In an isolated ion the five 3d orbitals have the same energy. Surround the ion with ligands and they split into two groups. An electron jumping the gap absorbs one part of visible light, and your eye sees what is left.
In a free ion the five 3d orbitals are degenerate (equal in energy). Ligand lone pairs repel the d electrons unequally, so the orbitals become non-degenerate: in an octahedral complex they split into three lower and two higher orbitals; in a tetrahedral complex into two lower and three higher, with a smaller gap. When white light passes through, an electron is promoted from a lower to a higher d orbital, absorbing a photon of energy ΔE = hf. The colour seen is the complementary colour of the light absorbed. Ions with an empty (d0) or full (d10) d subshell cannot make this transition, so they are colourless. A different ligand, oxidation state or coordination number changes ΔE and so changes the colour.
Use the figure: an electron moves up across ΔE, and the frequency of light absorbed matches the gap exactly.
| Light absorbed | Approx. wavelength / nm | Colour seen (complementary) | Example |
|---|---|---|---|
| red-orange | 600–800 | blue | [Cu(H2O)6]2+ |
| yellow-green | 500–580 | purple / violet | [Ti(H2O)6]3+ |
| green | 500–560 | pink / red | [Co(H2O)6]2+ |
| blue-violet | 400–480 | yellow / orange | Cr2O72−, [Fe(H2O)6]3+ (yellow-brown) |
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Into how many lower and higher energy orbitals do the 3d orbitals split in an octahedral complex?
An aqueous solution of a complex is blue. Which colour of light does it mainly absorb?
Why are zinc compounds white?
What does the term degenerate mean for d orbitals?
ΔE depends on the metal, its oxidation state, the ligands and the shape. Change any one and the light absorbed changes.
| Change | Example | Effect on ΔE | Colour |
|---|---|---|---|
| ligand H2O replaced by NH3 | [Cu(H2O)6]2+ → [Cu(NH3)4(H2O)2]2+ | larger (NH3 splits more strongly) | pale blue → deep blue |
| oxidation state raised | [Fe(H2O)6]2+ → [Fe(H2O)6]3+ | different (higher charge attracts ligands closer) | pale green → yellow-brown |
| octahedral to tetrahedral | [Co(H2O)6]2+ → [CoCl4]2− | smaller (tetrahedral split, different ligand) | pink → blue |
| d10 or d0 ion | [Zn(H2O)6]2+, [Sc(H2O)6]3+ | no d-d transition possible | colourless |
The colour you see is not the colour absorbed: a blue solution absorbs red-orange light. Octahedral splitting is 3 lower, 2 upper; tetrahedral is the reverse, 2 lower, 3 upper, with a smaller ΔE. Electrons are promoted within the 3d subshell, not from 3d to 4s. Colour needs a partly filled d subshell, so Zn2+ and Sc3+ compounds are white.
Unlocks once the four checks above are done. Worth more XP, written to A Level Paper 4 standard.
A complex absorbs light of wavelength 500 nm. What is ΔE in kJ mol−1? (h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, L = 6.02 × 1023 mol−1)
Why does replacing water ligands by ammonia change the colour of a copper(II) complex?
Which ion is colourless in aqueous solution?
Compared with an octahedral complex of the same ion, a tetrahedral complex has
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