Arrange the following in order of decreasing $\mathrm{N}-\mathrm{O} \mathrm{bond} \mathrm{length} : {\mathrm{NO}}_2^{+}, {\mathrm{NO}}_2^{-}, {\mathrm{NO}}_3^{-}$

N atom in ${\mathrm{NO}}_2^{+} \mathrm{is} \mathrm{sp} \mathrm{hybridised} \mathrm{while} \mathrm{in} {\mathrm{NO}}_2^{-} \mathrm{and} {\mathrm{NO}}_3^{-}, \mathrm{it} \mathrm{is} {\mathrm{sp}}^2 \mathrm{hybridised}.$

O  =  N  = O                                                                            Bond order = 2

Bond order= $\frac{2+1}{2}=1.5$

Bond order $=\frac{2+1+1}{3}=1.33$

Bond order $\propto \frac{1}{\mathrm{Bond} \mathrm{Length}}$

So, Bond length order is ${\mathrm{NO}}_3^{-} > {\mathrm{NO}}_2^{-} > {\mathrm{NO}}_2^{+}$