Why Do Transition Elements Have Variable Oxidation States?
Iron can be Fe²⁺ or Fe³⁺. Manganese runs from +2 all the way to +7. Copper does +1 and +2. Compare that with sodium, which is only ever +1, or magnesium, only ever +2. Why do the transition metals get to be so flexible? The whole explanation comes down to one idea: the 3d and 4s subshells are very close in energy.
The observation first
“Variable oxidation state” means an element forms stable compounds in more than one oxidation state. A few to have in mind:
- Iron: Fe²⁺ (as in FeSO₄) and Fe³⁺ (as in FeCl₃).
- Manganese: +2 in Mn²⁺, +4 in MnO₂, +6 in MnO₄²⁻, +7 in MnO₄⁻.
- Copper: +1 in Cu₂O, +2 in CuSO₄.
This is a defining property of transition elements, alongside coloured ions, catalytic behaviour and complex formation.
A quick definition worth knowing: a transition element is a d-block element that forms at least one stable ion with a partially filled d subshell. That is why scandium and zinc are often left out. Sc³⁺ is d⁰ and Zn²⁺ is d¹⁰, so neither is partially filled.
Why 3d and 4s being close in energy matters
When a transition metal is ionised, it loses its 4s electrons first, then its 3d electrons. The key point is that 3d and 4s sit at almost the same energy. So once you have removed the 4s electrons, pulling off the next 3d electron does not cost dramatically more energy.
Contrast that with a main-group metal. After you strip off its outer shell, you hit a full inner shell far below in energy, and the next electron is enormously harder to remove.
Small successive ionisation energy gaps
This is the heart of the exam answer. Look at how the successive ionisation energies climb.
- For a transition metal like iron, the 1st, 2nd, 3rd, 4th ionisation energies rise gradually, in small steps, because you are removing 4s then 3d electrons that are all close in energy. Several different numbers of electrons can therefore be removed without a sudden huge jump. The extra energy needed is repaid by the energy released when the ion bonds, hydrates or forms a lattice, so several oxidation states end up stable.
- For sodium, the jump from the 1st to the 2nd ionisation energy is massive (you go from removing a 3s electron to breaking into the filled 2p shell). That huge jump locks sodium into +1.
So it is not that transition metals “want” many states. It is that the energy cost of reaching them is small enough to be paid back, whereas for the s-block metals it is not.
Worked examples
Iron, [Ar] 3d⁶ 4s².
- Lose the two 4s electrons to give Fe²⁺, [Ar] 3d⁶.
- Lose one more, a 3d electron, to give Fe³⁺, [Ar] 3d⁵. That 3d⁵ arrangement is a stable half-filled subshell, which is part of why Fe³⁺ is so common.
Manganese, [Ar] 3d⁵ 4s².
- +2 at the low end (Mn²⁺, [Ar] 3d⁵, again a stable half-filled d subshell).
- Right up to +7 in MnO₄⁻, where manganese has, in effect, used all five 3d and both 4s electrons in its bonding. Seven accessible states, all because those electrons are close in energy.
Contrast with the s-block metals
| Metal | Outer config | Oxidation state(s) | Why |
|---|---|---|---|
| Na (Group 1) | 3s¹ | +1 only | Huge jump after the 1st IE (into 2p) |
| Mg (Group 2) | 3s² | +2 only | Huge jump after the 2nd IE (into 2p) |
| Fe (transition) | 3d⁶ 4s² | +2, +3 and more | 3d and 4s close in energy, IEs rise gently |
The s-block metals have one fixed state each because there is a cliff in their ionisation energies. The transition metals have a gentle slope instead, so several states are within reach.
How the exam asks it
The classic question is simply: “Explain why transition elements exhibit variable oxidation states.” A full-mark answer says:
- the 3d and 4s subshells are close in energy, so
- the successive ionisation energies increase only gradually / there is no large jump, so
- different numbers of electrons can be lost or used in bonding, and the states formed are stabilised by bonding or hydration.
Hit those three linked points and the marks are yours. A common slip is to stop at “3d and 4s are close in energy” without saying why that produces variable states (the gradual ionisation energies). Always finish the chain of reasoning.
Frequently asked questions
Why do transition elements have variable oxidation states? Their 3d and 4s subshells are close in energy, so successive ionisation energies rise only gradually. Different numbers of electrons can be removed without a large energy jump, so several oxidation states are stable.
Why does iron form both Fe²⁺ and Fe³⁺? Losing the two 4s electrons gives Fe²⁺ ([Ar] 3d⁶); losing one more 3d electron gives Fe³⁺ ([Ar] 3d⁵). Both are accessible because 3d and 4s are close in energy, and 3d⁵ is a stable half-filled subshell.
Why does sodium only form +1? After sodium loses its one 3s electron, the next must come from a full inner shell far lower in energy, so the 2nd ionisation energy is huge. That jump makes +1 its only feasible state.
Practise this
The Transition Elements topic (Topic 22) on ChemBank has targeted questions on oxidation states, the electronic configurations of ions, and exactly this reasoning.
Book a trial lesson with Mr Khemistry and make this reasoning automatic.