In simple terms
A friendly intro before the formal notes — no formulas yet.
The Rainbow of the d-block
Transition metal complexes are colourful because their d-orbitals are split into different energy levels by ligands. Electrons absorb specific colours (energies) of light to jump this energy gap, and we see the remaining colours that are transmitted.
Imagine a set of five identical bells. On their own, they all ring with the same note. Now, imagine you surround them with cushions. The cushions muffle some bells more than others, making them ring at different pitches. Similarly, ligands 'muffle' the d-orbitals, splitting their energy levels. An electron can then absorb a specific energy (a colour of light) to 'jump' from a lower-energy orbital to a higher-energy one. The colours not absorbed are what we see.
- 1
Split 3d subshell in ligand field — d–d transitions absorb light. | Sim hint: Complementary colour observed.
- 2
Different ligands → different ΔE → different colours. | Sim hint: Compare [Cu(H₂O)₆]²⁺ blue vs [CuCl₄]²⁻ yellow-green.
- 3
Degenerate d orbitals split into t₂g and e_g (octahedral). | Sim hint: Electron jumps between split levels.
- 4
Colourless: d⁰ or d¹⁰ — no d–d transitions (Sc³⁺, Zn²⁺). | Sim hint: Check d electron count.
Explore the concept
Use the live diagram and synced steps — play it or tap a step card to walk through.
Key formulas
Tap any symbol to reveal exactly what it means and its units.
Full topic notes
Formal explanation with the rigour you need for the exam.
The Origin of Colour: d-orbital Splitting
In an isolated, gaseous transition metal ion, the five d-orbitals are described as degenerate, meaning they all have exactly the same energy. However, in a complex, the central metal ion is surrounded by ligands. The lone pairs of electrons on the ligands repel the electrons in the d-orbitals of the metal ion. This repulsion is not uniform across all five d-orbitals because they have different spatial orientations. As a result, the d-orbitals split into two or more distinct energy levels. This phenomenon is known as crystal field splitting.
Splitting in Octahedral Complexes
In an octahedral complex, six ligands are positioned along the x, y, and z axes. The d-orbitals that point directly along these axes ( and ) experience the greatest repulsion from the ligand lone pairs and are raised to a higher energy level. This higher energy pair is called the e_g set. The three d-orbitals that point between the axes (, , and ) experience less repulsion and are stabilised at a lower energy level. This lower energy trio is called the t₂g set.
Degenerate d-orbitals: All five have the same energy in a free ion.
Octahedral field: Ligands approach along the axes.
e_g orbitals: and point at ligands, become higher energy.
t₂g orbitals: , , and point between ligands, become lower energy.
d-d Transitions and Absorption of Light
The energy difference between the t₂g and e_g sets is called the crystal field splitting energy, denoted as ΔE. When white light (which contains all wavelengths of the visible spectrum) passes through a solution of a transition metal complex, the complex can absorb a photon of light. If the energy of this photon exactly matches ΔE, it will be absorbed, and an electron will be promoted from a lower-energy t₂g orbital to a higher-energy e_g orbital. This process is called a d-d transition.
Here, is Planck's constant ( J s), is the speed of light ( m s⁻¹), is the frequency of light absorbed, and is its wavelength. The light that is not absorbed is transmitted through the solution and detected by our eyes. The colour we perceive is therefore the complementary colour to the one that was absorbed.
Factors Affecting ΔE and Colour
The magnitude of ΔE, and therefore the colour of the complex, is influenced by three main factors:
Identity of the Metal Ion: Different metals have different nuclear charges and d-electron counts, which alters the repulsion and thus ΔE.
Oxidation State of the Metal: A higher oxidation state increases the attraction for ligands, pulling them closer. This increases the repulsion between ligands and d-orbitals, leading to a larger ΔE. For example, [Fe(H₂O)₆]²⁺ (pale green) has a smaller ΔE than [Fe(H₂O)₆]³⁺ (yellow/brown).
Identity of the Ligand: Ligands vary in their ability to split the d-orbitals. The spectrochemical series ranks ligands from weak-field (small ΔE) to strong-field (large ΔE). A small part of the series is: Cl⁻ < H₂O < NH₃ < CN⁻. A larger ΔE means higher frequency (shorter wavelength) light is absorbed.
Remember the key relationship: Stronger field ligand → Larger ΔE → Higher frequency (shorter wavelength) light absorbed. This is a common source of exam questions, especially when explaining colour changes during ligand substitution reactions.
When are Complexes Colourless?
For a complex to be coloured, a d-d transition must be possible. This requires a partially filled d-subshell. If the d-subshell is completely full or completely empty, no d-d transition can occur, and the complex will be colourless.
d⁰ configuration: The metal ion has no d-electrons to be promoted. Examples include Sc³⁺ ([Ar] 3d⁰) and Ti⁴⁺ ([Ar] 3d⁰). Their complexes are colourless.
d¹⁰ configuration: The d-subshell is completely full. There are no vacant d-orbitals for an electron to be promoted into. Examples include Zn²⁺ ([Ar] 3d¹⁰) and Cu⁺ ([Ar] 3d¹⁰). Their complexes are colourless.
A common trap involves copper. Copper(II) compounds (Cu²⁺, d⁹) are almost always coloured (usually blue or green). However, copper(I) compounds (Cu⁺, d¹⁰) are colourless. Pay close attention to the oxidation state.
Worked examples
See the formulas applied — reveal one step at a time, like the exam.
The aqueous hexaaquacopper(II) ion, [Cu(H₂O)₆]²⁺, appears blue. It absorbs light with a maximum absorbance at a wavelength (λ) of 600 nm. Calculate the crystal field splitting energy, ΔE, in kJ mol⁻¹ for one mole of these ions. (Avogadro constant, mol⁻¹).
- 1
Calculate ΔE for one ion:
An aqueous solution of copper(II) sulfate is blue due to the [Cu(H₂O)₆]²⁺ ion. When concentrated hydrochloric acid is added, the solution turns a yellow-green colour as the [CuCl₄]²⁻ ion is formed. Explain this colour change with reference to the spectrochemical series and d-orbital splitting.
- 1
Identify ligands and complexes: The initial complex is [Cu(H₂O)₆]²⁺ with H₂O ligands. The final complex is [CuCl₄]²⁻ with Cl⁻ ligands.
How it all connects
The big idea sits in the middle — tap a linked idea to explore the link.
Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.
Glossary
Try to recall each definition before you reveal it.
Quick check
Answer in your head first — then tap to check. No pressure.
Revision flashcards
Flip the card. Test yourself before the exam.
What are degenerate orbitals?
Orbitals that have the same energy level. In an isolated gaseous transition metal ion, all five d-orbitals are degenerate.
Key takeaways
Review these before you close the topic — retrieval beats re-reading.
- ✓
Degenerate d-orbitals: All five have the same energy in a free ion.
- ✓
Octahedral field: Ligands approach along the axes.
- ✓
e_g orbitals: and point at ligands, become higher energy.
- ✓
t₂g orbitals: , , and point between ligands, become lower energy.
Practice — then mark it
The whole point: a real Cambridge question, marked mark-by-mark.
Practice Questions: Colour of Complexes
Practice Questions: Colour of Complexes
Extra simulations & links
PhET, GeoGebra and other curated tools — open in a new tab.
Frequently asked
Checkpoint
One marked question is worth ten re-reads — close the loop before you move on.
Reading it isn’t knowing it — prove it.
Before you move on: do Practice Questions: Colour of Complexes on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.