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Central Atom and Ligands

When Cu2+ ion...

Question

When $C u^{2 +}$ ion is treated with $K I$ , a white precipitate is formed. Explain the reaction with the help of a chemical equation.

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Solution

The reaction between $C u^{2 +}$ ion and $K I$ will take place as:

$\begin{matrix} 2 C u^{2 +} (a q) + 4 I^{-} (a q) \to & C u_{2} I_{2} (s) + I_{2} w h i t e p r e c i p i t a t e \end{matrix}$

In the aqueous solution, $K I$ dissociates to give $K^{+}$ and $I^{-}$ .

Since iodide ion $(I^{-})$ is a strong reducing
agent, it reduces $C u^{2 +}$ to $C u^{+}$ and a white precipitate of $C u_{2} I_{2}$ is formed.

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Central Atom and Ligands

Standard XII Chemistry

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