State the shape of (a) [Cu(H_2O)_6]^2+ and (b) [CuCl_4]^2-. [2 marks]

SingaporeChemistrySyllabus dot point

How do transition metals form coloured complex ions, and what controls the colour observed?

Describe the formation of complex ions with ligands, explain coordination number and shape, account for the origin of colour in terms of d orbital splitting and d-d transitions, and describe ligand exchange reactions

A focused answer to the H2 Chemistry learning outcome on transition-metal complexes. Ligands and dative bonding, coordination number and shape, the origin of colour from d orbital splitting and d-d transitions, and ligand exchange reactions with colour changes.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
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What this dot point is asking

SEAB wants you to describe how transition-metal ions form complex ions with ligands, explain coordination number and the resulting shape, account for the colour of complexes through $d$ orbital splitting and $d$ - $d$ transitions, and describe ligand-exchange reactions and their colour changes. The origin-of-colour explanation and ligand-exchange equations are signature transition-metal exam questions.

The answer

Ligands and complex ions

A ligand is a species with at least one lone pair that forms a dative (coordinate) bond to a central metal ion, giving a complex ion. Common ligands include $\text{H}_2\text{O}$ , $\text{NH}_3$ , $\text{Cl}^-$ , $\text{CN}^-$ and $\text{OH}^-$ .

The coordination number is the number of dative bonds from ligands to the central metal ion:

coordination number 6: octahedral (e.g. $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$ )
coordination number 4: tetrahedral (e.g. $[\text{CuCl}_4]^{2-}$ ) or, less commonly, square planar
coordination number 2: linear (e.g. $[\text{Ag}(\text{NH}_3)_2]^+$ )

The origin of colour

In an isolated transition-metal ion, the five $3d$ orbitals are degenerate (equal energy). When ligands approach, they repel the $d$ electrons unequally, splitting the $d$ orbitals into two sets separated by an energy gap $\Delta E$ .

An electron can absorb a photon of visible light whose energy equals $\Delta E$ and jump from the lower to the higher set, a $d$ - $d$ transition. The wavelength absorbed is removed from white light, so the complex appears the complementary colour of the light absorbed. For example, $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$ absorbs in the orange-red and looks blue.

Because Sc $^{3+}$ ( $d^0$ ) and Zn $^{2+}$ ( $d^{10}$ ) have no possible $d$ - $d$ transition (no partially filled $d$ ), their compounds are colourless.

What changes the colour

The size of $\Delta E$ , and hence the colour, depends on:

the ligand (different ligands cause different splitting),
the oxidation state of the metal,
the coordination number and geometry.

A larger $\Delta E$ means higher-energy (shorter-wavelength) light is absorbed, shifting the observed colour.

Ligand exchange

Ligands can be replaced by others, often with a striking colour change because $\Delta E$ changes:

Adding excess ammonia to copper(II): $[\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{NH}_3 \rightarrow [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+} + 4\text{H}_2\text{O}$ (pale blue to deep blue).
Adding concentrated chloride to copper(II): $[\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightarrow [\text{CuCl}_4]^{2-} + 6\text{H}_2\text{O}$ (blue to yellow-green, with a shape change from octahedral to tetrahedral).

Worked example

Explain why $[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}$ is a deeper, more intense blue than $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$ .

Both ions are coloured because of $d$ - $d$ transitions: an electron absorbs a photon of energy $\Delta E$ and jumps between the split $d$ orbital sets.

Ammonia causes a larger splitting ( $\Delta E$ ) than water. So the ammonia complex absorbs light of higher energy (shorter wavelength) than the aqua complex.

The change in the wavelength absorbed shifts the complementary colour seen, giving the deeper blue of the ammonia complex compared with the pale blue of the hexaaqua ion.

Try it: Complex ion colour predictor - choose a metal ion and ligand to see the predicted shape, coordination number and colour.

