The balanced equation for the reaction of fluorine with sodium iodide is:

F₂(g) + 2NaI(s) → 2NaF(s) + I₂(s)

Here's a breakdown of the reaction:

* Reactants:

* Fluorine (F₂): A diatomic, highly reactive gas.

* Sodium Iodide (NaI): A white, ionic solid.

* Products:

* Sodium Fluoride (NaF): A white, ionic solid.

* Iodine (I₂): A dark purple, solid.

Explanation:

Fluorine is a very strong oxidizing agent, meaning it readily gains electrons. Iodine is a weaker oxidizing agent than fluorine. In this reaction:

* Fluorine oxidizes iodide ions (I-) to form elemental iodine (I₂).

* Sodium ions (Na+) remain unchanged, forming sodium fluoride (NaF).

Important Note: This reaction is highly exothermic and can be explosive. It should only be performed by trained professionals in a controlled environment.