Chemistry of Transition Elements

0. What Examiners Want

This guide covers Topic 28 of Cambridge International AS & A Level Chemistry 9701 (2025–27) and follows the learning outcomes below, word for word. The right-hand column shows where each one is taught. It builds on Topic 24 (electrode potentials), Topic 26 (catalysis) and Topic 27 (the idea of polarisation).

1. What Is a Transition Element?

The transition elements are found in the d block, between Groups 2 and 13. Not every d-block element is a transition element.

The test is applied to the ion. Scandium only forms Sc³⁺ (no d electrons). Zinc only forms Zn²⁺ (a full d sub-shell). Copper is a transition element because Cu²⁺ is 3d⁹.

Electronic configurations. In atoms, the 4s sub-shell fills before the 3d sub-shell, except in chromium and copper, where the half-filled (3d⁵) and completely filled (3d¹⁰) arrangements are slightly more stable.

Ti to Cu: [Ar] 3d²4s², 3d³4s², 3d⁵4s¹, 3d⁵4s², 3d⁶4s², 3d⁷4s², 3d⁸4s², 3d¹⁰4s¹.

Ions. When a transition element forms a positive ion, the 4s electrons are lost first, then the 3d electrons:

• Fe: [Ar] 3d⁶4s² → Fe²⁺: [Ar] 3d⁶ → Fe³⁺: [Ar] 3d⁵
• Cu: [Ar] 3d¹⁰4s¹ → Cu⁺: [Ar] 3d¹⁰ and Cu²⁺: [Ar] 3d⁹
• Ti: [Ar] 3d²4s² → Ti³⁺: [Ar] 3d¹

Worked example. Write the full electronic configurations of Co²⁺ and Cr³⁺, and decide whether each is the ion of a transition element.

Co: 1s²2s²2p⁶3s²3p⁶3d⁷4s²; Co²⁺: 1s²2s²2p⁶3s²3p⁶3d⁷. Cr: …3d⁵4s¹; Cr³⁺: …3s²3p⁶3d³. Both have an incomplete 3d sub-shell, so cobalt and chromium are transition elements.

2. The 3d Orbitals

There are five 3d orbitals. In an isolated atom or ion they are degenerate, meaning they have the same energy. The syllabus asks you to sketch the shape of two of them:

• 3d_xy: four lobes lying between the x and y axes (pointing along the diagonals, in the xy plane);
• 3d_z²: two lobes along the z axis plus a doughnut-shaped ring around the middle, in the xy plane.

3d_xy has lobes between the axes; 3d_x²−y² has lobes on the x and y axes. 3d_xz and 3d_yz are 3d_xy rotated into the xz and yz planes.

When you sketch, draw and label the axes, make the lobes of equal size and say which orbital it is. A 3d_xy sketch with lobes on the axes would be the 3d_x²−y² orbital, which scores nothing.

3. Four Characteristic Properties

Transition elements are metals with high melting points and densities. Chemically, four properties set them apart, and each follows from the 3d sub-shell.

Learn each property together with its explanation: the question will ask for both.

(a) Variable oxidation states. The 3d and 4s sub-shells are similar in energy, so after the 4s electrons have been lost, the 3d electrons can also be lost with only a small extra energy cost. Successive ionisation energies rise steadily, with no huge jump, so several oxidation states are possible. Compare calcium, where the third ionisation energy is enormous because it breaks into a full inner shell.

Calcium shows a large jump after the 4s electrons; iron does not.

The highest state from Ti to Mn equals the total number of 4s + 3d electrons (Mn: +7 in MnO₄⁻). Across the row from iron, +2 becomes the commonest state as the 3d electrons are held more tightly.

Vanadium: VO₂⁺ (+5, yellow), VO²⁺ (+4, blue), V³⁺ (+3, green), V²⁺ (+2, violet). Different oxidation states usually give different colours.

