Colour in Transition-Metal Complexes

Transition-metal compounds are famous for their vivid colours: the deep blue of copper(II) sulfate, the purple of potassium manganate(VII), the rust-red of iron(III) chloride, the pink of cobalt(II) chloride hexahydrate. The origin of this colour is electronic. In an isolated gas-phase ion the five 3d orbitals are degenerate — equal in energy — but when ligands coordinate, the electrostatic field they generate raises some d orbitals and lowers others. The resulting energy gap, the crystal-field splitting energy Δ, typically corresponds to a photon of visible light. A d-electron is promoted from a lower-energy d orbital to a higher-energy one (a d–d transition), and the photon that supplied the energy is removed from the transmitted beam. What our eyes detect is the complementary part of the spectrum. This lesson develops the qualitative crystal-field model, the ΔE = hf relation, the spectrochemical series, the consequences of geometry and oxidation state, and the use of the Beer–Lambert law in quantitative colorimetric analysis. The story is firmly within AQA 7405 §3.2.5 but, as the synoptic links show, it reaches forward into UV-vis spectroscopy (§3.3.15) and backward into orbital energies (§3.1.1).

Spec mapping (AQA 7405): This lesson is anchored on §3.2.5 (colour in transition-metal complexes). It builds directly on L4 (general properties of transition metals), L5 (formation of complex ions), and L6 (substitution reactions of complex ions), where ligand swaps produced the colour changes that are quantitatively rationalised here. The forward link is §3.3.15 (UV-visible spectroscopy in the analytical course), where the qualitative ε from this lesson becomes a tabulated parameter for quantitative analysis. The electronic-structure basis — the spatial orientation of d orbitals and orbital energy ordering — is laid down in §3.1.1. Always refer to the official AQA specification document for the exact wording of each sub-section.

Assessment objectives: AO1 items include stating ΔE = hf (and ΔE = hc/λ), recognising the qualitative d-orbital splitting diagram for an octahedral and a tetrahedral field, and reproducing the spectrochemical series in the order H₂O < NH₃ < CN⁻ (and the broader sequence). AO2 is tested by predicting the direction of a colour change when the ligand, the oxidation state, or the coordination number changes — e.g. predicting whether the d–d band shifts to higher or lower energy when H₂O is replaced by NH₃ on Cu²⁺, and what that means for the observed colour. AO3 is tested by justifying why d⁰ and d¹⁰ ions are colourless, why MnO₄⁻ is so much more intense than [Ti(H₂O)₆]³⁺ (charge transfer vs d–d), and by interpreting calibration data to compute an unknown concentration with the Beer–Lambert law.

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Why Are Transition Metal Compounds Coloured?

The d-Orbital Splitting Model

In a free (gaseous) transition metal ion, the five 3d orbitals are degenerate (they all have the same energy). When ligands approach the metal ion to form a complex, the electrostatic interaction between the ligand lone pairs and the d electrons causes the d orbitals to split into two energy levels. The split arises because, in an octahedral arrangement, the six ligands sit on the Cartesian axes (±x, ±y, ±z) and the metal d orbitals point either directly at those positions or between them.

For an octahedral complex:

• Three d orbitals are lowered in energy: the t₂g set (d_xy, d_xz, d_yz). These lobes point between the ligand axes, so they avoid the ligand lone pairs and feel less electrostatic repulsion.
• Two d orbitals are raised in energy: the e_g set (d_x²–y² and d_z²). These lobes point directly along the ligand axes, into the ligand lone pairs, and are destabilised.
• The energy gap between the two sets is called Δ (delta) or Δ_oct — the octahedral crystal-field splitting energy.

For a tetrahedral complex:

• The splitting pattern is inverted: the two e orbitals (d_x²–y², d_z²) are now lower and the three t₂ orbitals (d_xy, d_xz, d_yz) are higher.
• The splitting energy is smaller: Δ_tet ≈ (4/9) × Δ_oct. Two reasons: there are only four ligands instead of six (less total interaction), and the ligands now sit between all five d orbitals so the directional matching with any one set is weaker.

The qualitative octahedral diagram you should be able to sketch is: a higher horizontal pair labelled e_g (containing d_x²–y² and d_z²), an arrow downward marked Δ_oct, and a lower horizontal triple labelled t₂g (containing d_xy, d_xz, d_yz).

