From left to right, sp^2, sp^2, sp, sp hybrid | vedclass

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$(A)$ To determine the hybridisation of carbon atoms, we count the number of sigma bonds and lone pairs (steric number). $1$. $sp^3$ hybridisation: $4

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$CH_2=CH-C\equiv N$

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$(A)$ To determine the hybridisation of carbon atoms, we count the number of sigma bonds and lone pairs (steric number). $1$. $sp^3$ hybridisation: $4$ sigma bonds. $2$. $sp^2$ hybridisation: $3$ sigma bonds. $3$. $sp$ hybridisation: $2$ sigma bonds. In option $A$, $CH_2=CH-C\equiv N$: - The first carbon $(CH_2=)$ has $3$ sigma bonds $\rightarrow sp^2$. - The second carbon $(-CH-)$ has $3$ sigma bonds $\rightarrow sp^2$. - The third carbon $(-C\equiv)$ has $2$ sigma bonds $\rightarrow sp$. - The fourth carbon $(\equiv N)$ is nitrogen, but the carbon atom attached to nitrogen has $2$ sigma bonds $\rightarrow sp$. Thus, the sequence is $sp^2, sp^2, sp, sp$.

From left to right, $sp^2, sp^2, sp, sp$ hybridisation is present in

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