Because the oxidation state varies, compound names include it: iron(II) sulfate, manganese(IV) oxide, chromium(VI) oxide.

(b) Catalysts. Transition elements and their compounds make good catalysts because they have more than one stable oxidation state (so they can accept and give back electrons in the course of a reaction) and vacant d orbitals that are energetically accessible and can form dative bonds with ligands (so they can bind reactants). Examples from Topic 26: Fe in the Haber process; V₂O₅ in the Contact process; Pt, Pd and Rh in catalytic converters; and Fe²⁺ / Fe³⁺ in the I⁻ / S₂O₈²⁻ reaction.

(c) Complex ions. Their ions have vacant d orbitals that are energetically accessible, which can accept lone pairs from ligands as dative bonds. They also have a high charge density (small, highly charged ions), which attracts ligands.

(d) Coloured compounds. Ligands split the d orbitals into two sets of different energy. An electron is promoted from the lower to the upper set, absorbing light of a particular frequency, and the complementary colour is seen (section 8).

Sc³⁺ (d⁰) and Zn²⁺ (d¹⁰) have no possible d–d transition, so their compounds are colourless. Cu⁺ (d¹⁰) is also colourless.

4. Ligands and Complex Ions

Ligand: a species that contains a lone pair of electrons that forms a dative covalent bond to a central metal atom / ion.

Complex: a molecule or ion formed by a central metal atom / ion surrounded by one or more ligands.

Coordination number: the number of coordinate (dative) bonds formed by the central metal ion.

Ligands are classified by the number of dative bonds each one forms:

The arrows show the lone pair donated by each donor atom. EDTA⁴⁻ uses two nitrogen atoms and four oxygen atoms.

The formula and charge of a complex. The charge is the sum of the charge on the metal ion and the charges on the ligands (neutral ligands such as H₂O and NH₃ contribute nothing).

To find the oxidation state of the metal from the complex, work backwards. In [Ni(CN)₄]²⁻: x + 4(−1) = −2, so x = +2.

Worked example. Predict the formula and charge of the complex formed by Cr³⁺ with six hydroxide ions, and state the coordination number.

Charge = +3 + 6(−1) = −3, so [Cr(OH)₆]³⁻. The coordination number is 6, because each OH⁻ forms one dative bond.

5. Shapes of Complexes

The shape depends on the coordination number (and, for four, on the metal and ligand).

Linear (180°), square planar (90°), tetrahedral (109.5°) and octahedral (90°).

Large ligands such as Cl⁻ give four-coordinate (tetrahedral) complexes, because only four fit around the metal ion. Small ligands such as H₂O and NH₃ give six-coordinate (octahedral) complexes. Silver(I), copper(I) and gold(I) form linear complexes.

Worked example. In the complex ion [Ag(NH₃)₂]⁺, name the shape and state the bond angles at Ag and at N.

Two ligands, so linear, with N–Ag–N = 180°. Each nitrogen has three N–H bonds and the dative bond to silver (four electron pairs, no lone pair), so the bond angle H–N–Ag is about 109.5°.

6. Ligand Exchange: Copper(II) and Cobalt(II)

In ligand exchange (ligand substitution), one ligand in a complex is replaced by another. Whenever you write Cu²⁺(aq) or Co²⁺(aq), the ion is really the hexaaqua complex, [Cu(H₂O)₆]²⁺ or [Co(H₂O)₆]²⁺. Water, ammonia and hydroxide ions are similar in size, so substituting them does not change the coordination number. Chloride ions are larger, so only four fit, and the shape changes from octahedral to tetrahedral.

Copper(II)

The pale blue [Cu(H₂O)₆]²⁺ is the starting point for all the copper reactions.

The hydroxide precipitate is also written Cu(OH)₂(s). The precipitate dissolves in excess ammonia: Cu(OH)₂(H₂O)₄ + 4NH₃ → [Cu(NH₃)₄(H₂O)₂]²⁺ + 2OH⁻ + 2H₂O. Add water to the yellow chloro complex and the equilibrium shifts back to pale blue.

