In the compound CH_2 = CH - CH_2 - CH_2 - C e | vedclass

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(D) To determine the hybridization of the carbon atoms,we number the chain: $C_1H_2 = C_2H - C_3H_2 - C_4H_2 - C_5 \equiv C_6H$. $C_1$ is $sp^2$ hybri

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$sp^2 - sp^3$

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(D) To determine the hybridization of the carbon atoms,we number the chain: $C_1H_2 = C_2H - C_3H_2 - C_4H_2 - C_5 \equiv C_6H$. $C_1$ is $sp^2$ hybridized. $C_2$ is $sp^2$ hybridized. $C_3$ is $sp^3$ hybridized. $C_4$ is $sp^3$ hybridized. $C_5$ is $sp$ hybridized. $C_6$ is $sp$ hybridized. The bond between $C_2$ and $C_3$ involves an $sp^2$ hybridized carbon and an $sp^3$ hybridized carbon. Therefore,the bond type is $sp^2 - sp^3$.

In the compound $CH_2 = CH - CH_2 - CH_2 - C \equiv CH$,the $C_2 - C_3$ bond is of the type

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