Question

The increasing order of $O-N-O$ bond angle in the species ${NO}_2,{NO}_2^{+}$ adn ${NO}_2^{-}$ is

• A. ${NO}_2^{+}<{NO}_2<{NO}_2^{-}$
• B. ${NO}_2<{NO}_2^{-}<{NO}_2^{+}$
• C. ${NO}_2^{+}<{NO}_2^{-}<{NO}_2$
• D. ${NO}_2<{NO}_2^{+}<{NO}_2^{-}$
• E. None of the above

Answer 1

Answer: (E)

None of the above

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 its central atom $(N^{+}$) undergoes sp-hybridisation. Hence, its bond angle is ${180}^o$.

In ${NO}_2^{-}$ ion, N-atom undergoes ${sp}^2$ hybridisation. The angle between hybrid orbital should be ${120}^o$ but one lone pair of electrons is lying on N-atom hence bond angle decreases to ${115}^o$

In ${NO}_2$ molecule N-atom has one unpaired electron in ${sp}^2$ hybrid orbital.The bond angle should be ${120}^o$ but actually, it is ${132}^o$. It may be due to one unpaired electron in ${sp}^2$ hybrid orbital.

Therefore the increasing order of bond angle is
$\underset{{115}^o}{{NO}_2^{-}}<\underset{{132}^o}{{NO}_2}<\underset{{180}^o}{{NO}_2^{+}}$