An inorganic Compound 'X' on treatment with concentrated H$$_2$$SO$$_4$$ produces brown fumes and gives dark brown ring with FeSO$$_4$$ in presence of concentrated H$$_2$$SO$$_4$$. Also Compound 'X' gives precipitate 'Y', when its solution in dilute HCl is treated with H$$_2$$S gas. The precipitate 'Y' on treatment with concentrated HNO$$_3$$ followed by excess of NH$$_4$$OH further gives deep blue coloured solution, Compound 'X' is:

We are given an inorganic compound 'X' that satisfies several chemical tests. First, it produces brown fumes with concentrated H$$_2$$SO$$_4$$, which indicates the presence of nitrate ions (NO$$_3^-$$) — concentrated H$$_2$$SO$$_4$$ reacts with nitrates to release brown NO$$_2$$ gas. Second, the dark brown ring test with FeSO$$_4$$ and concentrated H$$_2$$SO$$_4$$ is the classic brown ring test confirming the presence of nitrate ions.

These two tests confirm compound X is a nitrate salt. The options remaining are Cu(NO$$_3$$)$$_2$$, Co(NO$$_3$$)$$_2$$, and Pb(NO$$_3$$)$$_2$$.

Third, when a solution of X in dilute HCl is treated with H$$_2$$S gas, a precipitate Y forms. In the presence of dilute HCl (acidic conditions), only group II ions precipitate as sulfides with H$$_2$$S. Cu$$^{2+}$$ forms black CuS, Pb$$^{2+}$$ forms black PbS, while Co$$^{2+}$$ precipitates only in ammoniacal medium (group IV). This rules out Co(NO$$_3$$)$$_2$$.

Fourth, precipitate Y treated with concentrated HNO$$_3$$ followed by excess NH$$_4$$OH gives a deep blue coloured solution. The deep blue colour with excess ammonia is the characteristic reaction of the tetraamminecopper(II) complex, $$[\text{Cu(NH}_3)_4]^{2+}$$. Lead sulfide treated with concentrated HNO$$_3$$ gives Pb$$^{2+}$$, which does not form a deep blue complex with ammonia.

Therefore, the precipitate Y is CuS, and compound X is $$\text{Cu(NO}_3)_2$$, which is option 3.