Why "crystal field"? The model treats ligands as point negative charges that perturb the d-orbital energies purely electrostatically. It is a deliberate simplification of the full quantum-mechanical picture (ligand-field theory adds covalent character via molecular-orbital theory). Despite the simplification, crystal-field theory predicts the direction and rough magnitude of splittings remarkably well.

The d–d Transition

1. White light from the source passes through a solution of the complex.
2. A photon whose energy matches the splitting Δ promotes an electron from the lower set (t₂g) to the upper set (e_g).
3. The Planck–Einstein relation links the photon to the energy gap:
   • ΔE = h f = h c / λ
   • h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹.
   • For a typical visible photon (λ ≈ 500 nm), ΔE ≈ 4 × 10⁻¹⁹ J ≈ 240 kJ mol⁻¹.
4. The light that is not absorbed passes through. The transmitted beam appears as the complementary colour of the absorbed light.

This is the central mechanism: the colour we see is not the colour absorbed — it is what is left over.

The Colour Wheel

Colour absorbed	Approx. λ / nm	Colour observed (complementary)
Red	620–700	Green
Orange	590–620	Blue
Yellow	570–590	Violet / purple
Green	495–570	Red
Blue	450–495	Orange
Violet	380–450	Yellow

The opposite-pair memory aid: red↔green, yellow↔purple, blue↔orange.

Example

[Cu(H₂O)₆]²⁺ absorbs light in the red/orange region of the visible spectrum (λ_max ≈ 800 nm tail into 600 nm). The complementary colour is blue, which is why copper(II) sulfate solution appears pale blue.

Why d⁰ and d¹⁰ Ions Are Colourless

A d–d transition needs (a) at least one electron to promote and (b) at least one empty d orbital to promote into. Two configurations fail one or both tests:

• d⁰ (e.g. Sc³⁺ [Ar], Ti⁴⁺ [Ar]): no d electrons available — no transition is possible, so no visible photon is absorbed.
• d¹⁰ (e.g. Zn²⁺ [Ar] 3d¹⁰, Cu⁺ [Ar] 3d¹⁰): every d orbital is fully occupied — there is no empty d orbital to receive the promoted electron.

In both cases the solution is colourless. This is the standard explanation for the difference between blue Cu²⁺ (d⁹, one vacancy in e_g) and colourless Cu⁺ (d¹⁰, no vacancies), and between intensely-coloured Ti³⁺ (d¹, purple) and colourless Ti⁴⁺ (d⁰).

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Factors Affecting Colour

Four factors change the size of the d-orbital splitting Δ and therefore the wavelength absorbed and the colour observed.

1. The Ligand — The Spectrochemical Series

Different ligands cause different amounts of splitting. The empirical ranking from weak-field (small Δ) to strong-field (large Δ) is the spectrochemical series:

I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < CN⁻ < CO

• Weak-field ligands (e.g. I⁻, Cl⁻) generate small splittings, so a low-energy (long-wavelength, red end of the visible spectrum) photon is absorbed and a higher-energy complementary colour is observed.
• Strong-field ligands (e.g. CN⁻, CO) generate large splittings, so a high-energy (short-wavelength, blue/violet end) photon is absorbed and a lower-energy complementary colour is observed.

The order roughly tracks the σ-donor strength of the ligand and (for CO and CN⁻) their ability to π-back-bond into empty metal d orbitals. The detailed explanation belongs in ligand-field theory at undergraduate level (see Going Further); for A-Level you must know the order in the form H₂O < NH₃ < CN⁻ at minimum.

Example — chromium(III):

• [Cr(H₂O)₆]³⁺ is green/violet (water ligands; mid-size splitting).
• [Cr(NH₃)₆]³⁺ is purple (ammonia gives a larger Δ than water; the absorption shifts to shorter wavelength, so the complementary colour shifts).

Example — copper(II):

• [Cu(H₂O)₆]²⁺ — pale blue (water ligands, modest Δ).
• [Cu(NH₃)₄(H₂O)₂]²⁺ — deep royal blue (ammonia gives a larger Δ; absorption shifts from ~800 nm to ~600 nm, deepening the perceived blue).

