The hybridization of atomic orbitals of nitrogen in ${NO}_{2}^{+}, {NO}_{3}^{-}$ and ${NH}_{4}^{+}$ ions are respectively.

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${sp}^{2}, {sp}^{3} and sp$

${sp}^{2}, sp and {sp}^{3}$

$sp, {sp}^{3} and {sp}^{2}$

$sp, {sp}^{2} and {sp}^{3}$

answer is A.

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Detailed Solution

To predict the hybridization,

$no . of H . O = \frac{1}{2} n (o . of valance e^{-} s on central atom + no . of monovalent atoms - charge) For {NO}_{2}^{+}, no . of H . O = \frac{1}{2} (5 - 1) = 2 \to sp For {NO}_{3}^{-}, no . of H . O = \frac{1}{2} (5 + 1) = 3 \to {sp}^{2} For {NH}_{4}^{+}, no . of H . O = \frac{1}{2} (5 + 4 - 1) = 4 \to {sp}^{3}$