Question

Which of the following does not exist:
A. $CuC{l_2}$
B. $Cu{F_2}$
C. \[CuB{r_2}\]
D. $Cu{I_2}$

Answer 1

Hint: Copper mostly forms compounds with $ + 1$ and $ + 2$ oxidation states. The oxides, sulphides and halides of copper compounds are the simplest as they consist of only two elements. The copper compounds reacting with halogen can exist as $CuX{\text{ and Cu}}{{\text{X}}_2}$.

Complete step by step answer:
1. Copper belongs to the d-block elements and shows variable oxidation states due to the incomplete filling of d-orbitals.
2. Copper is relatively less reactive than the other members of the group.
3. $C{u^{2 + }}$ is more stable than $C{u^ + }$ because of the greater energy of hydration of $C{u^{2 + }}$.
4. Copper barely reacts with water and forms a brown black copper oxide when it reacts with oxygen.
5. All copper halides are known except the Iodide, because $C{u^{2 + }}$ oxidises ${I^ - }$ to ${I_2}$
6. Copper reacts with chlorine as depicted: $Cu + C{l_2} \to CuC{l_2}$
7. It reacts with fluorine as follows: $Cu + {F_2} \to Cu{F_2}$
8. With bromine it reacts in the following way: $Cu + B{r_2} \to CuB{r_2}$
9. But with iodide, copper does not react readily but the cupric ions oxidised the iodide ions into the molecular iodine and itself get reduced to cuprous ions. The following reaction explains: $2C{u^{2 + }} + 4{I^ - } \to 2CuI + {I_2}$ where the oxidation state of Copper ion in $CuI$ is $ + 1$.
10. Iodide ions are the strong reducing agents which reduces copper (II) iodide to copper (I) iodide by undergoing an internal redox reaction resulting in an insoluble compound.
11. Thus, all the three halogens except iodine react with copper and form their consecutive halides but $Cu{I_2}$ does not exist.

Hence, the right answer is (D).

Note:
Some roles of copper:
1. Copper plays a critical role in electron transport chains and transport of oxygen.
2. It is found in cytochrome c oxidase in mitochondria and helps in binding of oxygen.
3. It is also a major component in superoxide dismutases which disproportionate into oxygen and hydrogen peroxides.