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

s p, s p^{2}

and

s p^{3}

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s p^{2}, s p

and

s p^{3}

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s p, s p^{3}

and

s p^{2}

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

and

s p

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Solution

The correct option is A $s p, s p^{2}$ and $s p^{3}$
$N O_{2}^{+}$ contains 2 sigma bonds so hybridization is $s p$ and shape is linear.

$N O_{3}^{-}$ contains 3 sigma bonds so hybridization is $s p^{2}$ and shape is trigonal.

$N H_{4}^{+}$ contains 4 sigma bonds so hybridization is $s p^{3}$ and shape is tetrahedral.

Hence, option A is correct.

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

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

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

N H_{4}^{+}

N O_{3}^{-}, N O_{2}^{+} a n d N H_{4}^{+}

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

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