Part 3: Colour, light absorption and colorimetry

Explain d-orbital splitting, calculate photon energies and determine concentrations from calibrated absorbance measurements.

Colour starts with an energy gap

In an isolated ion the five d orbitals have equal energy in the simple model. A surrounding ligand field splits their energies because the orbitals interact differently with the ligands. A d electron can absorb a photon and move from a lower to a higher d-energy level when the photon energy matches an allowed gap.

For a coloured solution illuminated with white light, selected wavelengths are absorbed. The transmitted mixture of the remaining wavelengths produces the observed colour; for an opaque material, reflected light matters. Do not say a blue solution absorbs blue light as the reason it looks blue, or that its ordinary observed colour is emitted when electrons fall back down.

A change of ligand, coordination number/geometry or metal oxidation state can alter the splitting and hence the wavelengths absorbed. Diluting a complex normally makes its colour paler by lowering concentration; that alone does not mean the d-orbital energy gap changed. Dilution may additionally shift a ligand equilibrium in some systems.

Keep wavelength in metres

ΔE = hν = hc/λ, where h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹, ν is frequency in s⁻¹ and λ is wavelength in metres. ΔE from this equation is the energy per photon, not per mole. Multiply by the Avogadro constant only if the question asks for a molar energy.

Original worked example: a complex absorbs at 620 nm = 6.20 × 10⁻⁷ m. ν = c/λ = 4.84 × 10¹⁴ s⁻¹. ΔE = hc/λ = 3.21 × 10⁻¹⁹ J. With Nₐ = 6.022 × 10²³ mol⁻¹, this is 193 kJ mol⁻¹. A larger gap corresponds to higher frequency and shorter wavelength.

Do not force every colour into a d–d explanation

For the familiar aqua complexes, d⁰ ions have no d electron to excite and d¹⁰ ions have no appropriate vacancy within the full d sublevel. This helps explain colourless Sc³⁺ and Zn²⁺ aqua ions. It does not prove all d⁰/d¹⁰ compounds are colourless: charge-transfer transitions or coloured ligands can produce colour, as illustrated by purple manganate(VII), whose Mn(VII) is formally d⁰.

Not every transition-metal compound is coloured, and not every coloured compound is a transition-metal d–d system. Apply the splitting model to the species and transition the question actually asks about.

Measure concentration using a calibration curve

Prepare standards of known metal-ion concentration across a suitable range. If a ligand is added to intensify the colour, add enough to form the same coloured species in every standard and unknown; keep other reagents, pH and final volume controlled. Choose a wavelength strongly absorbed by that species, usually near its absorption maximum, or a suitable complementary-colour filter.

Zero the instrument using a reagent blank containing the solvent and added reagents but no analyte. Measure absorbance using clean, consistently orientated cuvettes with the same path length, wiped optical faces and no bubbles. Plot absorbance against concentration and measure the unknown under the same conditions. Dilute an unknown above the calibration range; do not extrapolate carelessly.

Over a validated linear range, absorbance is proportional to concentration at fixed path length. Percentage transmission is not itself linearly proportional to concentration. Use the measured calibration line rather than forcing every real dataset through zero. Repeat readings and consider scatter, blank error and any competing coloured species.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official AQA mark allocations.

Sources