From left to right,sp^2, sp^2, sp, sp hybridi | vedclass

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(A) To determine the hybridisation,we count the number of sigma bonds and lone pairs around each carbon atom.For $CH_2=CH-C \equiv N$:$1$. The first c

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$CH_2=CH-C \equiv N$

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(A) To determine the hybridisation,we count the number of sigma bonds and lone pairs around each carbon atom.For $CH_2=CH-C \equiv N$:$1$. The first carbon $(CH_2=)$ has $3$ sigma bonds,so it is $sp^2$ hybridised.$2$. The second carbon $(-CH=)$ has $3$ sigma bonds,so it is $sp^2$ hybridised.$3$. The third carbon $(-C \equiv)$ has $2$ sigma bonds,so it is $sp$ hybridised.$4$. The nitrogen atom is bonded to the carbon via a triple bond,and the carbon atom involved in the triple bond is $sp$ hybridised.Thus,the sequence $sp^2, sp^2, sp, sp$ corresponds to the structure $CH_2=CH-C \equiv N$.

From left to right,$sp^2, sp^2, sp, sp$ hybridisation is present in

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