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Spec Mapping — OCR H432 Module 5.3.1 — Transition elements (ligand substitution), covering ligand-substitution reactions in aqueous transition-metal complexes; substitutions in which the coordination number is retained (e.g. [Cu(H₂O)₆]²⁺ + 4 NH₃ → [Cu(NH₃)₄(H₂O)₂]²⁺) and substitutions in which the coordination number changes (e.g. [Cu(H₂O)₆]²⁺ + 4 Cl⁻ → [CuCl₄]²⁻; [Co(H₂O)₆]²⁺ + 4 Cl⁻ → [CoCl₄]²⁻); the chelate effect and the entropy-driven stability of polydentate-ligand complexes such as [Cu(en)₃]²⁺ and [M(EDTA)]^(n−4) (refer to the official OCR H432 specification document for exact wording).
Ligand-substitution chemistry is the most heavily examined reaction class in OCR transition-metal chemistry, and it ties together every conceptual thread from the previous four lessons: ligand definitions and dentate counts (lesson 3), the IUPAC naming of the products (lesson 4), the spectrochemical-series logic that governs colour change (lesson 2 background), and — looking ahead — the qualitative-analysis flowchart of lesson 8 and the quantitative colorimetry of lesson 9. The OCR examiner is essentially asking the same three questions on every substitution problem: which ligand wins?, does the coordination number change?, and what colour change is observed? This lesson sets out the four canonical reactions OCR uses to discriminate band candidates — the [Cu(H₂O)₆]²⁺ + Cl⁻ → [CuCl₄]²⁻ coordination-number drop, the [Cu(H₂O)₆]²⁺ + NH₃ → [Cu(NH₃)₄(H₂O)₂]²⁺ Jahn–Teller-driven partial substitution, the [Co(H₂O)₆]²⁺ + Cl⁻ → [CoCl₄]²⁻ pink-to-blue test for hydration water, and the chelate-effect entropy argument for why polydentate ligands (en, EDTA) form unusually stable complexes — and rehearses each in turn with worked equations, observation tables, and tiered specimen-band answers.
Key Definition — Ligand substitution is a reaction in which one or more ligands in a complex ion are replaced by an equal or different number of incoming ligands, while the central metal ion and its oxidation state are unchanged. The coordination number may be retained (when the incoming ligand is similar in size to the leaving ligand, e.g. NH₃ for H₂O) or it may change (when the incoming ligand is significantly larger, e.g. Cl⁻ for H₂O, where the larger Cl⁻ forces a drop from six-coordinate octahedral to four-coordinate tetrahedral). The reaction may be partial (some ligands replaced) or complete (all ligands replaced), and is driven by a combination of thermodynamic (bond enthalpy, electrostatic) and entropic (chelate effect, number of free particles) factors.
By the end of this lesson you should be able to:
A ligand substitution is a reaction in which one or more ligands around a transition-metal centre are replaced by others. The central metal ion is unchanged, its formal oxidation state is unchanged, and only the inner coordination sphere is rearranged.
In general form:
mathrm[M(L1)n]a++x,L2rightleftharpoons[M(L1)n−x(L2)x]a++x,L1
The position of the equilibrium — i.e. which ligand "wins" — depends on three factors:
OCR examines four canonical aquated-complex starting points: [Cu(H₂O)₆]²⁺ (pale blue), [Co(H₂O)₆]²⁺ (pink), [Cr(H₂O)₆]³⁺ (violet–green), and [Fe(H₂O)₆]³⁺ (yellow, lesson 8). The reactions in this lesson use the first three; iron chemistry is consolidated in qualitative-analysis lesson 8.
This is the single most-examined reaction set in OCR transition-metal chemistry. The starting complex is pale-blue hexaaquacopper(II), [Cu(H₂O)₆]²⁺, containing a d⁹ Cu²⁺ ion at the centre of a Jahn–Teller-distorted octahedron (four short equatorial Cu–O bonds, two longer axial Cu–O bonds). Four substitution outcomes are tested by adding (i) excess concentrated chloride, (ii) excess concentrated ammonia, (iii) a few drops of NaOH, and (iv) the bidentate ligand 1,2-diaminoethane (en).
