Question

Arrange $N{O}_2^{+},N{O}_2^{-}$ and $N{O}_3^{-}$ in the decreasing order of their $N-O$ bond lengths.

• 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 $N{O}_3^{-}>N{O}_2^{-}>N{O}_2^{+}$

$N$ atom in $N{O}_2^{+}$ is $sp$ hybridised whereas in $N{O}_2^{-}$ and $N{O}_3^{-}$, it is $s{p}^2$ hybridised.

$Bondorder=\frac{\text{No.of bond between two atoms in all possible structures}}{No.ofresonatingstructures}$

Bond order of $N{O}_2^{+}$ = 2

Bond order of $N{O}_2^{-}$ :

B.O = $\frac{2+1}{2}$=1.5

Bond order of $N{O}_3^{-}$ :


B.O = $\frac{2+1+1}{3}$=1.33

Since Bond length $\propto \frac{1}{Bondorder}$

So order of bond length is $N{O}_3^{-}>N{O}_2^{-}>N{O}_2^{+}$