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Potassium Permanganate

Reaction of a...

Question

Reaction of an inorganic sulphite X with dilute $H_{2} S O_{4}$ generates compound Y. Reaction of Y with $N a O H$ gives X. Further, the reaction of X with Y and water affords compound Z. Y and Z, respectively, are

A

S O_{2}

and

N a H S O_{3}

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B

S O_{2}

and

N a_{2} S O_{3}

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C

S O_{3}

and

N a H S O_{3}

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D

S

and

N a_{2} S O_{3}

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Solution

The correct option is A $S O_{2}$ and $N a H S O_{3}$
$X = N a_{2} S O_{3}$

The reactions are as follows,

$N a_{2} S O_{3} (X) + d i l . H_{2} S O_{4} \to N a_{2} S O_{4} + H_{2} + S O_{2} (Y)$

$S O_{2} + 2 N a O H \to N a_{2} S O_{3} (X) + H_{2} O$

$N a_{2} S O_{3} + S O_{2} + H_{2} O \to 2 N a H S O_{3}$

Thus, $Y = S O_{2}$ ; $Z = N a H S O_{3}$

Hence, option (b) is correct.

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