Which of the following reactions is an example of a redox reaction?

First, we recall the basic test for a redox (reduction-oxidation) reaction. A reaction is redox if at least one element shows an increase in its oxidation number (oxidation) while another shows a decrease (reduction). We shall therefore calculate the oxidation numbers of the key atoms on both sides of every given equation.

Option A

$$\text{XeF}_2 + \text{PF}_5 \; \rightarrow \; \text{XeF}^{+}\text{PF}_6^{-}$$

For $$\text{XeF}_2$$ we put $$\text{OS}_\text{F}=-1$$. Writing $$x$$ for the oxidation state of xenon, we have

$$x + 2(-1)=0 \;\;\Longrightarrow\;\; x=+2.$$

For $$\text{PF}_5$$, with $$\text{OS}_\text{F}=-1$$ and $$y$$ for phosphorus,

$$y + 5(-1)=0 \;\;\Longrightarrow\;\; y=+5.$$

Now consider $$\text{XeF}^{+}$$. If the oxidation state of Xe is $$x'$$, then

$$x' + (-1)=+1 \;\;\Longrightarrow\;\; x'=+2.$$

For $$\text{PF}_6^{-}$$ we set $$y'$$ for phosphorus, so

$$y' + 6(-1)=-1 \;\;\Longrightarrow\;\; y'=+5.$$

Thus Xe remains $$+2$$ and P remains $$+5$$. No oxidation state changes occur, so Option A is not redox.

Option B

$$\text{XeF}_6 + \text{H}_2\text{O} \; \rightarrow \; \text{XeOF}_4 + 2\text{HF}$$

In $$\text{XeF}_6$$, with $$x$$ for Xe,

$$x + 6(-1)=0 \;\;\Longrightarrow\;\; x=+6.$$

For $$\text{XeOF}_4$$, again letting $$x'$$ be the oxidation state of Xe, we use $$\text{OS}_\text{O}=-2$$ and obtain

$$x' + (-2) + 4(-1)=0 \;\;\Longrightarrow\;\; x'=+6.$$

The oxidation numbers of H and O in $$\text{H}_2\text{O}$$ and $$\text{HF}$$ remain $$+1$$ (for H) and $$-2$$ (for O) or $$-1$$ (for F) throughout. No element changes its oxidation state, hence Option B is not redox.

Option C

$$\text{XeF}_6 + 2\text{H}_2\text{O} \; \rightarrow \; \text{XeO}_2\text{F}_2 + 4\text{HF}$$

In $$\text{XeF}_6$$ we already found $$\text{OS}_\text{Xe}=+6$$. For $$\text{XeO}_2\text{F}_2$$ (put $$x$$ for Xe),

$$x + 2(-2) + 2(-1)=0 \;\;\Longrightarrow\;\; x=+6.$$

Again, the oxidation states of H, O and F do not change. Therefore Option C is not a redox reaction.

Option D

$$\text{XeF}_4 + \text{O}_2\text{F}_2 \; \rightarrow \; \text{XeF}_6 + \text{O}_2$$

For $$\text{XeF}_4$$ (let $$x$$ be Xe):

$$x + 4(-1)=0 \;\;\Longrightarrow\;\; x=+4.$$

For $$\text{XeF}_6$$ (let $$x'$$ be Xe):

$$x' + 6(-1)=0 \;\;\Longrightarrow\;\; x'=+6.$$

Thus Xe rises from $$+4$$ to $$+6$$, showing oxidation.

Now examine $$\text{O}_2\text{F}_2$$, which contains two oxygens and two fluorines. Putting $$y$$ for the average oxidation state of each O atom and $$\text{OS}_\text{F}=-1$$,

$$2y + 2(-1)=0 \;\;\Longrightarrow\;\; y=+1.$$

In $$\text{O}_2$$ the oxidation state of O is its elemental value $$0$$. Hence oxygen is reduced from $$+1$$ to $$0$$.

Because xenon is oxidised and oxygen is reduced, simultaneous oxidation and reduction occur; therefore Option D is a redox reaction.

Since only Option D satisfies the criterion for a redox change,

Hence, the correct answer is Option D.