A) \[{{I}_{2}}\]
B) \[S{{O}_{2}}\]
C) \[Hl\]
D) \[C{{l}_{2}}\]
Correct Answer: D
Solution :
Key Idea A more electronegative halogen displaces less electronegative halogen from its halides. The electronegativity order of halogen is as\[F>Cl>Br>I\]. Chlorine is more electro-negative than bromine so chlorine will liberate \[B{{r}_{2}}\] from KBr. \[C{{l}_{2}}+2KBr\xrightarrow{{}}2KCl+B{{r}_{2}}\]While iodine is less electronegative than bromine. So, it cannot displace bromine from potassium bromide. Note - A more electropositive halogen displaces less electropositive halogen from its halate. \[{{l}_{2}}+2KCl{{O}_{3}}\xrightarrow{{}}2KI{{O}_{3}}+C{{l}_{2}}\]
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