How does the oxidation state of O change in the following reaction? L(S)+ NaOH(aq) → LiOH(aq) + Na(s)

Oxygen Doesn't change

However, Li is oxidized (0 to +1), Na is reduced (+1 to 0)

Explanation:

On reactant side, Oxygen has -2 oxidation charge because we know common oxidation states such as oxygen -2, hydrogen +1 etc.

So NaOH, O is -2, H is +1, so Na has to be +1 to equal total charge of compound

In product side, LiOH, again O has to be -2, H is +1, so Li +1 as well..

We see that oxygen oxidation state doesn't change. However, for Li it becomes oxidized going from 0 to +1 whereas, Na is reduced going from +1 to 0.

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