When Cu ^ 2 + ion is treated with KI , a white precipitate, X...

The reaction between ${Cu}^{2 +}$ and $KI$ will takes place as:

${Cu}^{2 +} + 2KI ⟶ \mathop{{CuI}_{2} ↓}_{Unstable} + 2{K}^{+}$

${I}^{-}$ is a strong reducing agent, it reduces ${Cu}^{2 +}$ to ${Cu}^{+}$ and a white precipitate of ${Cu}_{2}{I}_{2}$ is formed.

$2{CuI}_{2} ⟶ \mathop{{Cu}_{2}{I}_{2} ↓}_{( White ) ' X '} + {I}_{2}$

Sodium thiosulphate is used as reducing agent to titrate iodine, with iodide being the product of the reaction.

$2{Na}_{2}{S}_{2}{O}_{3} + {I}_{2} \to 2NaI + \mathop{{Na}_{2}{S}_{4}{O}_{6}}_{( Y )}$

So, the compound $X\text{and}Y$ are ${Cu}_{2}{I}_{2}\text{and}{Na}_{2}{S}_{4}{O}_{6}$ respectively.