Cobalt(II)

Pink [Co(H₂O)₆]²⁺ becomes blue [CoCl₄]²⁻ in concentrated hydrochloric acid.

Blue anhydrous cobalt chloride paper (containing [CoCl₄]²⁻) turns pink in the presence of water, as water ligands replace the chloride ligands: it is a test for water.

Worked example. Concentrated aqueous ammonia is added a little at a time to aqueous copper(II) sulfate until it is in excess. Describe and explain what you see.

At first a pale blue precipitate of Cu(OH)₂(H₂O)₄ forms, because ammonia is a weak base and its hydroxide ions precipitate the hydroxide. With excess ammonia the precipitate dissolves to give a deep blue solution of [Cu(NH₃)₄(H₂O)₂]²⁺ by ligand exchange: four water ligands are replaced by four ammonia ligands.

7. Redox Reactions and Titrations

Predicting feasibility from E° values. A redox reaction is feasible when its standard cell potential is positive:

The higher (more positive) system is the stronger oxidising agent and drives the lower system in reverse.

Worked example. Is the oxidation of Fe²⁺ by acidified dichromate(VI) feasible? Is the oxidation of Cl⁻ by manganate(VII) feasible?

E°(Cr₂O₇²⁻/Cr³⁺) = +1.33 V and E°(Fe³⁺/Fe²⁺) = +0.77 V, so E°_cell = 1.33 − 0.77 = +0.56 V: feasible. E°(MnO₄⁻/Mn²⁺) = +1.52 V and E°(Cl₂/Cl⁻) = +1.36 V, so E°_cell = 1.52 − 1.36 = +0.16 V: feasible. This is why manganate(VII) titrations are done in sulfuric acid: in hydrochloric acid, MnO₄⁻ would also oxidise Cl⁻ to Cl₂, so the titre would be too large.

A positive E°_cell says the reaction is feasible, not that it is fast: a reaction with a large activation energy may not occur at a measurable rate.

Know the equation, the mole ratio and the colour change at the end-point for each titration.

(a) MnO₄⁻ / Fe²⁺. MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O and Fe²⁺ → Fe³⁺ + e⁻ give MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺. Potassium manganate(VII) (purple) goes in the burette. It acts as its own indicator: the end-point is the first permanent pink colour.

Worked example. 25.0 cm³ of an iron(II) solution needs 21.50 cm³ of 0.0200 mol dm⁻³ KMnO₄. Calculate the concentration of Fe²⁺.

Moles MnO₄⁻ = 0.0200 × 21.50 ÷ 1000 = 4.30 × 10⁻⁴. Moles Fe²⁺ = 5 × 4.30 × 10⁻⁴ = 2.15 × 10⁻³. Concentration = 2.15 × 10⁻³ ÷ 0.0250 = 0.0860 mol dm⁻³.

(b) MnO₄⁻ / C₂O₄²⁻. C₂O₄²⁻ → 2CO₂ + 2e⁻. Multiply by 5 and the manganate half-equation by 2: 2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 8H₂O + 10CO₂. The solution is warmed (about 60 °C) because the reaction is slow at first. It is then autocatalysed by the Mn²⁺ ions that form: the colour fades slowly at the start, then quickly.

Worked example. 25.0 cm³ of sodium ethanedioate solution reacts with 20.00 cm³ of 0.0200 mol dm⁻³ KMnO₄. Calculate the concentration of the ethanedioate.

Moles MnO₄⁻ = 0.0200 × 0.02000 = 4.00 × 10⁻⁴. Moles C₂O₄²⁻ = 4.00 × 10⁻⁴ × 5/2 = 1.00 × 10⁻³. Concentration = 1.00 × 10⁻³ ÷ 0.0250 = 0.0400 mol dm⁻³.

