Consider the species NO_3^-,NO_2^+,and NO_2^- | vedclass

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

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The hybridisation state of $N$ in $NO_3^-$ and $NO_2^-$ is the same.

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(C) To determine the hybridisation of the central atom $N$,we use the formula: $	ext{Hybridisation} = \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 $NO_3^-$: $N = \frac{1}{2} [5 + 0 - 0 + 1] = 3$. This corresponds to $sp^2$ hybridisation.$2$. For $NO_2^+$: $N = \frac{1}{2} [5 + 0 - 1 + 0] = 2$. This corresponds to $sp$ hybridisation.$3$. For $NO_2^-$: $N = \frac{1}{2} [5 + 0 - 0 + 1] = 3$. This corresponds to $sp^2$ hybridisation.Comparing these,both $NO_3^-$ and $NO_2^-$ have $sp^2$ hybridisation. Therefore,statement $C$ is correct.

Consider the species $NO_3^-$,$NO_2^+$,and $NO_2^-$. Pick the correct statement:

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