The hybridisations of N,C and O shown in the | vedclass

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(A) Hybridisation is determined from the steric number (number of atoms bonded to the central atom $+$ the number of lone pairs). The number of hybrid

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$sp^2, sp, sp^2$

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(A) Hybridisation is determined from the steric number (number of atoms bonded to the central atom $+$ the number of lone pairs). The number of hybrid orbitals must be equal to the steric number.From the Lewis structure of the isocyanate group $(R-N=C=O)$:$I$. For the $N$-atom: It is bonded to $2$ atoms ($R$ and $C$) and has $1$ lone pair. Steric number $= 2 + 1 = 3$. Therefore,hybridisation is $sp^2$.$II$. For the $C$-atom: It is bonded to $2$ atoms ($N$ and $O$) and has $0$ lone pairs. Steric number $= 2 + 0 = 2$. Therefore,hybridisation is $sp$.$III$. For the $O$-atom: It is bonded to $1$ atom $(C)$ and has $2$ lone pairs. Steric number $= 1 + 2 = 3$. Therefore,hybridisation is $sp^2$.Thus,the hybridisations are $sp^2, sp, sp^2$.

The hybridisations of $N$,$C$ and $O$ shown in the following compound are:

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