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

The inorganic sulphite is denoted by $$\text{X}$$. First we treat it with dilute sulphuric acid.

Whenever any sulphite or hydrogen-sulphite salt is treated with a dilute mineral acid, the following general reaction takes place

$$\text{SO}_3^{2-}+2\,\text{H}^+\;\longrightarrow\;\text{SO}_2\uparrow+\text{H}_2\text{O}$$

That is, sulphur dioxide gas is set free. Hence, in the present case

$$\text{X}+\,\text{dil.}\;\text{H}_2\text{SO}_4\;\longrightarrow\;\text{SO}_2\uparrow+\text{(other\;products)}$$

So the compound $$\textbf{Y}$$ produced in the first step is definitely

$$\boxed{\text{SO}_2}$$

Next we are told that the same compound $$\textbf{Y}$$ reacts with aqueous NaOH to regenerate $$\textbf{X}$$. Sulphur dioxide dissolves in alkali according to two possible stoichiometries:

With just one mole of base (limited alkali)

$$\text{SO}_2+\text{NaOH}\;\longrightarrow\;\text{NaHSO}_3$$

With two moles of base (excess alkali)

$$\text{SO}_2+2\,\text{NaOH}\;\longrightarrow\;\text{Na}_2\text{SO}_3+\text{H}_2\text{O}$$

The statement of the problem merely says “with NaOH”, it does not say “with excess NaOH”. In ordinary laboratory practice the product obtained from the 1 : 1 reaction is the acid (or hydrogen) sulphite, $$\text{NaHSO}_3$$. Therefore the species that is re-formed must be

$$\boxed{\text{X}=\text{NaHSO}_3}$$

We still have to check the third piece of information: “Reaction of X with Y and water affords compound Z”. Passing more $$\text{SO}_2$$ through an aqueous solution of sodium hydrogen-sulphite brings about the well-known condensation that gives sodium metabisulphite:

First write the bisulphite ion explicitly:

$$\text{HSO}_3^-+\text{SO}_2\;\rightleftharpoons\;[\text{SO}_2\cdot\text{SO}_3\text{H}]^-$$

Two such species then eliminate water to give the pyrosulphite (metabisulphite) ion, $$\text{S}_2\text{O}_5^{2-}$$. Combining the ionic steps and adding the sodium ions gives the overall equation in aqueous medium:

$$2\,\text{NaHSO}_3+\text{SO}_2\;\xrightarrow[\text{(aq.)}]{\text{presence of H}_2\text{O}}\;\text{Na}_2\text{S}_2\text{O}_5+\text{H}_2\text{O}$$

This is exactly of the form “X + Y + water → Z”, and it fits perfectly with $$\text{X}=\text{NaHSO}_3$$ and $$\text{Y}=\text{SO}_2$$.

Thus all three experimental facts are simultaneously satisfied only when

$$\boxed{\text{Y}=\text{SO}_2,\qquad\text{X}=\text{NaHSO}_3}$$

Looking at the given options, this pair corresponds to Option C.

Hence, the correct answer is Option C.