What is the hybridization of carbon atoms at | vedclass

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(C) The structure of the compound is $CH_3(6)-CH(5)=CH(4)-CH_2(3)-C(2) \equiv CH(1)$.$1$. Carbon at position $1$ is bonded to one hydrogen and triple-

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$sp, sp^3, sp^2$

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(C) The structure of the compound is $CH_3(6)-CH(5)=CH(4)-CH_2(3)-C(2) \equiv CH(1)$.$1$. Carbon at position $1$ is bonded to one hydrogen and triple-bonded to carbon $2$,so it has $2$ sigma bonds and $0$ lone pairs,resulting in $sp$ hybridization.$2$. Carbon at position $3$ is bonded to two hydrogens and two carbons,so it has $4$ sigma bonds,resulting in $sp^3$ hybridization.$3$. Carbon at position $5$ is double-bonded to carbon $4$ and bonded to one hydrogen and one carbon,so it has $3$ sigma bonds,resulting in $sp^2$ hybridization.Thus,the hybridization for positions $1, 3, 5$ is $sp, sp^3, sp^2$ respectively.

What is the hybridization of carbon atoms at positions $1$,$3$,and $5$ in the compound $CH_3-CH=CH-CH_2-C \equiv CH$?

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