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9701 · 28.3

Colour of complexes — practice questions

Practice and worked examples for 9701 Colour of complexes. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

The aqueous hexaaquacopper(II) ion, [Cu(H₂O)₆]²⁺, appears blue. It absorbs light with a maximum absorbance at a wavelength (λ) of 600 nm. Calculate the crystal field splitting energy, ΔE, in kJ mol⁻¹ for one mole of these ions. (Avogadro constant, L=6.02×1023L = 6.02 \times 10^{23} mol⁻¹).

Show solution outline
  1. Calculate ΔE for one ion: First, convert wavelength to metres: 600 nm=600×10−9 m600 \text{ nm} = 600 \times 10^{-9} \text{ m}. Use the formula ΔE=hc/λ\Delta E = hc/\lambda. ΔE=(6.63×10−34 J s)×(3.00×108 m s−1)600×10−9 m\Delta E = \frac{(6.63 \times 10^{-34} \text{ J s}) \times (3.00 \times 10^8 \text{ m s}^{-1})}{600 \times 10^{-9} \text{ m}} ΔE=3.315×10−19 J\Delta E = 3.315 \times 10^{-19} \text{ J} per ion.
  2. Calculate ΔE for one mole of ions: Multiply the energy per ion by the Avogadro constant, LL. ΔEmol=(3.315×10−19 J)×(6.02×1023 mol−1)\Delta E_{mol} = (3.315 \times 10^{-19} \text{ J}) \times (6.02 \times 10^{23} \text{ mol}^{-1}) ΔEmol=199563 J mol−1\Delta E_{mol} = 199563 \text{ J mol}^{-1}
  3. Convert to kJ mol⁻¹: Divide by 1000. ΔEmol=1995631000=199.6 kJ mol−1\Delta E_{mol} = \frac{199563}{1000} = 199.6 \text{ kJ mol}^{-1} Answer: 199.6 kJ mol⁻¹ (to 4 s.f.) or 200 kJ mol⁻¹ (to 3 s.f.)

Worked example 2

An aqueous solution of copper(II) sulfate is blue due to the [Cu(H₂O)₆]²⁺ ion. When concentrated hydrochloric acid is added, the solution turns a yellow-green colour as the [CuCl₄]²⁻ ion is formed. Explain this colour change with reference to the spectrochemical series and d-orbital splitting.

Show solution outline
  1. Identify ligands and complexes: The initial complex is [Cu(H₂O)₆]²⁺ with H₂O ligands. The final complex is [CuCl₄]²⁻ with Cl⁻ ligands.
  2. Compare ligand strength: According to the spectrochemical series, Cl⁻ is a weaker field ligand than H₂O.
  3. Relate ligand strength to ΔE: A weaker field ligand causes a smaller crystal field splitting energy (ΔE). Therefore, ΔE for [CuCl₄]²⁻ is smaller than ΔE for [Cu(H₂O)₆]²⁺.
  4. Relate ΔE to absorbed light: Since ΔE=hc/λ\Delta E = hc/\lambda, a smaller ΔE corresponds to the absorption of light with a longer wavelength (lower energy).
  5. Explain the colour change: [Cu(H₂O)₆]²⁺ absorbs orange light (shorter λ) and appears blue. [CuCl₄]²⁻ absorbs lower energy, longer wavelength light (e.g., blue/violet light) and so the transmitted light appears yellow-green.
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