In the reaction,the hybridisation states of c | vedclass

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(A) In the reactant $Br-\overset{1}{CH}=\overset{2}{CH}-Br$,both carbon atoms $1$ and $2$ are bonded to one hydrogen atom,one bromine atom,and one oth

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$1$ and $2$ are $sp^2$; $3$ and $4$ are $sp^3$

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(A) In the reactant $Br-\overset{1}{CH}=\overset{2}{CH}-Br$,both carbon atoms $1$ and $2$ are bonded to one hydrogen atom,one bromine atom,and one other carbon atom via a double bond,resulting in $sp^2$ hybridisation.
In the product $Br-\overset{3}{CH_2}-\overset{4}{CH_2}-Br$,both carbon atoms $3$ and $4$ are bonded to two hydrogen atoms,one bromine atom,and one other carbon atom via single bonds,resulting in $sp^3$ hybridisation.
Therefore,the correct hybridisation states are $1, 2$ as $sp^2$ and $3, 4$ as $sp^3$.

In the reaction,the hybridisation states of carbon atoms $1, 2, 3, 4$ are:
$Br-CH=CH-Br \xrightarrow[Catalyst]{H_2} Br-CH_2-CH_2-Br$

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In the reaction,the hybridisation states of carbon atoms $1, 2, 3, 4$ are: $Br-CH=CH-Br \xrightarrow[Catalyst]{H_2} Br-CH_2-CH_2-Br$