You are viewing a free preview of this lesson.
Subscribe to unlock all 10 lessons in this course and every other course on LearningBro.
One of the most visually striking properties of transition metal chemistry is colour. Solutions of copper(II) sulfate are blue, potassium permanganate is deep purple, and chromium(III) compounds are green. This lesson explains why transition metal complexes are coloured and what factors influence the colour observed.
In an isolated transition metal ion, all five d orbitals have the same energy — they are said to be degenerate. However, when ligands surround the metal ion to form a complex, the ligands' lone pairs create an electric field that interacts differently with the different d orbitals.
Some d orbitals point directly at the ligands and experience greater repulsion, raising their energy. Others point between the ligands and experience less repulsion. This creates an energy gap (ΔE) between two sets of d orbitals — the d orbitals have been split.
In an octahedral complex:
The energy difference between these two sets is called the crystal field splitting energy (Δoct or ΔE).
In a tetrahedral complex, the splitting is reversed (the three d(xy), d(xz), d(yz) orbitals are raised) and the splitting energy (Δtet) is smaller — typically about 4/9 of Δoct. This is why tetrahedral complexes often absorb different wavelengths from octahedral complexes of the same metal.
When white light passes through a solution of a transition metal complex, photons with energy exactly equal to ΔE are absorbed. This energy promotes an electron from a lower-energy d orbital to a higher-energy one — this is called a d-d transition.
The energy of the absorbed photon corresponds to a specific wavelength of visible light:
ΔE = hν = hc/λ
where h is Planck's constant, ν is frequency, c is the speed of light, and λ is wavelength.
The colour we see is the complementary colour of the wavelength absorbed. If a complex absorbs red light, it appears blue-green. If it absorbs blue light, it appears orange.
A colour wheel shows complementary colour pairs:
| Colour Absorbed | Approximate λ / nm | Complementary Colour (Observed) |
|---|---|---|
| Violet | 380–430 | Yellow |
| Blue | 430–480 | Orange |
| Blue-green | 480–500 | Red |
| Green | 500–560 | Magenta / red-purple |
| Yellow | 560–580 | Violet |
| Orange | 580–620 | Blue |
| Red | 620–700 | Blue-green / cyan |
For example:
flowchart LR
subgraph "Colour Wheel – Complementary Pairs"
A["Red<br>(620–700 nm)"] ---|"Complementary"| B["Cyan/Blue-green"]
C["Orange<br>(580–620 nm)"] ---|"Complementary"| D["Blue"]
E["Yellow<br>(560–580 nm)"] ---|"Complementary"| F["Violet"]
G["Green<br>(500–560 nm)"] ---|"Complementary"| H["Magenta"]
end
This table is essential for exam preparation. Memorise the colours and the associated ligand/geometry:
| Complex | Metal (Ox. State) | Geometry | Colour |
|---|---|---|---|
| [Fe(H₂O)₆]²⁺ | Fe(+2) | Octahedral | Pale green |
| [Fe(H₂O)₆]³⁺ | Fe(+3) | Octahedral | Yellow/brown |
| [Co(H₂O)₆]²⁺ | Co(+2) | Octahedral | Pink |
| [CoCl₄]²⁻ | Co(+2) | Tetrahedral | Blue |
| [Ni(H₂O)₆]²⁺ | Ni(+2) | Octahedral | Green |
| [Cu(H₂O)₆]²⁺ | Cu(+2) | Octahedral | Blue |
| [Cu(NH₃)₄(H₂O)₂]²⁺ | Cu(+2) | Octahedral | Deep blue |
| [CuCl₄]²⁻ | Cu(+2) | Tetrahedral | Yellow-green |
| [Cr(H₂O)₆]³⁺ | Cr(+3) | Octahedral | Green/violet |
| [Cr(NH₃)₆]³⁺ | Cr(+3) | Octahedral | Yellow |
| MnO₄⁻ | Mn(+7) | Tetrahedral | Deep purple |
| CrO₄²⁻ | Cr(+6) | Tetrahedral | Yellow |
| Cr₂O₇²⁻ | Cr(+6) | — | Orange |
The colour of a transition metal complex depends on the size of ΔE, which is influenced by several factors:
Different metals have different numbers of d electrons and different nuclear charges, giving different ΔE values. Even with the same ligand (H₂O) and the same geometry (octahedral), each transition metal gives a different colour.
Changing the oxidation state changes the number of d electrons and the charge density of the ion, both of which affect ΔE.
Different ligands cause different amounts of d-orbital splitting. This is summarised by the spectrochemical series, which lists ligands in order of increasing splitting:
I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < CN⁻ < CO
Weak-field ligands (left side) cause small splitting → absorb low-energy (long-wavelength) light. Strong-field ligands (right side) cause large splitting → absorb high-energy (short-wavelength) light.
For example, changing the ligands around Cr³⁺:
Tetrahedral complexes generally have a smaller splitting (Δtet ≈ 4/9 Δoct) than octahedral complexes. This affects which wavelengths are absorbed:
Common exam mistake: Students sometimes say "the colour changes because a different number of ligands are present." While true that the coordination number changes, the reason the colour changes is that the d-orbital splitting energy ΔE changes — always link back to ΔE and d-d transitions.
