JEE PYQ: P Block Elements Question 34
Question 34 - 2020 (06 Sep Shift 2)
Reaction of an inorganic sulphite X with dilute $\text{H}_2\text{SO}_4$ generates compound Y. Reaction of Y with NaOH gives X. Further, the reaction of X with Y and water affords compound Z. Y and Z, respectively, are:
(1) $\text{SO}_2$ and $\text{Na}_2\text{SO}_3$
(2) $\text{SO}_3$ and $\text{NaHSO}_3$
(3) $\text{SO}_2$ and $\text{NaHSO}_3$
(4) S and $\text{Na}_2\text{SO}_3$
Show Answer
Answer: (3)
Solution
$X = \text{Na}_2\text{SO}_3$. $\text{SO}_3^{2-} + \text{H}_2\text{SO}_4 \to [\text{H}_2\text{SO}_3] \to \text{H}_2\text{O} + \text{SO}_2$ (Y). $\text{SO}_2 + 2\text{NaOH} \to \text{Na}_2\text{SO}_3 + \text{H}_2\text{O}$. $\text{Na}_2\text{SO}_3 + \text{SO}_2 + \text{H}_2\text{O} \to 2\text{NaHSO}_3$ (Z).
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