Question
Find the oxidation state of manganese (Mn) in potassium permanganate, KMnO₄.
This is a classic NCERT Class 11 question and a guaranteed 1-marker in CBSE boards. JEE Main also uses KMnO₄ as a starting point for redox balancing problems, so knowing this cold saves you time.
Solution — Step by Step
Potassium (K) is a Group 1 alkali metal — its oxidation state is always +1. Oxygen in a non-peroxide compound is always −2. These are fixed rules; no exceptions for this compound.
The compound KMnO₄ is neutral, so all oxidation states must sum to zero.
Here, is the oxidation state of Mn.
The oxidation state of Mn in KMnO₄ is +7.
Why This Works
The oxidation number method is based on one rule: every neutral compound has charges that balance to zero. We assign known values first (K = +1, O = −2 are memorised facts), then treat the unknown as a variable and solve.
Manganese is special because it shows oxidation states from −3 all the way to +7 — the widest range among common transition metals. That’s why KMnO₄ is such a strong oxidising agent: Mn is at its highest possible state (+7) and desperately wants electrons, dropping down to Mn²⁺ (+2) in acidic medium or Mn⁴⁺ (+4) in neutral/basic medium.
This also explains why KMnO₄ is purple-violet. The intense colour comes from charge-transfer transitions, which are most dramatic when the metal is in a high oxidation state.
Alternative Method — Using Polyatomic Ion Rules
If you know that the permanganate ion is MnO₄⁻, you can find Mn’s oxidation state from the ion directly — no need to involve K at all.
The permanganate ion MnO₄⁻ carries a charge of −1. So:
Same answer, slightly faster. This approach is useful when questions give you the ion directly instead of the full salt.
Memorise the permanganate ion: MnO₄⁻ with Mn = +7. Dichromate Cr₂O₇²⁻ has Cr = +6. These two are the most asked oxidising agents in JEE and NEET redox chapters.
Common Mistake
The most common error is forgetting that there are four oxygen atoms, not one. Students write 1 + x + (−2) = 0 and get Mn = +1, which is wrong. Always multiply the oxidation state of each atom by the number of atoms present: 4 × (−2) = −8, not −2.
A related mistake: treating KMnO₄ as if it contains a peroxide linkage and assigning O = −1. It doesn’t — the oxygens here are ordinary oxide oxygens bonded to Mn. Only H₂O₂, Na₂O₂, and similar structures use O = −1.