(a) (i) Preparation of Potassium Permanganate ($\text{KMnO}_4$)

Step 1 (Fusion Reaction): [1 Mark]
Pyrolusite ore ($\text{MnO}_2$) is fused with potassium hydroxide ($\text{KOH}$) in the presence of atmospheric oxygen or an oxidizing agent like $\text{KNO}_3$ to form potassium manganate ($\text{K}_2\text{MnO}_4$).

Step 2 (Oxidation Reaction): [1 Mark]
The green potassium manganate ($\text{K}_2\text{MnO}_4$) is then subjected to electrolytic oxidation or treated with an acid/$\text{CO}_2$ to disproportionate into purple potassium permanganate ($\text{KMnO}_4$).

(a) (ii) Reasoning Questions

• (I) Sc salts are white: [1 Mark]
  Scandium ($Z=21$) has the electronic configuration $[\text{Ar}] 3d^1 4s^2$. In its salts, it exists as $\text{Sc}^{3+}$, which has a configuration of $[\text{Ar}] 3d^0$. Since there are no unpaired electrons in the $d$-orbital, $d-d$ transitions are not possible, rendering the salts colourless (white).
• (II) $E^\circ_{\text{Cu}^{2+}|\text{Cu}}$ vs $E^\circ_{\text{Mn}^{2+}|\text{Mn}}$: [1 Mark]
  The positive $E^\circ$ value for $\text{Cu}$ is due to its high enthalpy of atomization and high ionization energy, which are not compensated by its hydration enthalpy. Conversely, $\text{Mn}^{2+}$ has a stable $d^5$ configuration, making the reduction of $\text{Mn}^{2+}$ to $\text{Mn}$ energetically unfavourable, resulting in a negative $E^\circ$ value.
• (III) High Enthalpy of Atomisation: [1 Mark]
  Transition metals have a large number of unpaired electrons in their $(n-1)d$ orbitals. This allows for strong interatomic metallic bonding (covalent character) due to the overlapping of $d$-orbitals, requiring significant energy to break the metallic lattice.