Question

Mention the different steps for the preparation of potassium permanganate from pyrolusite ore and draw the structure of permanganate ion. (1+1=2)

Hint:

Fuse \( MnO_2 \) with KOH and air to get green \( K_2MnO_4 \), then oxidise electrolytically to purple \( KMnO_4 \); the \( MnO_4^- \) ion is tetrahedral.

Answer 1

Step 1: Fusion of pyrolusite (formation of manganate). Pyrolusite ore \( (MnO_2) \) is fused with potassium hydroxide (KOH) in the presence of air or an oxidising agent such as \( KNO_3 \). Manganese is oxidised from the +4 to the +6 state, giving the dark green potassium manganate:
\[ 2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O \]
Step 2: Oxidation of manganate to permanganate. The green manganate \( (MnO_4^{2-},\ Mn^{+6}) \) is then oxidised to the purple permanganate \( (MnO_4^{-},\ Mn^{+7}) \). This is done by electrolytic oxidation:
\[ MnO_4^{2-} \rightarrow MnO_4^{-} + e^{-} \]
or by chemical oxidation (disproportionation in neutral or slightly acidic medium):
\[ 3MnO_4^{2-} + 4H^{+} \rightarrow 2MnO_4^{-} + MnO_2 + 2H_2O \]
Step 3: Structure of the permanganate ion \( (MnO_4^{-}) \). The manganese is in the +7 oxidation state and is \( sp^3 \) hybridised. The ion is tetrahedral, with the Mn atom at the centre and the four oxygen atoms at the corners of a regular tetrahedron (O to Mn to O bond angle \( 109.5^{\circ} \)). All four Mn to O bonds are equivalent and have partial double bond character (\( \pi \) bonds formed by oxygen p and Mn d orbitals), and the overall ion carries a single negative charge.
\[\boxed{MnO_4^{-}\ \text{is tetrahedral, Mn in +7 state, four equivalent Mn to O bonds}}\]