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Reaction of an inorganic sulphite X with dilute $H_{2} {SO}_{4}$ 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:

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$S and {Na}_{2} {SO}_{3}$

${SO}_{3} and {NaHSO}_{3}$

${SO}_{2} and {NaHSO}_{3}$

${SO}_{2} and {Na}_{2} {SO}_{3}$

answer is D.

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Detailed Solution

$S + conc H_{2} {SO}_{4} \to \underset{X}{{SO}_{4 (g)}} \overset{NaOH}{\to} \underset{Y}{{Na}_{2} {SO}_{3}}$

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