Question

Amongst $Cu{{F}_{2}},CuC{{l}_{2}}$ and $CuB{{r}_{2}}$
(A) Only $Cu{{F}_{2}}$ is ionic
(B) Both $CuC{{l}_{2}}$ and $CuB{{r}_{2}}$ are covalent
(C) $Cu{{F}_{2}}$ and $CuC{{l}_{2}}$ are ionic but $CuB{{r}_{2}}$ is covalent
(D) $Cu{{F}_{2}},CuC{{l}_{2}}$ as well as $CuB{{r}_{2}}$ are ionic

Answer 1

Hint: Here you can apply Fazan's rule. According to it more the size of anion more will be the covalent character. In a periodic table if you move from top to bottom in a group the ionic character decreases and covalent character increases. In the halogen family, fluorine is more ionic and least covalent character. These questions contain multiple answers.

Complete step by step solution:
 From your chemistry lessons, you have learned about the Fazan rules, electronegativity, ionic as well as covalent character. Here we are going to deal with the halogen family. The above-given compounds contain both anions as well as a cation in them but the cation in all of them is the same that is copper ($Cu$).
According to Fazan’s rule, the size of the anion more will be the covalent character.
Halogens belong to group 17 of the periodic table. As the electronegativities of halogen is more so usually the compounds formed by them are ionic compounds but here the electronegativity of copper is nearly equal to that of halogens, so here it is not necessary that they all will make ionic compounds.
We know that in a periodic table ionic character decreases down the group and covalent character increases down the group. In the halogen family, the ionic character of Fluorine is more than bromine and covalent character of bromine more than fluorine.
Order of increasing ionic character (halogens)
ISo, in the compounds given above $Cu{{F}_{2}}$ will have ionic character because the size of Fluorine is less as a comparison to other halogens and it occupies the uppermost position in halogen family thus it will be more ionic.
Whereas $CuC{{l}_{2}}$ and $CuB{{r}_{2}}$ have the more covalent character in them instead of ionic because we know that as we move down the group size increases and covalent character also increases. $CuB{{r}_{2}}$ will contain more covalent character than $CuC{{l}_{2}}$.
So, among them only $Cu{{F}_{2}}$ will be ionic whereas $CuC{{l}_{2}}$ and $CuB{{r}_{2}}$ will be covalent.

Thus the correct option will be both (A) and (B).

Note: Electronegativity also decreases down the group as electronegativity decreases covalent character increases and ionic character decreases. Fluorine will have more electronegativity than other halogens. Electronegativity order in halogens: F>Cl>Br>I. Compounds like $Cu{{I}_{2}}$ do not exist because iodine ion acts as a strong reducing agent and it reduces $Cu(II)$ to $Cu(I)$ and makes it highly unstable.