A Level Chemistry: Transition Elements — Revision Notes

Condensed for the final weeks. For the full explanation, use the Transition Elements study guide. These notes focus on 28.1 and 28.2. For 28.3–28.5 (d-orbital splitting and colour, stereoisomerism and stability constants), use the colour, isomerism and Kstab revision notes.

Definition — get this exactly right

A transition element is a d-block element that forms one or more stable ions with a partially filled d subshell.

The wording earns the mark. It also explains the two exceptions:

• Scandium — forms only Sc³⁺, which is d⁰. Not a transition element.
• Zinc — forms only Zn²⁺, which is d¹⁰ (full). Not a transition element.

Both are d-block; neither is a transition element. That distinction is examined almost every series.

Electron configuration

4s fills before 3d, but 4s empties first on ionisation.

Fe   [Ar] 3d6 4s2      ->   Fe2+  [Ar] 3d6      (NOT 3d4 4s2)
Cr   [Ar] 3d5 4s1      <-   half-filled d is more stable
Cu   [Ar] 3d10 4s1     <-   filled d is more stable

Cr and Cu are the two anomalies. Learn them; they are not derivable from the filling order.

The four characteristic properties

Property	Cause
Variable oxidation states	4s and 3d are very close in energy, so varying numbers of electrons can be lost
Coloured compounds	d-orbitals split by ligands; d–d transition absorbs part of the visible spectrum
Catalytic activity	Variable oxidation states allow intermediate formation; d-orbitals allow adsorption
Complex ion formation	Small, highly charged ions with vacant d-orbitals accept lone pairs

Colour — the full chain of reasoning

1. Ligands approach and split the degenerate d-orbitals into two energy levels separated by ΔE.
2. An electron absorbs a photon of energy exactly ΔE and is promoted.
3. That frequency is removed from white light; the complementary colour is transmitted.
4. ΔE = hf, so the colour depends on the size of the split.

A d⁰ or d¹⁰ ion is colourless — no electron can be promoted (d⁰), or no vacancy exists (d¹⁰). This is why Sc³⁺, Zn²⁺, Cu⁺ and Ti⁴⁺ are all colourless.

The split size, and hence the colour, changes with ligand, oxidation state, and coordination number — which is why [Cu(H₂O)₆]²⁺ is pale blue but [Cu(NH₃)₄(H₂O)₂]²⁺ is deep blue.

Complex ions

Ligand — a species with a lone pair that forms a dative covalent bond to the central metal ion.

Type	Examples
Monodentate	H₂O, NH₃, Cl⁻, CN⁻, OH⁻
Bidentate	ethanedioate (C₂O₄²⁻), 1,2-diaminoethane (“en”)
Multidentate	EDTA⁴⁻ (hexadentate)

Coordination number	Shape
6	Octahedral
4, e.g. [CuCl₄]²⁻, [CoCl₄]²⁻	Tetrahedral
4, with some d⁸ ions such as Pt²⁺, e.g. [Pt(NH₃)₂Cl₂]	Square planar
2	Linear

Ligand exchange — the ligand exchange that actually occurs is the one giving the complex with the larger stability constant (Kstab) under the conditions used, not a single fixed “strength order” applied regardless of concentration. NH₃ and CN⁻ typically form more stable complexes than H₂O at ordinary concentrations, but a very high concentration of Cl⁻ can still drive out H₂O by mass action, as below — this is a concentration effect, not proof that Cl⁻ is intrinsically a stronger ligand.

Worked examples. [Cu(H₂O)₆]²⁺ (pale blue, octahedral) + excess NH₃(aq) → [Cu(NH₃)₄(H₂O)₂]²⁺ (deep blue, distorted octahedral) — only four of six waters are replaced. [Cu(H₂O)₆]²⁺ + excess concentrated HCl → [CuCl₄]²⁻ (yellow-green, tetrahedral) — coordination number falls from 6 to 4 because Cl⁻ is too large to fit six around the ion; the reaction needs a high Cl⁻ concentration (mass action), and adding water reverses it. [Co(H₂O)₆]²⁺ (pink) behaves the same way: + excess NH₃(aq) → [Co(NH₃)₆]²⁺ (yellow-brown); + excess concentrated HCl → [CoCl₄]²⁻ (blue, tetrahedral).

