Question

The $O-N-O$ bond angle is maximum in:

• A. $HN{O}_3$
• B. $N{O}_2^{+}$
• C. $HN{O}_2$
• D. $N{O}_2$

Answer 1

The correct option is B $N{O}_2^{+}$

In $HN{O}_3$, N atom is $s{p}^2$ hybridised having a shape of trigonal planar. Nitrogen has no lone pair of electrons. So, the bond angle will be ${120}^o$.

In $N{O}_2^{+}$, nitrogen has zero lone pair of electrons and has two bond pairs. So, it has a hybridisation of $sp$ and has a linear shape. So, the bond angle will be ${180}^o$.

In $HN{O}_2$, nitrogen has a hybridisation of $s{p}^2$ as it has 2 bond pairs of electrons and 1 lone pair of electrons. So, it has a bent or angular shape. But due to l.p - b.p repulsion, the bond angle is around ${111}^o$.

In $N{O}_2$, the hybridisation is $s{p}^2$. The shape is trigonal planar with a bond angle grater than ${120}^o$, due to only one unpaired electron (odd electron) on $N$.