MCQ
When $Cl_2$ is passed through cold dil. $NaOH$ the products are
  • $NaCl, \,NaOCl$
  • B
    $NaCl, \,NaClO_2$ 
  • C
    $ NaCl,\, NaClO_3$
  • D
    $ NaCl,\, NaClO_4$

Answer

Correct option: A.
$NaCl, \,NaOCl$
a
$I _{2( s )}+ H _2 O _{2( aq )}+2 OH _{\text {(aq) }}^{-} \rightarrow 2 I _{\text {(aq) }}^{-}+2 H _2 O _{\text {(l) }}+ O _{2( g )}$

Therefore, the product of the reaction of $I _2 I _2$ with $H _2 O _2. H _2 O _2$ in the basic medium is Iodine.

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