An inorganic compound 'X' on treatment with concentrated H2SO4 produces brown fumes and gives dark brown ring with…

Step 1: Analyze the anion responses Compound 'X' produces brown fumes exclusively upon heating with concentrated $\ce{H2SO4}$ and gives a positive dark brown ring test when treated with $\ce{FeSO4}$ and concentrated $\ce{H2SO4}$. These responses decisively confirm that the anion fundamentally present in 'X' is the Nitrate ion ($\ce{NO3-}$).

Step 2: Analyze the cation responses The solution of 'X' in dilute $\ce{HCl}$ reacts with $\ce{H2S}$ to form a precipitate 'Y'. Precipitation by $\ce{H2S}$ in an acidic medium firmly places the cation in Group II of qualitative analysis (e.g., $\ce{Pb^{2+}}$, $\ce{Cu^{2+}}$, $\ce{Bi^{3+}}$).

Step 3: Evaluate precipitate 'Y' The precipitate 'Y' (likely $\ce{PbS}$ or $\ce{CuS}$ based on options) dissolves in concentrated $\ce{HNO3}$. When excess $\ce{NH4OH}$ is added to this resulting solution, a deep blue coloured solution is obtained.

• If the metal was Lead ($\ce{Pb^{2+}}$), addition of excess ammonia would merely yield a white precipitate of $\ce{Pb(OH)2}$ that does not dissolve further.
• If the metal is Copper ($\ce{Cu^{2+}}$), the initially formed pale blue cupric hydroxide readily dissolves in excess $\ce{NH4OH}$ to form a stunning deep blue soluble complex, Tetraamminecopper(II) sulphate/nitrate ($\ce{[Cu(NH3)4]^{2+}}$). $\ce{Cu^{2+} + 4NH4OH -> [Cu(NH3)4]^{2+} + 4H2O}$

Step 4: Conclusion Since the anion is $\ce{NO3-}$ and the cation is beautifully confirmed as $\ce{Cu^{2+}}$, modifying compound 'X' structurally as $\ce{Cu(NO3)2}$.

Key Points to Remember:

• Nitrate ($\ce{NO3-}$) is the anion for the Brown Ring test with conc. $\ce{H2SO4}$.
• Nitrite ($\ce{NO2-}$) would give the brown ring test even with dilute $\ce{H2SO4}$.
• The uniquely defining deep blue solution with excess aqueous ammonia is the hallmark confirmatory test for Copper(II).