In the hydrocarbon,CH_3(1) - CH(2) = CH(3) - | vedclass

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(B) To determine the hybridization of carbon atoms,we count the number of sigma bonds and lone pairs attached to each carbon atom.$1$. Carbon $1$ $(CH

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$sp^3, sp^2, sp$

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(B) To determine the hybridization of carbon atoms,we count the number of sigma bonds and lone pairs attached to each carbon atom.$1$. Carbon $1$ $(CH_3)$: It is bonded to $3$ hydrogen atoms and $1$ carbon atom via single bonds. Total $4$ sigma bonds $\rightarrow sp^3$ hybridization.$2$. Carbon $3$ $(-CH=)$: It is bonded to $1$ hydrogen atom,$1$ carbon atom via a single bond,and $1$ carbon atom via a double bond. Total $3$ sigma bonds $\rightarrow sp^2$ hybridization.$3$. Carbon $5$ $(-C \equiv)$: It is bonded to $1$ carbon atom via a single bond and $1$ carbon atom via a triple bond. Total $2$ sigma bonds $\rightarrow sp$ hybridization.Thus,the sequence of hybridization for carbons $1, 3,$ and $5$ is $sp^3, sp^2, sp$.

In the hydrocarbon,$CH_3(1) - CH(2) = CH(3) - CH_2(4) - C(5) \equiv CH(6)$,the state of hybridization of carbons $1, 3$ and $5$ are in the following sequence:

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