A) \[C{{l}^{-}}+Cl{{O}^{-}}\]
B) \[C{{l}^{-}}+ClO_{2}^{-}\]
C) \[C{{l}^{-}}+ClO_{3}^{-}\]
D) \[C{{l}^{-}}+ClO_{4}^{-}\]
Correct Answer: A
Solution :
[a] \[2NaOH+C{{l}_{2}}\to NaCl+NaOCl+{{H}_{2}}O\] hence \[C{{l}^{-}}\] and \[OC{{l}^{-}}\]
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