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

N atom in ${\mathrm{NO}}_2^{+}$  is sp hybridized whereas in ${\mathrm{NO}}_2^{-}$ and ${\mathrm{NO}}_3^{-}$, it is sp² hybridized.

Bond order = $\frac{\mathrm{No}.\mathrm{of} \mathrm{bond} \mathrm{between} \mathrm{two} \mathrm{atoms} \mathrm{in} \mathrm{all} \mathrm{possible} \mathrm{structures}}{\mathrm{No}. \mathrm{of} \mathrm{resonating} \mathrm{structures}}$

Bond order of ${\mathrm{NO}}_2^{+}$ = 2

Bond order of ${\mathrm{NO}}_2^{-}$ :

B.O = 2+12=1.5

Bond order of ${\mathrm{NO}}_3^{-}$ :

 B.O = 2+1+13=1.33

Since$\mathrm{Bond} \mathrm{length} \propto \frac{1}{\mathrm{Bond} \mathrm{order}}$

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