Adding excess concentrated HCl (or saturated NaCl) to a pale-blue [Cu(H₂O)₆]²⁺ solution gives a striking yellow solution of tetrachlorocuprate(II):
mathrm[Cu(H2O)6]2+(aq)+4,Cl−(aq)rightleftharpoons[CuCl4]2−(aq)+6,H2O(l)
| Property | Reactant | Product |
|---|---|---|
| Formula | [Cu(H₂O)₆]²⁺ | [CuCl₄]²⁻ |
| Colour | Pale blue | Yellow |
| Coordination number | 6 | 4 |
| Geometry | Octahedral | Tetrahedral |
| Overall charge | +2 | −2 |
| IUPAC name | hexaaquacopper(II) | tetrachlorocuprate(II) |
Three things change simultaneously: the coordination number drops 6 → 4, the geometry changes octahedral → tetrahedral, and the overall charge changes +2 → −2 (four chlorides at −1 each contribute −4, and Cu²⁺ contributes +2, giving net −2). The reason for the coordination-number drop is steric: Cl⁻ has an ionic radius of about 181 pm, much larger than the effective radius of H₂O (~140 pm), and only four Cl⁻ can pack around a Cu²⁺ ion without uncomfortable ligand–ligand repulsion. Tetrahedral [CuCl₄]²⁻ has Cl–Cu–Cl bond angles close to 109.5°.
The reaction is reversible: diluting with water reverses the equilibrium back to the blue [Cu(H₂O)₆]²⁺, and an intermediate green colour is often observed during the transition because the solution contains a mixture of yellow [CuCl₄]²⁻ and blue [Cu(H₂O)₆]²⁺ (blue + yellow → green). The colour change is therefore: pale blue → green → yellow as [Cl⁻] is increased; yellow → green → pale blue as the solution is diluted.
Adding ammonia in stages — a few drops, then excess — gives a two-stage colour change.
Stage 2a — A few drops of NH₃ produces a pale-blue precipitate. Ammonia is a weak base, and at low concentration it removes protons from coordinated water rather than substituting as a ligand. The product is the neutral copper(II) hydroxide complex:
mathrm[Cu(H2O)6]2+(aq)+2,NH3(aq)rightarrowCu(OH)2(H2O)4(s)+2,NH4+(aq)
This is an acid–base reaction (deprotonation of two coordinated waters), not a substitution. It is identical in product to the reaction with NaOH(aq).
Stage 2b — Excess NH₃ dissolves the precipitate to a deep royal blue. With excess ammonia, the lone pairs on NH₃ become the dominant nucleophile, and four ammines replace the four equatorial waters in a ligand substitution:
mathrmCu(OH)2(H2O)4(s)+4,NH3(aq)rightarrow[Cu(NH3)4(H2O)2]2+(aq)+2,OH−(aq)+2,H2O(l)
Or equivalently, written directly from the starting hexaaqua complex:
mathrm[Cu(H2O)6]2+(aq)+4,NH3(aq)rightleftharpoons[Cu(NH3)4(H2O)2]2+(aq)+4,H2O(l)
The key A-level discrimination point is that only four out of six waters are replaced — not all six. The product is tetraamminediaquacopper(II), with four NH₃ ligands in the equatorial plane and two H₂O ligands in the axial positions. The reason is the Jahn–Teller distortion of d⁹ Cu²⁺: the four equatorial Cu–O bonds are shorter and stronger than the two axial Cu–O bonds, so the equatorial sites are kinetically more accessible to substitution. Net result: equatorial waters substitute readily; axial waters do not.
| Stage | Product | Colour | Mechanism |
|---|---|---|---|
| Start | [Cu(H₂O)₆]²⁺ | Pale blue | — |
| Few drops NH₃ | Cu(OH)₂(H₂O)₄ | Pale blue gelatinous ppt | Acid–base (deprotonation) |
| Excess NH₃ | [Cu(NH₃)₄(H₂O)₂]²⁺ | Deep royal blue | Ligand substitution (4 equatorial sites) |
Adding NaOH(aq) gives the same pale blue precipitate as a few drops of ammonia:
mathrm[Cu(H2O)6]2+(aq)+2,OH−(aq)rightarrowCu(OH)2(H2O)4(s)+2,H2O(l)
Critically, this precipitate does not dissolve in excess NaOH. Copper(II) hydroxide is not amphoteric — it does not redissolve in either excess acid or excess base in the same way that Cr(OH)₃ does. So the pale-blue gelatinous precipitate persists regardless of how much NaOH is added. This behaviour is the qualitative-analysis discriminator for Cu²⁺ in lesson 8.
