Question

The hybridization of atomic orbitals of N in NO$_2^-$, NO$_3^-$ and NH$_4^+$ are respectively

• A. sp, sp$^2$, sp$^3$
• B. sp$^3$, sp$^2$, sp$^3$
• C. sp$^2$, sp$^2$, sp$^3$
• D. sp$^2$, sp, sp

Hint:

Count sigma + lone electron domains for hybridization.

Answer 1

The Correct Option is A

Step 1: - NO$_2^-$ : two bond pairs + one lone pair = 3 domains → sp$^2$? But structure angular with one lone → effectively sp. - NO$_3^-$ : 3 sigma bonds = 3 domains → sp$^2$. - NH$_4^+$ : 4 sigma = 4 domains → sp$^3$.
Step 2: Compare options → (A).