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Potassium Permanganate (KMnO4): Preparation and Oxidising Reactions

Chemistry · D And F Block Elements · NEET

KMnO4 is made from the ore pyrolusite (MnO2). MnO2 is first fused with KOH and an oxidiser (air or KNO3) to give green potassium manganate K2MnO4 (Mn is +6). This K2MnO4 is then oxidised (by electrolysis or by passing chlorine/ozone) to purple KMnO4 (Mn is +7). As an oxidising agent KMnO4 changes based on the medium: in acid Mn goes +7 to +2, in neutral or faintly alkaline it goes +7 to +4 (MnO2), and in strong alkali it goes +7 to +6. Memory hook: "Acid Two, Neutral Four, Base Six" for the Mn oxidation state in each medium.
KMnO4 Preparation and Oxidation by MediumMnO2 (ore)pyrolusiteK2MnO4 (green)Mn = +6KMnO4 (purple)Mn = +7KOH,O2oxidiseAcidic mediumMn +7 -> +2 (5e-)colourless Mn2+Neutral / faint alkaliMn +7 -> +4 (3e-)brown MnO2Strong alkaliMn +7 -> +6 (1e-)green MnO4^2-
KMnO4 is built up from the ore MnO2 through green manganate (Mn +6) to purple permanganate (Mn +7). As an oxidiser, its Mn end-state depends on the medium: +2 in acid (5 electrons), +4 in neutral (3 electrons), +6 in strong alkali (1 electron).

Your doubts, answered

How is KMnO4 prepared from pyrolusite (MnO2)? What are the two steps?

There are two clear steps. Step 1 (fusion): MnO2 is fused with KOH in the presence of an oxidising agent like air (O2) or KNO3. This gives dark green potassium manganate, K2MnO4, where Mn is +6. Reaction: 2MnO2 + 4KOH + O2 -> 2K2MnO4 + 2H2O. Step 2 (oxidation): the green K2MnO4 is oxidised to purple permanganate KMnO4 (Mn +7), either by electrolytic oxidation of the alkaline solution, or chemically by passing Cl2 or O3. Reaction: 3K2MnO4 + 4H+ (or via electrolysis) -> 2KMnO4 + MnO2 + 2H2O + 2K+. So: ore -> manganate (+6, green) -> permanganate (+7, purple). This two-step route is a common NEET fill-in-the-blank.

What are the products when KMnO4 acts as an oxidiser in acidic, neutral and alkaline medium?

The Mn end-product depends on the medium. In ACIDIC medium (dilute H2SO4): MnO4^- + 8H+ + 5e^- -> Mn^2+ + 4H2O. Mn goes from +7 to +2, gaining 5 electrons (n-factor = 5). The purple colour disappears. In NEUTRAL or FAINTLY ALKALINE medium: MnO4^- + 2H2O + 3e^- -> MnO2 + 4OH^-. Mn goes +7 to +4, gaining 3 electrons (brown MnO2 forms). In STRONGLY ALKALINE medium: MnO4^- + e^- -> MnO4^2-. Mn goes +7 to +6, gaining only 1 electron (green manganate). Memory: 5-3-1 electrons for acid-neutral-base.

Why does KMnO4 oxidise iodide (I-) to iodate (IO3-) in neutral medium but only to I2 in acidic medium?

It is about oxidising strength in each medium. In acidic medium KMnO4 is a very strong oxidiser but is used up fast (Mn +7 to +2), so it only pushes I^- (-1) up to I2 (0). In neutral or faintly alkaline medium it acts as a milder but more sustained oxidiser (Mn +7 to +4), and there it oxidises iodide all the way to iodate IO3^- (I goes -1 to +5). Neutral/alkaline: 2MnO4^- + H2O + I^- -> 2MnO2 + 2OH^- + IO3^-. This exact fact was asked in NEET 2019 and NEET 2022.

What green paramagnetic compound forms when KMnO4 is heated? Is it decomposition or disproportionation?

On heating to 513 K, KMnO4 decomposes: 2KMnO4 -> K2MnO4 + MnO2 + O2. The green paramagnetic species is potassium manganate K2MnO4 (Mn is +6, d1, one unpaired electron, so it is coloured green and paramagnetic). Note: this thermal decomposition is NOT a disproportionation of a single element, because here both Mn (+7 -> +6 and +4) AND oxygen (-2 -> 0 in O2) change oxidation state. A NEET 2019 question used this to trap students. Contrast: 3MnO4^2- + 4H+ -> 2MnO4^- + MnO2 + 2H2O IS a true disproportionation (only Mn changes: +6 -> +7 and +4).

Why are both manganate MnO4^2- and permanganate MnO4^- tetrahedral, and which one is coloured due to d-d transition?

