$X_{2}$ (Halogen) would not be produced in:

Electrolysis of brine solution

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Reaction of

N a C l

with conc.

H_{2} S O_{4}

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N a I

and

N a I O_{3}

mixed in dil.

H_{2} S O_{4}

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Reaction of

N a B r

with conc.

H_{2} S O_{4}

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Solution

The correct option is B Reaction of $N a C l$ with conc. $H_{2} S O_{4}$
(A) Electrolysis of brine (aqueous solution of $N a C l$ ) yields $C l_{2}$ (at anode), $H_{2}$ (at cathode ) and $N a O H$ solution.

(B) Reaction of $N a C l$ with conc. $H_{2} S O_{4}$ is represented as-
$2 N a C l + H_{2} S O_{4} \to N a_{2} S O_{4} + H C l$

(C) Reaction of $N a B r$ with conc. $H_{2} S O_{4}$
$N a B r + H_{2} S O_{4} \to N a H S O_{4} + H B r$
$H_{2} S O_{4} + 2 H B r \to S O_{2} + B r_{2} + 2 H_{2} O$

(D) $N a I$ and $N a I O_{3}$ mixed in dil. $H_{2} S O_{4}$
$5 N a I + N a I O_{3} + 3 H_{2} S O_{4} \to 3 N a_{2} S O_{4} + 3 I_{2} + 3 H_{2} O$

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