Variable Oxidation States: Transition Metals Explained

30 September 2026Chemistry Try the interactive 3D model →

Most elements have just one or two common oxidation states: sodium is almost always +1, aluminium +3. Transition metals break this rule spectacularly. Iron appears as +2 in FeO and +3 in Fe2O3, manganese runs from +2 in MnO all the way to +7 in KMnO4, and chromium is found as +2, +3 and +6. A variable oxidation state means the same metal can lose different numbers of electrons in different compounds, and understanding why is central to d-block chemistry.

This matters far beyond exams. Variable oxidation states are what make transition metals coloured, magnetic and catalytically active. The purple of potassium permanganate, the orange of potassium dichromate, rust forming on iron and the catalysts in car exhausts all depend on a metal switching between oxidation states. Whenever a metal acts as an oxidizing or reducing agent, it is moving between these states.

Key Concepts

Why it happens: Transition metals have valence electrons in both the ns orbital and the (n-1)d orbitals, and the energy gap between these is small. Removing electrons from either set costs a similar amount of energy, so different totals of lost electrons give different stable states.
The +2 state is nearly universal: It forms by losing the two ns electrons first (for example Fe [Ar] 4s2 3d6 becomes Fe2+ [Ar] 3d6). Almost every first-row transition metal shows +2.
Higher states use d electrons too: After the ns electrons are gone, further oxidation removes (n-1)d electrons. Manganese reaches +7 in MnO4- by losing all seven of its 4s2 3d5 valence electrons.
Half-filled and full d configurations are extra stable: Mn2+ (d5) and Fe3+ (d5) resist further oxidation, while Fe2+ (d6) oxidizes readily to Fe3+. This explains why Mn2+ compounds are pale and stable whereas Fe2+ solutions darken in air.
Maximum oxidation state equals the group number: Sc shows +3, Ti +4, V +5, Cr +6, Mn +7, and in the heavier rows Os and Ru reach +8 in OsO4 and RuO4, the highest known oxidation state.
Stability pattern across the series: For 3d metals, high oxidation states become less stable toward the right (MnO4- and Cr2O7 2- are powerful oxidizers), while for 4d and 5d metals high states are more stable (MoO3 and WO3 are ordinary stable oxides).
Disproportionation: Some intermediate states are unstable and split into a higher and a lower state, for example 3 MnO4 2- + 2 H2O gives 2 MnO4- + MnO2 + 4 OH-.

Formulas

Oxidation states are found with a simple charge-balance equation rather than a fixed formula:

Oxidation state of the metal x: sum of (oxidation state x subscript) over all atoms = overall charge of the species. Standard values: alkali metals +1, alkaline earth metals +2, fluorine -1, oxygen -2 (except peroxides -1 and OF2 +2), hydrogen +1 (except metal hydrides -1), halogens -1 unless bonded to oxygen or a more electronegative halogen.
Example setup for KMnO4: K(+1) + Mn(x) + 4 x O(-2) = 0, so 1 + x - 8 = 0 and x = +7. The same arithmetic works for ions: in MnO4 2-, x + 4(-2) = -2 gives x = +6.
d-electron count: d electrons remaining = group number - oxidation state (for groups 3-7). Fe3+ in group 8 is handled as d5: 8 - 3 = 5. This count predicts colour and magnetism.

Worked Example

Question: Find the oxidation state of chromium in K2Cr2O7 and explain why this compound is a strong oxidizing agent.

Write the balance equation: 2 x K(+1) + 2 x Cr(x) + 7 x O(-2) = 0, since the compound is neutral.
Solve for x: 2(1) + 2x + 7(-2) = 0, which gives 2 + 2x - 14 = 0, so 2x = 12 and x = +6. Chromium is in the +6 state.
Check the d-electron count: Chromium is in group 6, so d electrons = 6 - 6 = 0. Cr(VI) has an empty d shell.
Explain the oxidizing power: With no d electrons left, Cr(VI) strongly tends to gain electrons and drop to the far more stable Cr3+ (d3) state, which is why potassium dichromate oxidizes Fe2+ to Fe3+, ethanol to ethanoic acid, and SO2 to sulfate in classic test-tube reactions.

Common Misconceptions

Transition metals lose d electrons before s electrons: Wrong. The ns electrons are removed first even though they filled first. Fe2+ is [Ar] 3d6, never [Ar] 4s2 3d4, because once the atom is ionized the 3d orbitals drop below 4s in energy.
Higher oxidation state always means stronger compound: Not so. For 3d metals the highest states are often unstable oxidizers (Mn(VII), Cr(VI)), whereas the heavier 4d and 5d metals stabilize high states (W(VI) in WO3 is perfectly ordinary). Stability depends on the row as well as the number.
Oxidation state equals actual charge on the atom: It is a bookkeeping number, not a measured charge. In MnO4- the manganese is not literally missing seven electrons; the +7 is the charge it would have if every bond were fully ionic, which is a useful fiction for tracking electron transfer.

Real-World Applications

Potassium permanganate (KMnO4): Uses the Mn(VII) to Mn(II) change as a self-indicating oxidizer in redox titrations, water disinfection and organic synthesis; the purple colour vanishes as it reacts.
Iron and steel: The Fe2+/Fe3+ pair drives rusting, and blast-furnace chemistry is essentially the managed reduction of Fe(III) ores to Fe(0) metal.
Catalysts: The contact process for sulfuric acid cycles vanadium between +4 and +5 (V2O5 catalyst), and car catalytic converters cycle platinum-group metals through oxidation states to clean exhaust gases.
Batteries: Rechargeable cells work by shuttling a metal between oxidation states, such as Co3+/Co4+ in lithium-ion cathodes and Pb/PbO2 in lead-acid batteries.

Variable oxidation states arise because ns and (n-1)d electrons sit at similar energies, letting transition metals lose different numbers of electrons. The +2 state is almost always available, higher states peak at the group number, and half-filled d5 configurations act as stability islands. Learn to compute the state with charge balance and you can predict colours, oxidizing power and reactivity across the whole d block.