Question

The decreasing order of $N-O$ bond lengths in $N{O}_2^{+}$, $N{O}_2^{-}$ and $N{O}_3^{-}$ is:

• 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 $N{O}_3^{-}>N{O}_2^{-}>N{O}_2^{+}$

By finding out the bond order of the molecules, we can predict the bond length order.
For $N{O}_2^{+}$,
B.O=2
$Bondorder=\frac{No.ofbonds}{No.ofresonatingstructures}$,
For $N{O}_2^{-}$,
B.O=$\frac{3}{2}=1.5$
For $N{O}_3^{-}$,
B.O=$\frac{4}{3}=1.33$
The order of bond order is:
$N{O}_2^{+}>N{O}_2^{-}>N{O}_3^{-}$
$Bondorder\propto \frac{1}{Bondlength}$
Thus, the order of bond length is just reverse.