Animated Solution for Chemistry - Redox Reactions: Which of the following reactions is an example of a redox reaction?

Imagine a cosmic dance where elements constantly trade their most prized possessions: electrons. This beautiful exchange is the heart of a redox reaction. The term "redox" is a portmanteau of two fundamental chemical processes: reduction and oxidation.

In any chemical reaction, if one species loses electrons (oxidation), another must gain them (reduction). They are two sides of the same coin; you simply cannot have one without the other. But how do we spot this invisible exchange of electrons? We use a powerful bookkeeping tool called the oxidation state.

Oxidation States

To determine if a reaction is redox, we must act like chemical detectives and track the oxidation states of every element from the reactant side to the product side.

- If an element's oxidation state increases, it has lost electrons. This is oxidation. - If an element's oxidation state decreases, it has gained electrons. This is reduction.

Let's put our detective hats on and analyze the first option provided in the question:

We need to assign oxidation states to Xenon ($\mathrm{Xe}$) and Oxygen ($\mathrm{O}$) before and after the reaction. We know that Fluorine ($\mathrm{F}$) is the most electronegative element on the periodic table, so it stubbornly holds onto a $-1$ oxidation state in all its compounds.

On the Reactant Side: - In Xenon tetrafluoride (${\mathrm{XeF}}_4$), there are four fluorine atoms, contributing a total charge of $-4$. For the molecule to be neutral, Xenon must be in a $+4$ oxidation state. - In dioxygen difluoride (${\mathrm{O}}_2{\mathrm{F}}_2$), the two fluorine atoms contribute $-2$. To balance this, the two oxygen atoms must share a $+2$ charge, meaning each oxygen atom is in a rare $+1$ oxidation state.

On the Product Side: - In Xenon hexafluoride (${\mathrm{XeF}}_6$), the six fluorine atoms contribute $-6$. Thus, Xenon is forced into a $+6$ oxidation state. - The oxygen gas (${\mathrm{O}}_2$) is in its elemental, uncombined form. By definition, any element in its free state has an oxidation state of $0$.

Let's summarize the changes we just discovered:

1. Xenon goes from $+4$ to $+6$. Its oxidation state increased, meaning it underwent oxidation. 2. Oxygen goes from $+1$ to $0$. Its oxidation state decreased, meaning it underwent reduction.

Because both oxidation and reduction are occurring simultaneously in this single chemical equation, it is the textbook definition of a redox reaction!

If you were to analyze the other options, you would find that they are either simple adduct formations or hydrolysis reactions. In those processes, the oxidation states of Xenon, Oxygen, and Fluorine remain completely unchanged from left to right. Therefore, option (a) stands alone as the correct answer.