When a salt is treated with sodium hydroxide solution it gives gas X. On passing gas X through reagent Y a brown coloured precipitate is formed. X and Y respectively, are

When an ammonium salt (for example $$NH_4Cl$$) is warmed with aqueous $$NaOH$$, the hydroxide ion removes a proton from the ammonium ion.

$$NH_4^+ + OH^- \rightarrow NH_3 \uparrow + H_2O$$   $$-(1)$$

Thus the evolved gas $$X$$ is ammonia, $$NH_3$$.

To identify ammonia, Nessler’s reagent is commonly used. Nessler’s reagent is a strongly alkaline solution of $$K_2HgI_4$$ in excess $$KOH$$. In the presence of ammonia, it undergoes a redox-complex reaction to give a brown precipitate of Millon’s base $$\left(Hg_2O\;.\;NH_2I\right)$$.

Overall (simplified) reaction with excess alkali:

$$2\,K_2HgI_4 + NH_3 + 3\,KOH \rightarrow Hg_2O.NH_2I \downarrow\;(\text{brown}) + 7\,KI + 2\,H_2O$$   $$-(2)$$

Therefore, when gas $$X$$ ($$NH_3$$) is passed through reagent $$Y$$ (Nessler’s reagent: $$K_2HgI_4 + KOH$$), a brown coloured precipitate forms.

Hence:

$$X = NH_3, \qquad Y = K_2HgI_4 + KOH$$

Option B is correct.