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?