In the compound ^1CH_2 = ^2CH - ^3CH_2 - ^4CH | vedclass

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(D) $1$. Identify the hybridization of $C_2$: It is attached to one double bond and two single bonds, so it has $3$ sigma bonds and $0$ lone pairs, ma

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$sp^2 - sp^3$

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(D) $1$. Identify the hybridization of $C_2$: It is attached to one double bond and two single bonds, so it has $3$ sigma bonds and $0$ lone pairs, making it $sp^2$ hybridized.$2$. Identify the hybridization of $C_3$: It is attached to four single bonds, so it has $4$ sigma bonds and $0$ lone pairs, making it $sp^3$ hybridized.$3$. Therefore, the $C_2 - C_3$ bond is formed by the overlap of an $sp^2$ orbital and an $sp^3$ orbital, which is of the $sp^2 - sp^3$ type.

In the compound $^1CH_2 = ^2CH - ^3CH_2 - ^4CH_2 - ^5C \equiv ^6CH$, the $C_2 - C_3$ bond is of which type?

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