Coordination Compounds · Lesson 9 of 12
Colour and Ligand Effects
“Explain observed colours through electronic absorption while respecting geometry, selection rules and competing transitions.”
• Connect absorbed light with electronic energy gaps and observed colour. • Calculate a photon energy and approximate molar splitting. • Explain ligand-substitution colour changes. • Recognise limitations of simple d–d colour explanations.
A coloured solution selectively removes light
White light contains a range of visible wavelengths. A coloured solution absorbs some wavelengths more strongly than others, so the transmitted mixture has a different colour. What we see is the light that remains, not the colour of the absorbed photons. In a simple one-band picture the observed colour is approximately complementary to the absorbed colour. Actual spectra contain broad bands, so a single wavelength cannot fully specify the appearance of every sample.
The boundaries of named colour regions are approximate. Around 498 nm, light is usually described as blue-green; removing a substantial band in that region can leave a violet appearance for [Ti(H₂O)₆]³⁺. Concentration, path length and additional absorption bands affect perceived colour. A table that labels 535 nm as yellow or 600 nm as red should be treated cautiously: those are better described as green and orange-region wavelengths, respectively.
| Approximate wavelength | Typical spectral description | Illustrative context |
|---|---|---|
| 475 nm | Blue | Absorption band listed for [Co(NH₃)₆]³⁺ |
| 498–500 nm | Blue-green | Ti(III) aqua transition / cobalt aqua-ammine band |
| 535 nm | Green | Band listed for [CoCl(NH₃)₅]²⁺ |
| 600 nm | Orange region | Simplified copper aqua absorption example |
| 310 nm | Ultraviolet | Outside the visible range |
| Entity | Representative band / nm | Reported appearance |
|---|---|---|
| [CoCl(NH₃)₅]²⁺ | 535 | Violet |
| [Co(H₂O)(NH₃)₅]³⁺ | 500 | Red |
| [Co(NH₃)₆]³⁺ | 475 | Yellow-orange |
| [Co(CN)₆]³⁻ | 310 (UV) | Pale yellow |
| Hydrated Cu(II) | 600 (simplified band reference) | Blue |
| [Ti(H₂O)₆]³⁺ | 498 | Violet |
These representative band values do not constitute complete spectra. In particular, a UV maximum alone does not explain a pale visible tint; weaker bands or absorption tails can contribute. The hydrated copper entry is a simplified band reference for a distorted, solvated environment rather than proof of a unique four-coordinate aqua structure.
A d¹ ion makes the energy-level argument clear
Titanium(III) has one d electron. In octahedral [Ti(H₂O)₆]³⁺ it occupies t₂g in the ground state. Absorbing a suitable photon promotes it to e_g, leaving t₂g⁰e_g¹ in the excited configuration. In the simple d¹ model the photon energy equals Δo. The electron later relaxes through allowed pathways; the transmitted beam has already lost the absorbed part of its spectrum. The resulting solution is violet.
Problem
A representative absorption is at 498 nm. Estimate Δo per photon, per mole and as wavenumber.
- 1.Convert 498 nm to 4.98×10⁻⁷ m. Compute hc/λ = 3.99×10⁻¹⁹ J per photon.
- 2.Multiply by 6.02214076×10²³ mol⁻¹ to obtain 2.40×10⁵ J mol⁻¹, or about 240 kJ mol⁻¹.
- 3.For wavenumber, use λ = 4.98×10⁻⁵ cm, so 1/λ = 20 080 cm⁻¹, approximately 2.01×10⁴ cm⁻¹.
- 4.This estimates the d¹ transition gap. The absorbed photon is blue-green; the observed violet is the remaining-light effect, not a claim that 498 nm itself is violet.
Change the ligand, change the spectrum
For a fixed metal oxidation state and comparable geometry, different ligands change the orbital energy separation. A larger relevant gap requires a higher-energy photon and hence a shorter wavelength. For octahedral Ni(II) complexes, the qualitative field-strength order H₂O < NH₃ < N-bound NO₂⁻ suggests the reverse absorption-wavelength order: λ(nitrito) < λ(ammine) < λ(aqua) for comparable transitions. Nickel has several electronic states and bands, so this is a qualitative comparison rather than a formula for every spectral maximum.
Adding en progressively to [Ni(H₂O)₆]²⁺ replaces two waters per bidentate ligand while retaining six donors. The sequence is [Ni(H₂O)₆]²⁺ ⇌ [Ni(en)(H₂O)₄]²⁺ ⇌ [Ni(en)₂(H₂O)₂]²⁺ ⇌ [Ni(en)₃]²⁺, with each substitution consuming one en and releasing two H₂O. Representative appearances progress from green through pale blue and blue-purple to violet. At intermediate compositions, equilibrium mixtures and isomer populations can make colours less sharply separated than the ideal formula sequence implies.
