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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?

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