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(C) To determine the hybridisation of carbon atoms,we count the number of sigma bonds and lone pairs (if any) attached to each carbon atom.$1$. The st

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$3$

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(C) To determine the hybridisation of carbon atoms,we count the number of sigma bonds and lone pairs (if any) attached to each carbon atom.$1$. The structure is $CH_2 = CH - C \equiv C - CH_2 - C \equiv N$.$2$. Let's analyze each carbon atom:- $CH_2$ (terminal): $sp^2$ hybridised ($3$ sigma bonds).- $CH$ (second carbon): $sp^2$ hybridised ($3$ sigma bonds).- $C$ (third carbon): $sp$ hybridised ($2$ sigma bonds,one triple bond).- $C$ (fourth carbon): $sp$ hybridised ($2$ sigma bonds,one triple bond).- $CH_2$ (fifth carbon): $sp^3$ hybridised ($4$ sigma bonds).- $CN$ (cyanide carbon): $sp$ hybridised ($2$ sigma bonds,one triple bond).$3$. The carbon atoms with $sp$ hybridisation are the third,fourth,and the carbon in the cyano group $(-CN)$.$4$. Therefore,there are $3$ $sp$ hybridised carbon atoms in the given compound.

Number of $sp$ hybridised carbon atoms in the given compound is:
$CH_2 = CH - C \equiv C - CH_2 - CN$

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Number of $sp$ hybridised carbon atoms in the given compound is:$CH_2 = CH - C \equiv C - CH_2 - CN$