1. Transition Metal Definition & Characteristic Properties
- Scandium (Sc): Electronic configuration is [Ar] 3d¹ 4s². It forms only one stable ion, Sc³⁺, which has the configuration [Ar] 3d⁰. Because the 3d sub-level is completely empty, scandium does not satisfy the definition.
- Zinc (Zn): Electronic configuration is [Ar] 3d¹⁰ 4s². It forms only the Zn²⁺ ion, with configuration [Ar] 3d¹⁰. Because the 3d sub-level is completely full, zinc does not satisfy the definition.
Four Characteristic Properties of Transition Metals
- Variable oxidation states: e.g. Fe exists as Fe²⁺ and Fe³⁺; Mn exists from +2 to +7. (The 3d and 4s electrons have very similar energies, allowing differing numbers of electrons to be lost). See Variable Oxidation States and Catalysts.
- Formation of complex ions: Readily coordinate with lone-pair donor ligands.
- Formation of coloured compounds: Absorption of visible light due to d-d electronic transitions.
- Catalytic activity: Act as heterogeneous and homogeneous catalysts due to variable oxidation states and vacant d orbitals.
2. Complex Ions, Ligands & Coordination Number
- Complex Ion: A central transition metal ion surrounded by coordinately bonded ligands.
- Ligand: a molecule or ion that forms a co-ordinate bond with a transition metal ion by donating a pair of electrons.
- Coordination Number: The total number of coordinate bonds formed between the central metal ion and its surrounding ligands.
Classification of Ligands by Denticity
| Denticity | Definition | Representative Examples | Typical Coordination Number & Complex |
|---|---|---|---|
| Monodentate | Donates ONE lone pair per ligand molecule/ion. | :H₂O, :NH₃, :Cl⁻, :CN⁻, :OH⁻ | Coord 6: [Fe(H₂O)₆]²⁺, [Cu(H₂O)₆]²⁺
Coord 4: [CuCl₄]²⁻ (Cl⁻ is large and charged) |
| Bidentate | Donates TWO lone pairs from different atoms in the same molecule. | 1,2-diaminoethane (en, H₂NCH₂CH₂NH₂)
Ethanedioate ion (C₂O₄²⁻) |
Coord 6: [Fe(C₂O₄)₃]³⁻, [Cr(en)₃]³⁺ |
| Multidentate | Donates THREE OR MORE lone pairs from different atoms in the same molecule. | EDTA⁴⁻ (hexadentate: donates 6 lone pairs from 4 carboxylate oxygens and 2 amine nitrogens) | Coord 6: [Cu(EDTA)]²⁻, [Fe(EDTA)]⁻ |
3. Ligand Substitution Reactions
In a ligand substitution reaction, one ligand replaces another around the metal ion. Learn the equation and the colour change for each reaction below.
Ligands of similar size: NH₃ replaces H₂O
NH₃ and H₂O are similar in size and both are uncharged. So the co-ordination number stays 6, and the complex stays octahedral. These equations show the overall change with excess concentrated ammonia.
| Metal ion | Equation | Colour change |
|---|---|---|
| Cu²⁺ | [Cu(H₂O)₆]²⁺ + 4NH₃ → [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O | Blue solution to deep blue solution |
| Co²⁺ | [Co(H₂O)₆]²⁺ + 6NH₃ → [Co(NH₃)₆]²⁺ + 6H₂O | Pink solution to pale brown (straw) solution. It darkens in air, as Co²⁺ is oxidised to Co³⁺. |
With copper, the substitution is incomplete. Only four of the six water ligands are replaced, and the co-ordination number stays 6.
A larger ligand: Cl⁻ replaces H₂O
Add concentrated hydrochloric acid. The Cl⁻ ligand is larger than H₂O, so only four fit around the metal ion. The co-ordination number changes from 6 to 4, and the shape changes from octahedral to tetrahedral.
| Metal ion | Equation | Colour change |
|---|---|---|
| Cu²⁺ | [Cu(H₂O)₆]²⁺ + 4Cl⁻ → [CuCl₄]²⁻ + 6H₂O | Blue to yellow/green |
| Co²⁺ | [Co(H₂O)₆]²⁺ + 4Cl⁻ → [CoCl₄]²⁻ + 6H₂O | Pink to blue |
| Fe³⁺ | [Fe(H₂O)₆]³⁺ + 4Cl⁻ → [FeCl₄]⁻ + 6H₂O | Changes to yellow |
Bidentate and multidentate ligands
- Bidentate: [Cu(H₂O)₆]²⁺ + 3H₂NCH₂CH₂NH₂ → [Cu(H₂NCH₂CH₂NH₂)₃]²⁺ + 6H₂O. Each H₂NCH₂CH₂NH₂ forms two co-ordinate bonds, so three of them replace six water ligands.
