When Cu²⁺ ion is treated with KI, a white precipitate, X appears in solution. The solution is titrated with sodium thiosulphate, the compound Y is formed. X and Y respectively are :

Solution

Chemistry ma’am · 4 steps

Copper(II) oxidises iodide and drops to white Cu₂I₂; thiosulphate ends up as tetrathionate.

1. Copper meets iodide. Cu²⁺ is a strong enough oxidising agent to oxidise I⁻ to I₂, and it is reduced to Cu⁺ in the process. So CuI₂ never survives. What precipitates is copper(I) iodide, written Cu₂I₂.

   2Cu²⁺ + 4I⁻ → Cu₂I₂ v + I₂

2. Why it is white. Cu⁺ has a 3d¹⁰ configuration, a full d subshell. No d–d transition is possible, so the solid is white. That white solid is X = Cu₂I₂.
3. Titrate the iodine. The free I₂ is titrated with sodium thiosulphate. Iodine oxidises thiosulphate to tetrathionate: I₂ + 2Na₂S₂O₃ → Na₂S₄O₆ + 2NaI. The average oxidation state of sulphur goes from +2 to +2.5.
4. So. X is Cu₂I₂ and Y is sodium tetrathionate, Na₂S₄O₆. That is option (b).

The idea to keep. Cu²⁺ plus I⁻ always gives Cu₂I₂ plus I₂, never CuI₂. In iodometry, thiosulphate always ends up as tetrathionate, S₄O₆²⁻.

Answer from the NTA final answer key. Solution by PrepWiser.