Question

The increasing order of $O-N-O$ bond angle in the species $NO_2,\, NO_2^ +$ and $NO_2 ^-$ is

• A. $NO_{2}^+< NO_{2}< NO_{2}^{-}$
• B. $NO_{2}< NO_{2}^{-}< NO_{2}^{+}$
• C. $NO_{2}^{+}< NO_{2} < NO_{2}^{-}$
• D. $NO_{2}< NO_{2}^{+}< NO_{2}^{-}$

Answer 1

Answer: (C)

$NO_{2}^{+}< NO_{2} < NO_{2}^{-}$

Answer 2

No option is correct. As the number of lone pair of electrons increases, bond angle decreases $NO _{2}^{+}$ion is isoelectronic with $CO _{2}$ molecule. It is a linear ion and its central atom $\left( N ^{+}\right)$undergoes sp-hybridisation. Hence, its bond angle is $180^{\circ}$. $\ln NO _{2}^{-}$ion, $N$-atom undergoes $sp ^{2}$ hybridisation. The angle between hybrid orbital should be $120^{\circ}$ but one lone pair of electrons is lying on $N$-atom, hence bond angle decreases to $115^{\circ}$. In $NO _{2}$ molecule, $N$-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 angle is $\underset{115^{\circ}}{NO^{-}_{2}} < \underset{132^{\circ}}{NO_{2}} < \underset{180^{\circ}}{NO^{+}_{2}}$