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Cu^{2+} salt reacts with potassium iodide to give
Option: 1 \mathrm{Cu_{2}I_{2}}
Option: 2 \mathrm{Cu_{2} I_{3}}
Option: 3 \mathrm{C u I}
Option: 4 \mathrm{C u\left(I_{3}\right)_{2}}

Answers (1)

best_answer

Cu^{2+} reacts with I^{-} to form precipitate of Cu_{2}I_{2}.

Cu^{2+} is reduced to Cu^{+} while Iodide ions are oxidised to Iodine as shown below

\mathrm{Cu}^{2+}+\mathrm{I}^{-} \rightarrow \mathrm{Cu}_{2} \mathrm{I}_{2}+\mathrm{I}_{2}

The I_{2} complexes with I^{-} to give tri iodide ions \left(I_{3}^{-}\right) which forms a reddish solution.

Hence, the correct answer is option (1).

Posted by

sudhir.kumar

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