1. Identify the reagent conditions

   The solution is treated with:

   • conc. $HCl$
   • then $H_2S$

   In qualitative salt analysis, passing $H_2S$ in the presence of acid precipitates only those metal ions whose sulfides are very insoluble, i.e. Group II cations.

2. Effect of concentrated $HCl$

   Concentrated $HCl$ makes the medium strongly acidic, so the concentration of $S^{2-}$ from $H_2S$ remains very low: $H_2S \rightleftharpoons 2H^+ + S^{2-}$ High $[H^+]$ suppresses ionization of $H_2S$, therefore only very insoluble sulfides precipitate.

3. Check each given cation

   Given ions are: $Al^{3+},\ Zn^{2+},\ Ca^{2+},\ Fe^{3+},\ Ni^{2+},\ Ba^{2+},\ Cu^{2+}$

   Now test which of these form sulfides precipitable in acidic medium:

   • $Al^{3+}$: does not precipitate with $H_2S$ in acidic medium.
   • $Zn^{2+}$: $ZnS$ precipitates in alkaline/ammoniacal medium, not in acidic medium.
   • $Ca^{2+}$: does not precipitate.
   • $Fe^{3+}$: does not precipitate as sulfide in acidic medium under this group analysis condition.
   • $Ni^{2+}$: $NiS$ precipitates in basic/ammoniacal medium, not in acidic medium.
   • $Ba^{2+}$: does not precipitate.
   • $Cu^{2+}$: forms highly insoluble $CuS$, which does precipitate in acidic medium.

4. Count the precipitated cations

   Only: $Cu^{2+} \rightarrow CuS \downarrow$

   Hence, the total number of cations precipitated is: $1$

5. Option matching

   • A: $1$ ✅
   • B: $3$ ❌
   • C: $4$ ❌
   • D: $2$ ❌

Final Answer: A (1)