(c) Cu²⁺ / I⁻ / S₂O₃²⁻. Excess iodide reduces copper(II) to copper(I), forming a white precipitate of CuI, and iodine (brown): 2Cu²⁺ + 4I⁻ → 2CuI + I₂. The iodine is titrated with sodium thiosulfate: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻. Starch is added when the iodine colour has faded to pale yellow; the end-point is the loss of the blue-black colour. From the two equations, 2Cu²⁺ ≡ I₂ ≡ 2S₂O₃²⁻, so moles Cu²⁺ = moles S₂O₃²⁻.

Worked example. 25.0 cm³ of copper(II) sulfate solution is added to excess potassium iodide. The iodine formed needs 18.60 cm³ of 0.100 mol dm⁻³ sodium thiosulfate. Calculate the concentration of the copper(II) sulfate.

Moles S₂O₃²⁻ = 0.100 × 0.01860 = 1.86 × 10⁻³ = moles Cu²⁺. Concentration = 1.86 × 10⁻³ ÷ 0.0250 = 0.0744 mol dm⁻³.

(d) Other redox systems. Any redox titration can be handled the same way: write the two half-equations, balance the electrons, find the mole ratio, then calculate. For example, dichromate(VI) oxidises iron(II): Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺, in the ratio 1 : 6, and it is orange → green. Dichromate needs an indicator for the end-point.

8. Why Complexes Are Coloured

White light contains all visible frequencies. A solution looks coloured when it absorbs part of the spectrum; the colour you see is the complementary colour of the light that is absorbed.

Opposite colours on the wheel are complementary: a complex that absorbs red light looks cyan.

Degenerate and non-degenerate orbitals. In an isolated ion the five d orbitals have the same energy: they are degenerate. When ligands approach, their lone pairs repel the d electrons. The repulsion is not the same for every d orbital, so the d orbitals split into two sets of different energy: they are now non-degenerate.

Octahedral: two orbitals higher, three lower. Tetrahedral: three orbitals higher, two lower. ΔE is the energy gap.

• Octahedral: the d_x²−y² and d_z² orbitals point towards the six ligands, so they are repelled more and are higher in energy; d_xy, d_xz and d_yz point between the ligands and are lower.
• Tetrahedral: the reverse; d_xy, d_xz and d_yz are nearer the ligands (higher), and d_x²−y² and d_z² are lower. The splitting is smaller than in octahedral complexes.

Origin of colour. An electron in the lower set absorbs light whose energy equals ΔE and is promoted to the upper set (a d–d transition). The frequency absorbed is given by ΔE = hν. The remaining light, which has had that frequency removed, is the colour seen.

Worked example. A complex absorbs light of wavelength 600 nm. Calculate ΔE in J and in kJ mol⁻¹. (h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹, L = 6.022 × 10²³ mol⁻¹.)

ν = c / λ = 3.00 × 10⁸ ÷ 6.00 × 10⁻⁷ = 5.00 × 10¹⁴ Hz. ΔE = 6.63 × 10⁻³⁴ × 5.00 × 10¹⁴ = 3.32 × 10⁻¹⁹ J per ion. Per mole: 3.32 × 10⁻¹⁹ × 6.022 × 10²³ = 2.00 × 10⁵ J = 200 kJ mol⁻¹. Light of 600 nm is orange, so the complex looks blue.

Effect of the ligand (qualitative). The size of ΔE depends on the metal ion, its oxidation state and the ligands. Different ligands split the d orbitals by different amounts. A change in ΔE changes the frequency absorbed and so the complementary colour seen. For copper(II), the aqua complex [Cu(H₂O)₆]²⁺ is pale blue, but when four NH₃ replace four H₂O the splitting is larger and the solution is deep blue. A change of shape (octahedral to tetrahedral with Cl⁻) also changes ΔE, giving yellow [CuCl₄]²⁻.

Larger ΔE means light of higher frequency is absorbed, so a different complementary colour is seen.

