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 :
Solution
$2 \mathrm{Cu}^{2+}+4 \mathrm{KI} \longrightarrow \underset{\text { White ppt. }}{\mathrm{Cu}_{2} \mathrm{I}_{2}}+\mathrm{I}_{2}$
<br/><br/>$\mathrm{I}_{2}+\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \longrightarrow 2 \mathrm{Nal}+\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}$
<br/><br/>$\mathrm{X}=\mathrm{Cu}_{2} \mathrm{I}_{2}$
<br/><br/>$\mathrm{Y}=\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}$
About this question
Subject: Chemistry · Chapter: d and f Block Elements · Topic: Properties of Transition Metals
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