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Q.

When $C u S O_{4}$ 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.

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a

$x = C u_{2} I_{2} y = N a_{2} S_{2} O_{5}$

b

$x = C u_{2} I_{2} y = N a_{2} S_{4} O_{6}$

c

$x = C u I_{2} y = N a_{2} S_{4} O_{3}$

d

$x = C u I_{2} y = N a_{2} S_{4} O_{6}$

answer is B.

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Detailed Solution

$C u S O_{4} + 2 K I \overset{}{\to} \underset{U n s t a b l e}{C u I_{2}} + K_{2} S O_{4} 2 C u I_{2} \overset{}{\to} \underset{W h i t e}{C u_{2} I_{2}} + I_{2} K I + I_{2} \overset{}{\to} K I_{3} B r o w n s o l u t i o n K I_{3} + 2 N a_{2} S_{2} O_{3} \overset{}{\to} K I + N a_{2} S_{4} O_{6} + 2 N a I$

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When CuSO4 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.