The hybridization of atomic orbitals of nitrogen in $N O_{2}^{+}, N O_{2}^{-}$ and $N H_{4}^{+}$ is:

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

and

s p^{2}

respectively

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s p, s p^{2}

and

s p^{3}

respectively

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s p^{2}, s p

and

s p^{3}

respectively

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

and

s p

respectively

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Solution

The correct option is B $s p, s p^{2}$ and $s p^{3}$ respectively
Hybridization of $N$ in:
1. $N O_{2}^{+}$ = $s p$
2. $N O_{2}^{-}$ = $s p^{2}$
3. $N H_{4}^{+}$ = $s p^{3}$
Therefore, option $B$ is correct.

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Similar questions

The hybridisation of atomic orbitals of nitrogen in ${N O_{2}}^{+}$ , ${N O_{3}}^{-}$ and ${N H_{4}}^{+}$ are:

Question 03

The types of hybrid orbitals of nitrogen in $N O_{2}^{+}, N O_{3}^{-} a n d N H_{4}^{+}$ respectively are expected to be

(a) $s p, s p^{3} a n d s p^{2}$

(b) $s p, s p^{2} a n d s p^{3}$

(d) $s p^{2}, s p^{3} a n d s p$

N O_{2}^{+}, N O_{3}^{-}

N H_{4}^{+}

C H_{4}, C_{2} H_{6}, a n d C_{2} H_{4}

s p, s p^{2} a n d s p^{3}

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