NaOH(aq) precipitates the metal hydroxide at any concentration, limited or excess — e.g. Cu²⁺(aq) + 2OH⁻(aq) → Cu(OH)₂(s), pale blue — written as a complex, [Cu(H₂O)₆]²⁺ + 2OH⁻ → [Cu(OH)₂(H₂O)₄](s) + 2H₂O, where two OH⁻ ligands replace two water ligands to give an uncharged, insoluble complex — and Cu(OH)₂ does not redissolve in excess NaOH(aq). NH₃(aq) gives the same precipitate with a little NH₃ (NH₃ acting as a base, supplying the OH⁻), and for Cu²⁺ that precipitate does redissolve in excess NH₃(aq), forming the deep blue ammine complex by ligand exchange. These are not general rules. From the 9701 qualitative-analysis table: Cr(OH)₃ (grey-green) does dissolve in excess NaOH(aq), giving a dark green solution, but not in excess NH₃(aq); the Fe²⁺, Fe³⁺ and Mn²⁺ hydroxides are insoluble in excess of either reagent. Learn each ion from the table rather than applying one rule.

(Background, beyond 9701’s 28.1/28.2 outcomes.) The chelate effect: multidentate ligands displace monodentate ones because the reaction increases the number of free particles, so ΔS is positive and ΔG becomes more negative. It is an entropy effect, not an enthalpy one.

Key colours

Ion	Colour
[Cu(H₂O)₆]²⁺	pale blue
[Cu(NH₃)₄(H₂O)₂]²⁺	deep blue
[CuCl₄]²⁻	yellow-green
[Fe(H₂O)₆]²⁺	pale green
[Fe(H₂O)₆]³⁺	yellow-brown
MnO₄⁻	purple
Cr₂O₇²⁻	orange
CrO₄²⁻	yellow

Redox titrations

Multiple accessible oxidation states make transition elements central to redox titrations.

MnO₄⁻/Fe²⁺: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. Worked example: 25.0 cm³ of Fe²⁺(aq) needed 23.50 cm³ of 0.0200 mol dm⁻³ KMnO₄(aq). Moles MnO₄⁻ = 0.02350 × 0.0200 = 4.70 × 10⁻⁴ mol; moles Fe²⁺ = 5 × that = 2.35 × 10⁻³ mol; [Fe²⁺] = 2.35 × 10⁻³ ÷ 0.0250 = 0.0940 mol dm⁻³.

MnO₄⁻/C₂O₄²⁻: 2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 10CO₂ + 8H₂O. Purple MnO₄⁻ reacting with colourless C₂O₄²⁻/Mn²⁺ makes this titration self-indicating — the end point is the first permanent pink tinge.

Cu²⁺/I⁻ (iodometry): 2Cu²⁺ + 4I⁻ → 2CuI(s) + I₂. Cu²⁺ oxidises I⁻ to I₂ while being reduced to Cu⁺, which precipitates as white CuI; the liberated I₂ is then titrated against standardised thiosulfate — an indirect method, since Cu²⁺ has no sharp end point of its own. Note: E°(Cu²⁺/Cu⁺) = +0.15 V and E°(I₂/I⁻) = +0.54 V give E°cell = 0.15 − 0.54 = −0.39 V, which looks infeasible from standard-conditions data alone — but the reaction proceeds because the very low solubility of CuI removes Cu⁺ from solution as fast as it forms, shifting the equilibrium and driving the reaction forward (Le Chatelier), a case where product removal matters more than the standard-conditions calculation.

Exam traps

• Calling Sc and Zn transition elements.
• Writing Fe²⁺ as 3d⁴4s² — the 4s electrons go first.
• Saying the colour seen is the colour absorbed. It is the complementary colour.
• Explaining the chelate effect by enthalpy instead of entropy.
• Forgetting the overall charge on a complex ion = metal charge + sum of ligand charges.

Self-test

1. Define a transition element and explain why zinc is excluded.
2. Give the electron configuration of Cr and of Fe²⁺.
3. Explain colour in four steps.
4. Why is [Zn(H₂O)₆]²⁺ colourless?
5. (Background, beyond 9701’s 28.1/28.2 outcomes.) Why does EDTA⁴⁻ displace six water ligands?

Answers: 1. A d-block element that forms at least one stable ion with a partially filled d subshell; zinc forms only Zn²⁺, which is 3d¹⁰ — full, not partially filled. 2. Cr = [Ar]3d⁵4s¹; Fe²⁺ = [Ar]3d⁶. 3. Ligands split the d-orbitals; an electron absorbs a photon of energy equal to ΔE and is promoted; that frequency is removed from white light; the complementary colour is transmitted. 4. Zn²⁺ is d¹⁰ — the d subshell is full, so no d–d transition is possible. 5. The chelate effect: one EDTA⁴⁻ replaces six H₂O, increasing the number of free particles, so ΔS is positive and ΔG more negative.