The big idea: A transition element is a d-block metal that forms at least one stable ion with a partially filled d sub-shell.
That 'partially filled d sub-shell' (not empty, not full) is the source of almost everything special about these metals — their variable oxidation states, their coloured ions, and their catalytic activity.
The transition elements are the d-block — the central block of metals (Sc → Zn in the first row). Their chemistry comes from a partially filled 3d sub-shell.
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Define the terms: - d-block — the central block of the periodic table (groups 3–12); the highest-energy electrons go into d orbitals. - Sub-shell — a set of orbitals of the same type (the 3d sub-shell holds up to 10 electrons in 5 orbitals). - Partially filled — between 1 and 9 electrons in the d sub-shell (so not d⁰ and not d¹⁰).
Why Sc and Zn are often excluded: Scandium and zinc are in the d-block, but by the strict definition they are not transition elements:
- Scandium forms only Sc³⁺, which is [Ar] 3d⁰ — an empty d sub-shell. - Zinc forms only Zn²⁺, which is [Ar] 3d¹⁰ — a full d sub-shell.
Neither forms an ion with a partially filled d sub-shell, so neither shows the typical transition-metal behaviour (variable states, coloured ions). That is why they are usually left out.
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Across the first row (Sc → Zn) electrons fill the 3d sub-shell. The 4s sub-shell fills before 3d (it is slightly lower in energy when empty), so most atoms end in 3dx 4s².
| Atom | Z | Electron configuration | Note |
|---|---|---|---|
| Sc | 21 | [Ar] 3d¹ 4s² | forms only Sc³⁺ → [Ar] (empty d): often excluded |
| Ti | 22 | [Ar] 3d² 4s² | |
| V | 23 | [Ar] 3d³ 4s² | |
| Cr | 24 | [Ar] 3d⁵ 4s¹ | anomaly — a half-full 3d⁵ is extra stable |
| Mn | 25 | [Ar] 3d⁵ 4s² | shows the widest range of oxidation states (up to +7) |
| Fe | 26 | [Ar] 3d⁶ 4s² | +2 and +3 both common |
| Co | 27 | [Ar] 3d⁷ 4s² | |
| Ni | 28 | [Ar] 3d⁸ 4s² | |
| Cu | 29 | [Ar] 3d¹⁰ 4s¹ | anomaly — a full 3d¹⁰ is extra stable |
| Zn | 30 | [Ar] 3d¹⁰ 4s² | forms only Zn²⁺ → 3d¹⁰ (full d): often excluded |
The Cr and Cu anomalies: Two atoms break the simple pattern because a half-full or completely full d sub-shell is extra stable:
- Chromium is [Ar] 3d⁵ 4s¹ (not 3d⁴ 4s²) — one 4s electron shifts into 3d to give a stable half-full 3d⁵. - Copper is [Ar] 3d¹⁰ 4s¹ (not 3d⁹ 4s²) — one 4s electron shifts into 3d to give a stable full 3d¹⁰.
Forming the ions — remove 4s electrons FIRST: This is the key HL rule. The 4s sub-shell fills first but is emptied first when the atom is ionised.
To write a transition-metal ion: take electrons from 4s before 3d.
Example — iron, Fe = [Ar] 3d⁶ 4s²: - Fe²⁺: remove the two 4s electrons → [Ar] 3d⁶ - Fe³⁺: then remove one 3d electron → [Ar] 3d⁵
Fe²⁺ (1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶): both electrons are removed from 4s first, leaving a partially filled 3d sub-shell — so iron(II) is a true transition-metal ion.
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Worked example — deduce an ion's configuration
Deduce the full electron configuration of the cobalt(II) ion, Co²⁺. (Co has Z = 27.)
Solution
- Write the atom first. Cobalt (Z = 27) follows the normal pattern:
- Remove the 4s electrons first. Co²⁺ means remove 2 electrons, both from 4s:
- Check: the 3d sub-shell holds 7 electrons — it is partially filled, so Co²⁺ is a genuine transition-metal ion.
