Question

The correct order of increasing bond angles in the following triatomic species is

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

Answer 1

The correct option is C $N{O}_2^{-}<N{O}_2<N{O}_2^{+}$

As the number of lone pairs of electrons on the central atom increases, the bond angle decreases.

In $N{O}_2^{+}$ ion, it is isoelectronic with the $C{O}_2$ molecule. It is a linear ion and its central atom $\left(N^{+}\right)$ undergoes $sp$-hybridisation, hence bond angle is ${180}^{\circ }$
In $N{O}_2$ molecule, $N$- atom undergoes $s{p}^2$ hybridisation. So, it adopts bent geometry. The angle between hybrid orbital should be ${120}^{\circ }$ but actually, it is ${132}^{\circ }$. It may be due to one unpaired electron in a $s{p}^2$ hybrid orbital.
In $N{O}_2^{-}$ ion, $N$- atom undergoes $s{p}^2$ hybridisation. So, it adopts bent geometry. For $N{O}_2^{-}$, actual $O-N-O$ bond angle is ${115}^{\circ }$. It's slightly deviated from expected ${120}^{\circ }$ because of repulsion between the interacting bond pairs and lone pair of electrons.