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

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

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

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

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

answer is B.

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Detailed Solution

Count $σ - b o n d,$ lone pairs and unpaired electrons or count number of atoms directly attached, lone pairs and unpaired electrons.

This is = 2 then sp , = 5, then $s p^{3} d$

= 3 then sp² , $= 6 then {sp}^{3} d^{2}, = 4 then {sp}^{3}$

	$σ - b o n d$	Lone pair	Unpaired electron	Total
I.	3	0	0	$3 (s p^{2})$
II.	2	0	0	2(sp)
III.	4	0	0	$4 (s p^{3})$

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