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
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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).
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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.
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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.