When Cu2+ ion is treated with Kl, a white precipitate is formed. Explain the reaction with the help of a chemical equation.

When copper (II) ion (Cu2+) reacts with iodide ion (I-), a white precipitate of copper (II) iodide (CuI2) is formed. This is a precipitation reaction, which can be represented by the following chemical equation:

2KI(aq) + CuSO4(aq) -> CuI2(s) + K2SO4(aq)

In this equation, KI is potassium iodide, CuSO4 is copper sulfate, CuI2 is copper iodide, and K2SO4 is potassium sulfate. The (aq) denotes substances that are in aqueous solution (dissolved in water), and the (s) denotes a solid substance.

The reaction occurs because copper (II) ions and iodide ions combine to form copper (II) iodide, which is insoluble in water and therefore forms a solid precipitate. This is a common type of reaction in chemistry known as a precipitation reaction.