The correct option is A True
The given statement is True.
(A) is Hg2Cl2 (calomel). It makes part of calomel electrode and is insoluble in water.
(A) is blackened by ammonia forming (B).
Hg2Cl2 (A)+2NH3→[Hg+Hg(NH2)Cl] (B)↓+NH4Cl
(B) is soluble in aqua regia forming (C).
[Hg+Hg(NH2)Cl] (B)+3Clfrom aqua regia→2HgCl2(C)
(C) gives orange ppt with KI but ppt dissolves in excess of KI forming (D).
HgCl2(C) +2KI→HgI2(D) ↓(orange) +2KCl