Q33 / 180 · KCET 2022 · Chemistry

All Cu(II) halides are known, except the iodide, the reason for it is that

  • Correct answer
    A
    $\mathrm{Cu}^{2+}$ oxidises iodide to iodine
  • B
    $\mathrm{Cu}^{2+}$ has much more negative hydration enthalpy
  • C
    $\mathrm{Cu}^{2+}$ ion has smaller size.
  • D
    iodide is bulky ion.

Solution

$2 \mathrm{Cu}^{2+}+4 \mathrm{I}^{-} \longrightarrow 2 \mathrm{CuI}_2$ This $\mathrm{Cu}$ (II) iodide immediately decomposes to liberate $\mathrm{I}_2$ and insoluble copper (I) iodide. $$ 2 \mathrm{CuI}_2 \longrightarrow 2 \mathrm{CuI}+\mathrm{I}_2 $$

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

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