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Chemistry Question on Methods of Preparation of Haloalkanes

The reaction condition leading to the best yield of C2H5Cl_2H_5 Cl are-

A

C2H6(excess)+CI2    UV light→C_2H_6(excess)+CI_2 \, \, \, \, \underrightarrow{UV \, light}

B

C2H6+CI2(excess)   →room tempdarkC_2H_6+CI_2 (excess) \, \, \, \xrightarrow[room \, temp]{dark}

C

C2H6+CI2(excess)    UV light→C_2H_6+CI_2(excess) \, \, \, \, \underrightarrow{UV \, light}

D

C2H6+CI2    UV light→C_2H_6+CI_2 \, \, \, \, \underrightarrow{UV \, light}

Answer

C2H6(excess)+CI2    UV light→C_2H_6(excess)+CI_2 \, \, \, \, \underrightarrow{UV \, light}

Explanation

Solution

During chlorination of alkane, if exebss of alkane is treated with Cl2_2(g) in presence of light or heat, chance of mono-chlorination predominate.
C2H6(excess)+CI2    UV light→ CH3CH2CI+HCIC_2H_6(excess)+CI_2 \, \, \, \, \underrightarrow{UV \, light} \, CH_3CH_2CI+HCI