A) \[AlC{{l}_{3}}\]
B) sun light
C) \[ZnC{{l}_{2}}\]
D) dark
Correct Answer: B
Solution :
\[{{C}_{6}}{{H}_{6}}+3C{{l}_{2}}\xrightarrow[{}]{sun\,light}\underset{BHC}{\mathop{{{C}_{6}}{{H}_{6}}C{{l}_{6}}}}\,\] BHC is formed is presence of bright sun light.
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