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CHEM20018 · Chemistry: Reactions And Synthesis

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Chapter 7 of 9 · CHEM20018

Coordination Chemistry: Stability, HSAB & Substitution

Coordination chemistry quantifies and explains the stability and reactivity of metal complexes. Stability is captured by stepwise formation constants Kf1…Kfn and the overall βn, linked to free energy by ΔG° = −RT ln K = ΔH° − TΔS°. Trends are rationalised by the Irving–Williams series (a stability maximum at Cu²⁺), the entropy-driven chelate effect, and the HSAB principle of hard/soft acid–base matching. Reactivity is about kinetics — labile vs inert complexes and associative vs dissociative mechanisms — with square-planar Pt(II) substitution (and the medicinal example of cisplatin) as the showcase.

In this chapter

What this chapter covers

  • 01Stepwise formation constants Kf1…Kfn and the overall stability constant βn = Kf1·Kf2·…·Kfn
  • 02Linking stability to thermodynamics: ΔG° = −RT ln K = ΔH° − TΔS°
  • 03Irving–Williams series Mn < Fe < Co < Ni < Cu > Zn and its electrostatic + CFSE + Jahn–Teller explanation
  • 04Chelate effect: multidentate ligands bind more strongly, driven by favourable entropy; 5- and 6-membered rings best
  • 05HSAB principle: hard acids prefer hard bases, soft acids prefer soft bases
  • 06Substitution kinetics: labile vs inert (a kinetic, not thermodynamic, distinction)
  • 07Associative (A/Ia) vs dissociative (D/Id) mechanisms and rate-law evidence
  • 08Square-planar Pt(II) substitution, the trans effect, and the cisplatin mechanism
Worked example · free

Overall stability constant β₃ to ΔG° of complex formation

Q [6 marks]. For the complex [M(en)₃]²⁺ (en = ethylenediamine, a bidentate ligand), the stepwise formation constants at 298 K are Kf1 = 4.0×10⁷, Kf2 = 1.5×10⁶ and Kf3 = 5.0×10⁴. (a) Deduce the likely coordination number and geometry. (b) Calculate the overall ΔG° of formation. (R = 8.314 J K⁻¹ mol⁻¹.)
  • 1 mark — CN 6, octahedral(a) Each en is bidentate (two N-donors). Three en ligands therefore supply 3 × 2 = 6 donor atoms, so the coordination number is 6 and the geometry is octahedral.
  • 2 marks — correct β₃(b) The overall stability constant is the product of the stepwise constants: β₃ = Kf1 × Kf2 × Kf3 = (4.0×10⁷)(1.5×10⁶)(5.0×10⁴) = 3.0×10¹⁸.
  • 1 mark — ln β₃Take the natural log: ln β₃ = ln(3.0×10¹⁸) = ln 3.0 + 18 ln 10 = 1.099 + 41.45 = 42.55.
  • 2 marks — substitution + final value with correct signApply ΔG° = −RT ln β₃ = −(8.314)(298)(42.55) = −1.054×10⁵ J mol⁻¹ ≈ −105 kJ mol⁻¹.
(a) Coordination number 6, octahedral. (b) β₃ = 3.0×10¹⁸ and ΔG° = −RT ln β₃ ≈ −105 kJ mol⁻¹ (formation is strongly favourable).
Sia tip — βn is always the PRODUCT of the stepwise Kf values, so ln β is the SUM of their logs — handy if you are given individual log K values. Keep R in J (8.314) for an answer in J mol⁻¹, then convert to kJ. A large negative ΔG° is expected because each Kf is large; a positive answer means a sign slip.
Glossary

Key terms

Formation (stability) constant
The equilibrium constant for binding a ligand to a metal; stepwise constants Kf1…Kfn multiply to give the overall βn.
Irving–Williams series
The general order of stability for high-spin divalent first-row metal complexes, Mn²⁺ < Fe²⁺ < Co²⁺ < Ni²⁺ < Cu²⁺ > Zn²⁺, peaking at Cu²⁺.
Chelate effect
The greater stability of complexes formed by multidentate ligands compared with equivalent monodentate ligands, arising mainly from a favourable entropy change.
HSAB principle
Hard (small, high-charge) acids bind hard bases and soft (large, polarisable) acids bind soft bases most strongly; 'hard–hard' and 'soft–soft' combinations are the most stable.
Labile vs inert
A kinetic classification: labile complexes undergo ligand substitution rapidly, inert complexes slowly — independent of how thermodynamically stable the complex is.
FAQ

Coordination Chemistry: Stability, HSAB & Substitution FAQ

Why does the Irving–Williams series peak at copper?

Three effects combine across the row: the M²⁺ ionic radius contracts, strengthening electrostatic metal–ligand attraction; crystal-field stabilisation energy (CFSE) adds and is large near d⁸ (Ni²⁺); and Cu²⁺ (d⁹) gains extra Jahn–Teller stabilisation. Zn²⁺ (d¹⁰) has no CFSE and is larger, so stability drops after the Cu²⁺ maximum.

Is the chelate effect driven by enthalpy or entropy?

Mainly entropy. Replacing several monodentate ligands with one multidentate ligand releases more free particles into solution, increasing the entropy of the system, which makes ΔG° more negative. Enthalpy differences are usually small, so the chelate effect is described as entropy-driven.

Why does cisplatin work and its trans isomer not?

cis-[Pt(NH₃)₂Cl₂] aquates to lose chloride and then binds two adjacent guanine bases on DNA, forming a cis-1,2 intrastrand cross-link that distorts the helix and blocks replication. The trans isomer cannot bridge adjacent bases in the same way, so it lacks the same cytotoxic activity — geometry, set by square-planar Pt(II) substitution chemistry, is decisive.

Study strategy

Exam move

Split your revision into stability (thermodynamics) and reactivity (kinetics). On the stability side, practise the βn = ∏Kf → ΔG° = −RT ln β pipeline numerically, and be able to explain the Irving–Williams maximum, the entropy origin of the chelate effect, and HSAB matching in one or two crisp sentences each — these are guaranteed short-answer marks. On the kinetics side, distinguish labile/inert (kinetic) from stable/unstable (thermodynamic) carefully, since mixing them up is a classic error. Learn the associative mechanism of square-planar Pt(II), the trans effect ordering, and the cisplatin DNA-binding story as a worked case that ties the whole section together. Donnelly's section rewards qualitative reasoning, so prioritise being able to argue trends over rote facts.

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