1. Reaction on heating potassium permanganate

Potassium permanganate decomposes on heating:

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

At about $513\,\text{K}$, one of the products formed is potassium manganate, $K_2MnO_4$.

2. Colour of $K_2MnO_4$

The manganate ion $MnO_4^{2-}$ is green in colour.

So the required product is green.

3. Magnetic nature of $K_2MnO_4$

In $MnO_4^{2-}$, let oxidation state of Mn be $x$:

$x + 4(-2) = -2$ $x - 8 = -2$ $x = +6$

So manganese is in the $+6$ oxidation state.

Atomic number of Mn = 25

Ground-state electronic configuration of Mn:

$[Ar] \, 3d^5 4s^2$

For $Mn^{6+}$, remove 6 electrons:

$Mn^{6+} : [Ar] \, 3d^1$

A $d^1$ system has one unpaired electron, hence it is paramagnetic.

4. Match with options

• A: paramagnetic and colourless $\rightarrow$ wrong (not colourless)
• B: diamagnetic and green $\rightarrow$ wrong (it is paramagnetic)
• C: diamagnetic and colourless $\rightarrow$ wrong
• D: paramagnetic and green $\rightarrow$ correct

Therefore, the correct answer is:

$\boxed{D}$