Answer:
The correct answer is option D.
Explanation:
All the cations given belong to the group-1 cation and will give precipitate of their respective chlorides when HCl is added to their aqueous solution.
[tex]Ag^+(aq)+HCl(aq)\rightarrow AgCl(s)+H^+(aq)[/tex]
[tex]Cu^{2+}+(aq)+2HCl(aq)\rightarrow CuCl_2(s)+2H^+(aq)[/tex]
[tex]Hg_{2}^{2+}+(aq)+2HCl(aq)\rightarrow Hg_2Cl_2(s)+2H^+(aq)[/tex]
[tex]Pb^{2+}+(aq)+2HCl(aq)\rightarrow PbCl_2(s)+2H^+(aq)[/tex]
All the precipitates are insoluble in cold water but out of all these only lead(II) chloride is soluble in warm water.