Formation of coloured ionsAQA A-Level Chemistry: Revision notes
Section 1
Why transition metal ions are coloured
Transition metal ions can be identified by their colour, for example blue Cu²⁺(aq), pale green Fe²⁺(aq) and pink Co²⁺(aq). White light contains all visible wavelengths. Colour arises when some wavelengths of visible light are absorbed and the remaining wavelengths are transmitted or reflected. The colour seen is the mixture of the light that is not absorbed.
A solution that appears blue has absorbed light at the red and orange end of the spectrum.
Do not say the ion absorbs the colour you see. It absorbs other wavelengths, and the light left over is the colour you see.
Section 2
Electron promotion and ΔE
In a transition metal complex the d orbitals are not all at the same energy. When light is absorbed, d electrons move from the ground state to an excited state (a higher d energy level). The energy absorbed equals the energy difference:
ΔE = hν = hc/λ
where h is the Planck constant, ν the frequency, c the speed of light and λ the wavelength. A larger ΔE means a higher frequency and shorter wavelength absorbed.
Worked example: light of wavelength 510 nm: ΔE = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (510 × 10⁻⁹) = 3.90 × 10⁻¹⁹ J. Multiply by L = 6.02 × 10²³ mol⁻¹ for 235 kJ mol⁻¹.
Convert wavelength from nm to m (× 10⁻⁹) before using ΔE = hc/λ. Multiply by L and divide by 1000 to get kJ mol⁻¹.
Section 3
Why some ions are colourless
If the d sub-level is empty (Sc³⁺) or full (Zn²⁺, Cu⁺, [Ar]3d¹⁰), no d–d electron promotion is possible. No visible light is absorbed, so the compounds are white or colourless. Copper(II) sulfate is blue because Cu²⁺ is [Ar]3d⁹.
Section 4
What changes the colour
Changes in oxidation state, co-ordination number and ligand all alter ΔE and so change which wavelengths are absorbed:
- Oxidation state: Fe²⁺(aq) is pale green; Fe³⁺(aq) is yellow-brown
- Co-ordination number and ligand: pink [Co(H₂O)₆]²⁺ becomes blue [CoCl₄]²⁻
- Ligand: pale blue [Cu(H₂O)₆]²⁺ becomes deep blue [Cu(NH₃)₄(H₂O)₂]²⁺
Worked example: [Cu(H₂O)₆]²⁺ absorbs at 800 nm, ΔE = 2.49 × 10⁻¹⁹ J (150 kJ mol⁻¹). The ammine complex absorbs at 600 nm, ΔE = 3.32 × 10⁻¹⁹ J (200 kJ mol⁻¹). The ammonia ligands give a larger ΔE.
Section 5
Spectroscopy and colorimetry
The absorption of visible light is used in spectroscopy. A simple colorimeter measures the absorbance of a coloured solution at a chosen wavelength, selected with a filter of the colour that is absorbed most strongly. Absorbance is proportional to concentration.
To find an unknown concentration: make standards of known concentration, measure their absorbances, plot a calibration curve (or find absorbance per mol dm⁻³), then read off the concentration of the sample. If the sample was diluted before measuring, multiply by the dilution factor.
A pale solution can be converted into a more intensely coloured complex, for example with ammonia for Cu²⁺, to improve sensitivity.
State the dilution factor and multiply last: concentration measured × (final volume ÷ volume of original sample).
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Formation of coloured ions
- Aqueous copper(II) sulfate is blue and aqueous zinc sulfate is colourless. The copper(II) ion has the electron configuration [Ar]3d⁹ and the zinc ion has the configuration [Ar]3d¹⁰.Explain why aqueous copper(II) sulfate is coloured but aqueous zinc sulfate is colourless.2 marks
- A solution of [Ti(H₂O)₆]³⁺ is purple. A student is investigating how the colour of a complex ion is related to the light it absorbs. Use h = 6.63 × 10⁻³⁴ J s and c = 3.00 × 10⁸ m s⁻¹.The purple solution absorbs green light with a wavelength of 510 nm. Calculate the energy difference, ΔE, in J for one ion.2 marks
- A student uses a colorimeter to find the concentration of copper(II) ions in a waste solution. Standard solutions of copper(II) sulfate with concentrations of 0.020, 0.040, 0.060 and 0.080 mol dm⁻³ are each treated with excess ammonia solution, and their absorbances at the chosen wavelength are 0.14, 0.28, 0.42 and 0.56 respectively.Explain why the student adds excess ammonia solution to each solution, and why a coloured filter is used in the colorimeter.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).