- The ethanedioate ion, C₂O₄²⁻, is also a bidentate ligand.
- Multidentate: [Cu(H₂O)₆]²⁺ + EDTA⁴⁻ → [CuEDTA]²⁻ + 6H₂O. One EDTA⁴⁻ ion forms six co-ordinate bonds.
For the reactions of these aqua ions with sodium hydroxide, ammonia and carbonate, see Reactions of Aqueous Ions.
4. The Chelate Effect & Thermodynamic Stability
When monodentate ligands (such as H₂O or NH₃) are replaced by bidentate or multidentate ligands (such as EDTA⁴⁻ or 1,2-diaminoethane), the resulting complex is vastly more stable. This phenomenon is known as the chelate effect.
Thermodynamic Justification of the Chelate Effect
Consider the ligand substitution of hexaaquacopper(II) by EDTA⁴⁻:
[Cu(H₂O)₆]²⁺(aq) + EDTA⁴⁻(aq) → [Cu(EDTA)]²⁻(aq) + 6H₂O(l)
- Enthalpy change (ΔH): Metal-ligand coordinate bond strengths are very similar (Cu−O and Cu−N bonds broken vs formed), so ΔH is approximately zero.
- Entropy change (ΔS): On the left side of the equation, there are 2 particles in solution (one complex ion and one EDTA⁴⁻). On the right side, there are 7 particles (one chelated complex ion and six released water molecules). The increase in independent particles in solution creates substantial disorder, producing a large positive entropy change (ΔS > 0).
- Gibbs free energy (ΔG): Since ΔG = ΔH − TΔS, the large positive ΔS term makes ΔG highly negative, driving the reaction decisively forward and rendering the chelate complex exceptionally stable.
In short: 2 particles become 7, so ΔS is positive. ΔH is about 0, because similar bonds (Cu−N for Cu−O, similar in number) are broken and made. So ΔG = ΔH − TΔS is negative, and the reaction is feasible.
Haem: oxygen transport and carbon monoxide
- Haem contains an iron(II) ion held by a multidentate ligand called a porphyrin.
- Haem is part of haemoglobin. O₂ forms a co-ordinate bond to the Fe(II), so oxygen is carried in the blood.
- CO also forms a co-ordinate bond to the Fe(II). It binds more strongly than O₂ and does not readily come off. So less O₂ is carried. This is why carbon monoxide is toxic.
5. Geometries of Transition Metal Complexes
- Octahedral (Coordination Number 6, 90° bond angles): Formed with small, uncharged ligands like H₂O and NH₃. Examples: [Fe(H₂O)₆]²⁺, [Cu(H₂O)₆]²⁺, [Fe(CN)₆]³⁻, [Cr(NH₃)₆]³⁺.
- Tetrahedral (Coordination Number 4, 109.5° bond angles): Formed when ligands are large and charged, such as chloride ions (Cl⁻). Steric hindrance and electrostatic repulsions prevent 6 chloride ions from packing around the metal. Examples: [CuCl₄]²⁻ (yellow-green), [CoCl₄]²⁻ (deep blue).
- Square Planar (Coordination Number 4, 90° bond angles): Formed by platinum(II) and nickel(II) complexes. The most famous example is the anticancer chemotherapy drug cisplatin, [Pt(NH₃)₂Cl₂].
- Linear (Coordination Number 2, 180° bond angles): Formed by silver(I) complexes, notably Tollens' reagent: [Ag(NH₃)₂]⁺.
In the diagrams, a solid wedge shows a bond coming towards you and a dashed line shows a bond going away from you.
6. Isomerism in Complex Ions
Cis-trans isomerism in square planar complexes
[Pt(NH₃)₂Cl₂] is square planar. The two Cl⁻ ligands can be next to each other (cis, 90°) or opposite each other (trans, 180°). Cisplatin is the cis isomer. It is used as an anticancer drug: see Amino Acids, Proteins and Cisplatin in Unit 4.
Cis-trans isomerism in octahedral complexes
Octahedral complexes with two monodentate ligands of one kind and four of another show cis-trans isomerism. In [Co(NH₃)₄Cl₂]⁺, the two Cl⁻ ligands are at 90° in the cis isomer and at 180° in the trans isomer.
