The species in which the nitrogen atom is in | vedclass

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(C) The hybridisation of an atom can be determined by calculating the steric number $(SN)$: $SN = \frac{1}{2} [V + M - C + A]$ Where $V$ is the number

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$NO_2^{+}$

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(C) The hybridisation of an atom can be determined by calculating the steric number $(SN)$: $SN = \frac{1}{2} [V + M - C + A]$ 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. For $NO_2^{+}$: $V = 5$ (nitrogen), $M = 0$, $C = 1$, $A = 0$. $SN = \frac{1}{2} [5 + 0 - 1 + 0] = \frac{4}{2} = 2$. $A$ steric number of $2$ corresponds to $sp$ hybridisation. Therefore, the nitrogen atom in $NO_2^{+}$ is $sp$ hybridised.

The species in which the nitrogen atom is in a state of $sp$ hybridisation is

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