The copper (II) halide which does not exist is

Question

The copper (II) halide which does not exist is (A) CuF2 (B) CuBr2 (C) CuI2 (D) CuCl2

Tardigrade Admin · Accepted Answer

CuI 2 is not stable and thus does not exist iodide ion will act as reducing agent that converts Cu(II) to Cu(I) and part of iodide ion will be oxidised to iodine. The first reaction involved is as follows: 2 Cu 2+(a q)+4 I -(a q) longrightarrow 2 operatornameCuI(s)+ I 2(a q)

Q. The copper (II) halide which does not exist is

$C u F_{2}$

$C u B r_{2}$

$C u I_{2}$

$C u C l_{2}$

$C u I_{2}$ is not stable and thus does not exist iodide ion will act as reducing agent that converts $C u (II)$ to $C u (I)$ and part of iodide ion will be oxidised to iodine. The first reaction involved is as follows:

$2 C u^{2 +} (a q) + 4 I^{-} (a q) ⟶ 2 CuI (s) + I_{2} (a q)$

Q. The copper (II) halide which does not exist is

CuF2​

CuBr2​

CuI2​

CuCl2​

CuI2​ is not stable and thus does not exist iodide ion will act as reducing agent that converts Cu(II) to Cu(I) and part of iodide ion will be oxidised to iodine. The first reaction involved is as follows: 2Cu2+(aq)+4I−(aq)⟶2CuI(s)+I2​(aq)

$C u F_{2}$

$C u B r_{2}$

$C u I_{2}$

$C u C l_{2}$

$C u I_{2}$ is not stable and thus does not exist iodide ion will act as reducing agent that converts $C u (II)$ to $C u (I)$ and part of iodide ion will be oxidised to iodine. The first reaction involved is as follows:

$2 C u^{2 +} (a q) + 4 I^{-} (a q) ⟶ 2 CuI (s) + I_{2} (a q)$