Key principle: precipitation with $H_2S$ in acidic medium

When $H_2S$ is passed in presence of dilute $HCl$, the concentration of $S^{2-}$ ions is very low because ionization of $H_2S$ is suppressed in acidic medium.

Therefore, only those metal ions whose sulfides have very small solubility product precipitate in this medium.

These are mainly Group II cations such as: $Cu^{2+},\ Pb^{2+},\ Hg^{2+},\ Bi^{3+},\ Cd^{2+},\ As^{3+},\ Sb^{3+},\ Sn^{2+}/Sn^{4+}$

Ions like $Fe^{3+}$, $Zn^{2+}$, $Ba^{2+}$ do not precipitate with $H_2S$ in dilute acid.

Check each option

Option A: $Ba^{2+},\ Zn^{2+}$

• $Ba^{2+}$ does not form sulfide precipitate in acidic medium.
• $Zn^{2+}$ belongs to later group precipitation; $ZnS$ precipitates in ammoniacal/basic medium, not in dilute $HCl$.

Hence, both are not precipitated.

Option A: Incorrect

Option B: $Bi^{3+},\ Fe^{3+}$

• $Bi^{3+}$ precipitates as $Bi_2S_3$ with $H_2S$ in acidic medium.
• $Fe^{3+}$ does not precipitate under these conditions because $Fe_2S_3$ is not precipitated from dilute acid medium by $H_2S$.

Hence, both are not precipitated.

Option B: Incorrect

Option C: $Cu^{2+},\ Pb^{2+}$

• $Cu^{2+}$ precipitates as $CuS$.
• $Pb^{2+}$ precipitates as $PbS$.

Both sulfides are sufficiently insoluble to precipitate in acidic medium.

Option C: Correct

Option D: $Hg^{2+},\ Bi^{3+}$

• $Hg^{2+}$ precipitates as $HgS$.
• $Bi^{3+}$ precipitates as $Bi_2S_3$.

Both are Group II ions and precipitate with $H_2S$ in dilute $HCl$.

Option D: Correct

Final answer

The pairs in which both ions precipitate on passing $H_2S$ in presence of dilute $HCl$ are: $\boxed{C,\ D}$

Comparison with stored correct answer

Stored correct answer: C, D

Our derived answer matches exactly.