The products obtained from a reaction of hydrogen peroxide and acidified potassium permanganate are

We need to find the products when hydrogen peroxide reacts with acidified potassium permanganate ($$KMnO_4$$).

In acidic medium, $$KMnO_4$$ acts as a strong oxidising agent and $$H_2O_2$$ acts as a reducing agent (it gets oxidised).

Since $$MnO_4^-$$ is reduced in acidic medium, the reduction half-reaction is $$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$$. Meanwhile, $$H_2O_2$$ is oxidised according to $$H_2O_2 \rightarrow O_2 + 2H^+ + 2e^-$$.

Substituting and balancing electrons by multiplying the reduction half-reaction by 2 and the oxidation half-reaction by 5 gives the balanced half-reactions: $$2MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O$$ and $$5H_2O_2 \rightarrow 5O_2 + 10H^+ + 10e^-$$.

Combining these half-reactions yields the overall reaction $$2MnO_4^- + 5H_2O_2 + 6H^+ \rightarrow 2Mn^{2+} + 8H_2O + 5O_2$$.

Therefore, the products are $$Mn^{2+}$$, $$H_2O$$, and $$O_2$$. In acidic medium, $$KMnO_4$$ is reduced to $$Mn^{2+}$$ (not $$Mn^{4+}$$). $$Mn^{4+}$$ as $$MnO_2$$ is formed only in neutral or weakly basic medium.

Therefore, the correct answer is Option D: $$Mn^{2+}$$, $$H_2O$$, $$O_2$$ only.