Problem
Compare [Fe(CN)₆]⁴⁻ with [Fe(H₂O)₆]²⁺.
- 1.Both contain Fe(II), d⁶, but the ligand sets produce different splitting and covalent interactions.
- 2.The cyanido entity is low spin and the aqua entity high spin, changing both orbital occupancy and possible electronic transitions.
- 3.Their absorption spectra and therefore observed colours differ. Equal oxidation state does not imply equal colour, and an overall anion/cation sign is not a sufficient colour explanation.
Why “unpaired” does not mean “coloured”
[Ni(H₂O)₆]²⁺ is green and paramagnetic; square-planar [Ni(CN)₄]²⁻ is diamagnetic and often described as colourless or pale under dilute conditions. Pairing alone does not cause colourlessness. The cyanido entity has a different geometry, splitting pattern and set of accessible transitions, with strong absorptions often outside the visible range or only weak visible absorption. Many paired complexes are coloured, and d⁰ or d¹⁰ complexes can be coloured by charge-transfer or ligand-centred transitions.
A d–d transition changes the d-electron state; a charge-transfer transition moves substantial electron density between ligand and metal. Selection rules also affect intensity. These distinctions explain why “no unpaired electrons, therefore no colour” and “no ligands, therefore colourless” are invalid universal rules. Solids can have lattice effects and charge-transfer absorption, and removing a ligand may change composition rather than merely erase splitting.
Water, minerals and the environment of the metal
Anhydrous copper(II) sulfate is white, while its common pentahydrate is blue. Hydration changes the electronic environment around copper; it is an illustrative comparison, not proof that every dehydrated transition-metal salt is colourless. Aqueous copper(II) is more realistically represented by a distorted hydrated environment, commonly [Cu(H₂O)₆]²⁺, than by treating four equatorial waters as a complete universal coordination sphere.
Ruby and emerald both contain Cr(III) at approximately octahedral sites, but their surrounding lattices differ. Ruby is Cr-doped Al₂O₃; emerald is Cr-bearing beryl, Be₃Al₂Si₆O₁₈. Different host bonding and local fields shift absorption bands, leaving ruby red and emerald green. The same metal and oxidation state can therefore generate different colours when the surrounding chemical environment changes.
Keep model and observation connected
To explain a colour, identify the species actually present, establish geometry and d count, consider the relevant transitions, then relate absorption to transmitted or reflected light. CFT provides a first model for d–d transitions but does not capture covalency, every intensity rule or the complete spectrum. An apparently simple colour test is evidence about a chemical environment, not a unique molecular fingerprint by itself.
Quick Quiz
Quiz
A complex absorbs blue-green light and appears violet. Which light reaches the observer?
Observed colour comes from wavelengths not removed as strongly.
If a comparable transition energy increases, what happens to its absorption wavelength?
E = hc/λ makes wavelength inversely related to transition energy.
What is the ground d occupancy of octahedral Ti(III)?
Ti³⁺ is d¹ and occupies the lower octahedral level.
Which statement about colour is defensible?
Charge-transfer and other transitions can give colour independently of unpaired-electron counting.
How many water ligands are replaced by one bidentate en in the stated nickel sequence?
en supplies two N donors in place of two O donors while CN stays six.
Practice Problems
- Calculate the per-photon and molar energy of a 500 nm absorption.
- Predict the qualitative wavelength order for comparable octahedral Ni(II) aqua, ammine and N-bound nitrito transitions.
- Write the three balanced nickel/en substitution steps and check CN.
- Explain why [Ni(H₂O)₆]²⁺ and [Ni(CN)₄]²⁻ differ in colour without claiming pairing is the only cause.
- Why are ruby and emerald different colours despite both containing Cr(III)?
- Correct “removing all water ligands makes every transition-metal compound colourless.”
1. λ = 5.00×10⁻⁷ m. 2. E = hc/λ = 3.9729×10⁻¹⁹ J per photon. 3. Multiplying by NA gives 2.3925×10⁵ J mol⁻¹ = 239 kJ mol⁻¹ to three significant figures. 4. This conversion gives a transition energy, not automatically a unique oxidation state.
Key Takeaways
• Observed colour is associated with light remaining after absorption. • Photon energy equals hc/λ. • The Ti(III) d¹ example links a simple d–d transition to Δo. • Ligand substitution changes electronic energies and spectra. • Compare like transitions when using wavelength orders. • Diamagnetism does not imply colourlessness. • Charge-transfer and lattice effects extend beyond simple CFT. • A metal’s environment matters as much as its oxidation state for colour.