Animated Solution for Chemistry - d and f-Block Elements: Excess of $\text{KI}$ reacts with ${\text{CuSO}}_4$ solution and then ${\text{Na}}_2{\text{S}}_2{\text{O}}_3$ solution is added to it. Which of the following statements is incorrect for this reaction?

Imagine you are standing in a chemistry laboratory, holding a test tube filled with a beautiful, bright blue solution of copper sulphate (${\text{CuSO}}_4$). Now, you start adding an excess of colorless potassium iodide ($\text{KI}$) solution into it. What happens next is a fascinating sequence of redox reactions that forms the very foundation of a famous analytical technique known as iodometric titration.

When these two solutions mix, you might initially expect a simple double displacement reaction to occur, forming cupric iodide and potassium sulphate:

However, there is a major catch here. Cupric iodide (${\text{CuI}}_2$) is highly unstable. Why? Because the copper(II) ion (${\text{Cu}}^{2+}$) is a relatively strong oxidizing agent, and the iodide ion (${\text{I}}^{-}$) is a good reducing agent. They cannot coexist peacefully. The ${\text{Cu}}^{2+}$ immediately oxidizes the ${\text{I}}^{-}$ into elemental iodine (${\text{I}}_2$), while reducing itself to the more stable copper(I) state.

This causes the transient ${\text{CuI}}_2$ to rapidly decompose:

As a result, you will observe a white precipitate of cuprous iodide (${\text{Cu}}_2{\text{I}}_2$) suspended in a solution that turns deep brown due to the liberated aqueous iodine.

Now, the problem states that sodium thiosulphate (${\text{Na}}_2{\text{S}}_2{\text{O}}_3$), commonly known as hypo, is added to this brown mixture. Hypo is a classic reagent used to titrate free iodine. When it is added, it reacts specifically with the liberated iodine:

In this step, the brown iodine (${\text{I}}_2$) is reduced back to colorless iodide ions (${\text{I}}^{-}$), and the thiosulphate ion (${\text{S}}_2{\text{O}}_3^{2-}$) is oxidized to the tetrathionate ion (${\text{S}}_4{\text{O}}_6^{2-}$). As the reaction proceeds, the brown color fades away, leaving behind the white precipitate of ${\text{Cu}}_2{\text{I}}_2$ clearly visible.

Let's evaluate the given options based on our chemical journey:

Option (a): ${\text{Cu}}_2{\text{I}}_2$ is formed. This is True, as it precipitates out. Option (b): ${\text{CuI}}_2$ is formed. This is False. Although it might form transiently, it is highly unstable and immediately decomposes. It is not considered a final product of this reaction. Option (c): ${\text{Na}}_2{\text{S}}_2{\text{O}}_3$ is oxidised. This is True, it oxidizes to sodium tetrathionate. Option (d): Evolved ${\text{I}}_2$ is reduced. This is True, it reduces back to sodium iodide.

Therefore, the incorrect statement is that ${\text{CuI}}_2$ is formed.