The problem asks to identify a mixture of two salts based on a series of qualitative analysis tests shown in a flowchart.

Step 1: Analysis of the first reaction and the precipitate P

1. The mixture of salts forms a solution S.
2. When dilute HCl is added to solution S, a white precipitate (P) is formed. This indicates the presence of a Group I cation, which forms an insoluble chloride. The Group I cations are $Ag^+$, $Pb^{2+}$, and $Hg_2^{2+}$.
3. The precipitate P is found to dissolve in hot water. This is a characteristic test for lead(II) chloride ($PbCl_2$). $AgCl$ and $Hg_2Cl_2$ are insoluble in hot water.
4. Therefore, one of the cations present in the mixture must be $Pb^{2+}$.
5. Based on this conclusion, we can evaluate the options:
   • A: $Pb(NO_3)_2$ and $Zn(NO_3)_2$ - Contains $Pb^{2+}$. Possible.
   • B: $Pb(NO_3)_2$ and $Bi(NO_3)_3$ - Contains $Pb^{2+}$. Possible.
   • C: $AgNO_3$ and $Bi(NO_3)_3$ - Contains $Ag^+$, not $Pb^{2+}$. The precipitate $AgCl$ would not dissolve in hot water. Incorrect.
   • D: $Pb(NO_3)_2$ and $Hg(NO_3)_2$ - Contains $Pb^{2+}$. Possible.

Step 2: Analysis of the second reaction and the precipitate R

1. The filtrate (Q) obtained after removing the precipitate P contains the second cation.
2. When $H_2S$ gas is passed through the filtrate Q, a white precipitate (R) is formed. Note that filtrate Q is acidic because dilute HCl was added in the previous step.
3. We need to check the reaction of the second cation from the possible options (A, B, D) with $H_2S$ in an acidic medium.

Step 3: Evaluation of the remaining options

• Option A: $Pb(NO_3)_2$ and $Zn(NO_3)_2$

  • The filtrate Q contains $Zn^{2+}$ ions.
  • The reaction is $Zn^{2+} + H_2S \rightarrow ZnS \downarrow + 2H^+$.
  • Zinc sulfide ($ZnS$) is a white precipitate. Although $Zn^{2+}$ is a Group IV cation (precipitates in basic medium), its precipitation can occur in weakly acidic solution if the concentration of $Zn^{2+}$ is high. Given the observation, this option is consistent with the results.
  • Therefore, Option A is correct.

• Option B: $Pb(NO_3)_2$ and $Bi(NO_3)_3$

  • The filtrate Q contains $Bi^{3+}$ ions.
  • The standard reaction is $2Bi^{3+} + 3H_2S \rightarrow Bi_2S_3 \downarrow + 6H^+$.
  • Bismuth(III) sulfide ($Bi_2S_3$) is a black precipitate. This contradicts the observation that R is a white precipitate.
  • Alternative Interpretation: All the salts given are nitrates. In an acidic solution (from HCl), nitrate ions ($NO_3^−$) can act as an oxidizing agent and oxidize $H_2S$ to elemental sulfur, which is a white or pale-yellow colloidal precipitate. $3H_2S(g) + 2NO_3^-(aq) + 2H^+(aq) \rightarrow 3S(s) \downarrow + 2NO(g) + 4H_2O(l)$
  • If this reaction is considered to be the one producing the precipitate, then the observation of a white precipitate would be explained. This is a known interference in qualitative analysis. Assuming this is the intended reaction for this option.
  • Therefore, Option B is also considered correct under this interpretation.

• Option D: $Pb(NO_3)_2$ and $Hg(NO_3)_2$

  • The filtrate Q contains $Hg^{2+}$ ions.
  • The reaction is $Hg^{2+} + H_2S \rightarrow HgS \downarrow + 2H^+$.
  • Mercury(II) sulfide ($HgS$) is a black precipitate. This contradicts the observation.
  • Even if the oxidation of $H_2S$ by nitrate is considered, the precipitation of $HgS$ is extremely rapid and complete due to its very low solubility product ($K_{sp} \approx 10^{-52}$), so the observed precipitate would be black $HgS$.
  • Therefore, Option D is incorrect.

Conclusion Based on the analysis, Option A is clearly correct. Option B can be considered correct if the white precipitate R is interpreted as elemental sulfur formed by the oxidation of $H_2S$ by nitrate ions. This is a plausible scenario in the context of a multiple-choice question designed to test deeper knowledge of inorganic qualitative analysis. Both A and B are correct.