The big idea: In an isolated transition-metal ion the five d orbitals all have the same energy (they are degenerate).
When ligands bond to the metal ion to form a complex, they push on those d orbitals unevenly. This raises some d orbitals in energy and lowers others, so the five d orbitals split into two energy levels separated by a gap.
That energy gap is called the splitting energy, Δ (delta). The whole of this micro comes down to the size of Δ.
Define the terms: - Ligand — a molecule or ion (e.g. H2O, NH3, Cl⁻, CN⁻) that bonds to a central metal ion by donating a lone pair of electrons. - Complex (ion) — a central metal ion surrounded by ligands, e.g. [Cu(H2O)6]2+. - d-orbital splitting — the breaking of the five d orbitals into a lower set and an upper set when ligands approach. - Δ (splitting energy) — the energy gap between the two split d-orbital levels.
| Energy level | Which d orbitals | Relative energy |
|---|---|---|
| upper set | the higher-energy d orbitals (e.g. 2 of them in an octahedral complex) | higher — top of the gap Δ |
| the gap | no orbitals — this is the splitting energy Δ (delta) | the size of Δ sets the colour |
| lower set | the lower-energy d orbitals (e.g. 3 of them in an octahedral complex) | lower — bottom of the gap Δ |
The five d orbitals split into a lower set and an upper set; the gap is the splitting energy Δ. A d electron absorbs a photon of energy Δ and jumps to the upper set (a d-d transition). Shown for the d⁹ ion Cu²⁺.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
Free preview
This is the free notes preview
You're reading the free notes. Aimnova Pro unlocks the full study experience — and you can try it with your first topic free to keep:
- FlashcardsLock in vocabulary and key terms with spaced repetition.
- Practice questionsAnswer exam-style questions and get instant AI marking.
- Mock exams & past-paper vaultSit full mocks and see exactly how examiners award marks.
- Personalised study planA daily plan built around your exam date and weak areas.
Because the d orbitals are now at two energy levels, a d electron can absorb a photon of light and jump from the lower level to the upper level. This is a d-d transition.
The photon must match Δ: An electron can only make the jump if the photon's energy is exactly equal to Δ.
For many transition-metal complexes, Δ happens to match the energy of visible light — so the complex absorbs part of the visible spectrum. That is why so many transition-metal compounds are coloured.
The visible spectrum (≈ 400–700 nm). A coloured complex absorbs one band of this light; the colour you SEE is what is left over (the complement).
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
You see the COMPLEMENT of what is absorbed: White light contains all the visible colours. The complex removes (absorbs) the band that matches Δ; the colour you see is the left-over light — the complementary colour of the absorbed band.
Use the colour wheel: the colour seen is the one opposite the absorbed colour. For example, a complex that absorbs red light looks green; one that absorbs blue light looks orange.
| Colour ABSORBED | Approx. wavelength absorbed | Colour SEEN (complement) |
|---|---|---|
| red | ≈ 700 nm | green |
| orange | ≈ 620 nm | blue |
| yellow | ≈ 580 nm | violet |
| green | ≈ 530 nm | red |
| blue | ≈ 470 nm | orange |
| violet | ≈ 410 nm | yellow |
IB-style question — why a complex is coloured
Explain why an aqueous solution of [Cu(H2O)6]2+ is coloured. [3]
How to score the marks
- Mark 1 — splitting. The water ligands split the five d orbitals of the Cu2+ ion into two energy levels separated by a gap Δ.
- Mark 2 — d-d transition. A d electron absorbs a photon of visible light whose energy equals Δ and is promoted from the lower level to the upper level (a d-d transition).
- Mark 3 — complement. The complex absorbs red/orange light, so the colour seen is the complementary colour — blue — i.e. the light that is not absorbed.
Final answer
Ligands split the d orbitals by Δ → a d electron absorbs visible light of energy Δ (d-d transition) → the colour seen is the complement of the absorbed light.
