Question

Arrange the following in order of decreasing $N-O$ bond length :

$N{O}_2^{+},N{O}_2{}^{-},N{O}_3{}^{-}$.

• A. $N{O}_3{}^{-}>N{O}_2^{+}>N{O}_2{}^{-}$
• B. $N{O}_3{}^{-}>N{O}_2{}^{-}>N{O}_2{}^{+}$
• C. $N{O}_2{}^{+}>N{O}_3{}^{-}>N{O}_2{}^{-}$
• D. $N{O}_2{}^{-}>N{O}_3{}^{-}>N{O}_2{}^{+}$

Answer 1

The correct option is B $N{O}_3{}^{-}>N{O}_2{}^{-}>N{O}_2{}^{+}$

To answer this question, firstly we have to calculate the bond orders. The higher the bond order is, shorter the bond length.

Nitrogen atom has $2+5=7$ electrons

Oxygen atom has $2+6=8$ electrons

Now $NO$ has 6 electrons in the bonding and 1 in the antibonding.

${NO}_2^{+}$ has 2 electrons lesser than NO; So 6 electrons in the bonding and no electrons in the antibonding.

Thus, $N{O}_2^{+}$,Bond Order=$\frac{6-0}{2}=3$

For ${NO}_2^{-}$, Band order=$\frac{6-3}{2}=1.5$

For ${NO}_3^{-}$, Band order=$\frac{6-4}{2}=1$

The order of Band order is :

${NO}_3^{-}$> ${NO}_2^{-}$> ${NO}_2^{+}$

Hence the correct option is B.