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

Question

All Cu (II) halides are known except the iodide. The reason for it is that (A) iodide is a bulky ion (B) Cu2+ oxidizes iodide to iodine (C) Cu2+(aq) has much more negative hydration enthalpy (D) Cu2+ ion has smaller size

Tardigrade Admin · Accepted Answer

2Cu2+ + 41- → 2CuI2 The Cul2 immediately decomposes to liberate I2 and insoluble copper(I) iodide. 2CuI2 → 2CuI + I2

Q. All $C u$ (II) halides are known except the iodide. The reason for it is that

iodide is a bulky ion

$C u^{2 +}$ oxidizes iodide to iodine

$C u_{(a q)}^{2 +}$ has much more negative hydration enthalpy

$C u^{2 +}$ ion has smaller size

$2 C u^{2 +} + 4 1^{-} \to 2 C u I_{2}$
The $C u l_{2}$ immediately decomposes to liberate $I_{2}$ and insoluble copper $(I)$ iodide.
$2 C u I_{2} \to 2 C u I + I_{2}$

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

iodide is a bulky ion

Cu2+ oxidizes iodide to iodine

Cu(aq)2+​ has much more negative hydration enthalpy

Cu2+ ion has smaller size

2Cu2++41−→2CuI2​ The Cul2​ immediately decomposes to liberate I2​ and insoluble copper(I) iodide. 2CuI2​→2CuI+I2​

Q. All $C u$ (II) halides are known except the iodide. The reason for it is that

$C u^{2 +}$ oxidizes iodide to iodine

$C u_{(a q)}^{2 +}$ has much more negative hydration enthalpy

$C u^{2 +}$ ion has smaller size

$2 C u^{2 +} + 4 1^{-} \to 2 C u I_{2}$
The $C u l_{2}$ immediately decomposes to liberate $I_{2}$ and insoluble copper $(I)$ iodide.
$2 C u I_{2} \to 2 C u I + I_{2}$