$$KMnO_4$$ decomposes on heating at $$513$$ K to form $$O_2$$ along with

We need to identify the products formed when $$KMnO_4$$ is heated at 513 K (along with $$O_2$$).

Write the thermal decomposition reaction.

When potassium permanganate is heated, it undergoes the following decomposition:

$$2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2$$

Verify the balancing.

Left side: K: 2, Mn: 2, O: 8

Right side: K: 2, Mn: 1+1=2, O: 4+2+2=8. Balanced.

Understand the chemistry.

In $$KMnO_4$$, manganese is in the +7 oxidation state. During thermal decomposition:

- One Mn goes from +7 to +6 in $$K_2MnO_4$$ (potassium manganate) - this is a reduction

- Another Mn goes from +7 to +4 in $$MnO_2$$ (manganese dioxide) - this is also a reduction

- Oxygen goes from -2 to 0 in $$O_2$$ - this is an oxidation

This is a disproportionation-type reaction where the oxygen is oxidised while manganese is reduced to two different lower oxidation states.

The products (along with $$O_2$$) are $$K_2MnO_4$$ and $$MnO_2$$.

The correct answer is Option (4): $$K_2MnO_4$$ and $$MnO_2$$.