d- and f-Block Elements · Lesson 8 of 12
Preparing Permanganate and Understanding Manganate
“Track manganese from dioxide through manganate to permanganate and explain structure, colour and disproportionation.”
• Describe alkaline oxidative fusion and subsequent conversion to permanganate. • Balance manganate disproportionation and electrolytic oxidation. • Relate oxoanion charge to manganese state, d count and magnetic behaviour. • Explain thermal decomposition and a peroxodisulfate preparation route.
The green intermediate is a different oxidation state
Starting with MnO₂, manganese is +4. Alkaline fusion with an oxidant first makes green manganate, MnO₄²⁻, in which Mn is +6. Further oxidation makes purple permanganate, MnO₄⁻, in which Mn is +7. The extra negative charge on manganate means the metal is less oxidised, even though both ions contain four oxygen atoms. Formula charge is therefore essential information.
With oxygen as oxidant, a balanced fusion representation is 2MnO₂+4KOH+O₂→2K₂MnO₄+2H₂O. Each Mn loses two electrons; one O₂ accepts the four electrons. Nitrate can be another oxidant in an alkaline fusion, but the oxygen equation should not be labelled as though nitrate were its reactant. The melt is strongly alkaline and hot; its chemistry is different from mixing the same ingredients in cold water.
Problem
Identify the electron transfer in forming manganate from manganese dioxide.
- 1.MnO₂ has x−4=0, giving Mn(+4). K₂MnO₄ has 2(+1)+x−8=0, giving Mn(+6). Each Mn loses two electrons.
- 2.For two Mn atoms, four electrons are released. O₂ becomes oxygen in oxidation state −2 and accepts four electrons in total.
- 3.The full equation has K 4, Mn 2, O 10 and H 4 on both sides. Oxidation-state changes and ordinary atom counts agree.
Two routes turn green manganate into purple permanganate
A commercial route oxidises manganate electrolytically in alkaline solution. At the anode the essential half-reaction is MnO₄²⁻→MnO₄⁻+e⁻. The manganese state rises by one, and oxidation occurs at the anode regardless of the electrode sign convention. This is not a complete cell equation: a corresponding cathode process consumes the electrons.
On reducing alkalinity or acidifying, manganate can disproportionate: 3MnO₄²⁻+4H⁺→2MnO₄⁻+MnO₂(s)+2H₂O. Two Mn(VI) atoms are oxidised to Mn(VII) and a third is reduced to Mn(IV). Strongly alkaline conditions stabilise manganate relative to this pathway. The equation shows why some of the manganese appears as brown dioxide rather than all becoming permanganate.
Problem
Why does manganate disproportionation require three manganese centres?
- 1.Oxidation +6→+7 releases one electron per Mn. Reduction +6→+4 consumes two electrons per Mn.
- 2.Two oxidised Mn atoms supply the two electrons needed by one reduced Mn atom. Thus three initial Mn atoms give two permanganates and one dioxide.
- 3.Balancing oxygen and hydrogen in acid then gives 3MnO₄²⁻+4H⁺→2MnO₄⁻+MnO₂+2H₂O. Charge is −2 on both sides.
- 4.This internal electron exchange explains why no separate oxidant is required in the disproportionation step.
Problem
Compare 0.300 mol manganate converted by the two idealised routes.
- 1.Complete electrolytic one-electron oxidation can give 0.300 mol permanganate from 0.300 mol manganate if no side reaction or material loss occurs.
- 2.Disproportionation gives two permanganates from three manganates: 0.300×2/3=0.200 mol permanganate and 0.100 mol MnO₂.
- 3.These are stoichiometric limits, not claims of measured industrial yield. Recycling dioxide can alter the overall process material economy.
Tetrahedral shape does not make the ions identical
Both oxoanions are approximately tetrahedral, with four oxygen atoms around manganese. Formal d counts are group 7 minus oxidation state: manganate(VI) is d¹ and paramagnetic because it has one unpaired electron; permanganate(VII) is d⁰ and is diamagnetic in the usual electronic-ground-state description. Weak temperature-dependent effects in real solids do not create a hidden d electron or invalidate the formal count.