Optical isomerism in octahedral complexes
Octahedral complexes with three bidentate ligands show optical isomerism. [Ni(H₂NCH₂CH₂NH₂)₃]²⁺ exists as two non-superimposable mirror images.
7. D-Orbital Splitting Diagram
The diagram below displays how the five degenerate 3d orbitals split into two distinct energy levels under the electrostatic influence of octahedral ligands:
Beyond the OxfordAQA specification: the labels t2g and eg, and which d-orbitals go up or down in energy, are not required. You only need to know that ligands split the d-orbitals into levels separated by ΔE.
8. Origin of Colour in Complex Ions
The distinctive, vibrant colours of transition metal complexes arise from electronic transitions between split d-orbitals:
- In an isolated gaseous transition metal atom or ion, all five 3d orbitals are degenerate (identical in energy).
- When ligands bond to the metal ion, the 3d orbitals no longer all have the same energy. Beyond the OxfordAQA specification: in an octahedral complex, ligand lone pairs repel electrons in the orbitals that point along the axes (dx²−y² and dz²) more strongly than those pointing between the axes (dxy, dyz, dxz).
- This unequal repulsion splits the 3d sub-level into two non-degenerate energy levels separated by an energy gap, ΔE.
- When white light shines on the complex, an electron absorbs a specific frequency of visible light and is promoted from the lower d-level to the higher d-level. The energy absorbed is governed by Planck's equation:
ΔE = hν = hc / λ - The frequencies of light that are NOT absorbed are transmitted or reflected through the solution. The human eye perceives the complementary colour to the absorbed wavelength (e.g. hydrated copper(II) absorbs red and orange light (most strongly near 800 nm) and transmits blue light, so it appears blue).
For a d-d transition to occur, there must be an electron in a lower d-orbital capable of promotion, AND a vacancy in a higher d-orbital to accept it.
- Sc³⁺ ([Ar] 3d⁰) has no 3d electrons to promote: colourless.
- Zn²⁺ ([Ar] 3d¹⁰) has all 3d orbitals completely filled; there is no vacant orbital to receive an excited electron: colourless.
9. Changing the Colour, and Colorimetry
What changes ΔE
A change in oxidation state, co-ordination number or ligand changes the size of ΔE. A different ΔE means a different frequency of light is absorbed, so the colour changes. Examples:
- Ligand: [Cu(H₂O)₆]²⁺ is blue, but [Cu(NH₃)₄(H₂O)₂]²⁺ is deep blue.
- Ligand and co-ordination number: [Co(H₂O)₆]²⁺ is pink, but [CoCl₄]²⁻ is blue.
- Oxidation state: the four oxidation states of vanadium each have a different colour (see Vanadium).
(h = 6.63 × 10⁻³⁴ J s; c = 3.00 × 10⁸ m s⁻¹; L = 6.02 × 10²³ mol⁻¹)
Step 1: ΔE = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (600 × 10⁻⁹) = 3.32 × 10⁻¹⁹ J
Step 2: per mole: 3.32 × 10⁻¹⁹ × 6.02 × 10²³ = 2.00 × 10⁵ J mol⁻¹ = 200 kJ mol⁻¹
Measuring concentration with a colorimeter
Visible light spectroscopy measures how much light a coloured solution absorbs. A colorimeter does this for one colour of light. The more concentrated the solution, the higher the absorbance.
- Choose a filter of the colour that is absorbed by the solution (the complementary colour to the colour you see).
- Make standard solutions of known concentration. Measure the absorbance of each, and plot a calibration curve of absorbance against concentration.
- Measure the absorbance of the unknown solution and read its concentration off the calibration curve.
- If the colour is pale, add a ligand that forms a more intensely coloured complex first.
10. Catalysts
Catalysts (spec 3.2.5.6) and variable oxidation states (spec 3.2.5.5) now have their own page: Transition Metals: Variable Oxidation States and Catalysts.
It covers heterogeneous catalysts (Contact process, Haber process, support media and poisoning), homogeneous catalysis by Fe²⁺, and autocatalysis by Mn²⁺.
11. Worked Calculations
Calculate the energy gap (ΔE) between the split d-orbitals in J and in kJ mol⁻¹.
(Planck constant h = 6.63 × 10⁻³⁴ J s; speed of light c = 3.00 × 10⁸ m s⁻¹; Avogadro constant L = 6.022 × 10²³ mol⁻¹).
