CH_4 + Cl_2 xrightarrow{hv} CH_3Cl + HClTo ob | vedclass

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(A) The correct answer is $A$. To maximize the yield of the monohalogenated product $(CH_3Cl)$,the reaction should be carried out with an excess of th

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(A) The correct answer is $A$. To maximize the yield of the monohalogenated product $(CH_3Cl)$,the reaction should be carried out with an excess of the alkane $(CH_4)$.By keeping the concentration of $CH_4$ high relative to $Cl_2$,the probability of the chlorine radical $(Cl^{\bullet})$ colliding with a $CH_4$ molecule is significantly higher than its collision with a $CH_3Cl$ molecule.If $Cl_2$ were in excess,the chlorine radicals would react with the already formed $CH_3Cl$ to produce polyhalogenated products like $CH_2Cl_2$ via the following steps:$Cl^{\bullet} + CH_3Cl 	o {^{\bullet}}CH_2Cl + HCl$${^{\bullet}}CH_2Cl + Cl_2 	o CH_2Cl_2 + Cl^{\bullet}$Therefore,a high ratio of $CH_4$ to $Cl_2$ is required.

$CH_4 + Cl_2 \xrightarrow{hv} CH_3Cl + HCl$
To obtain high yields of $CH_3Cl$,the ratio of $CH_4$ to $Cl_2$ must be

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$CH_4 + Cl_2 \xrightarrow{hv} CH_3Cl + HCl$To obtain high yields of $CH_3Cl$,the ratio of $CH_4$ to $Cl_2$ must be