The hybridization of orbitals of N atom in NO | vedclass

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(B) The hybridization of the central atom can be calculated using the formula: $H = \frac{1}{2} [V + M - C + A]$,where $V$ is the number of valence el

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$sp^{2}, sp, sp^{3}$

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(B) The hybridization of the central atom can be calculated using the formula: $H = \frac{1}{2} [V + M - C + A]$,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 $NO_{3}^{-}$: $H = \frac{1}{2} [5 + 0 - 0 + 1] = 3$,which corresponds to $sp^{2}$ hybridization. For $NO_{2}^{+}$: $H = \frac{1}{2} [5 + 0 - 1 + 0] = 2$,which corresponds to $sp$ hybridization. For $NH_{4}^{+}$: $H = \frac{1}{2} [5 + 4 - 1 + 0] = 4$,which corresponds to $sp^{3}$ hybridization. Thus,the hybridization states are $sp^{2}, sp, sp^{3}$ respectively.

The hybridization of orbitals of $N$ atom in $NO_{3}^{-}$,$NO_{2}^{+}$ and $NH_{4}^{+}$ are respectively $:$

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