The hybridisation of N in NO_2^{+},NO_3^{-} a | vedclass

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(A) To determine the hybridisation,we use the formula: $	ext{Steric Number} = 	ext{Number of bond pairs} + 	ext{Number of lone pairs}$.For $NO_2^{+

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$sp, sp^2, sp^3$

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(A) To determine the hybridisation,we use the formula: $	ext{Steric Number} = 	ext{Number of bond pairs} + 	ext{Number of lone pairs}$.For $NO_2^{+}$:$	ext{Steric Number} = 2 + 0 = 2$. Hybridisation is $sp$.For $NO_3^{-}$:$	ext{Steric Number} = 3 + 0 = 3$. Hybridisation is $sp^2$.For $NH_4^{+}$:$	ext{Steric Number} = 4 + 0 = 4$. Hybridisation is $sp^3$.Thus,the hybridisation states are $sp, sp^2, sp^3$ respectively.

The hybridisation of $N$ in $NO_2^{+}$,$NO_3^{-}$ and $NH_4^{+}$ respectively is

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