Electron Configurations of Transition Metals

Spec Mapping — OCR H432 Module 5.3.1 — Transition elements, covering the electronic configuration of atoms and ions of the first-row d-block elements (Sc to Zn), the anomalous configurations of Cr [Ar] 3d⁵ 4s¹ and Cu [Ar] 3d¹⁰ 4s¹ explained by the extra stability of half-filled and fully-filled 3d sub-shells, the rule that 4s electrons are removed before 3d when a positive ion forms, and the use of these configurations to rationalise variable oxidation states and the colourless/non-transition status of Sc³⁺ (d⁰) and Zn²⁺ (d¹⁰) (refer to the official OCR H432 specification document for exact wording).

Electron configuration is the bedrock upon which the whole of transition-metal chemistry rests: every observed property — colour, variable oxidation state, complex-ion shape, catalysis — can be traced back to how many d-electrons the cation carries and how the 3d and 4s sub-shells are arranged. Lesson 1 established the OCR definition of a transition element as a d-block element forming at least one ion with a partially filled d sub-shell. This lesson supplies the electronic mechanics that justify the exclusions of Sc and Zn, that account for the Cr and Cu anomalies, and that explain why Fe²⁺ is [Ar] 3d⁶ rather than the naïve [Ar] 3d⁴ 4s² that simple Aufbau reversal would predict. The single most counter-intuitive idea — that 4s fills before 3d in neutral atoms but ionises first when cations form — is the source of more lost marks at A-Level than any other concept in Module 5.3.1, and the bulk of this lesson is devoted to embedding the rule until it becomes second nature.

Key Definitions:

• Aufbau principle — electrons fill atomic orbitals from lowest energy upwards.
• Hund's rule — within a degenerate sub-shell, electrons singly occupy each orbital with parallel spins before any pairing.
• Pauli exclusion principle — no two electrons in an atom may share all four quantum numbers; an orbital therefore holds at most two electrons of opposite spin.
• Shorthand [Ar] notation — replaces the 1s² 2s² 2p⁶ 3s² 3p⁶ core of any d-block element with the symbol [Ar].
• Exchange energy — the quantum-mechanical stabilisation arising from indistinguishable parallel-spin electrons swapping orbitals; maximised at d⁵ (five parallel spins) and d¹⁰ (five parallel pairs).

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The first-row d-block at a glance

The OCR specification focuses entirely on the first row of the d-block — atomic numbers 21 (Sc) to 30 (Zn). All ten elements share the [Ar] = 1s² 2s² 2p⁶ 3s² 3p⁶ noble-gas core, and they differ only in how 3d and 4s are populated. The Aufbau principle says: fill 4s before 3d. For eight of the ten elements this works exactly; for Cr and Cu it does not.

Z	Element	Atomic configuration	Anomalous?	Why
21	Sc	[Ar] 3d¹ 4s²	No	—
22	Ti	[Ar] 3d² 4s²	No	—
23	V	[Ar] 3d³ 4s²	No	—
24	Cr	[Ar] 3d⁵ 4s¹	Yes	Half-filled 3d is exchange-stabilised
25	Mn	[Ar] 3d⁵ 4s²	No	—
26	Fe	[Ar] 3d⁶ 4s²	No	—
27	Co	[Ar] 3d⁷ 4s²	No	—
28	Ni	[Ar] 3d⁸ 4s²	No	—
29	Cu	[Ar] 3d¹⁰ 4s¹	Yes	Fully-filled 3d is exchange-stabilised
30	Zn	[Ar] 3d¹⁰ 4s²	No	—

The pattern is straightforward except for Cr and Cu, both of which "borrow" a single electron from 4s and place it in 3d to achieve a half-filled (d⁵) or fully-filled (d¹⁰) 3d sub-shell. These two anomalies are recurring exam fodder and must be memorised verbatim.

graph LR
  A[Atom-building 4s before 3d] --> B{Anomalous?}
  B -->|Cr Cu only| C["Promote 4s into 3d<br/>achieves d5 or d10<br/>extra exchange stability"]
  B -->|other eight| D["Standard 3dn 4s2<br/>Aufbau works"]
  C --> E["Cr is 3d5 4s1<br/>Cu is 3d10 4s1"]
  D --> F["Sc 3d1 4s2 through Zn 3d10 4s2"]

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Why are Cr and Cu anomalous?

