Part 2: Ligand substitution, chelation and haem

Follow changes in ligand, coordination number, charge and colour without confusing substitution with redox.

Follow the ligands and keep the metal oxidation state

In ligand substitution, an incoming ligand replaces another at the metal centre. When neutral H₂O is replaced by neutral NH₃, the overall charge is unchanged. Their similar sizes allow the coordination number to remain six in the cobalt and copper examples. These are equilibria: adding a high concentration of the incoming ligand can favour the substituted complex.

Adding a little ammonia to many aqueous metal ions first causes acid–base precipitation because ammonia accepts protons from water ligands. Excess ammonia can then dissolve some precipitates by complex formation. The overall substitution equations below describe ligand exchange, not the intermediate precipitate stage.

Copper and cobalt do not give identical formulae

For copper(II), four ammonia ligands replace four waters; two waters remain. The solution becomes deep blue, and the coordination number stays six. Writing [Cu(NH₃)₆]²⁺ is not the required product.

For cobalt(II), six ammonia ligands can replace the six waters. The pink aqua complex gives an initially straw/yellow-brown ammine solution in the usual school description. In air, oxidation to cobalt(III) ammine species can rapidly darken/change the colour. This later redox step is distinct from the ligand substitution; observed colours depend on conditions and time.

Chloride substitution can change coordination number

Chloride is a larger ligand than water. High chloride concentrations can produce tetrahedral four-coordinate complexes from six-coordinate aqua ions. Adding water lowers chloride concentration and can reverse the change. The metal oxidation state remains +2 for cobalt/copper and +3 for iron in these examples.

The cobalt change is pink to blue. [CuCl₄]²⁻ is commonly described as yellow; mixtures with blue aqua copper(II) often look green. A green solution is not proof of a pure green complex. Iron(III) chloride solutions are yellow, but hydrolysis and mixed complexes mean colour alone does not establish a unique formula.

Why a multidentate ligand often binds so effectively

The chelate effect is the enhanced stability commonly found when several monodentate ligands are replaced by suitable multidentate ligands. For one aqua complex plus one EDTA⁴⁻, the model equation produces one metal–EDTA complex and six released waters. Releasing more independently moving species often gives a favourable entropy contribution.

The full criterion is ΔG = ΔH − TΔS. Breaking six metal–water bonds and forming six donor bonds can give a relatively modest enthalpy change if their strengths are similar, allowing the entropy contribution to favour chelation. Bond strengths are not always equal, and solvent ordering, protonation and conditions also affect stability. Do not claim every EDTA reaction has exactly zero ΔH or that every chelate is harmless.

Original worked example: in a constructed substitution with ΔH = +8.0 kJ mol⁻¹ and ΔS = +60 J K⁻¹ mol⁻¹ at 298 K, ΔG = 8.0 − 298 × 0.060 = −9.88 kJ mol⁻¹. The positive entropy term more than offsets the endothermic enthalpy change.

The same ligand idea operates in haemoglobin

The iron in haem is conventionally described as Fe(II). A large multidentate porphyrin ligand supplies four nitrogen donor atoms around it; a protein histidine donor occupies another site, leaving a site for reversible oxygen binding. This is a coordination-bond example, not a claim that oxygen is merely trapped in a cavity.

Carbon monoxide competes strongly for the oxygen-binding site and can displace oxygen, reducing blood’s oxygen-carrying ability. At this level, explain the danger using stronger competitive coordination and fewer available oxygen sites. Do not imply ordinary oxygen transport is simply permanent oxidation of Fe²⁺ to Fe³⁺.

Ammonia also explains silver-halide solubility

AgCl dissolves in dilute ammonia through formation of the linear [Ag(NH₃)₂]⁺ complex; AgBr requires concentrated ammonia and AgI remains insoluble under the usual test conditions. Lowering free Ag⁺ concentration through complex formation draws more AgCl into solution. This connects the halide tests to ligand chemistry.

Original back-titration example: 20.00 cm³ of 0.0150 mol dm⁻³ EDTA contains 0.300 mmol. If a metal sample consumes some of it and the remaining EDTA requires 8.00 cm³ of 0.0100 mol dm⁻³ Zn²⁺, 0.0800 mmol remains. For 1:1 metal:EDTA binding, the sample contained 0.220 mmol of the metal. Multiply by any sampling dilution factor only after this subtraction.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official AQA mark allocations.

Sources