Cu 2 + salt reacts with potassium iodide to give

{Cu}^{2 +}

salt reacts with potassium iodide to give

${Cu}_{2} I_{2}$
${Cu}_{2} I_{3}$
$Cul$
$Cu {(I_{3})}_{2}$

Solution

$2 {Cu}^{+ 2} + 4 I^{-} ⟶ {Cu}_{2} I_{2} (s) + I_{2}$

$2 {Cu}^{+ 2} + 3 I^{-} ⟶ 2 CuI + I_{2}$

Asked in: JEE Main 2021 (20 Jul Shift 2)

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