1. Criterion for a redox reaction

A reaction is redox if at least one element undergoes oxidation number change.

So, we check oxidation states in each option.

2. Option A

$XeF_6 + H_2O \to XeOF_4 + 2HF$

Let oxidation state of Xe in $XeF_6$ be $x$:

$x + 6(-1)=0 \Rightarrow x=+6$

Now in $XeOF_4$:

$x + (-2) + 4(-1)=0$ $x-2-4=0$ $x=+6$

Xe remains $+6$.

Hydrogen, oxygen, fluorine also keep their usual oxidation states.

So this is not a redox reaction.

3. Option B

$XeF_6 + 2H_2O \to XeO_2F_2 + 4HF$

In $XeF_6$, Xe is:

$x+6(-1)=0 \Rightarrow x=+6$

In $XeO_2F_2$:

$x+2(-2)+2(-1)=0$ $x-4-2=0$ $x=+6$

Again, Xe remains $+6$.

Hence this is not redox.

4. Option C

$XeF_4 + O_2F_2 \to XeF_6 + O_2$

First, Xe in $XeF_4$:

$x+4(-1)=0 \Rightarrow x=+4$

Xe in $XeF_6$:

$x+6(-1)=0 \Rightarrow x=+6$

So Xe is oxidized from $+4$ to $+6$.

Now check oxygen in $O_2F_2$.

Let oxidation state of O be $x$:

$2x+2(-1)=0$ $2x-2=0$ $x=+1$

In $O_2$, oxygen is $0$.

So O changes from $+1$ to $0$, hence oxygen is reduced.

Therefore, oxidation and reduction both occur.

So Option C is a redox reaction.

5. Option D

$XeF_2 + PF_5 \to [XeF]^+ PF_6^-$

In $XeF_2$, Xe is:

$x+2(-1)=0 \Rightarrow x=+2$

In $[XeF]^+$:

$x+(-1)=+1$ $x=+2$

Xe remains $+2$.

For phosphorus:

In $PF_5$:

$x+5(-1)=0 \Rightarrow x=+5$

In $PF_6^-$:

$x+6(-1)=-1$ $x-6=-1$ $x=+5$

P also remains $+5$.

Hence this is not redox.

6. Conclusion

Only Option C involves change in oxidation numbers, so it is the redox reaction.

7. Comparison with stored correct answer

Stored correct answer: C

My derived answer: C

They agree.