the right side $\mathrm{p}_z$ orbital and no electrons in the nitrogen $\mathrm{p}_z$ orbital. But the left side oxygen (in the first structure, $\ce{O-}$) has 2 electrons in its $\mathrm{p}_z$ orbital. Then the nitrogen $\mathrm{p}_z$ and oxygen $\mathrm{p}_z$ overlap each other and make a π bond.

I have said before "There should only be one electron for each orbital." But, according to my understanding above, the π bond is made by $\mathrm{p}_z$ with 2 electrons (oxygen) and $\mathrm{p}_z$ with no electrons (nitrogen) overlapping each other.

So, is there any mistake in my imagination?
Yes. I think that there is a mistake in your imagination. Nitrogen is evidently $sp^2$-hybridised in this case. Nitrogen has 5 valence electrons. Two electrons are used to make 2 $\sigma$ bonds with two hybrid orbitals. Another two electrons become the lone pair and reside also in a hybrid orbital. This leaves us with one electron in one unhybridised $p$ orbital. This $p$ orbital is then used to make the $\pi$ bond to either oxygen atom, contributing one electron to the $\pi$ bond while the oxygen atom also contributes one electron from its $p$ orbital.