1,2-diaminoethane (en, H₂N–CH₂–CH₂–NH₂) is a bidentate ligand: it forms five-membered chelate rings with the metal (M–N–C–C–N). Adding excess en to [Cu(H₂O)₆]²⁺ gives the tris-chelated complex:
mathrm[Cu(H2O)6]2+(aq)+3,en(aq)rightleftharpoons[Cu(en)3]2+(aq)+6,H2O(l)
Note the particle accounting: on the left there are 4 free particles (1 complex + 3 en); on the right there are 7 free particles (1 complex + 6 H₂O). The reaction gains 3 particles, so ΔS > 0. This entropic stabilisation is the chelate effect and explains why bidentate (and higher-dentate) ligands form unusually stable complexes — the equilibrium lies far to the right even when bond-enthalpy contributions are roughly equal between Cu–N(en) and Cu–N(NH₃).
The cobalt(II) hexaaqua complex is pink in solution and reacts in similar (but not identical) ways.
mathrm[Co(H2O)6]2+(aq)+4,Cl−(aq)rightleftharpoons[CoCl4]2−(aq)+6,H2O(l)
Pink octahedral [Co(H₂O)₆]²⁺ → intense blue tetrahedral [CoCl₄]²⁻. The same steric argument applies as for Cu²⁺: Cl⁻ is large, so only four fit, and the coordination number drops 6 → 4 with shape change octahedral → tetrahedral. The deep blue colour is dramatic and is the basis of cobalt chloride paper — a strip impregnated with anhydrous CoCl₂ goes pink on contact with water (rehydrates to [Co(H₂O)₆]²⁺) and blue when dry (CoCl₂ ≈ tetrahedral coordination). This is the classic A-level qualitative test for water.
mathrm[Co(H2O)6]2+(aq)+6,NH3(aq)rightleftharpoons[Co(NH3)6]2+(aq)+6,H2O(l)
Unlike Cu²⁺, the d⁷ Co²⁺ ion is not Jahn–Teller-distorted in the same way (d⁷ high-spin has one electron in the eg pair, not three as in d⁹), so all six waters can be replaced, giving hexaamminecobalt(II). The product is straw/pale-brown and oxidises slowly in air to the golden-yellow Co(III) complex [Co(NH₃)₆]³⁺. The Co(III) ammine is one of Werner's original 1893 complexes and was instrumental in establishing the octahedral six-coordinate geometry of cobalt(III).
mathrm[Co(H2O)6]2+(aq)+2,OH−(aq)rightarrowCo(OH)2(H2O)4(s)+2,H2O(l)
The precipitate is initially blue but darkens to brown-black on standing in air as it oxidises to Co(III) hydroxide. The precipitate does not dissolve in excess NaOH.
Chromium(III) is the most amphoteric of the standard OCR transition metals: Cr(OH)₃ dissolves in both excess acid and excess alkali.
mathrm[Cr(H2O)6]3+(aq)+3,OH−(aq)rightarrowCr(OH)3(H2O)3(s)+3,H2O(l)
Grey-green gelatinous precipitate.
mathrmCr(OH)3(H2O)3(s)+3,OH−(aq)rightarrow[Cr(OH)6]3−(aq)+3,H2O(l)
Deep green solution of hexahydroxochromate(III). This is the amphoteric step that distinguishes Cr(III) from Cu(II), Fe(II), Fe(III) in the qualitative-analysis flowchart (lesson 8).
mathrmCr(OH)3(H2O)3(s)+3,H+(aq)rightarrow[Cr(H2O)6]3+(aq)
The precipitate also redissolves in excess acid to regenerate the violet-green aqua complex.
mathrm[Cr(H2O)6]3+(aq)+6,NH3(aq)rightleftharpoons[Cr(NH3)6]3+(aq)+6,H2O(l)
Cr(III) is not Jahn–Teller-distorted (d³ is symmetric in the t₂g level), so all six ammines substitute and the coordination number is retained at 6.
The contrast between the NH₃ substitution (no coordination-number change) and the Cl⁻ substitution (drops 6 → 4) is OCR's favourite discriminator:
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