Question

The copper (II) halide which does not exist is

• A. $CuF_2$
• B. $CuBr_2$
• C. $CuI_2$
• D. $CuCl_2$

Answer 1

Answer: (C)

$CuI_2$

Answer 2

$CuI _{2}$ 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:

$2 Cu ^{2+}(a q)+4 I ^{-}(a q) \longrightarrow 2 \operatorname{CuI}(s)+ I _{2}(a q)$