Identify the anion from the first tests

The compound $X$:

• on treatment with concentrated $H_2SO_4$ produces brown fumes
• gives a dark brown ring with $FeSO_4$ in presence of concentrated $H_2SO_4$

These are characteristic tests for the nitrate ion, $NO_3^-$.

• Brown fumes are due to $NO_2$.
• Brown ring test confirms nitrate.

Therefore, $X$ must be a nitrate, so options containing nitrite are rejected.

Hence, eliminate:

• B: $Pb(NO_2)_2$

Possible options left:

• A: $Co(NO_3)_2$
• C: $Cu(NO_3)_2$
• D: $Pb(NO_3)_2$

Identify the cation from the $H_2S$ test

The solution of $X$ in dilute $HCl$ is treated with $H_2S$, giving a precipitate $Y$.

In acidic medium, $H_2S$ precipitates cations of Group II as sulfides.

Among the remaining options:

• $Cu^{2+}$ gives black precipitate of $CuS$
• $Pb^{2+}$ gives black precipitate of $PbS$
• $Co^{2+}$ is not precipitated in acidic medium in this way (it belongs to later group behavior)

So A is eliminated.

Possible options now:

• C: $Cu(NO_3)_2$
• D: $Pb(NO_3)_2$

Use the confirmatory test for precipitate $Y$

The precipitate $Y$ on treatment with concentrated $HNO_3$ followed by excess $NH_4OH$ gives a deep blue coloured solution.

This is the characteristic test for $Cu^{2+}$:

• $CuS$ dissolves in oxidizing acid like concentrated $HNO_3$ to form $Cu^{2+}$
• then with excess $NH_4OH$, copper forms the deep blue tetraammine complex:

$Cu^{2+} + 4NH_3 \rightarrow [Cu(NH_3)_4]^{2+}$

This deep blue solution is diagnostic of copper.

Lead does not give a deep blue solution with excess ammonia.

Conclusion

Therefore, the compound $X$ is:

$\boxed{Cu(NO_3)_2}$

Option check

• A: $Co(NO_3)_2$ — incorrect
• B: $Pb(NO_2)_2$ — incorrect
• C: $Cu(NO_3)_2$ — correct
• D: $Pb(NO_3)_2$ — incorrect