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Question

Salt (A) makes part of electrode and is insoluble in water. (A) is blackened by NH3 forming (B). (B) is soluble in aqua regia forming (C). (C) gives orange precipitate with KI but the precipitate dissolves in excess of KI forming (D). Here,
(A) : Hg2Cl2 (calomel)
(B) : (HgNH2Cl+Hg)
(C) : HgCl2
(D) : K2HgI4
If true enter 1, else enter 0.

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Solution

The salt (A) is Hg2Cl2. It is a part of calomel electrode. It is water insoluble.
It reacts with ammonia to form (B) which is (HgNH2Cl+Hg). It is black in colour.
Hg2Cl2+2NH4OHH2NHgCl+Hgblack+NH4Cl+2H2O
(B) dissolves in aqua regia to form (C) which is HgCl2.
(C) gives orange precipitate with KI which dissolves in excess of KI to form (D) which is K2HgI4.
HgCl2+2KIHgI2orangeprecipitate+2KCl
HgI2+2KI(excess)K2[HgI4]

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