When $\mathrm{Cu}^{2+}$ ion is treated with $\mathrm{KI}$, a white precipitate, $\mathrm{X}$ appears in solution. The solution is titrated with sodium thiosulphate, the compound $\mathrm{Y}$ is formed. $\mathrm{X}$ and $\mathrm{Y}$ respectively are :

Question

$\mathrm{X=CuI_2}$	$\mathrm{Y=Na_2S_4O_6}$

$\mathrm{X=Cu_2I_2}$	$\mathrm{Y=Na_2S_4O_6}$

$\mathrm{X=CuI_2}$	$\mathrm{Y=Na_2S_2O_3}$

$\mathrm{X=Cu_2I_2}$	$\mathrm{Y=Na_2S_4O_5}$

Solution

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