Animated Solution for Chemistry - d and f-Block Elements: Fusion of ${\text{MnO}}_2$ with $\text{KOH}$ in presence of ${\text{O}}_2$ produces a salt $\text{W}$. Alkaline solution of $\text{W}$ upon eletrolytic oxidation yields another salt $\text{X}$. The manganese containing ions present in $\text{W}$ and $\text{X}$, respectively, are $\text{Y}$ and $\text{Z}$. Correct statement(s) is (are)

Imagine you are in an industrial chemistry lab, staring at a pile of black pyrolusite ore, which is essentially manganese dioxide ($Mn{O}_2$). Our goal? To extract the beautiful, highly reactive chemistry hidden within it.

When we fuse this black powder with potassium hydroxide ($KOH$) and introduce a stream of oxygen gas ($O_2$), a fascinating transformation occurs. The oxygen acts as an oxidizing agent, pulling the manganese from its $+4$ state up to a $+6$ state. The result is a vibrant green salt: potassium manganate ($K_{2}Mn{O}_4$).

This gives us our first crucial piece of the puzzle. Salt $W$ is $K_{2}Mn{O}_4$, and the manganese-containing ion $Y$ is the manganate ion, $Mn{O}_4^{2-}$.

But we don't stop there. The manganate ion is just a stepping stone. To reach the pinnacle of manganese's oxidation states, we take the alkaline solution of potassium manganate and subject it to electrolytic oxidation.

At the anode, the manganate ion is stripped of one more electron. It transitions from the green $Mn{O}_4^{2-}$ to the deep, majestic purple of the permanganate ion, $Mn{O}_4^{-}$. Here, manganese achieves its maximum possible oxidation state of $+7$.

Thus, salt $X$ is potassium permanganate ($KMn{O}_4$), and the ion $Z$ is $Mn{O}_4^{-}$.

Now, let's put these ions under the microscope and evaluate Option (A).

In the manganate ion ($Mn{O}_4^{2-}$), manganese sits at a $+6$ oxidation state. The electron configuration of neutral manganese is $[Ar]4s^{2}3d^5$. Stripping away six electrons leaves us with exactly one lonely electron in the $3d$ subshell ($[Ar]3d^1$). Because of this unpaired electron, the manganate ion is paramagnetic.

Conversely, in the permanganate ion ($Mn{O}_4^{-}$), manganese is at a $+7$ oxidation state. All valence electrons have been removed, leaving a completely empty $3d$ subshell ($[Ar]3d^0$). With zero unpaired electrons, permanganate is strictly diamagnetic.

Option (A) claims the exact opposite, so we can confidently cross it out.

Let's look at Option (B). If permanganate has no d-electrons, how can it be so intensely purple? Shouldn't it be colorless?

This is where a beautiful phenomenon called Ligand-to-Metal Charge Transfer (LMCT) comes into play. Even though the d-orbitals of manganese are empty, an electron from the oxygen atom's filled p-orbital can temporarily absorb a photon of visible light and jump into the empty d-orbital of manganese. This transition is what gives both manganate and permanganate their striking colors.

Furthermore, in both ions, the central manganese atom is $s{p}^3$ hybridized, leading to a perfect tetrahedral geometry. Option (B) is absolutely correct.

Moving to Option (C), we need to understand how the double bonds in these tetrahedral structures are formed.

The single bonds are standard sigma bonds formed by $s{p}^3$ hybridized orbitals. However, the pi bonds are formed by the sideways overlap of the filled $2p$ orbitals of the oxygen atoms with the empty $3d$ orbitals of the manganese atom.

Because this interaction involves a d-orbital and a p-orbital, it is formally known as $d\pi -p\pi$ bonding. Option (C) hits the nail on the head.

Finally, let's examine Option (D). What happens if we take our green manganate solution and make it acidic?

The manganate ion is highly unstable in the presence of $H^{+}$ ions. It undergoes a rapid disproportionation reaction. This means it acts as both an oxidizing and a reducing agent simultaneously.

Three moles of manganate will react to form two moles of permanganate (oxidized to $+7$) and one mole of solid manganese dioxide (reduced to $+4$).

This confirms that Option (D) is also correct.

By systematically breaking down the preparation, electronic structure, bonding, and chemical stability of these fascinating manganese complexes, we have found our answers. The correct statements are (B), (C), and (D).