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Ligand substitution is the reaction in which one ligand in the coordination sphere of a complex ion is replaced by another. It is one of the most fertile areas of transition-metal chemistry because the products are usually distinguishable by eye — substituting water with ammonia, chloride, EDTA⁴⁻, or thiocyanate changes the d-orbital splitting and therefore the absorbed wavelength. In this lesson we develop the thermodynamic drivers that make a substitution favourable (the equilibrium constant K, dissected into ΔH° and TΔS° contributions, with the chelate effect quantified explicitly), then catalogue the canonical AQA examples for copper(II), cobalt(II) and iron(III). We close with Required Practical 12, the preparation and purification of a transition-metal complex, and a practical-skills box for the dropwise NH₃ addition to Cu²⁺ that every A-Level chemist must be able to describe from memory.
Spec mapping (AQA 7405): This lesson anchors §3.2.5 (transition metals), specifically the ligand-substitution sub-section and the associated stability-constant treatment. It builds on lesson 5 (complex-ion foundations — coordination number, dentate count, geometry, the chelate-effect derivation), feeds forward into lesson 7 (colour change accompanying substitution, spectrochemical series, Beer–Lambert), lesson 8 (reactions of aqueous metal ions with NaOH, NH₃ and Na₂CO₃), and provides the chemistry behind Required Practical 12 (preparation of a transition-metal complex). Stability-constant K_stab expressions reuse the equilibrium-law machinery from §3.1.6 (Kc). Refer to the official AQA 7405 specification for verbatim wording.
Assessment objectives: AO1 (recall) — typical colour changes for the Cu²⁺/Co²⁺/Fe³⁺ aqua complexes with NH₃, Cl⁻, EDTA⁴⁻ and SCN⁻, and the coordination-number/geometry change in each. AO2 (apply) — write balanced substitution equations with correct overall charges, predict coordination number from ligand size, predict colour direction from d-electron count and ligand-field strength. AO3 (analyse) — apply the chelate-effect entropy argument quantitatively, rationalise the observed sequence of intermediate colours when NH₃ is added dropwise to Cu²⁺, and use stability-constant data to predict the displacement of one ligand by another.
A ligand substitution is the replacement of one ligand in a complex by a different ligand:
[ML₆]^(n+) + 6L′ ⇌ [ML′₆]^(n+) + 6L
If the incoming and outgoing ligands carry the same charge (e.g. H₂O for NH₃, both neutral) the overall complex charge is preserved. If the charges differ — for instance H₂O (0) being replaced by Cl⁻ (−1) — the overall complex charge changes. So [Cu(H₂O)₆]²⁺ (charge +2) becomes [CuCl₄]²⁻ (charge +2 from Cu plus 4×(−1) from Cl⁻, net −2) when four chloride ions replace six waters. Writing the equation correctly is therefore a charge-balance exercise as well as a counting exercise.
A second distinction matters: substitution can be simple (no change in coordination number, both ligands of similar size) or size-driven (a switch from CN 6 to CN 4 because the incoming ligand is significantly larger). NH₃ and H₂O are almost identical in size — both occupy ~3 Å of metal–ligand bond length — so substitutions of H₂O by NH₃ preserve coordination number. Cl⁻ is much larger (~3.3 Å when bound), so only four can pack around the metal: chloride substitutions therefore collapse the geometry from octahedral to tetrahedral.
Key point: Three things change in a substitution: the identity of the ligand, sometimes the coordination number (and shape), and almost always the colour. Each is a marking point in an exam answer.
The pale-blue hexaaqua copper(II) complex is the most heavily examined starting point for ligand substitution.
Adding ammonia dropwise initially produces a pale-blue precipitate of Cu(OH)₂ (an acid–base reaction in which NH₃ acts as a base, deprotonating two of the coordinated waters):
[Cu(H₂O)₆]²⁺(aq) + 2NH₃(aq) → Cu(OH)₂(H₂O)₄(s) + 2NH₄⁺(aq)
This is not a substitution — it is a deprotonation that produces a neutral hydroxide species. On further addition of NH₃, the precipitate redissolves in a true ligand substitution that replaces four of the six waters (the two axial waters remain):
[Cu(H₂O)₆]²⁺(aq) + 4NH₃(aq) ⇌ [Cu(NH₃)₄(H₂O)₂]²⁺(aq) + 4H₂O(l)
pale blue → deep royal blue
Why only four of the six waters? Copper(II) is a d⁹ ion and is therefore subject to a Jahn–Teller distortion: the two axial bonds of the octahedron are elongated and held weakly, while the four equatorial bonds are shorter and stronger. The four strong equatorial sites are the ones substituted by NH₃; the two long axial waters are essentially spectators. The product retains coordination number 6 but is described as tetragonally distorted octahedral rather than perfectly octahedral — a detail that elevates an A* answer.
