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

  • A
    $X = Cu_2I_2, Y = Na_2S_4O_5$
  • B
    $X = Cu_2I_2, Y = Na_2S_4O_6$
  • C
    $X = CuI_2, Y = Na_2S_4O_3$
  • D
    $X = CuI_2, Y = Na_2S_4O_6$

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