A colorless aqueous solution contains nitrates of two metals, X and Y. When it was added to an aqueous solution of NaCl, a white precipitate was formed. This precipitate was found to be partly soluble in hot water to give a residue P and a solution Q. The residue P was soluble in aq. $$NH_{3}$$ and also in excess sodium thiosulfate. The hot solution Q gave a yellow precipitate with KI. The metals X and Y, respectively, are

The original colourless solution contains the nitrates $$X(NO_3)_m$$ and $$Y(NO_3)_n$$.

Step 1 - Action of $$NaCl$$
Adding aqueous $$NaCl$$ produces a white precipitate. White chlorides that precipitate from water are mainly $$AgCl$$, $$PbCl_2$$ and $$Hg_2Cl_2$$. Hence the cations X and Y must come from this group.

Step 2 - Heating the chloride precipitate
The precipitate is “partly soluble” in hot water, giving:
  • Residue $$P$$ (insoluble portion)
  • Solution $$Q$$ (hot filtrate)

Solubility behaviour of the suspected chlorides in hot water:
  • $$AgCl$$ - practically insoluble, even on heating.
  • $$PbCl_2$$ - markedly more soluble on heating, dissolves to give $$Pb^{2+}$$ in the filtrate.
  • $$Hg_2Cl_2$$ - does not dissolve but disproportionates; this is not mentioned.
Therefore, the chloride that remains as residue $$P$$ is $$AgCl$$ and the chloride that enters the hot solution $$Q$$ is $$PbCl_2$$.

Step 3 - Confirming residue $$P$$
Residue $$P$$ is stated to dissolve in aqueous $$NH_3$$ as well as in excess $$Na_2S_2O_3$$. These are the classic confirmatory tests for $$Ag^+$$:
$$AgCl + 2\,NH_3 \rightarrow [Ag(NH_3)_2]^+ + Cl^-$$
$$AgCl + 2\,S_2O_3^{2-} \rightarrow [Ag(S_2O_3)_2]^{3-} + Cl^-$$
Thus residue $$P = AgCl$$ and X = Ag.

Step 4 - Confirming solution $$Q$$
The hot solution $$Q$$ (which contains the ions that dissolved) gives a yellow precipitate with $$KI$$:
$$Pb^{2+} + 2\,I^- \rightarrow PbI_2 \downarrow\ (\text{yellow})$$
This is the confirmatory test for $$Pb^{2+}$$, so Y = Pb.

Conclusion
X is $$Ag$$ and Y is $$Pb$$.

Option A which is: Ag and Pb