Answer:
The correct answer is option C. The unknown solution definitely has Hg22+ present.
Step-by-step explanation:
In the analysis of group 1 metal cation, the unknown solution is treated with sufficient quantity of 6 M HCl solution and if group 1 metal cations are present then white precipitate of Agcl, PbCl2 or Hg2Cl2 is formed. The precipitate of PbCl2 is soluble in hot water but the other two remains insoluble after treating with hot water. Precipitate of AgCl disappears upon treatment of NH3 solution but Hg2Cl2 becomes black in the reaction with NH3. The black Colour appears due to the formation of metallic Hg.
Balanced chemical equation of the reation is -
Hg2Cl2 + 2NH3 ---------> HgNH2Cl (white ppt.) + Hg (black ppt.) + NH4Cl
Therefore, from the given information the conclusion which can be drawn is that the unknown solution definitely has Hg22+ present.