[Cu(H₂O)₆]²⁺(aq) + 4Cl⁻(aq) ⇌ [CuCl₄]²⁻(aq) + 6H₂O(l)
pale blue → yellow
Coordination number drops from 6 to 4 because Cl⁻ is a much larger ligand than H₂O. The shape becomes tetrahedral. The overall charge of the complex switches from +2 to −2 (Cu²⁺ plus four Cl⁻ contributes net −2). In practice the solution often appears green during the addition — this is the mixed-colour appearance of unreacted blue [Cu(H₂O)₆]²⁺ together with newly-formed yellow [CuCl₄]²⁻, not a separate green species.
[Cu(H₂O)₆]²⁺(aq) + EDTA⁴⁻(aq) → [Cu(EDTA)]²⁻(aq) + 6H₂O(l)
EDTA⁴⁻ is hexadentate — it binds through two nitrogen atoms and four carboxylate oxygens, wrapping itself around the metal centre and occupying all six coordination sites with a single molecule. The product is deep blue. The reaction goes essentially to completion because of the chelate effect, quantified below.
The pink hexaaqua cobalt(II) complex provides the second canonical AQA example.
The first step parallels copper — small additions of NH₃ produce a pale-blue/buff precipitate of Co(OH)₂(H₂O)₄ via deprotonation. In excess ammonia the precipitate redissolves with full substitution of all six waters (NH₃ and H₂O are size-matched):
[Co(H₂O)₆]²⁺(aq) + 6NH₃(aq) ⇌ [Co(NH₃)₆]²⁺(aq) + 6H₂O(l)
pink → pale yellow/straw brown
Cobalt(II) ammine complexes are mildly air-sensitive: O₂ from the atmosphere oxidises Co(II) to Co(III), producing the much more stable yellow-brown [Co(NH₃)₆]³⁺. This oxidation is one of the few cases where a transition-metal substitution product noticeably changes colour on standing in air, and is a classic demonstration of how complexation can shift the standard electrode potential of a metal couple.
[Co(H₂O)₆]²⁺(aq) + 4Cl⁻(aq) ⇌ [CoCl₄]²⁻(aq) + 6H₂O(l) ΔH° > 0
pink → blue
Coordination number drops from 6 to 4; the shape changes from octahedral to tetrahedral; overall charge changes from +2 to −2. The reaction is endothermic in the forward direction, so heating drives the equilibrium to the right (blue dominates) and cooling drives it back (pink dominates). This is the basis of cobalt(II) chloride moisture-testing paper: anhydrous CoCl₂ contains the blue [CoCl₄]²⁻-like environment, and on exposure to water it reverts to pink [Co(H₂O)₆]²⁺, giving a vivid blue-to-pink colour change used in school laboratories and silica-gel desiccants.
The pale-yellow/violet hexaaqua iron(III) complex is the third examined system.
[Fe(H₂O)₆]³⁺(aq) + SCN⁻(aq) ⇌ [Fe(SCN)(H₂O)₅]²⁺(aq) + H₂O(l)
pale yellow → deep blood-red
This is the qualitative test for Fe³⁺ in solution. The intensity of the red colour is so high that even trace iron(III) — micromolar concentrations — gives a perceptible colour. The complex is a 1:1 monothiocyanato species (only one of the six waters is substituted in the test conditions); higher complexes [Fe(SCN)₂(H₂O)₄]⁺ and so on form at higher SCN⁻ concentrations.
[Fe(H₂O)₆]³⁺(aq) + EDTA⁴⁻(aq) → [Fe(EDTA)]⁻(aq) + 6H₂O(l)
pale yellow → pale yellow (visually less dramatic, but K_stab ≈ 10²⁵ for Fe(III) EDTA — among the largest stability constants tabulated for first-row metals).
The position of the equilibrium [ML₆] + 6L′ ⇌ [ML′₆] + 6L depends on the stability constant K_stab. For the substitution to proceed essentially to completion we need K_stab ≫ 1, i.e. ΔG° ≪ 0.
ΔG° = ΔH° − TΔS°
Two distinct routes can make ΔG° favourable:
Enthalpic route — the new metal–ligand bonds are stronger than the old ones. Replacing H₂O (a weak-field, σ-only donor) with NH₃ (a slightly stronger σ donor) gives a modest enthalpic gain. Replacing H₂O with CN⁻ or CO (strong π-acceptors) gives a much larger gain. This route dominates when the incoming ligand is much higher up the spectrochemical series than the outgoing ligand (lesson 7 develops this fully).
Entropic route — the chelate effect. When one polydentate ligand replaces several monodentate ligands, the total number of free particles in solution rises and entropy increases.
Consider replacing six monodentate H₂O ligands by three bidentate ethane-1,2-diamine (en) ligands:
[M(H₂O)₆]^(n+) + 3en ⇌ [M(en)₃]^(n+) + 6H₂O
Particle count: left-hand side 1 + 3 = 4 species; right-hand side 1 + 6 = 7 species. Net gain of three free particles per reaction. ΔS° for this kind of swap is typically about +88 J K⁻¹ mol⁻¹ at 298 K. The entropic contribution to ΔG° is therefore:
TΔS° = 298 × 0.088 = +26.2 kJ mol⁻¹ of favourable free-energy gain from entropy alone.
