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Electrolysis of aqueous sodium chloride (X) forms three products A,B and C of which B and C are gases. B cannot react with A but C can react with A in cold condition to give X and Y in hot condition give X and Z. The difference in the oxidation states of C in Y and Z is

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answer is D.

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Detailed Solution

$6 NaOH + \underset{(hot & conc .)}{3 {Cl}_{2}} \to 5 NaCl + {NaClO}_{3} + H_{2} O$

$2 NaOH + \underset{(cold, dil .)}{{Cl}_{2}} \to NaCl + NaOCI + H_{2} O$

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