Question
Why are transition metal compounds coloured? How does crystal field theory (CFT) explain the origin of colour? When would a transition metal compound be colourless?
(JEE Main and CBSE 12 — colour prediction using CFT is a high-frequency conceptual question)
Solution — Step by Step
In a free metal ion, all five d-orbitals have the same energy. In a coordination compound, the ligand field splits these orbitals into two sets (e.g., and in octahedral geometry) with an energy gap called crystal field splitting energy ().
When visible light hits the complex, a d-electron absorbs a photon with energy equal to and jumps from the lower set to the higher set. This is a d-d transition. The remaining (transmitted or reflected) wavelengths give the compound its observed colour.
The colour we see is the complementary colour of the absorbed wavelength.
| Factor | Effect on |
|---|---|
| Ligand strength | Strong field ligands (CN-, CO) increase ; weak field (I-, Br-, Cl-) decrease it |
| Oxidation state | Higher charge → larger |
| Geometry | Octahedral has larger than tetrahedral |
Spectrochemical series (increasing field strength): I^- < Br^- < Cl^- < F^- < H_2O < NH_3 < en < CN^- < CO
A larger absorbs higher energy (shorter wavelength) light. A smaller absorbs lower energy (longer wavelength) light.
A compound is colourless when d-d transitions cannot occur:
- d0 configuration: No d-electrons to excite. Examples: ( is colourless),
- d10 configuration: All d-orbitals are fully occupied — no empty orbital to jump to. Examples: , ,
- If the falls outside the visible spectrum range (UV or IR), the compound appears colourless to our eyes
So: and complexes are typically colourless. All others ( to ) are generally coloured.
graph TD
A["d-electrons in complex"] --> B{"d0 or d10?"}
B -->|"Yes"| C["Colourless - No d-d transition"]
B -->|"No: d1-d9"| D["d-d transition possible"]
D --> E["Light absorbed at wavelength = Delta"]
E --> F["Observed colour = Complementary"]
G["Factors affecting Delta"] --> H["Ligand: Spectrochemical series"]
G --> I["Oxidation state"]
G --> J["Geometry"]Why This Works
Crystal field theory provides a simple model: ligands are treated as negative point charges that repel d-electrons, splitting the d-orbital energies. The energy gap between split levels happens to fall in the visible light range (400-700 nm) for most transition metal complexes — which is why they are coloured.
The complementary colour relationship: if a complex absorbs red light (~700 nm, low energy, small ), it appears green (the complementary colour). If it absorbs violet light (~400 nm, high energy, large ), it appears yellow. The spectrochemical series tells us which ligands push higher.
Example: absorbs yellow-green light ( ~ 500 nm) and appears purple. Changing ligands changes , which changes the absorbed wavelength, which changes the colour.
Alternative Method
For JEE, remember these colourless species: Zn2+ compounds (d10), Cu+ compounds (d10), Sc3+ compounds (d0), Ti4+ compounds (d0). If a question asks “which is colourless?” and lists transition metal ions, check for d0 or d10 first.
Common Mistake
The most common error: saying ” is blue because copper is a transition metal.” The colour does not come from being a transition metal — it comes from d-d transitions enabled by the ligand field. (d10) is colourless despite being a transition metal ion. The d-electron configuration and ligand environment determine colour, not just the metal identity.
Also, students confuse the absorbed colour with the observed colour. If a question says “a complex absorbs at 600 nm (orange)” — the observed colour is the complementary, which is blue. Always convert absorbed to complementary before answering.