Manganate solutions are green and permanganate solutions purple. The intense permanganate colour is largely associated with oxygen-to-metal charge-transfer excitation, not promotion of a nonexistent d electron within a d⁰ shell. A useful tetrahedral drawing communicates connectivity and geometry; it should not be interpreted as four permanently isolated O⁻ charges bonded to a bare metal ion. Metal–oxygen bonding is more delocalised than such a drawing suggests.
KMnO₄ forms dark purple, nearly black-looking crystals. At 293 K, the stated solubility is about 6.4 g per 100 g water. The denominator is mass of solvent, not volume of solution, so this number cannot be directly treated as 6.4 mol L⁻¹ or as a mass percentage of the whole solution. Its solid structure is related to that of KClO₄; this does not mean the two salts have identical redox chemistry.
Other transformations preserve the same electron logic
In a laboratory preparation route, peroxodisulfate oxidises Mn²⁺ to permanganate: 2Mn²⁺+5S₂O₈²⁻+8H₂O→2MnO₄⁻+10SO₄²⁻+16H⁺. Each Mn rises from +2 to +7, a five-electron loss. Each peroxodisulfate accepts two electrons when its peroxide linkage is reduced to sulfate, so five are needed for two Mn. The equation states the overall reaction; kinetic conditions can be important.
On heating to around the stated decomposition temperature of 513 K, permanganate undergoes 2KMnO₄→K₂MnO₄+MnO₂+O₂. Two Mn(VII) centres end as Mn(VI) and Mn(IV), accepting four electrons in total; oxygen oxidation from −2 to 0 supplies the matching electrons when considered across the whole balanced reaction. Count all oxygen atoms before assigning the redox changes: eight on the left become four plus two plus two on the right.
Problem
For 0.0200 mol KMnO₄ decomposed by the ideal equation, calculate O₂ and MnO₂ amounts.
- 1.The balanced equation uses two mol permanganate for one mol each of O₂ and MnO₂.
- 2.Each product amount is 0.0200/2=0.0100 mol. Oxygen mass is 0.0100×32.00=0.320 g.
- 3.The stoichiometric amounts require complete decomposition by this pathway; actual gas collection can be affected by incomplete conversion and measurement conditions.
Quick Quiz
Quiz
What is the Mn state in manganate?
Explanation: x−8=−2 gives +6.
What occurs at the anode during manganate conversion?
Explanation: Anodic oxidation changes Mn(VI) to Mn(VII).
Which oxoanion is formally d¹ and paramagnetic?
Explanation: Group 7 minus oxidation state 6 leaves one formal d electron.
How much permanganate comes from 3 mol manganate by disproportionation?
Explanation: The balanced ratio is three manganates to two permanganates plus one dioxide.
What gas forms in the usual thermal decomposition equation?
Explanation: 2KMnO₄→K₂MnO₄+MnO₂+O₂ conserves all elements.
Practice Problems
- Write the preparation sequence from MnO₂ and label the changes.
- Balance manganate disproportionation in acid and check charge.
- Predict colour and magnetism for manganate and permanganate.
- How many mol peroxodisulfate are required for 0.0100 mol Mn²⁺ oxidation to permanganate?
- If 0.0600 mol KMnO₄ decomposes completely, find amounts of manganate, MnO₂ and O₂.
- Convert 6.4 g KMnO₄ per 100 g water into saturated-solution mass percentage.
Alkaline oxidative fusion: 2MnO₂+4KOH+O₂→2K₂MnO₄+2H₂O changes Mn(+4) to (+6). Electrolytic oxidation: MnO₄²⁻→MnO₄⁻+e⁻ changes (+6) to (+7). Alternatively acid-promoted disproportionation gives 2Mn(VII) and one Mn(IV) from three Mn(VI).
Key Takeaways
• Manganese dioxide first becomes green manganate in alkaline oxidative fusion. • Anodic one-electron oxidation turns manganate into permanganate. • Manganate disproportionates into Mn(VII) and Mn(IV) when the medium permits. • Both oxoanions are tetrahedral but differ in charge, d count, colour and magnetism. • Intense permanganate colour is compatible with a d⁰ metal through charge transfer. • Peroxodisulfate oxidation and thermal decomposition obey electron as well as atom balance. • Stoichiometric limits must be distinguished from measured process yields.