Reaction of an inorganic sulphite $X$ with dilute ${\mathrm{H}}_2{\mathrm{SO}}_4$ generates compound Y. Reaction of $\mathrm{Y}$ with $\mathrm{NaOH}$ gives $\mathrm{X}$. Further, the reaction of $\mathrm{X}$ with $\mathrm{Y}$ and water affords compound $\mathrm{Z}$. Y and $\mathrm{Z}$, respectively, are:

Given: We need to find X, Y and Z.

Concept Used: P-Block

Detailed Explanation: - The reaction of Sodium sulphite $\left({\mathrm{Na}}_2{\mathrm{SO}}_3\right)$ with dilute Sulphuric $acid\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)$ generates a compound of Sulphur dioxide $\left({\mathrm{SO}}_2\right)$.

${\mathrm{Na}}_2{\mathrm{SO}}_3( \mathrm{s})+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)\to {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+{\mathrm{SO}}_2( \mathrm{g})$

Sodium sulphite dil. Sulphuric acid Water Sulphur dioxide

The reaction of Sulphur dioxide $\left({\mathrm{SO}}_2\right)$ with $\left(\mathrm{NaOH}\right)$ gives Sodium sulphite $\left({\mathrm{Na}}_2{\mathrm{SO}}_3\right)$.

${\mathrm{SO}}_2( \mathrm{g})+\mathrm{NaOH}\left(\mathrm{aq}\right)\to {\mathrm{Na}}_2{\mathrm{SO}}_3\left(\mathrm{aq}\right)$

Sulfur dioxide Sodium hydroxide Sodium sulphite

The reaction of Sodium sulphite $\left({\mathrm{Na}}_2{\mathrm{SO}}_3\right)$ with Sulphur dioxide( $\left.{\mathrm{SO}}_2\right)$ and water afford the compound Sodium bisulfite $\left({\mathrm{NaHSO}}_3\right)$.

${\mathrm{SO}}_2\left(\mathrm{ }\mathrm{g}\right)+{\mathrm{Na}}_2{\mathrm{SO}}_3\left(\mathrm{ }\mathrm{s}\right)\to {\mathrm{NaHSO}}_3\left(\mathrm{ }\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Sulfur dioxide Sodium sulphite Sodium bisulphate Water

Therefore, $\mathrm{X}$ is ${\mathrm{Na}}_2{\mathrm{SO}}_3,\mathrm{Y}$ is ${\mathrm{SO}}_2$ and $\mathrm{Z}$ is ${\mathrm{NaHSO}}_3$.