Which of the following compounds shows the hy | vedclass

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(A) In the molecule $H_2C = CH - C \equiv N$:$1$. The first carbon atom $(CH_2)$ is bonded to two hydrogens and one double bond,so it is $sp^2$ hybrid

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$H_2C = CH - C \equiv N$

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(A) In the molecule $H_2C = CH - C \equiv N$:$1$. The first carbon atom $(CH_2)$ is bonded to two hydrogens and one double bond,so it is $sp^2$ hybridized.$2$. The second carbon atom $(CH)$ is bonded to one hydrogen,one double bond,and one single bond,so it is $sp^2$ hybridized.$3$. The third carbon atom $(C)$ is bonded to one single bond and one triple bond,so it is $sp$ hybridized.$4$. The nitrogen atom $(N)$ is bonded to one triple bond and has one lone pair,so it is $sp$ hybridized.Thus,the hybridization sequence is $sp^2 - sp^2 - sp - sp$.

Which of the following compounds shows the hybridization mode $sp^2 - sp^2 - sp - sp$ from left to right?

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