Chemistry - STRUCTURAL ISOMERISM Question with Solution | TestHub

Key Idea : As the number of lone pair of electrons increases, bond angle decreases. ${\mathrm{NO}}_2^{+}$ion is isoelectronic with ${\mathrm{CO}}_2$ molecule. It is a linear ion and its central atom $\left({\mathrm{N}}^{+}\right)$ undergoes $sp$-hybridisation, hence bond angle is $180^{\circ }$. In ${\mathrm{NO}}_2^{-}$ion, $\mathrm{N}$-atom undergoes $s{p}^2$ hybridisation. The angle between hybrid orbital should be $120^{\circ }$ but one lone pair of electrons is lying on $\mathrm{N}$-atom, hence bond angle decreases to $115^{\circ }$. In ${\mathrm{NO}}_2$ molecule, $\mathrm{N}$-atom has one unpaired electron in $s{p}^2$-hybrid orbital. The bond angle should be $120^{\circ }$ but actually it is $132^{\circ }$. It may be due to one unpaired electron in $s{p}^2$-hybrid orbital. Therefore, the increasing order of bond angles is : $$\underset{\left(115^{\circ }\right)}{{\mathrm{NO}}_2^{-}}<\underset{\left(132^{\circ }\right)}{{\mathrm{NO}}_2}<\underset{\left(180^{\circ }\right)}{{\mathrm{NO}}_{+}^{+}}$$