The hybridization of nitrogen atoms in NO^{+} | vedclass

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

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$sp, sp^2, sp^3$

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(A) To determine the hybridization,we use the formula: $	ext{Steric Number} = \frac{1}{2} [V + M - C + A]$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.$1$. For $NO^{+}_2$: $V=5, M=0, C=1, A=0$. $	ext{Steric Number} = \frac{1}{2} [5 + 0 - 1 + 0] = 2$. Hybridization is $sp$.$2$. For $NO^{-}_3$: $V=5, M=0, C=0, A=1$. $	ext{Steric Number} = \frac{1}{2} [5 + 0 - 0 + 1] = 3$. Hybridization is $sp^2$.$3$. For $NH^{+}_4$: $V=5, M=4, C=1, A=0$. $	ext{Steric Number} = \frac{1}{2} [5 + 4 - 1 + 0] = 4$. Hybridization is $sp^3$.Thus,the hybridization is $sp, sp^2, sp^3$.

The hybridization of nitrogen atoms in $NO^{+}_2$,$NO^{-}_3$,and $NH^{+}_4$ respectively is:

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