Potassium permanganate on heating at 513 K gives a product which is:

We have to analyse what happens when solid potassium permanganate, whose formula is $$KMnO_4$$, is heated to about $$513\;{\rm K}$$. Potassium permanganate is well known to undergo thermal decomposition. First, let us write the balanced thermal decomposition reaction that is found experimentally:

$$2\,KMnO_4 \;\xrightarrow{513\;{\rm K}}\; K_2MnO_4 \;+\; MnO_2 \;+\; O_2$$

Now we look at the individual products one by one. The first product on the right‐hand side is $$K_2MnO_4$$, which is called potassium manganate. The second is $$MnO_2$$ (manganese dioxide) and the third is molecular oxygen $$O_2$$.

The question asks about “the product”, and in the context of colour and magnetism the visually dominant solid that forms is $$K_2MnO_4$$, because $$MnO_2$$ is a dark brown or black solid that often masks nothing, while $$K_2MnO_4$$ imparts a distinct green colour in the melt or in aqueous solution. So we examine the properties of $$K_2MnO_4$$ in detail.

First, let us determine the oxidation state of manganese in $$K_2MnO_4$$. We know the rule:

Sum of oxidation states in a neutral compound $$= 0.$$

In $$K_2MnO_4$$:

$$2\,(+1)\;+\; x\;+\;4\,(-2)\;=\;0,$$

where $$x$$ is the oxidation state of Mn. Simplifying step by step,

$$+2 \;+\; x \;-\; 8 \;=\; 0,$$

$$x \;-\; 6 \;=\; 0,$$

$$x \;=\; +6.$$

So manganese is in the $$+6$$ oxidation state in $$K_2MnO_4$$.

Next we need its magnetic behaviour. For that we recall the electronic configuration concept. The ground-state electronic configuration of a neutral manganese atom (atomic number $$25$$) is

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

If six electrons are removed (because the oxidation state is $$+6$$), we remove the two $$4s$$ electrons first and four of the $$3d$$ electrons afterwards, giving

$$[Ar]\,3d^1.$$

There is one unpaired $$d$$ electron present. A substance having one or more unpaired electrons shows paramagnetism. Therefore $$K_2MnO_4$$ is paramagnetic.

Finally, the colour. Textbook data and laboratory observation both state that potassium manganate is green. The presence of the $$MnO_4^{2-}$$ ion (manganate ion) is responsible for this green colour.

Putting together the two properties we have just derived, $$K_2MnO_4$$ is paramagnetic (because of the single unpaired electron) and green (because of the electronic transitions within $$MnO_4^{2-}$$).

Among the four options supplied, the only one that matches “paramagnetic and green” is:

Option D.

Hence, the correct answer is Option D.