Question

${\mathrm{Cu}}^{2+}$ salts react with $\mathrm{KI}$ and forms:

• A. ${\mathrm{Cu}}_2{\mathrm{I}}_2$
• B. $\mathrm{CuI}$
• C. $\mathrm{Cu}{\left({\mathrm{I}}_3\right)}_2$
• D. ${\mathrm{CuI}}_3$

Answer 1

The correct option is A

${\mathrm{Cu}}_2{\mathrm{I}}_2$

Explanation for the correct answer:

1. The oxidation state of iodine in Potassium Iodide ($\mathrm{KI}$) is $-1$which is its lowest possible oxidation state.
2. Since it is present in the lowest oxidation state, it will participate in the reaction as a reducing agent.
3. Reduction of copper from ${\mathrm{Cu}}^{2+}$to ${\mathrm{Cu}}^{+}$will occur.
4. The following reaction will take place-

$2{\mathrm{CuSO}}_4\left(\mathrm{aq}\right)+4\mathrm{KI}\left(\mathrm{aq}\right)\to 2{\mathrm{K}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)+{\mathrm{Cu}}_2{\mathrm{I}}_2\left(\mathrm{s}\right)+{\mathrm{I}}_2$

Thus, ${\mathrm{Cu}}^{2+}$ salts react with $\mathrm{KI}$ and form ${\mathrm{Cu}}_2{\mathrm{I}}_2$.

Hence, option A is the correct answer.