A) \[C{{l}_{2}}\]
B) Conc. \[HCl\]
C) \[HBr\]
D) \[{{H}_{2}}S\]
Correct Answer: A
Solution :
\[C{{l}_{2}}+2B{{r}^{-}}\,\to \,2C{{l}^{-}}+B{{r}_{2}}\]
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