Which of the following is formed in solution when [Cu(NH3)4]2+ is treated with KCN till the colour o — Qualitative and Quantitative Analysis Chemistry Question

💡 Solution & Explanation

Step 1: Identify white sublimable chlorides among the options, which include mercurous chloride (Hg2Cl2) and mercuric chloride (HgCl2). Step 2: When mercurous chloride (Hg2Cl2, calomel) is treated with ammonia (NH3), it undergoes a disproportionation reaction to form a mixture of black finely divided metallic mercury (Hg) and white mercuric amido chloride: Hg2Cl2 + 2NH3 -> Hg↓ (black) + Hg(NH2)Cl↓ (white) + NH4Cl. Step 3: This mixture appears dark grey or black, whereas HgCl2 forms a white precipitate with ammonia. Thus, Hg2Cl2 is the substance, matching option (a).