When Cu$$^{2+}$$ 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

When $$Cu^{2+}$$ is treated with excess KI:

$$ 2Cu^{2+} + 4I^- \rightarrow Cu_2I_2 \downarrow + I_2 $$

$$Cu^{2+}$$ is reduced to $$Cu^+$$ which forms the white precipitate of cuprous iodide ($$Cu_2I_2$$), and iodide is oxidized to iodine ($$I_2$$).

Therefore, $$X = Cu_2I_2$$ (white precipitate).

The liberated $$I_2$$ is then titrated with sodium thiosulphate:

$$ I_2 + 2Na_2S_2O_3 \rightarrow Na_2S_4O_6 + 2NaI $$

The product sodium tetrathionate ($$Na_2S_4O_6$$) is compound Y.

Therefore, $$X = Cu_2I_2$$ and $$Y = Na_2S_4O_6$$.