# Bonding in the nitrate anion [duplicate]

Hi so I was studying chemical bonding where i encountered a problem which is stated below.

When we talk about $${NO_3}^-$$ we draw its structure as following But the thing which I dont understand is that in two of the O atoms there are three lone pairs(ie: $$6 e^-$$) but as we know from O's $$e^-$$ config. (which is $$1s^2 2s^2 2p^4$$) there are two unpaired $$e^-$$ which will take part in bonding as there is no vacant orbitals for $$e^-$$ to get excited to . Now if in this O atom only one sigma bond is present then technically we are left with $$5e^-$$ and not $$6e^-$$ which is shown in the figure.

Similarly in the N atom (whose $$e^-$$ is $$1s^2 2s^2 2p^3$$ ) does not have any other vacant orbital to excite its $$e^-$$ thus showing that it can make only 3 bonds then why is it making 4 bonds( 1 pi and 3 sigma) ?