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(A) To determine the hybridisation of carbon atoms,we count the number of sigma bonds and lone pairs attached to each carbon atom:$1$. $sp^3$ hybridis

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$H_2C=CH-C\equiv N$

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(A) To determine the hybridisation of carbon atoms,we count the number of sigma bonds and lone pairs attached to each carbon atom:$1$. $sp^3$ hybridisation: $4$ sigma bonds (tetrahedral geometry).$2$. $sp^2$ hybridisation: $3$ sigma bonds (trigonal planar geometry).$3$. $sp$ hybridisation: $2$ sigma bonds (linear geometry).Let's analyze option $A$: $H_2C=CH-C\equiv N$- The first carbon $(CH_2=)$ has $3$ sigma bonds $\rightarrow sp^2$.- The second carbon $(-CH-)$ has $3$ sigma bonds $\rightarrow sp^2$.- The third carbon $(-C\equiv)$ has $2$ sigma bonds $\rightarrow sp$.- The nitrogen atom is attached to the third carbon with a triple bond,and the carbon itself is $sp$ hybridised. The nitrogen atom in $C\equiv N$ is also $sp$ hybridised due to the triple bond and one lone pair.Thus,the sequence $sp^2-sp^2-sp-sp$ corresponds to $H_2C=CH-C\equiv N$.

Which of the following represents the given mode of hybridisation $sp^2-sp^2-sp-sp$ from left to right?

Similar Questions

Which of the following carbon atoms is the most electronegative?

Which of the following bonds is the most acidic?

The hybridisation of each carbon in the following compound respectively is:
$CH_3-C(=O)-CH_2-CN$

Amongst the given options,the compound$(s)$ in which all the atoms are in one plane in all the possible conformations (if any),is (are):
$(A)$ $CH_2=CH-CH=CH_2$
$(B)$ $HC\equiv C-CH=CH_2$
$(C)$ $H_2C=C=O$
$(D)$ $H_2C=C=CH_2$

In the compound $CH_2 = CH - CH_2 - CH_2 - C \equiv CH$,the $C_2 - C_3$ bond is of the type

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