A white powder "A" on heating gave a non-combustible gas and a white residue. The residue on heating turns yellow. The residue dissolves in dil. HCl and the solution gives a white ppt. with K4[Fe(CN)6

Question

A white powder "A" on heating gave a non-combustible gas and a white residue. The residue on heating turns yellow. The residue dissolves in dil. HCl and the solution gives a white ppt. with K4[Fe(CN)6]. "A" would be:

Prashant Kotian · Accepted Answer

Correct answer: B. Step 1: Cu2+ ions react with iodide (I-) to form unstable copper(II) iodide (CuI2), which spontaneously decomposes to precipitate white copper(I) iodide (CuI, precipitate 'P') and liberate free iodine (I2). Step 2: In the presence of excess KI, the liberated iodine (I2) dissolves by reacting with iodide to form soluble triiodide ions (KI3) in the liquor ('M'). Step 3: Thus, precipitate 'P' is CuI and liquor 'M' is KI3 solution, matching option (c).

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