Answer:
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 they will selectively precipitates as white precipitate of chlorides. These cations are respectevely: Ag+,Pb2+,Hg22+ . If we consider the whole periodic table, the only elements whose chlorides are insoluble are those of silver, lead (II) and mercury (I), while chlorides of the other elements are soluble.
However, the precipitate of PbCl2 is soluble in hot water but the other two remains insoluble after treating with hot water. The precipitate of AgCl disappears upon treatment of NH3 solution but Hg2Cl2 becomes gray--black in the reaction with NH3. The black Colour appears due to the formation of metallic Hg.
Balanced chemical equation of the reaction is :
Hg2Cl2 + 2NH3 ---------> HgNH2Cl (white ppt.) + Hg (black ppt.) + NH4Cl
Therefore, the unknown solution definitely has Hg22+ present.