Question

The hybridization of atomic orbitals of nitrogen in ${\mathrm{NO}}_2^{+}$ ,${\mathrm{NO}}_3^{-}$and ${\mathrm{NH}}_4^{+}$are:

• A. $\mathrm{sp},{\mathrm{sp}}^2\mathrm{and}{\mathrm{sp}}^3$ respectively
• B. ${\mathrm{sp}}^2,{\mathrm{sp}}^3\mathrm{and}\mathrm{sp}$ respectively
• C. ${\mathrm{sp}}^3,{\mathrm{sp}}^2\mathrm{and}\mathrm{sp}$ respectively
• D. ${\mathrm{sp}}^3,\mathrm{sp}\mathrm{and}{\mathrm{sp}}^2$ respectively

Answer 1

The correct option is A

$\mathrm{sp},{\mathrm{sp}}^2\mathrm{and}{\mathrm{sp}}^3$ respectively

Explanation for the correct option:

(a) $\mathrm{sp},{\mathrm{sp}}^2\mathrm{and}{\mathrm{sp}}^3$ respectively

• Since the hybridizations are $\mathrm{sp},{\mathrm{sp}}^2\mathrm{and}{\mathrm{sp}}^3$ respectively
• so this option is correct.

Explanation for the incorrect option :

(b)${\mathrm{sp}}^2,{\mathrm{sp}}^3\mathrm{and}\mathrm{sp}$ respectively

• Since the hybridizations of ${\mathrm{NO}}_2^{+}$ ,${\mathrm{NO}}_3^{-}$and ${\mathrm{NH}}_4^{+}$are $\mathrm{sp},{\mathrm{sp}}^2\mathrm{and}{\mathrm{sp}}^3$ respectively
• But this option states hybridization to be ${\mathrm{sp}}^2,{\mathrm{sp}}^3\mathrm{and}\mathrm{sp}$ respectively
• Hence this option is not correct

(c ) ${\mathrm{sp}}^3,{\mathrm{sp}}^2\mathrm{and}\mathrm{sp}$ respectively

• Since the hybridizations of ${\mathrm{NO}}_2^{+}$ ,${\mathrm{NO}}_3^{-}$and ${\mathrm{NH}}_4^{+}$are $\mathrm{sp},{\mathrm{sp}}^2\mathrm{and}{\mathrm{sp}}^3$ respectively
• But this option states hybridization to be ${\mathrm{sp}}^3,{\mathrm{sp}}^2\mathrm{and}\mathrm{sp}$ respectively
• Hence this option is not correct

(d) ${\mathrm{sp}}^3,\mathrm{sp}\mathrm{and}{\mathrm{sp}}^2$ respectively

• Since the hybridizations of ${\mathrm{NO}}_2^{+}$ ,${\mathrm{NO}}_3^{-}$and ${\mathrm{NH}}_4^{+}$are $\mathrm{sp},{\mathrm{sp}}^2\mathrm{and}{\mathrm{sp}}^3$ respectively
• But this option states hybridization to be ${\mathrm{sp}}^3,\mathrm{sp}\mathrm{and}{\mathrm{sp}}^2$ respectively
• Hence this option is not correct

Hence (a) $\mathrm{sp},{\mathrm{sp}}^2\mathrm{and}{\mathrm{sp}}^3$ respectively is the correct option