Animated Solution for Chemistry - d and f-Block Elements: Potassium permanganate on heating at 513 K gives a product which is

The world of transition metals is nothing short of a magical canvas, painted with vibrant colors and governed by the profound laws of quantum mechanics. Today, we are going to dive deep into a classic problem that beautifully intertwines chemical reactivity, oxidation states, and magnetic properties.

Imagine you are in a laboratory, holding a test tube filled with deep purple crystals of potassium permanganate ($KMn{O}_4$). This compound is a legendary oxidizing agent, but it has a breaking point.

When you heat potassium permanganate to a scorching 513 K, it can no longer hold itself together. The intense thermal energy causes the molecule to undergo a fascinating internal redox reaction.

In this auto-redox process, the manganese atom is partially reduced, while the oxygen atoms are oxidized to form free oxygen gas. The purple crystals break down, yielding three distinct products: potassium manganate ($K_{2}Mn{O}_4$), manganese dioxide ($Mn{O}_2$), and oxygen gas ($O_2$).

The balanced chemical equation for this transformation is a cornerstone of inorganic chemistry:

This single equation is the key to unlocking our problem. But now, we face a new challenge: which of these products is the question asking about?

To find our culprit, we must play the role of a chemical detective and look closely at the options provided. The options describe the mystery product as either "green" or "colourless".

Let's examine our solid products. Manganese dioxide ($Mn{O}_2$) is a well-known, dark brown or black powder. It is often used as a catalyst, but it clearly doesn't fit the color description in our options.

On the other hand, potassium manganate ($K_{2}Mn{O}_4$) is famous for its striking, brilliant green color. This immediately tells us that the question is directing our focus entirely onto $K_{2}Mn{O}_4$.

Now that we have identified our target, we need to determine its magnetic nature. The first step in this quantum journey is to calculate the oxidation state of the central manganese atom.

Let's set up a simple algebraic equation. We know that potassium, being an alkali metal, always has an oxidation state of $+1$. Oxygen, in most of its compounds, carries an oxidation state of $-2$.

Let the oxidation state of manganese be $x$. Since the overall molecule is electrically neutral, the sum of all oxidation states must equal zero:

Solving this simple linear equation, we get:

Manganese is sitting at a $+6$ oxidation state in this green compound. This is a highly oxidized state, but it is one step lower than the $+7$ state it held in the original reactant.

To understand how this molecule interacts with a magnetic field, we must look at its electronic configuration. Manganese, with an atomic number of 25, has a ground state configuration of $[Ar]3d^{5}4s^2$.

To form the $M{n}^{+6}$ ion, it must strip away six of its outermost electrons. According to the rules of quantum mechanics, electrons are always removed from the outermost principal quantum shell first.

Therefore, it first loses the two electrons from the higher-energy $4s$ orbital. After that, it loses four electrons from the $3d$ orbital.

This leaves us with a final electronic configuration of:

Notice that solitary electron sitting in the $3d$ subshell? That single, unpaired electron is the absolute secret to the molecule's magnetic identity.

According to the principles of magnetism, any chemical species with one or more unpaired electrons will be weakly attracted to an external magnetic field. This fundamental property is known as paramagnetism.

Because our $M{n}^{+6}$ ion has exactly one unpaired electron ($n=1$), it is definitively paramagnetic.

A Bonus Insight

You might be wondering, why does the color change so drastically from deep purple to bright green? This is where the physics of light and electrons becomes truly beautiful.

In the original potassium permanganate ($KMn{O}_4$), manganese is in a $+7$ state, meaning it has zero d-electrons ($d^0$). Its intense purple color does not come from d-d transitions, but rather from a phenomenon called Ligand-to-Metal Charge Transfer (LMCT), where an electron temporarily jumps from the oxygen atom to the empty d-orbital of manganese.

However, in our green product ($K_{2}Mn{O}_4$), the manganese is in a $+6$ state with a $d^1$ configuration. Because it now possesses a d-electron, it can undergo classic d-d transitions.

When white light hits the molecule, this single d-electron absorbs specific wavelengths of light to jump to a higher energy d-orbital. The light that is not absorbed is reflected back to our eyes, and in this case, it corresponds to the green region of the visible spectrum!

We have successfully decoded every layer of this problem, from the macroscopic color changes to the microscopic quantum states.

The thermal decomposition of potassium permanganate yields potassium manganate ($K_{2}Mn{O}_4$). This compound is characterized by its beautiful green color.

Furthermore, because its central manganese atom possesses exactly one unpaired d-electron, it is definitively paramagnetic.

Therefore, the product is both paramagnetic and green, which perfectly aligns with our correct option.

This problem is a brilliant reminder of how macroscopic observable properties, like color and magnetism, are directly dictated by the invisible, quantum dance of electrons!