Both ions are tetrahedral because the Mn-O bonds have partial double-bond character from p-pi to d-pi bonding: filled p-orbitals of oxygen overlap with empty d-orbitals of manganese (asked in NEET 2019). Their colour is different in origin: permanganate MnO4^- has Mn +7 = d0, so it has NO d electrons; its intense purple colour comes from charge transfer (ligand-to-metal), not d-d transition, and it is diamagnetic. Manganate MnO4^2- has Mn +6 = d1, so it CAN show a real d-d transition and is paramagnetic (one unpaired electron), and it is green. This d0 vs d1 point is a classic NEET trap.

⚠️ The NEET trap
✗ KMnO4 is intensely purple, so it must be paramagnetic and its colour comes from a d-d transition.
✓ KMnO4 has Mn in +7 = d0, so it has NO unpaired d electrons: it is DIAMAGNETIC and its purple colour is from ligand-to-metal charge transfer, not a d-d transition. It is the green manganate MnO4^2- (Mn +6, d1) that is paramagnetic and shows a genuine d-d transition.
🧠 Colour does NOT mean paramagnetic. Check the d-electron count: d0 = charge-transfer colour + diamagnetic; d1 or more = possible d-d transition + paramagnetic.

Real NEET questions

NEET 2022

In the neutral or faintly alkaline medium, KMnO4 oxidises iodide into iodate. The change in oxidation state of manganese in this reaction is from:

A · +7 to +4 ✓
B · +6 to +4
C · +7 to +3
D · +6 to +5
Solution: In neutral or faintly alkaline medium the reaction is 2MnO4^- + H2O + I^- -> 2MnO2 + 2OH^- + IO3^-. Manganese starts as Mn +7 in MnO4^- and is reduced to Mn +4 in MnO2. Iodide (-1) is oxidised to iodate (+5). So Mn changes from +7 to +4. Answer (A).
ReNEET 2026

The green paramagnetic species formed by heating KMnO4 at 513 K is:

A · K2MnO4 ✓
B · Mn3O4
C · MnO
D · KO2
Solution: On heating: 2KMnO4 -> K2MnO4 + MnO2 + O2. Potassium manganate K2MnO4 has Mn in +6 (d1, one unpaired electron), so it is green and paramagnetic. MnO2 is not the green species asked for. Answer (A).
NEET 2019

Which of the following reactions are disproportionation reactions? (a) 2Cu+ -> Cu2+ + Cu (b) 3MnO4^2- + 4H+ -> 2MnO4^- + MnO2 + 2H2O (c) 2KMnO4 -> K2MnO4 + MnO2 + O2 (d) 2MnO4^- + 3Mn2+ + 2H2O -> 5MnO2 + 4H+

A · (a) and (b) only ✓
B · (a), (b) and (c)
C · (a), (c) and (d)
D · (a) and (d) only
Solution: Disproportionation = the SAME element in one oxidation state is both oxidised and reduced. (a) Cu +1 -> Cu +2 and Cu 0: yes. (b) Mn +6 -> +7 and +4: yes. (c) here Mn (+7 -> +6 and +4) AND oxygen (-2 -> 0) both change, so it is thermal decomposition, not disproportionation. (d) Mn +7 reduced to +4 and Mn +2 oxidised to +4 is comproportionation, not disproportionation. So only (a) and (b). Answer (A).

Solved D And F Block Elements NEET PYQs

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Frequently asked

What is the colour of KMnO4 and K2MnO4?

KMnO4 (permanganate, Mn +7) is deep purple. K2MnO4 (manganate, Mn +6) is dark green. The colour change purple to green is a clue that Mn was reduced from +7 to +6.

What is the n-factor of KMnO4 in acidic medium?

In acidic medium Mn goes from +7 to +2, gaining 5 electrons, so the n-factor (equivalent factor) of KMnO4 is 5. In neutral medium it is 3 (+7 to +4) and in strongly alkaline medium it is 1 (+7 to +6).

Why is acidified KMnO4 used and not neutral for titrations?

In acidic medium KMnO4 is a strong 5-electron oxidiser giving colourless Mn2+, so the purple to colourless end point is sharp and self-indicating. Dilute H2SO4 is used (not HCl, because HCl is itself oxidised by KMnO4).

Is KMnO4 paramagnetic or diamagnetic?

KMnO4 is diamagnetic because Mn is +7 (d0), with no unpaired electrons. Its colour comes from charge transfer, not from d electrons.

Related concepts:
Potassium Dichromate (K2Cr2O7): Preparation and OxidationWhat Is Disproportionation (Cu+, MnO4^2-) With ExamplesWhy Transition Metals Show Variable Oxidation StatesWhy Transition Metal Ions Are Coloured (d-d Transitions)How to Find Oxidation State of Chromium in CrO5
Practise Chemistry for NEET:
Chemistry MCQs (free, chapter-wise)D And F Block Elements PYQs