Step 1: Calculate ΔE for a single photon:
ΔE = hc / λ
ΔE = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (5.00 × 10⁻⁷) = 3.978 × 10⁻¹⁹ J
Step 2: Convert to energy per mole (kJ mol⁻¹):
ΔEmolar = ΔE × L = (3.978 × 10⁻¹⁹ J) × (6.022 × 10²³ mol⁻¹) = 2.396 × 10⁵ J mol⁻¹
ΔEmolar = 2.396 × 10⁵ / 1000 = 240 kJ mol⁻¹
Step 1: Write the balanced equation:
[Ni(H₂O)₆]²⁺(aq) + 3en(aq) ⇌ [Ni(en)₃]²⁺(aq) + 6H₂O(l)
Step 2: Explain using thermodynamics:
4 particles on the reactant side (1 complex + 3 en) form 7 particles on the product side (1 complex + 6 water molecules). The substantial increase in particles in solution produces a large positive entropy change (ΔS > 0). Since ΔH is close to zero, ΔG = ΔH − TΔS becomes strongly negative, making the equilibrium position lie almost entirely to the right.
Go deeper
These UK AQA pages cover the same ideas in more detail. They may include content that is not in 9620.
Exam-Style Practice Questions
Test your understanding of these core syllabus concepts with targeted questions and detailed explanations.
Question 1: Which of the following elements is classified as a d-block element but NOT a transition metal?
Show Answer & Explanation
Correct Answer: D
Zinc forms only the Zn²⁺ ion, which has a full 3d¹⁰ sub-level. A transition metal must form at least one stable ion with a partially filled d sub-level (3d¹ to 3d⁹).
Question 2: Why does the ligand exchange of [Cr(H₂O)₆]³⁺ with EDTA⁴⁻ proceed with a high equilibrium constant?
Show Answer & Explanation
Correct Answer: B
This is the chelate effect: 1 [Cr(H₂O)₆]³⁺ + 1 EDTA⁴⁻ → 1 [Cr(EDTA)]⁻ + 6H₂O. An increase from 2 to 7 independent particles creates significant disorder, driving the reaction via positive ΔS.
Question 3: What is the coordination number and shape of the complex ion [CuCl₄]²⁻?
Show Answer & Explanation
Correct Answer: B
Chloride ligands (Cl⁻) are relatively large and carry a negative charge. Due to steric crowding and electrostatic repulsion, only four chloride ligands can fit around copper, forming a tetrahedral complex.
Question 4: In the reaction between peroxodisulfate ions and iodide ions, why can Fe²⁺ ions act as an effective homogeneous catalyst?
Show Answer & Explanation
Correct Answer: B
Fe²⁺ readily oxidises to Fe³⁺ by reducing S₂O₈²⁻, and Fe³⁺ subsequently oxidises I⁻ to I₂ while regenerating Fe²⁺. Alternating between Fe²⁺ and Fe³⁺ bypasses the high-repulsion barrier between like-charged anions.
Written Questions
Write your answer, then open the mark scheme. Each point marked (1) is worth one mark.
Question 5 (2 marks): Write an equation for the reaction of [Co(H₂O)₆]²⁺ with an excess of concentrated hydrochloric acid. State the colour change.
Show Answer & Mark Scheme
- [Co(H₂O)₆]²⁺ + 4Cl⁻ → [CoCl₄]²⁻ + 6H₂O (1)
- Pink to blue (1)
Question 6 (3 marks): [Co(NH₃)₄Cl₂]⁺ shows cis-trans isomerism. State the shape of the complex, and explain how the cis and trans isomers are different.
Show Answer & Mark Scheme
- Octahedral (1)
- Cis: the two Cl⁻ ligands are at 90° to each other (1)
- Trans: the two Cl⁻ ligands are at 180° to each other (opposite) (1)
Question 7 (2 marks): A complex absorbs light of wavelength 450 nm. Calculate the energy absorbed, in kJ mol⁻¹. (h = 6.63 × 10⁻³⁴ J s; c = 3.00 × 10⁸ m s⁻¹; L = 6.02 × 10²³ mol⁻¹)
Show Answer & Mark Scheme
- ΔE = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (450 × 10⁻⁹) = 4.42 × 10⁻¹⁹ J (1)
- × 6.02 × 10²³ = 266 kJ mol⁻¹ (1)
Question 8 (3 marks): Explain why carbon monoxide is toxic.
Show Answer & Mark Scheme
- CO forms a co-ordinate bond to the Fe(II) in haem (1)
- CO binds more strongly than O₂ and does not readily come off (1)
- So less O₂ is carried in the blood (1)