Question

When $C{u}^{2+}$ 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. $X=Cu{I}_{2}Y=N{a}_2{S}_2{O}_3$
• B. $X=Cu{I}_{2}Y=N{a}_2{S}_4{O}_6$
• C. $X=C{u}_2{I}_{2}Y=N{a}_2{S}_4{O}_6$
• D. $X=C{u}_2{I}_{2}Y=N{a}_2{S}_4{O}_5$

Answer 1

Answer: (C)

$C{u}^{2+}+2KI⟶\underset{\text{ Unstable }}{Cu{I}_2}\downarrow +2K^{+}$
$I^{-}$is strong R.A it reduces $C{u}^{2+}$ to $C{u}^{+}$
$2Cu{I}_2⟶\underset{\text{ (White)’X’ }}{C{u}_2{I}_2\downarrow }+I_2$
$KI+I_2⟶K^{+}{I}_3^{-}\text{(Brown solution) }$
$I_3^{-}\rightleftharpoons I_2+I^{-}$
$K{I}_3+N{a}_2{S}_2{O}_3\to KI+\underset{(Y)}{N{a}_2{S}_4{O}_6}$