For the EDTA⁴⁻ swap [M(H₂O)₆]^(n+) + EDTA⁴⁻ → [M(EDTA)]^((n−4)) + 6H₂O the particle change is 2 → 7 — a net gain of five — and the entropic free-energy gain rises to roughly +40 to +50 kJ mol⁻¹ at 298 K. This is why EDTA complexes have stability constants so much higher than the corresponding ammine complexes despite the metal–N bond enthalpies being similar.
Key point: The chelate effect is entropic, not enthalpic. ΔH° for replacing 6 NH₃ by 3 en is close to zero (the M–N bonds being made and broken are essentially identical). The full ~26 kJ mol⁻¹ advantage comes from TΔS°. State this explicitly in any exam answer asking why chelate complexes are more stable than equivalent monodentate complexes.
For [M(H₂O)₆]^(n+) + 6L ⇌ [ML₆]^(n+) + 6H₂O:
K_stab = [ML₆^(n+)] / ([M(H₂O)₆^(n+)] × [L]⁶)
Water does not appear in the expression because it is the solvent and its activity is fixed near unity. K_stab is reported as log₁₀ K_stab to make the numbers manageable. Indicative values for some copper(II) systems:
| Complex | log K_stab |
|---|---|
| [Cu(NH₃)₄(H₂O)₂]²⁺ | ~13 |
| [Cu(en)₂(H₂O)₂]²⁺ | ~20 |
| [Cu(EDTA)]²⁻ | ~18.8 |
The 5–7-order-of-magnitude gap between the NH₃ and en/EDTA complexes is the chelate effect. A complex with a larger K_stab will displace one with a smaller K_stab — for example, addition of EDTA⁴⁻ to a deep-blue solution of [Cu(NH₃)₄(H₂O)₂]²⁺ produces the pale-blue [Cu(EDTA)]²⁻ complex with release of NH₃.
Example 1 — write the balanced equation. Concentrated HCl is added to aqueous CuSO₄. Write the equation for the substitution that produces the yellow species.
[Cu(H₂O)₆]²⁺(aq) + 4Cl⁻(aq) ⇌ [CuCl₄]²⁻(aq) + 6H₂O(l)
Check charge: left (+2) + (−4) = −2; right (−2) + 0 = −2. Balanced.
Example 2 — predict the shape change. What is the geometry of the product in Example 1, and why does it differ from the reactant?
The reactant [Cu(H₂O)₆]²⁺ is six-coordinate, octahedral (Jahn–Teller distorted). The product [CuCl₄]²⁻ is four-coordinate, tetrahedral, because Cl⁻ is a larger ligand than H₂O and only four can fit around the smaller central ion.
Example 3 — predict the colour from electron count. Cu²⁺ is d⁹. Both the aqua and the ammine complexes have d–d transitions, but the ammine is deeper in colour. Suggest a reason.
NH₃ is a stronger-field ligand than H₂O (it is higher in the spectrochemical series — lesson 7). It produces a larger d-orbital splitting Δₒct, so the absorbed photon has higher energy (shorter wavelength). For Cu²⁺, the absorption shifts from the red region (giving pale blue) towards the yellow-orange region (giving deep royal blue). The intensity also rises slightly because of changes in the symmetry environment around Cu(II) on going from six H₂O to four NH₃ plus two H₂O.
RP12 in the AQA practical handbook is the synthesis of a transition-metal complex (typically a cobalt(II) or chromium(III) ammine, or hexaammine nickel(II) chloride). The generic procedure is:
A spot of product solution is placed on a TLC plate alongside a spot of an authentic reference (e.g. the starting salt). The plate is developed in a suitable solvent — often a polar/non-polar mixture — and the relative Rf values reveal whether the product is a single complex or a mixture (e.g. unreacted starting material plus product). A pure product shows a single spot at an Rf distinct from the reactant. TLC is preferred over column chromatography for routine purity checks because it is fast, cheap, and uses tiny sample volumes — the principles transfer directly to the analytical-chemistry course (paper, TLC, gas, and HPLC chromatography all share the same partition-equilibrium underpinning).
A staple A-Level demonstration. Set up ~10 cm³ of 0.1 mol dm⁻³ aqueous CuSO₄ in a small beaker. Add dilute (0.1 mol dm⁻³) aqueous ammonia dropwise from a teat pipette with continuous swirling. Observe:
The sequence pale blue → milky pale-blue precipitate → deep blue solution is one of the most memorable colour sequences in inorganic chemistry and is frequently tested as a sequence-of-observations question. The two distinct chemical events — acid–base deprotonation, then substitution proper — must be named separately to score full marks.
Question 1. [13 marks total]
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