Identify reagent $Y$ from the ferro/ferricyanide tests

We are told that aqueous solution of $Y$ behaves as follows:

• With potassium hexacyanoferrate(II), an intense blue precipitate forms, which dissolves in excess reagent.
• With potassium hexacyanoferrate(III), a brown coloration appears due to formation of $Z$.

These are characteristic reactions of $\mathrm{Fe^{3+}}$ salts.

• $\mathrm{Fe^{3+}}$ with ferrocyanide $[\mathrm{Fe(CN)_6}]^{4-}$ gives Prussian blue / ferric ferrocyanide.
• $\mathrm{Fe^{3+}}$ with ferricyanide $[\mathrm{Fe(CN)_6}]^{3-}$ gives a brown coloration.

Hence, $Y$ is a solution containing $\mathrm{Fe^{3+}}$, typically $\mathrm{FeCl_3}$.

Use the methylene blue test

The question states:

• $p$-Amino-$N,N$-dimethylaniline is added to a strongly acidic solution of $X$.
• The resulting solution is treated with a few drops of aqueous solution of $Y$.
• A blue colour appears due to formation of methylene blue.

This is the well-known qualitative test for sulfide ion $\mathrm{S^{2-}}$ / $\mathrm{H_2S}$.

In acidic medium, sulfide gives $\mathrm{H_2S}$, which reacts with $p$-amino-$N,N$-dimethylaniline in presence of an oxidizing agent such as $\mathrm{Fe^{3+}}$ to form methylene blue.

Therefore, $X$ must be a sulfide salt.

Match with options

The options are:

• A: $\mathrm{NaNO_3}$
• B: $\mathrm{NaCl}$
• C: $\mathrm{Na_2SO_4}$
• D: $\mathrm{Na_2S}$

Only $\mathrm{Na_2S}$ provides sulfide ion.

Check consistency of all statements

• Strongly acidic solution of $\mathrm{Na_2S}$ produces $\mathrm{H_2S}$.
• $\mathrm{H_2S}$ with $p$-amino-$N,N$-dimethylaniline and $\mathrm{Fe^{3+}}$ gives methylene blue.
• The identified $Y$ as $\mathrm{Fe^{3+}}$ also matches the ferrocyanide/ferricyanide tests.

So all clues are internally consistent.

Final answer

$X = \mathrm{Na_2S}$

Therefore, the correct option is D.