The conventional explanation, accepted by every A-Level mark scheme, is that a half-filled sub-shell (d⁵) and a fully-filled sub-shell (d¹⁰) carry an extra increment of stability beyond what simple orbital-energy ordering would predict. The quantum-mechanical source of this stabilisation is exchange energy — a purely quantum effect arising from the indistinguishability of parallel-spin electrons. The more parallel-spin pairs that can be drawn between electrons in the same sub-shell, the larger the exchange stabilisation.

For five degenerate d-orbitals, the number of exchange pairs with parallel spin is:

Configuration	Exchange pairs (parallel spin)
d¹	0
d²	1
d³	3
d⁴	6
d⁵ (all parallel)	10
d⁶	10 (5 in spin-up sub-set, 5 cross-pairs lost)
d⁹	10 + 6 = 16
d¹⁰	20 (10 in spin-up sub-set + 10 in spin-down sub-set)

The jump in exchange-pair count at d⁵ and d¹⁰ explains the extra stability that drives Cr to adopt [Ar] 3d⁵ 4s¹ rather than the naïve [Ar] 3d⁴ 4s², and Cu to adopt [Ar] 3d¹⁰ 4s¹ rather than [Ar] 3d⁹ 4s².

For an A-Level mark-scheme answer you do not need the exchange-pair counting; you need the outcome:

"Chromium adopts [Ar] 3d⁵ 4s¹ because a half-filled 3d sub-shell is particularly stable. The promotion of one 4s electron into 3d is energetically favourable. Copper adopts [Ar] 3d¹⁰ 4s¹ for the analogous reason — a fully filled 3d sub-shell is particularly stable."

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The rule for ions: remove 4s first

This is the most counter-intuitive — and most heavily examined — rule in d-block chemistry. State it explicitly in your exam answers:

When a transition metal forms a positive ion, the 4s electrons are removed before the 3d electrons, even though 4s was filled before 3d during atom-building.

The physical reason: in a neutral atom, 4s lies very slightly below 3d in energy (because the 4s orbital penetrates the [Ar] core more effectively). Once electrons populate 3d, however, the additional 3d–3d electron repulsions and the increased effective nuclear charge experienced by 3d cause the 3d energy to drop below 4s. In the resulting cation, 4s is now the higher-energy, more easily removed orbital.

Worked examples — write these out before every exam

Atom	Configuration	M²⁺	M³⁺
Fe (Z=26)	[Ar] 3d⁶ 4s²	[Ar] 3d⁶	[Ar] 3d⁵
Mn (Z=25)	[Ar] 3d⁵ 4s²	[Ar] 3d⁵	[Ar] 3d⁴
Co (Z=27)	[Ar] 3d⁷ 4s²	[Ar] 3d⁷	[Ar] 3d⁶
Ni (Z=28)	[Ar] 3d⁸ 4s²	[Ar] 3d⁸	[Ar] 3d⁷ (rare)
Cu (Z=29)	[Ar] 3d¹⁰ 4s¹	[Ar] 3d⁹	—
Cr (Z=24)	[Ar] 3d⁵ 4s¹	[Ar] 3d⁴	[Ar] 3d³
Sc (Z=21)	[Ar] 3d¹ 4s²	—	[Ar] (d⁰)
Zn (Z=30)	[Ar] 3d¹⁰ 4s²	[Ar] 3d¹⁰	—

Two cases trap candidates:

• Cr²⁺ = [Ar] 3d⁴. The neutral atom is already [Ar] 3d⁵ 4s¹ (anomalous), so to lose two electrons we remove the 4s electron plus one 3d electron.
• Cu²⁺ = [Ar] 3d⁹. The neutral atom is already [Ar] 3d¹⁰ 4s¹ (anomalous), so to lose two electrons we remove the 4s electron plus one 3d electron.

For very high oxidation states (Mn(VII), Cr(VI)), all valence electrons — both 3d and 4s — are stripped to leave bare [Ar]. So MnO₄⁻ contains Mn(VII) = [Ar], and K₂Cr₂O₇ contains Cr(VI) = [Ar].