Common traps

Saying the colour seen is the colour absorbed: The complex shows the complementary colour of the light absorbed, not the absorbed colour itself.
Forgetting why Sc and Zn compounds are colourless: No partially filled $d$ subshell means no $d$ - $d$ transition, so no visible-light absorption.
Confusing coordination number with charge: Coordination number counts dative bonds; it is not the same as the overall charge of the complex.
Writing the wrong copper-ammonia complex: Only four water ligands are replaced; the product is $[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}$ , not $[\text{Cu}(\text{NH}_3)_6]^{2+}$ .
Ignoring the shape change in chloride exchange: Going from six water to four chloride ligands changes the geometry from octahedral to tetrahedral.

Examples in context

Example 1. Haemoglobin and ligand exchange. The iron(II) centre in haemoglobin binds oxygen reversibly as a ligand, but carbon monoxide binds far more strongly, displacing oxygen and preventing transport. SEAB uses this biological ligand-exchange example to show why carbon monoxide is toxic, linking the abstract idea to physiology.

Example 2. Testing for copper(II) and identifying complexes. Adding ammonia dropwise then in excess to an unknown solution gives a pale blue precipitate that dissolves to a deep blue solution, a characteristic confirmation of copper(II). The sequence of colour changes, explained by precipitation then ligand exchange, is exactly what qualitative-analysis questions reward.

Try this

Q1. Define the terms ligand and coordination number. [2 marks]

Cue. A ligand donates a lone pair in a dative bond to a metal ion; coordination number is the number of such dative bonds to the central ion.

Q2. State the shape of (a) $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$ and (b) $[\text{CuCl}_4]^{2-}$ . [2 marks]

Cue. (a) Octahedral. (b) Tetrahedral.

Q3. Explain why a solution of zinc sulfate is colourless. [2 marks]

Cue. $\text{Zn}^{2+}$ is $[\text{Ar}]\,3d^{10}$ ; the full $d$ subshell allows no $d$ - $d$ transition, so no visible light is absorbed.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Specimen (9729)4 marksExplain, in terms of d orbital splitting, why aqueous copper(II) ions are blue, and predict how the colour might change if the water ligands were replaced by a ligand causing a larger splitting.

Show worked answer →

In an isolated ion the five 3d orbitals are degenerate (equal energy). When ligands approach, they split the d orbitals into two sets of slightly different energy, separated by an energy gap delta E.

An electron can absorb a photon of visible light whose energy equals delta E and jump from the lower to the higher set (a d-d transition).

The light absorbed is removed from white light; the colour seen is the complementary colour of the absorbed light. The hexaaqua copper(II) ion absorbs in the orange-red region and so appears blue.

A ligand causing a larger splitting (larger delta E) means light of higher energy (shorter wavelength) is absorbed, so the complementary colour seen shifts, typically toward a deeper blue or green-blue.

Markers reward the splitting of degenerate d orbitals, the d-d transition absorbing a photon of energy delta E, the complementary-colour idea, and the effect of a larger delta E.

2022 (style)4 marksWhen concentrated aqueous ammonia is added slowly to aqueous copper(II) sulfate, a pale blue precipitate forms first, which then dissolves to give a deep blue solution. Identify the species formed at each stage and write equations.

Show worked answer →

Stage 1: a small amount of ammonia acts as a base, precipitating copper(II) hydroxide:
Cu2+ + 2OH- -> Cu(OH)2 (pale blue precipitate). (Ammonia provides OH- from NH3 + H2O equilibrium.)

Stage 2: with excess concentrated ammonia, ligand exchange occurs. Ammonia ligands replace four of the water ligands to form the deep blue tetraamminediaquacopper(II) complex:
[Cu(H2O)6]2+ + 4NH3 -> [Cu(NH3)4(H2O)2]2+ + 4H2O.

The deep blue colour confirms the complex.

Markers reward the pale blue Cu(OH)2 precipitate with equation, the ligand-exchange step with the correct complex, and the deep blue final colour.