Complexes of Sc³⁺ (3d⁰) and Zn²⁺ (3d¹⁰) are colourless because:
In both cases, d-d transitions are impossible, so no visible light is absorbed and the compounds are colourless.
Cu⁺ is also colourless: Cu⁺ has the configuration [Ar] 3d¹⁰ (full d sub-shell), so its complexes are colourless. Cu²⁺ (3d⁹) has an incomplete d sub-shell and forms coloured complexes.
The colour of transition metal complexes has practical applications in analysis:
To use colourimetry effectively:
Question: Concentrated HCl is added to a blue CuSO₄ solution. The solution turns yellow-green. Explain the colour change.
Answer:
Transition metal complexes are coloured because d-orbital splitting allows d-d transitions that absorb specific wavelengths of visible light. The observed colour is the complementary colour of the absorbed light. The colour depends on the metal, its oxidation state, the ligand (spectrochemical series), and the geometry of the complex. Sc³⁺ and Zn²⁺ complexes are colourless because d-d transitions cannot occur. Learn the reference table of complex colours — it appears in almost every inorganic exam paper.
Edexcel 9CH0 specification Topic 15 — Transition Metals, sub-topic 15.3 covers d-orbital splitting in octahedral and tetrahedral ligand fields, the energy gap Δ_oct (Δ_t = ⁴⁄₉ Δ_oct), the origin of colour through electron promotion (d–d transitions), the relationship E = hf = hc/λ between absorbed wavelength and observed (complementary) colour, and how factors (ligand identity, oxidation state, coordination number) affect Δ and therefore colour (refer to the official specification document for exact wording). Examined in Paper 1 (9CH0/01) and Paper 3 (9CH0/03). Synoptic links: Topic 1 (Atomic Structure) for d-orbital identity, Topic 2 (Bonding) for shape, Topic 15.2 for ligand exchange, and to UV-vis spectroscopy concepts (calculation of ε via Beer-Lambert law occasionally).
Question (8 marks):
(a) Explain why [Cu(H₂O)₆]²⁺ is pale blue but [Cu(NH₃)₄(H₂O)₂]²⁺ is deep royal blue, in terms of d-orbital splitting and the visible spectrum. (5)
(b) Predict and explain the colour change observed when conc. HCl is added to a Cu²⁺ aqueous solution. (3)
Solution with mark scheme:
(a) Step 1 — d-orbital splitting in octahedral field.
The five degenerate d orbitals of Cu²⁺ split in an octahedral ligand field into a lower set t₂g (d_xy, d_xz, d_yz) and an upper set e_g (d_x²−y², d_z²) separated by Δ_oct.
B1 — splitting diagram or t₂g/e_g labels.
Step 2 — d–d transition.
For d⁹ Cu²⁺, the configuration is t₂g⁶ e_g³ (one hole in e_g). A photon of energy E = Δ_oct can promote an electron from t₂g to e_g; the wavelength of light absorbed is λ = hc/Δ_oct.
M1 — d–d transition mechanism described, E = hc/λ used or stated.
Step 3 — spectrochemical series.
NH₃ is higher in the spectrochemical series than H₂O, so NH₃ produces a larger Δ_oct. The wavelength absorbed is therefore shorter (higher energy). The complementary colour shifts accordingly.
M1 — spectrochemical comparison correct.
Step 4 — colour outcome.
[Cu(H₂O)₆]²⁺ absorbs in the orange/red region (~800 nm), so the transmitted/observed colour is pale blue.
[Cu(NH₃)₄(H₂O)₂]²⁺ has 4 NH₃ + 2 H₂O ligands; the average Δ is larger, absorption shifts to ~600 nm (yellow/orange), so the complementary colour is deeper, more intense blue (more of the red/orange absorbed, less of the blue).
A1 — explicit complementary-colour reasoning.
A1 — quantitative colour shift (absorbed wavelength shorter → observed colour darker / more intense blue).
(b) Adding conc. HCl drives the equilibrium [Cu(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CuCl₄]²⁻ + 6H₂O. [CuCl₄]²⁻ is tetrahedral with Δ_t = ⁴⁄₉ Δ_oct (smaller). Cl⁻ is also a weaker field ligand than H₂O. Both effects push absorption to longer wavelength (lower energy); λ_max shifts from ~800 nm into the near-IR / longer visible. The complementary colour observed is yellow / yellow-green.
M1 — tetrahedral identified, Δ_t < Δ_oct stated.
M1 — Cl⁻ weaker-field stated.
A1 — observed colour yellow / yellow-green from longer-wavelength absorption.
Total: 8 marks (M3 A3 B1 + 1 description).
Question (6 marks): A solution of Cr(III) is treated successively with H₂O, NH₃, and CN⁻.
(a) Explain how the colour changes through this sequence in terms of the spectrochemical series. (3)
(b) [V(H₂O)₆]²⁺ is purple-violet; [V(H₂O)₆]³⁺ is green. Discuss how oxidation state affects observed colour. (3)
Mark scheme decomposition by AO:
Subscribe to continue reading
Get full access to this lesson and all 10 lessons in this course.