The oxidation state of manganese in the product obtained in a reaction of potassium permanganate and hydrogen peroxide in basic medium is

We need to find the oxidation state of manganese in the product formed when potassium permanganate reacts with hydrogen peroxide in basic medium.

In basic medium, $$KMnO_4$$ reacts with $$H_2O_2$$ as follows: $$2KMnO_4 + 3H_2O_2 \rightarrow 2MnO_2 + 2KOH + 3O_2 + 2H_2O$$.

Since the permanganate ion is reduced in this reaction, the manganese-containing product is manganese dioxide, $$MnO_2$$, which appears as a brown precipitate.

Let the oxidation state of Mn in $$MnO_2$$ be $$x$$. Substituting into the charge balance gives $$x + 2(-2) = 0$$.

From the above equation, we have $$x - 4 = 0$$, which leads to $$x = +4$$.

Therefore, the oxidation state of manganese changes from +7 in $$KMnO_4$$ to +4 in $$MnO_2$$.

Hence, the answer is 4.