2. Oxidation State of the Metal

A higher oxidation state means:

• Greater positive charge on the metal ion.
• A smaller cation (less electron screening of the nucleus).
• Stronger electrostatic attraction for the ligand lone pairs.
• Ligands held closer to the metal centre, increasing their perturbation of the d orbitals.
• Result: larger Δ, different wavelength absorbed, different colour.

Examples:

• [Fe(H₂O)₆]²⁺ is pale green (Fe in +2; modest charge density).
• [Fe(H₂O)₆]³⁺ is yellow/brown (Fe in +3; greater charge density, larger Δ).
• [Cr(H₂O)₆]²⁺ is sky-blue (Cr +2); [Cr(H₂O)₆]³⁺ is green/violet (Cr +3) — same metal, same ligand, different oxidation state, different colour.

3. Coordination Number / Geometry

Changing the number and arrangement of ligands changes both the splitting pattern and its magnitude. The two A-Level cases are:

• Octahedral (6 ligands): pattern (3 down, 2 up), splitting Δ_oct.
• Tetrahedral (4 ligands): pattern inverted (2 down, 3 up), splitting Δ_tet ≈ (4/9)Δ_oct — substantially smaller.

A smaller Δ absorbs a lower-energy (longer-wavelength) photon, which usually red-shifts the absorption. The complementary colour is therefore visibly different.

Example — cobalt(II):

• [Co(H₂O)₆]²⁺ is pink (octahedral; coordination number 6).
• [CoCl₄]²⁻ is blue (tetrahedral; coordination number 4).

Two factors are at work here: the geometry change (octahedral → tetrahedral) and the ligand change (H₂O → Cl⁻, weaker field). Both reduce Δ; both shift the absorption to longer wavelength.

4. Charge Density on the Metal (Combined)

The first three factors all act, ultimately, through the strength of the metal–ligand interaction. A useful unifying mantra: anything that pushes the ligand closer to a more highly-charged metal centre increases Δ and blue-shifts the absorption. Anything that loosens or weakens the interaction decreases Δ and red-shifts the absorption.

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Predicting Colour Changes — Worked Examples

The exam style is: "State and explain the colour change when reagent X is added to solution Y." You need to identify (1) what changed (ligand, oxidation state, coordination number), (2) whether Δ increases or decreases, and (3) the resulting shift in absorption wavelength and observed colour.

Worked Prediction 1 — Adding NH₃ to Cu²⁺

Starting solution: [Cu(H₂O)₆]²⁺ (pale blue). Excess concentrated NH₃ is added. After loss of two H₂O molecules and substitution by four NH₃:

[Cu(H₂O)₆]²⁺ + 4 NH₃ → [Cu(NH₃)₄(H₂O)₂]²⁺ + 4 H₂O

Analysis:

• Ligand has changed (water replaced by ammonia for four of the six sites).
• NH₃ is higher in the spectrochemical series than H₂O.
• Therefore Δ increases; the absorption shifts to shorter wavelength (higher energy); the complementary colour (the observed colour) shifts.
• Observation: pale blue → deep royal blue. Correct.

Worked Prediction 2 — Adding HCl to [Co(H₂O)₆]²⁺

[Co(H₂O)₆]²⁺ + 4 Cl⁻ ⇌ [CoCl₄]²⁻ + 6 H₂O

Analysis:

• Ligand has changed (water → chloride) — Cl⁻ is lower in the spectrochemical series.
• Coordination number has changed (6 → 4) and geometry has changed (octahedral → tetrahedral).
• Both effects decrease Δ; the absorption shifts to longer wavelength.
• Observation: pink (octahedral hexaaqua) → blue (tetrahedral tetrachloro).

Worked Prediction 3 — Why Cu⁺ is colourless but Cu²⁺ is blue

Both ions can in principle form aqua complexes. The difference is electronic:

• Cu²⁺ is [Ar] 3d⁹: one vacancy in the e_g set; a d–d transition is allowed; visible photons absorbed; blue colour observed.
• Cu⁺ is [Ar] 3d¹⁰: every d orbital filled; no d–d transition possible; visible light passes through unabsorbed; solution is colourless.

