Animated Solution for Chemistry - d and f-Block Elements: Given below are two statements. Statement I Potassium permanganate on heating at $573\text{ K}$ forms potassium manganate. Statement II Both potassium permanganate and potassium manganate are tetrahedral and paramagnetic in nature. In the light of the above statements, choose the most appropriate answer from the options given below

Welcome to another exciting journey into the world of inorganic chemistry! Today, we are tackling a classic JEE Main problem that beautifully intertwines chemical reactions, molecular geometry, and the magnetic properties of transition metal complexes. This question tests your ability to critically evaluate two distinct statements about potassium permanganate ($KMn{O}_4$) and potassium manganate ($K_{2}Mn{O}_4$). Let's break it down step by step and uncover the fascinating chemistry behind these vibrant compounds.

Let's start by dissecting Statement I, which claims that potassium permanganate forms potassium manganate upon heating at $573\text{ K}$. Is this true? Absolutely! Potassium permanganate is a powerful oxidizing agent, but it is thermally unstable at elevated temperatures.

When you heat solid $KMn{O}_4$ to about $573\text{ K}$, it undergoes a thermal decomposition reaction. The intense purple crystals break down to form green potassium manganate ($K_{2}Mn{O}_4$), black manganese dioxide ($Mn{O}_2$), and oxygen gas ($O_2$). The balanced chemical equation for this transformation is:

Because this reaction perfectly matches the description in Statement I, we can confidently conclude that Statement I is true.

Moving on to Statement II, we are presented with two claims: first, that both compounds are tetrahedral, and second, that both are paramagnetic. Let's evaluate the geometry first.

In both the permanganate ion ($Mn{O}_4^{-}$) and the manganate ion ($Mn{O}_4^{2-}$), the central manganese atom is bonded to four oxygen atoms. Despite the presence of double bonds, the geometry is determined by the number of sigma ($\sigma$) bonds. Manganese forms exactly four $\sigma$ bonds with the surrounding oxygen atoms and has zero lone pairs on the central metal atom.

According to VSEPR theory, four bonding pairs and zero lone pairs correspond to an $s{p}^3$ hybridization state. This hybridization inherently leads to a tetrahedral geometry. Therefore, the first half of Statement II is correct—both ions are indeed tetrahedral.

Now comes the most crucial part: evaluating the magnetic nature of these compounds. To determine whether a complex is paramagnetic or diamagnetic, we must find the oxidation state of the central metal and write its electronic configuration.

Let's calculate the oxidation state of manganese in potassium permanganate ($KMn{O}_4$). Let the oxidation state of Mn be $x$. Potassium is an alkali metal with a $+1$ charge, and each oxygen atom carries a $-2$ charge. Setting the sum of oxidation states to zero gives us:

Next, let's do the same for potassium manganate ($K_{2}Mn{O}_4$). Here, we have two potassium ions:

The neutral manganese atom (atomic number $Z=25$) has the ground-state electronic configuration $[Ar]4s^{2}3d^5$.

In $KMn{O}_4$, manganese is in the $+7$ oxidation state. This means it has lost all seven of its valence electrons (two from the $4s$ orbital and five from the $3d$ orbital). The resulting configuration is $[Ar]4s^{0}3d^0$. Because there are zero unpaired electrons, potassium permanganate is strictly diamagnetic.

In $K_{2}Mn{O}_4$, manganese is in the $+6$ oxidation state. It has lost six valence electrons, leaving exactly one electron in the $d$-subshell. The configuration is $[Ar]3d^1$. The presence of this one unpaired electron makes potassium manganate paramagnetic.

Bringing it all together, Statement II claims that both compounds are paramagnetic. However, our rigorous quantum mechanical check proved that $KMn{O}_4$ is diamagnetic. Therefore, Statement II is false.

Since Statement I is true and Statement II is false, the correct choice is Option (a).

A Quick Bonus Concept: You might be wondering, if $KMn{O}_4$ has a $d^0$ configuration with no unpaired electrons, how can it exhibit such a brilliant, intense purple color? Normally, transition metals are colored due to $d-d$ electron transitions. The secret here is a phenomenon called Ligand-to-Metal Charge Transfer (LMCT). Electrons from the oxygen ligands are temporarily excited into the empty $d$-orbitals of the manganese atom, absorbing specific wavelengths of light and reflecting that iconic purple hue. Keep this in mind, as it is a highly tested concept in competitive exams!