Final answer
Co²⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷ (or [Ar] 3d⁷).
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The headline property: a transition metal can exist in several different oxidation states. Iron is +2 or +3; manganese ranges from +2 all the way to +7.
Why several oxidation states?: In a transition metal the 4s and 3d sub-shells are very close in energy.
That means a similar, small amount of energy is needed to remove a 4s electron or a 3d electron. So once the 4s electrons are lost (giving the common +2 state), the 3d electrons can also be removed in ones and twos — each costing roughly the same — giving +3, +4 … and so on.
By contrast, a group-1 or group-2 metal has only its outer s electrons available, so it shows one fixed oxidation state.
| Metal | Common oxidation states | Example species |
|---|---|---|
| iron, Fe | +2, +3 | Fe²⁺ (in FeSO4), Fe³⁺ (in FeCl3) |
| copper, Cu | +1, +2 | Cu⁺ (in Cu2O), Cu²⁺ (in CuSO4) |
| manganese, Mn | +2 up to +7 | Mn²⁺ (pale pink), MnO4⁻ (Mn is +7) |
| chromium, Cr | +3, +6 | Cr³⁺, Cr2O7²⁻ (Cr is +6) |
Finding the oxidation state of the metal: Use the rule that the charges in a species add up to its overall charge. In the permanganate ion MnO4⁻: each O is −2 (4 × −2 = −8) and the whole ion is −1, so Mn must be +7 (because +7 − 8 = −1).
Common states to know: - +2 is common for nearly all of them (the two 4s electrons are removed). - Iron: +2 and +3 (Fe³⁺ has a stable half-full 3d⁵). - Copper: +1 and +2 (the only first-row metal with a stable +1 in solution chemistry you meet). - Manganese: +2 up to +7 (the widest range; Mn is +7 in MnO4⁻).
How this is tested: This HL micro shows up in two formats:
- Paper 1A (MCQ): deduce the configuration of a given transition-metal ion, or pick the metal/ion with a partially filled d sub-shell, or work out the oxidation state of a metal in a species (e.g. Cr in Cr2O7²⁻). - Paper 2: an explain mark — why transition metals show variable oxidation states — usually paired with writing the configuration of an ion.
The trap on Paper 2 is answering 'because of the d electrons' without naming why that matters.
Score the explain mark: For variable oxidation states, the marking point is the energy gap: the 4s and 3d sub-shells are close in energy, so successive electrons can be removed using similar (small) amounts of energy. Name that cause, then state the effect (several stable states).
IB-style question — deduce the ion and its state (a)
(a) Deduce the full electron configuration of the manganese(II) ion, Mn²⁺, and state the oxidation state of manganese in it. (Mn has Z = 25.) [2]
How to score the marks
- Mark 1 — the configuration. Mn atom = [Ar] 3d⁵ 4s². Remove the two 4s electrons first to give the +2 ion: Mn²⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ (= [Ar] 3d⁵).
- Mark 2 — the oxidation state. The ion carries a 2+ charge, so manganese is in oxidation state +2.
Final answer
Mn²⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ ([Ar] 3d⁵); oxidation state = +2.
IB-style question — explain variable oxidation states (b)
(b) Explain why iron can form both the +2 and +3 ions, whereas calcium forms only the +2 ion. [3]
How to score the marks
- Mark 1 — the energy gap. In iron (a transition metal) the 4s and 3d sub-shells are close in energy, so a similar (small) amount of energy is needed to remove electrons from either.
- Mark 2 — so more than one state. After losing the two 4s electrons (→ Fe²⁺), a 3d electron can also be removed using a similar amount of energy, giving Fe³⁺; the resulting [Ar] 3d⁵ is a stable half-full sub-shell.
- Mark 3 — contrast with calcium. Calcium has only its two outer 4s electrons available; removing a core electron would need far too much energy, so it forms only Ca²⁺ (one fixed state).
Final answer
Fe: 4s and 3d are close in energy → 3d electrons can also be removed for a similar cost → +2 and +3. Ca has only 4s electrons available → only +2.