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The hybridization of N in ${NO}_{2}^{+}, {NO}_{3}^{-}$ and ${NH}_{4}^{+}$ respectively are

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a

sp, {sp}^{2}, {sp}^{3}

b

sp, {sp}^{3}, {sp}^{2}

c

{sp}^{2}, sp, {sp}^{3}

d

{sp}^{3}, {sp}^{2}, sp

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detailed solution

Correct option is A

${NO}_{2}^{+}$ has 16 valence electrons. There arrangement will be $: o_{. .}^{. .} - N = o_{. .}^{. .}$ Two pairs of electrons around N implies linear geometry involving sp hybridization. NO3 has 24 (= 5 x 3 x 6 + 1) valence electrons. These are arranged as

There are three pairs of electrons around N. Its geometry will be trigonal planar and will result from ${sp}^{2}$ hybridization. ${NH}_{4}^{+}$ has ${sp}^{3}$ hybridization.

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