Question

The hybridization of orbitals of $N$ atom in $N{O}_3^{-},N{O}_2^{+}$ and $N{H}_4^{+}$ are respectively:

• A. $sp,s{p}^2,s{p}^3$
• B. $s{p}^2,sp,s{p}^3$
• C. $sp,s{p}^3,s{p}^2$
• D. $s{p}^2,sp,sp$

Answer 1

Answer: (B)

$s{p}^2,sp,s{p}^3$

Maximum covalency of nitrogen is $5$.

In $N{O}_3^{-}$, there are $3$ bond pairs and hence, $s{p}^2$ hybridization.

In $N{O}_2^{+}$, there are $2$ bond pairs and hence, $sp$ hybridization.

In $N{H}_4^{+}$, there are $4$ bond pairs and hence, $s{p}^3$ hybridization.

Hence, option B is correct.