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5.2.3Unit 5. Transition Metals and Organic Nitrogen Chemistry

Colour in Aqueous ions — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions

5 min read 5 parts Lesson 15 of 43 Unit quiz

📚 IAL Edexcel Linear (Chemistry) Hub / Topical Revision
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Part 1 of 5The Origin of Colour

5.2.3 Colour in Aqueous ions
Splitting
  • In a complex the ligands split the five d orbitals into two groups with a small energy gap between them.
Absorption
  • An electron can be promoted from the lower to the higher group by absorbing light of exactly matching energy.
Transmitted colour
  • The colour seen is made of the wavelengths not absorbed, which is the complementary colour of the light removed.
Requirement
  • The d subshell must be partially filled, so there is a vacancy in the upper group to receive the electron.

Part 2 of 5Energy Relationships

Energy Relationships
Equation
  • ΔE = hν = hc/λ, where h is the Planck constant and ν is the frequency of the absorbed light.
Larger gap
  • A bigger energy gap means light of higher frequency and shorter wavelength is absorbed.
Colourless ions
  • Sc³⁺ (3d⁰) and Zn²⁺ (3d¹⁰) have no possible d to d transition, so their solutions are colourless.

Part 3 of 5What Changes the Colour

1
The identity of the metal affects the splitting, so Cu²⁺ and Co²⁺ ions with the same ligand give different colours.
2
The oxidation state alters the gap, as shown by Fe²⁺(aq) being pale green and Fe³⁺(aq) being yellow-brown.
3
The type of ligand changes the splitting: [Cu(H₂O)₆]²⁺ is pale blue while [Cu(NH₃)₄(H₂O)₂]²⁺ is deep blue.
4
The coordination number and shape matter, since [CuCl₄]²⁻ is yellow-green compared with the blue aqua ion.
5
A colorimeter can measure absorbance at the wavelength of maximum absorption to determine concentration.

Part 4 of 5Worked Example

Question

  • A complex absorbs light of wavelength 600 nm.
  • Calculate ΔE in J per ion and in kJ mol⁻¹.
  • (h = 6.63 x 10⁻³⁴ J s, c = 3.00 x 10⁸ m s⁻¹, L = 6.02 x 10²³ mol⁻¹.)

  1. λ = 600 x 10⁻⁹ m = 6.00 x 10⁻⁷ m.
  2. ΔE = hc/λ = (6.63 x 10⁻³⁴ x 3.00 x 10⁸) / 6.00 x 10⁻⁷ = 3.32 x 10⁻¹⁹ J.
  3. Per mole3.32 x 10⁻¹⁹ x 6.02 x 10²³ = 1.99 x 10⁵ J mol⁻¹.

Answer ΔE = 3.32 x 10⁻¹⁹ J per ion, about 1.99 x 10² kJ mol⁻¹

Part 5 of 5Exam Checklist & Tips

5.2.3 – Key Points for the Exam

  • d orbital splitting Separation of d orbitals into two energy levels by the ligands.
  • Complementary colour The colour seen when other wavelengths of visible light are absorbed.
  • Planck constant 6.63 x 10⁻³⁴ J s, linking energy and frequency.
  • Absorbance A measure of how much light a solution absorbs.

Common Mistakes

  • Saying that the colour seen is the colour absorbed.
  • Claiming Zn²⁺ is coloured because zinc is a d-block element.
  • Forgetting to convert wavelength from nm to m.
  • Attributing colour to electrons moving between ligands rather than d orbitals.

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Chemistry 0620 & Edexcel OL Product Prediction Rules

Aqueous Electrolysis Rules (Inert Electrodes)

  • At Cathode (-): Least reactive cation discharged. If metal is more reactive than hydrogen (Na+, K+, Mg2+), H2 gas is produced: 2H+ + 2e- → H2. If Cu2+ or Ag+, the metal coats the cathode.
  • At Anode (+): If concentrated halide present (Cl-, Br-, I-), halogen is formed (e.g. 2Cl- → Cl2 + 2e-). If sulfate (SO42-) or nitrate (NO3-), O2 gas is discharged: 4OH- → O2 + 2H2O + 4e-.