The species in which the N atom is in a state | vedclass

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(C) To determine the hybridization of the $N$ atom,we use the formula: $	ext{Steric Number} = \frac{1}{2} [V + M - C + A]$,where $V$ is the number of

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$NO_{2}^{+}$

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(C) To determine the hybridization of the $N$ atom,we use the formula: $	ext{Steric Number} = \frac{1}{2} [V + M - C + A]$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.For $NO_{2}^{+}$: $	ext{Steric Number} = \frac{1}{2} [5 + 0 - 1 + 0] = \frac{4}{2} = 2$. $A$ steric number of $2$ corresponds to $sp$ hybridization.For $NO_{3}^{-}$: $	ext{Steric Number} = \frac{1}{2} [5 + 0 - 0 + 1] = 3$ ($sp^2$ hybridization).For $NO_{2}$: The $N$ atom has one unpaired electron and is $sp^2$ hybridized.For $NO_{2}^{-}$: $	ext{Steric Number} = \frac{1}{2} [5 + 0 - 0 + 1] = 3$ ($sp^2$ hybridization).Thus,the $N$ atom is $sp$ hybridized in $NO_{2}^{+}$.Therefore,the correct option is $C$.

The species in which the $N$ atom is in a state of $sp$ hybridization is:

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