Stop wasting time on topics you know
Our AI identifies your weak areas and focuses your study time where it matters. No more overstudying easy topics.
Change the size of Δ and you change the colour, because a different Δ matches a different band of visible light. Three things control Δ:
1. The metal and its oxidation state: A different metal, or the same metal in a different oxidation state, gives a different Δ.
A higher oxidation state pulls the ligands in more tightly, which usually increases Δ — so the colour changes. For example Fe2+ and Fe3+ complexes are different colours.
2. The identity of the ligand (spectrochemical series): Different ligands split the d orbitals by different amounts. Ranking ligands by the size of Δ they cause gives the spectrochemical series:
weak-field (small Δ): I⁻ < Br⁻ < Cl⁻ < F⁻ < H2O < NH3 < CN⁻ ≈ CO (strong-field, large Δ).
Swapping a weak-field ligand for a strong-field one increases Δ — the complex then absorbs shorter-wavelength (higher-energy) light, and its colour shifts.
| Ligand (weak → strong field) | Effect on Δ |
|---|---|
| I⁻ < Br⁻ < Cl⁻ < F⁻ | small Δ (weak-field ligands) |
| H2O < NH3 | medium Δ |
| CN⁻, CO | large Δ (strong-field ligands) |
3. The number and geometry of the ligands: How many ligands there are and how they are arranged (the geometry — e.g. octahedral with 6 ligands vs tetrahedral with 4) changes the pattern and size of the splitting. A tetrahedral complex, for instance, has a smaller Δ than the equivalent octahedral one, so it absorbs different light and shows a different colour.
Same metal, different ligand → different colour: A neat illustration: adding concentrated ammonia to pale-blue [Cu(H2O)6]2+ replaces water with the stronger-field ligand NH3, giving a deep-blue ammine complex. NH3 causes a larger Δ, so a different band is absorbed and the colour deepens.
How this is tested: This HL micro is examined two ways:
- Paper 1A (MCQ): match an absorbed colour/wavelength to the observed colour using complements, or rank ligands by Δ from the spectrochemical series. - Paper 2: an explain mark — why a named complex is coloured (splitting → d-d transition → complement) — or a predict mark — how the colour changes when the ligand or oxidation state changes.
For the explain mark you must link Δ → absorption of visible light → complementary colour seen, not just say 'd electrons move'.
Relate Δ to the wavelength absorbed: A larger Δ needs a higher-energy photon, which is shorter-wavelength light (E and λ are inversely related). So strong-field ligands → larger Δ → shorter wavelength absorbed → the observed colour shifts. State this direction explicitly to earn the predict mark.
IB-style question — changing the ligand (a)
(a) When the water ligands in a pale-coloured aqua complex are replaced by cyanide ions, CN⁻, the splitting energy Δ increases. State and explain the effect on the wavelength of light absorbed. [2]
How to score the marks
- Mark 1 — the effect. The wavelength of light absorbed decreases (a shorter wavelength is absorbed).
- Mark 2 — the reason. CN⁻ is a strong-field ligand, so Δ is larger; a larger Δ needs a higher-energy photon, and higher energy means shorter wavelength.
Final answer
Shorter wavelength absorbed — CN⁻ is a strong-field ligand → larger Δ → higher-energy (shorter-λ) photon needed.
IB-style question — changing the oxidation state (b)
(b) A solution containing the Mn2+ complex is very pale, but on oxidation to a Mn3+ complex it becomes more strongly coloured. Suggest, in terms of Δ, why changing the oxidation state changes the colour. [2]
How to score the marks
- Mark 1 — Δ changes. Changing the oxidation state changes the charge on the metal ion, which changes how strongly it pulls the ligands and therefore changes the size of Δ.
- Mark 2 — colour changes. A different Δ means a different band of visible light matches the d-d transition, so a different colour is absorbed and the complementary colour seen changes.
Final answer
Different oxidation state → different Δ → a different wavelength of visible light is absorbed → a different complementary colour is seen.