Why some compounds are colourless. Sc³⁺ (3d⁰) has no d electrons to promote, and Zn²⁺ and Cu⁺ (3d¹⁰) have no vacant place in the d orbitals for the electron to move to. Their compounds are colourless (or white).

9. Stereoisomerism and Stability Constants

These two topics are in the coursebook chapter and are examined in the 2023 and 2024 Paper 4 questions on complexes, although they are not listed among the 28.1–28.3 outcomes in the 2025–27 syllabus above. Treat this section as a useful extension.

Stereoisomerism. Complexes with the same formula can differ in the arrangement of ligands in space.

Cis and trans isomers. cis-platin is the anti-cancer drug.

• Geometric (cis / trans) isomerism: in square planar [MA₂B₂] and octahedral [MA₄B₂] complexes, the two B ligands are either next to each other (cis) or opposite (trans).
• Optical isomerism: octahedral complexes with bidentate ligands, such as [Ni(en)₃]²⁺, exist as two non-superimposable mirror images. cis-[Ni(en)₂(H₂O)₂]²⁺ is also optically active, but the trans isomer is not.

The two optical isomers of [Ni(en)₃]²⁺: they rotate the plane of polarised light in opposite directions.

Stability constants. The stability constant, K_stab, of a complex ion is the equilibrium constant for the formation of the complex from the aqueous metal ion and the ligand(s). Water is left out of the expression because it is in large excess. For [Cu(H₂O)₆]²⁺ + 4NH₃ ⇄ [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O:

The larger the value of K_stab, the more stable the complex, so a ligand with a larger K_stab replaces one with a smaller value. For copper(II): log₁₀K_stab is about 5.6 for Cl⁻ and 13.1 for NH₃, so ammonia displaces chloride from [CuCl₄]²⁻. Chelating (bidentate and polydentate) ligands usually give much larger values than monodentate ones, because replacing several water ligands by one chelating ligand releases more molecules (an increase in entropy). That is why EDTA⁴⁻ removes toxic metal ions from the body.

Worked example. The equilibrium [Fe(H₂O)₆]³⁺ + SCN⁻ ⇄ [Fe(SCN)(H₂O)₅]²⁺ + H₂O has equilibrium concentrations [Fe(H₂O)₆]³⁺ = 1.0 × 10⁻³, SCN⁻ = 2.0 × 10⁻⁴ and complex = 3.0 × 10⁻⁴ mol dm⁻³. Calculate K_stab.

K_stab = 3.0 × 10⁻⁴ ÷ (1.0 × 10⁻³ × 2.0 × 10⁻⁴) = 1.5 × 10³ dm³ mol⁻¹ (the units are 1 / concentration).

10. Quick Sheet and Checklist

Checklist. Before the exam you should be able to:

• [ ] define a transition element and explain why scandium and zinc are not transition elements
• [ ] write the electronic configurations of the atoms and ions of Ti to Cu, including the Cr and Cu exceptions
• [ ] sketch 3d_xy and 3d_z² with labelled axes
• [ ] explain variable oxidation states, catalysis, complex formation and colour in terms of the 3d sub-shell
• [ ] define ligand, complex, coordination number, monodentate, bidentate and polydentate, with examples
• [ ] predict the formula and charge of a complex and give its shape and bond angle
• [ ] give the formulae, colours and equations for the ligand exchange of copper(II) and cobalt(II) with water, ammonia, hydroxide and chloride ions
• [ ] use E° values to decide whether a redox reaction is feasible
• [ ] write equations and do titration calculations for MnO₄⁻ with Fe²⁺ and C₂O₄²⁻, and for Cu²⁺ with I⁻ and S₂O₃²⁻
• [ ] define degenerate and non-degenerate d orbitals and draw the splitting for octahedral and tetrahedral complexes
• [ ] explain why complexes are coloured, and how a change of ligand changes the colour
• [ ] draw cis / trans and optical isomers of simple complexes, and write a K_stab expression with its units