Let us write the oxidation number of each atom involved in the given reaction above its symbol as:
`overset(+1)(H_(2)) overset(-2)(O) + overset(0)(F_(2)) to overset(+1)(H) overset(-1)(F) + overset(-1)(H)overset(-2)(O)overset(+1)(F)`
Here , we have observed that the oxidation number of F increases from 0 in `F_(2)` to `+1` in HOF .
Also, the oxidation number decreases from 0 in `F_(2)` to `–1` in HF. Thus, in the above reaction, F is both oxidized and reduced. Hence, the given reaction is a redox reaction.