(Note: Cu⁺(aq) is also unstable to disproportionation — 2 Cu⁺ → Cu + Cu²⁺ — but that is a separate stability argument from the colour argument.)

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Specific Colour Examples — Reference List

Complex	d-config	Geometry	Colour observed	Approx λ_max absorbed
[Ti(H₂O)₆]³⁺	d¹	Octahedral	Purple/violet	~500 nm (yellow-green)
[V(H₂O)₆]²⁺	d³	Octahedral	Violet	~500–600 nm
[V(H₂O)₆]³⁺	d²	Octahedral	Green	~600 nm (red)
[Cr(H₂O)₆]³⁺	d³	Octahedral	Green/violet	~580 nm
[Cr(NH₃)₆]³⁺	d³	Octahedral	Purple	~460 nm (shorter λ)
[Mn(H₂O)₆]²⁺	d⁵	Octahedral	Very pale pink	spin-forbidden, weak
[Fe(H₂O)₆]²⁺	d⁶	Octahedral	Pale green	~700 nm
[Fe(H₂O)₆]³⁺	d⁵	Octahedral	Yellow/brown	mostly charge-transfer
[Co(H₂O)₆]²⁺	d⁷	Octahedral	Pink	~500 nm
[CoCl₄]²⁻	d⁷	Tetrahedral	Blue	~600–700 nm
[Ni(H₂O)₆]²⁺	d⁸	Octahedral	Green	~700 nm
[Cu(H₂O)₆]²⁺	d⁹	Octahedral (Jahn–Teller distorted)	Pale blue	~800 nm (red tail)
[Cu(NH₃)₄(H₂O)₂]²⁺	d⁹	Octahedral (distorted)	Deep royal blue	~600 nm
[CuCl₄]²⁻	d⁹	Tetrahedral	Yellow / green-yellow	~450 nm
Sc³⁺, Ti⁴⁺, Cu⁺, Zn²⁺	d⁰ or d¹⁰	various	Colourless	no d–d transition
MnO₄⁻	(formally d⁰)	Tetrahedral	Intense purple	~525 nm, charge transfer

Note on Mn²⁺: The d⁵ configuration has every d orbital singly occupied with parallel spins. The only available d–d transitions would require flipping one of those spins to occupy the upper set, because the upper-set orbitals already contain parallel-spin partners. Spin-forbidden transitions still occur, just with vanishingly small probability, giving the famously pale pink of Mn(II) salts.

Charge-Transfer Bands — Why MnO₄⁻ is So Intense

The intense purple of MnO₄⁻ is not a d–d transition (the formal d-count is d⁰). Instead it is a charge-transfer transition: an electron is promoted from a filled oxygen-based ligand orbital onto an empty metal-based orbital. Charge-transfer transitions are allowed by quantum-mechanical selection rules, so their molar absorptivities ε are typically 10³–10⁴ times larger than those of d–d transitions. Hence the same molar concentration of MnO₄⁻ gives a vastly more intense colour than [Cu(H₂O)₆]²⁺. This is why permanganate is used as a titrant — even 10⁻⁵ mol dm⁻³ solutions are visibly coloured. Charge-transfer also explains the deep yellow/brown of Fe³⁺(aq), the orange of dichromate, and the yellow of chromate.

Signposted, not examined at A-Level: Charge-transfer is named and identified at A-Level; the full molecular-orbital justification belongs to UV-vis spectroscopy at undergraduate level.

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Why Some Compounds Are Colourless — Summary Table

Ion	d electrons	Colour	Explanation
Sc³⁺	d⁰	Colourless	No d electrons to promote
Ti⁴⁺	d⁰	Colourless	No d electrons to promote
Ti³⁺	d¹	Purple	Single d–d transition possible
Cu²⁺	d⁹	Blue	d–d transition possible (one vacancy in e_g)
Cu⁺	d¹⁰	Colourless	Full d sub-shell, no empty orbital
Zn²⁺	d¹⁰	Colourless	Full d sub-shell, no empty orbital
Mn²⁺	d⁵	Very pale pink	d–d transitions are spin-forbidden; very low transition probability

Exam Tip: Mn²⁺ solutions appear almost colourless because the d–d transition requires a spin flip, which is "spin-forbidden" and has a very low probability. The very pale pink colour is due to this weak absorption.

