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Which has the greater N–O bond length, $\ce{NO_2^{-}}$ or $\ce{NO^_3^{-}}$?
I don't understand why the N-O bonds should be of different lengths in these two molecules (don't the same bonds have the same length)? I didn't quite know how to determine this and so I've tried googling the answer, but none of the answers already provided helped clear things up for me... Why does $\ce{NO_3^{-}}$ have a greater N-O bond length?
There are two ways to think of it. First: draw out the (major) contributing Lewis structures for both ions. Each iin has one pi bond, but it's shared between two linkages in $\ce{NO2^-}$ ion versus three linkages in the $\ce{NO3^-}$ ion. So the $\ce{NO3^-}$ ion has less pi bonding in each linkage making those bonds weaker overall.
Second: In all of those Lewis structuresthere are nonbinding pairs of electrons on the oxygen atoms, plus a pair on the nitrogen in $\ce{NO2^-}$. There are more such pairs in $\ce{NO3^-}$ than in $\ce{NO2^-}$. These pairs tend to repel each other and the bonding pairs, and $\ce{NO3^-}$ requires more room, thus longer bonds, to accommodate the greater number of closely packed nonbonding electron pairs.
These two factors tend to occur together when the molecules have nonnetalluc atoms that fill their votes, as here.
You will need to draw the Lewis structure to determine if the bonds are single, double or triple. Remember that single bonds are the longest, triple bonds are the shortest.
For the nitrite ion (NO2-), there is one single bond and one double bond between the nitrogen and oxygen. This leads to a resonance structure that effectively makes the bond lengths 1.5 each
For the nitrate ion (NO3-), there is one double bond that leads to resonance amongst all three of the nitrogen-oxygen bonds. This effectively gives each bond the characteristic of 1.3 bonds, and slightly longer than in the nitrite ion.
