Number of sp hybridised carbon atoms in the g | vedclass

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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.

List-$I$	List-$II$
$A$. Ethane	$1$. $2 \ sp$ carbons
$B$. Ethylene	$2$. $6 \ sp^2$ carbons
$C$. Acetylene	$3$. $2 \ sp^3$ carbons
$D$. Benzene	$4$. $2 \ sp^2$ carbons
	$5$. $1 \ sp$ and $1 \ sp^2$ carbons

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

Similar Questions

In which of the following species is the underlined carbon having $sp^3$ hybridisation?

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

Match the following lists:
List-$I$ List-$II$
$A$. Ethane $1$. $2 \ sp$ carbons
$B$. Ethylene $2$. $6 \ sp^2$ carbons
$C$. Acetylene $3$. $2 \ sp^3$ carbons
$D$. Benzene $4$. $2 \ sp^2$ carbons
$5$. $1 \ sp$ and $1 \ sp^2$ carbons

Explain how the electronegativity of carbon atoms is related to their state of hybridization in an organic compound.

Which of the following compounds do not have $sp^3$ carbon atom$(s)$?
$I$. Acetone
$II$. Acetic acid
$III$. Buta-$1,3$-diene
$IV$. Propyne
$V$. Naphthalene

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