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Colorimetry and the Beer–Lambert Law

Once we accept that a coloured solution absorbs a specific portion of the visible spectrum, we can use that absorption to quantify concentration. This is colorimetry, and the underlying mathematical law is the Beer–Lambert law, named after the two nineteenth-century scientists whose names label it.

Principle of a Colorimeter

1. A beam of white light is passed through a filter that selects a narrow band of wavelengths near the solution's absorption maximum.
2. The filtered light passes through the coloured solution held in a transparent cuvette of known path length l (usually 1.00 cm).
3. A detector measures the intensity of the transmitted light I, compared to the intensity I₀ of the incident beam.
4. The absorbance is defined as A = log₁₀(I₀ / I). It is dimensionless and additive — convenient for plotting.

Beer–Lambert Law

A = ε c l

Where:

• A = absorbance (dimensionless).
• ε = molar absorptivity (or molar extinction coefficient), units dm³ mol⁻¹ cm⁻¹.
• c = concentration of the absorbing species (mol dm⁻³).
• l = path length of the cuvette (cm).

ε depends on the species and the wavelength but not on c. The product ε l is the gradient of an A-vs-c calibration plot at fixed l.

Using a Calibration Curve — Step by Step

1. Choose the wavelength. Run a preliminary scan across the visible region (or pick the filter whose colour is complementary to the solution) and identify the wavelength of maximum absorbance, λ_max.
2. Prepare standards. Make up a series of solutions of known concentrations — typically 5–7 standards spanning the expected range of the unknown.
3. Measure absorbances. Record the absorbance of each standard at λ_max. Always zero the colorimeter with a blank (pure solvent in an identical cuvette).
4. Plot A vs c. The graph should be a straight line through the origin, with gradient ε l.
5. Determine the unknown. Measure the absorbance of the unknown solution at the same λ_max in the same cuvette. Read the concentration off the graph, or compute c = A / (ε l).

Selecting the Filter / Wavelength

• The optimum filter passes the colour that the solution absorbs most strongly — i.e. the complementary colour of the solution as observed.
• For a blue solution (e.g. CuSO₄(aq)), the absorbed light is in the orange/red region; use an orange or red filter (or set λ ≈ 600–650 nm on a spectrophotometer).
• For a purple solution (e.g. KMnO₄), absorbed light is in the green region; use a green/yellow filter (λ ≈ 525 nm).
• For a yellow solution (e.g. dichromate), absorbed light is in the violet/blue region; use a blue/violet filter.

Common Misconception: Students sometimes pick a filter the same colour as the solution. Wrong — the filter must transmit the colour the solution most strongly absorbs, which is the complementary colour. Using the wrong filter gives a small absorbance and a noisy calibration plot.

Worked Example — Cu²⁺ Concentration

Question: A series of standard copper(II) sulfate solutions was prepared and the absorbance of each measured at 635 nm in a 1.00 cm cuvette:

Concentration / mol dm⁻³	0.020	0.040	0.060	0.080	0.100
Absorbance A	0.13	0.26	0.39	0.52	0.65

An unknown CuSO₄ solution has absorbance 0.45 at the same wavelength and path length. Determine its concentration.

Solution:

• Plot A against c. The five points lie on a straight line through the origin.
• Gradient = (0.65 − 0) / (0.100 − 0) = 6.5 dm³ mol⁻¹ (this is ε l for l = 1.00 cm, so ε = 6.5 dm³ mol⁻¹ cm⁻¹).
• For A = 0.45: c = A / (gradient) = 0.45 / 6.5 = 0.069 mol dm⁻³ (2 s.f.).

Equivalently, c = A / (ε l) = 0.45 / (6.5 × 1.00) = 0.069 mol dm⁻³.

Forward link (§3.3.15, UV-vis): This colorimetric procedure is the same logic that underpins UV-visible spectroscopy in the analytical course. The colorimeter is a low-cost monochromator using filters; a UV-vis spectrophotometer uses a diffraction grating to scan continuously. The Beer–Lambert law is identical in both.

Practical-Skills Box — Determining Cu²⁺ in an Unknown

Aim: Determine the concentration of Cu²⁺ in an unknown solution by colorimetry.