Topic 4 of 6
Ligands, complexes and observations
Follow copper colours, QA evidence and haemoglobin exchange.
A-Level 9476 (2026-2027)
Ligands donate electron pairs to a metal centre
A complex can be positive, neutral or negative.
A ligand is an ion or molecule that donates a lone pair to a central metal atom or ion to form a coordinate bond. A complex contains that central metal bonded to its surrounding ligands. H2O and NH3 are neutral donors through O and N respectively; Cl- is a negatively charged donor. The metal accepts the pair: it does not supply both electrons in that coordinate bond.
| Complex | Ligands and coordination | Overall charge |
|---|---|---|
| [Cu(H2O)6]2+ | Six water donors; coordination number 6. | +2: all water ligands are neutral. |
| [Cu(NH3)4(H2O)2]2+ | Four ammonia and two water donors; coordination number 6. Often abbreviated [Cu(NH3)4]2+ in equations. | +2: ammonia and water are neutral. |
| [CuCl4]2- | Four chloride donors; coordination number 4. | +2 + 4(-1) = -2. |
Coordination number counts donor atoms directly bonded to the metal. For these one-donor ligands it equals the number of ligands. Determine oxidation state separately from overall complex charge: the negative chloride complex still contains copper in the +2 state.
Two ligand changes, the same copper oxidation state
Water gives the blue copper(II) aqua complex. Excess ammonia produces a deep-blue ammine complex; concentrated chloride produces a yellow chloro complex. Both retain copper(II).
Excess ammonia: [Cu(H2O)6]2+(aq) + 4NH3(aq) ⇌ [Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l). Ammonia replaces four coordinated water molecules; the solution becomes deep blue.
Concentrated chloride: [Cu(H2O)6]2+(aq) + 4Cl-(aq) ⇌ [CuCl4]2-(aq) + 6H2O(l). High chloride concentration favours the yellow complex. Dilution favours the blue aqua form. These are ligand exchanges, not redox reactions.
Check your understandingWhat is chromium's oxidation state in [Cr(OH)6]3-, and why does its overall negative charge not answer that question?Think it through, then reveal the answer
Separate precipitation, ligand exchange and oxidation
The sequence of additions explains the qualitative-analysis observation.
A few drops of aqueous ammonia supply OH- through NH3 + H2O ⇌ NH4+ + OH-. Copper(II) first forms blue Cu(OH)2(s): Cu2+(aq) + 2OH-(aq) → Cu(OH)2(s). In excess ammonia, complex formation removes dissolved copper ions and the precipitate dissolves to the deep-blue ammine solution. NaOH does not provide that ammonia ligand, so the precipitate remains in excess NaOH.
| Ion | NaOH(aq), then excess | NH3(aq), then excess |
|---|---|---|
| Cr3+ | Grey-green Cr(OH)3 precipitate; dissolves in excess to dark-green [Cr(OH)6]3-. | Grey-green precipitate; insoluble in excess. |
| Cu2+ | Pale-blue Cu(OH)2 precipitate; insoluble in excess. | Blue precipitate; dissolves in excess to a deep-blue complex. |
| Fe2+ | Green Fe(OH)2 precipitate; insoluble in excess; turns brown in air. | Same precipitate and excess-reagent behaviour. |
| Fe3+ | Red-brown Fe(OH)3 precipitate; insoluble in excess. | Same precipitate and excess-reagent behaviour. |
| Mn2+ | Off-white Mn(OH)2 precipitate; insoluble in excess; rapidly turns brown in air. | Same precipitate and excess-reagent behaviour. |
Chromium(III) hydroxide dissolves in excess hydroxide by forming a soluble hydroxo complex: Cr(OH)3(s) + 3OH-(aq) ⇌ [Cr(OH)6]3-(aq). Chromium remains +3. In contrast, the browning of freshly formed iron(II) or manganese(II) hydroxide in air involves oxidation. Record the initial colour and subsequent change separately.
In haemoglobin, oxygen binds reversibly to an iron-containing centre. Carbon monoxide competes for the binding site and binds much more strongly, displacing oxygen and reducing oxygen transport. A simplified exchange is HbO2 + CO ⇌ HbCO + O2. This illustrates ligand competition; it is not a reaction in which CO burns with oxygen.