The reaction of $$H_2O_2$$ with potassium permanganate in acidic medium leads to the formation of mainly

The reaction of $$H_2O_2$$ with $$KMnO_4$$ in acidic medium is a well-known redox reaction.

The balanced equation is:

$$2KMnO_4 + 5H_2O_2 + 3H_2SO_4 \rightarrow 2MnSO_4 + K_2SO_4 + 5O_2 + 8H_2O$$

Analysis of oxidation states:

- In $$KMnO_4$$, manganese is in the $$+7$$ oxidation state.

- In the product $$MnSO_4$$, manganese is in the $$+2$$ oxidation state.

In acidic medium, $$KMnO_4$$ acts as a strong oxidising agent and is reduced from $$Mn^{7+}$$ to $$Mn^{2+}$$. Meanwhile, $$H_2O_2$$ acts as a reducing agent and is oxidised to $$O_2$$.

Note: In neutral or weakly acidic medium, $$KMnO_4$$ is reduced to $$Mn^{4+}$$ (as $$MnO_2$$), but in strongly acidic medium, the reduction goes all the way to $$Mn^{2+}$$.

Therefore, the correct answer is Option A: $$Mn^{2+}$$.