Q46 / 225 · JEE Main 2004 · Chemistry

Excess of $\mathrm{KI}$ reacts with $\mathrm{CuSO}_4$ solution and then $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ solution is added to it. Which of the statements is incorrect for this reaction?

  • A
    $\mathrm{Cu}_2 \mathrm{I}_2$ is reduced
  • B
    Evolved $\mathrm{I}_2$ is reduced
  • C
    $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ is oxidized
  • Correct answer
    D
    $\mathrm{Cul}_2$ is formed

Solution

$2 \mathrm{CuSO}_4+4 \mathrm{KI}$ (excess) $\longrightarrow 2 \mathrm{~K}_2 \mathrm{SO}_4+\mathrm{Cu}_2 \mathrm{I}_2+\mathrm{I}_2 \uparrow$ $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3+\mathrm{I}_2 \longrightarrow \mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6+2 \mathrm{Nal}$

Hence, the correct answer is option: 'D'.

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