The hybridization of atomic orbitals of nitro | vedclass

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

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$sp$,$sp^2$,and $sp^3$ respectively

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(B) To determine the hybridization,we use the formula: $	ext{Hybridization} = \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.$1$. For $NO_2^+$: $	ext{Hybridization} = \frac{1}{2} [5 + 0 - 1 + 0] = 2$,which corresponds to $sp$ hybridization.$2$. For $NO_3^-$: $	ext{Hybridization} = \frac{1}{2} [5 + 0 - 0 + 1] = 3$,which corresponds to $sp^2$ hybridization.$3$. For $NH_4^+$: $	ext{Hybridization} = \frac{1}{2} [5 + 4 - 1 + 0] = 4$,which corresponds to $sp^3$ hybridization.Thus,the correct order is $sp$,$sp^2$,and $sp^3$ respectively.

The hybridization of atomic orbitals of nitrogen in $NO_2^+$,$NO_3^-$,and $NH_4^+$ are:

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