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Why Sc³⁺ and Zn²⁺ are not transition-metal ions

The OCR exclusions of Sc and Zn fall out directly from electron configuration:

• Sc: atom [Ar] 3d¹ 4s². The only stable ion is Sc³⁺ = [Ar] (lose 4s², lose 3d¹). Sc³⁺ has d⁰ — no d-electrons — and therefore cannot undergo d–d transitions, has no variable oxidation state, and shows no transition-metal character. Sc³⁺ is colourless and diamagnetic.
• Zn: atom [Ar] 3d¹⁰ 4s². The only stable ion is Zn²⁺ = [Ar] 3d¹⁰ (lose 4s²). Zn²⁺ has d¹⁰ — a fully filled d sub-shell — and again cannot undergo d–d transitions (no vacant $e_g$eg​ to promote into). Zn²⁺ is colourless, diamagnetic, and shows only one oxidation state.

Both ions sit at the boundaries of the partially-filled-d test: Sc³⁺ at d⁰, Zn²⁺ at d¹⁰. A partially-filled d sub-shell has between 1 and 9 d-electrons inclusive, so neither qualifies.

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Orbital-box (arrow-in-box) diagrams

OCR mark schemes routinely ask for orbital-box diagrams alongside the [Ar] shorthand. The conventions are: one box per orbital, half-arrow up for spin-up, half-arrow down for spin-down, electrons enter degenerate orbitals singly with parallel spins (Hund's rule) before pairing.

Notice how Mn atom and Fe³⁺ have identical 3d arrangements (five unpaired electrons in five separate boxes) — Fe³⁺ is isoelectronic with Mn atom for the 3d sub-shell alone. This is a recurring synoptic prompt in OCR Paper 3.

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Worked Example 1 — Build Fe²⁺ from scratch

Question: Write the full electron configuration of Fe²⁺ and explain each step.

Working:

1. Fe atomic number = 26. Total electrons = 26.
2. Neutral Fe configuration follows Aufbau (no anomaly for Fe): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s², shortened to [Ar] 3d⁶ 4s².
3. Fe²⁺ has 26 − 2 = 24 electrons.
4. Apply the "remove 4s first" rule: lose both 4s electrons, not two 3d electrons.
5. Resulting configuration: [Ar] 3d⁶, with no 4s electrons remaining.

Common error: a student tempted by Aufbau-reversal might write [Ar] 3d⁴ 4s² for Fe²⁺. This is wrong by two electrons in the wrong sub-shell and would lose marks.

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Worked Example 2 — Build Cu²⁺ correctly

Question: Write the configuration of Cu²⁺ and identify whether copper is a transition element by reference to this configuration.

Working:

1. Cu atomic number = 29; anomalous neutral configuration [Ar] 3d¹⁰ 4s¹.
2. Cu²⁺ has 27 electrons.
3. Apply remove-4s-first: lose the single 4s electron. That accounts for one of the two electrons lost.
4. The second electron must come from 3d: lose one 3d electron.
5. Resulting configuration: [Ar] 3d⁹ (nine d-electrons; one unpaired).
6. Cu²⁺ has d⁹ — partially filled. By the OCR definition, copper is a transition element because it forms at least one stable ion (Cu²⁺) with a partially filled d sub-shell. ✓

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Worked Example 3 — V across its four oxidation states

Vanadium famously cycles through four colours when reduced by zinc in acidic solution: yellow → blue → green → violet. The configurations are:

Species	Notation	Configuration	Colour
V atom	V	[Ar] 3d³ 4s²	(silvery metal)
V(V)	VO₂⁺	[Ar] d⁰	Yellow
V(IV)	VO²⁺	[Ar] 3d¹	Blue
V(III)	[V(H₂O)₆]³⁺	[Ar] 3d²	Green
V(II)	[V(H₂O)₆]²⁺	[Ar] 3d³	Violet

Each one-electron addition lowers the oxidation state by one and changes the d-count by one. The colour change is direct evidence that d-electron count and colour are intrinsically linked (lesson 1 covered the d–d transition mechanism).

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