3.1.11Additional lessons
Colour in Transition Metal Complexes (HL) — IB Chemistry HL Revision Notes & Topical Questions
5 min read 5 parts Lesson 58 of 172
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Section 1 of 5Origin of Colour
d-orbital splitting
- d-orbital splitting Ligands approaching a metal ion repel its d electrons unequally and split the d sub-level into two sets of different energy.
Absorption
- Absorption A d electron in the lower set absorbs light with energy equal to the gap ΔE and moves into the higher set.
Colour seen
- Colour seen The complex transmits the unabsorbed wavelengths, so the observed colour is complementary to the one absorbed.
Relationship
- Relationship The energy gap is linked to frequency by ΔE = hν, where h is Planck's constant.
No colour
- No colour Ions with d⁰ or d¹⁰ configurations, such as Sc³⁺ and Zn²⁺, cannot undergo d-d transitions and are colourless.
Section 2 of 5Factors Affecting Colour
1
A different metal ion changes the nuclear charge and the d electron count, so it changes the size of ΔE.
2
A change of oxidation state of the same metal changes the splitting, so Fe²⁺ and Fe³⁺ differ in colour.
3
Different ligands split the d sub-level by different amounts, as shown by the spectrochemical series.
4
Strong-field ligands such as CN⁻ give large ΔE, while weak-field ligands such as Cl⁻ give small ΔE.
5
The shape of the complex, such as octahedral versus tetrahedral, also changes the splitting pattern and the colour.
| Complex | Colour |
|---|---|
| [Cu(H₂O)₆]²⁺ | Pale blue |
| [Cu(NH₃)₄(H₂O)₂]²⁺ | Deep blue |
| [CuCl₄]²⁻ | Yellow-green |
Section 3 of 5Interpreting Spectra
Interpreting Spectra
Absorption spectrum
- Absorption spectrum A graph of absorbance against wavelength shows a peak at the wavelength absorbed.
Colour wheel
- Colour wheel The absorbed colour and the observed colour sit on opposite sides of the colour wheel.
Ligand order
- Ligand order Spectrochemical series: I⁻ < Br⁻ < Cl⁻ < OH⁻ < H₂O < NH₃ < CN⁻, in order of increasing splitting.
Colorimetry
- Colorimetry Absorbance at a chosen wavelength is proportional to concentration, so colour can be used for quantitative analysis.
Section 4 of 5Worked Example
Question
- A complex absorbs light of wavelength 600 nm.
- Calculate the energy gap per mole in kJ mol⁻¹ using h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹ and Nₐ = 6.02 × 10²³ mol⁻¹.
- ΔE = hc ÷ λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (600 × 10⁻⁹).
- ΔE = 3.315 × 10⁻¹⁹ J per ion.
- Per mole3.315 × 10⁻¹⁹ × 6.02 × 10²³ = 1.995 × 10⁵ J mol⁻¹.
- Divide by 1000 to give kJ mol⁻¹.
Answer ΔE ≈ 200 kJ mol⁻¹
Section 5 of 5Exam Checklist & Tips
3.1.11 – Key Points for the Exam
- d-d transition Movement of an electron between split d orbitals on absorbing light.
- Ligand field splitting The energy difference ΔE between the two sets of d orbitals in a complex.
- Complementary colour The colour seen when another part of the visible spectrum is absorbed.
- Spectrochemical series Ligands arranged by the size of splitting they cause.
Common Mistakes
- Saying that the complex emits the colour we see rather than transmitting or reflecting it.
- Stating that the observed colour is the same as the colour absorbed.
- Assuming all d-block ions are coloured, including Zn²⁺.
- Forgetting that ligand identity affects colour as much as the metal.
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