13.6 Substitution in Complexes: Labile and Inert, Trans Effect, Stability Constants, Chelate Effect

13.6.1 Thermodynamic stability and kinetic inertness ★★★

• Thermodynamic stability is about the equilibrium: how far a complex forms, measured by the formation constant β (or ΔG° = −RT ln β). Kinetic stability is about speed: how fast its ligands are replaced, measured by a rate constant or half-life.SN
• Labile: substitution complete within about 1 min at 25 °C (0.1 M, Taube's working rule); inert: slower. These are kinetic words; “stable” and “unstable” are thermodynamic words Geo 2023 Q1(b) 5m.SN
• The two are independent. [Ni(CN)₄]²⁻ is very stable (log β₄ ≈ 30) yet labile (¹⁴CN⁻ exchanges in seconds). [Co(NH₃)₆]³⁺ is thermodynamically unstable in acid (K ≈ 10²⁵ for hydrolysis to [Co(H₂O)₆]³⁺) yet survives for days in 6 M HCl: it is inert Geo 2025 Q7(a) 10m.SN

13.6.2 Which complexes are labile? ★★★

• Water-exchange rate constants span 16 powers of ten: [Cu(H₂O)₆]²⁺ ≈ 10⁹ s⁻¹, [Ni(H₂O)₆]²⁺ ≈ 10⁴, [Cr(H₂O)₆]³⁺ ≈ 10⁻⁶, [Ir(H₂O)₆]³⁺ ≈ 10⁻¹⁰. Higher charge, smaller ion and higher LFSE make a complex inert.SN
• VBT (Taube): outer-orbital (sp³d²) complexes are labile; inner-orbital (d²sp³) complexes are labile only if one t₂g orbital is empty (d⁰–d²). So d³ and low-spin d⁴–d⁶ are inert.SN
• CFT: in a dissociative step the octahedron becomes a square pyramid; the loss of LFSE is the crystal-field activation energy (CFAE). Large positive CFAE = inert.SN

13.6.3 Substitution in octahedral complexes ★★

• Mechanisms (Langford–Gray): D (dissociative, five-coordinate intermediate), A (associative, seven-coordinate intermediate), I (interchange, no intermediate: I_d or I_a). Most octahedral substitutions are I_d: rate depends on the leaving group much more than on the entering group.SN
• Eigen–Wilkins mechanism: the complex and the entering ligand first form an outer-sphere encounter pair (fast pre-equilibrium K_E), then the inner-sphere ligand is lost (rate constant k):

• At low [Y] the reaction looks second order; at high [Y] it saturates at rate = k[complex]. Base hydrolysis of ammine complexes, rate = k[complex][OH⁻], goes by the S_N1CB path (OH⁻ removes an N–H proton; the amido conjugate base loses Cl⁻ fast).SN

13.6.4 Substitution in square-planar complexes ★★★

• Square-planar d⁸ complexes (16 e, empty p_z) react by an associative path: Y attacks above the plane, a five-coordinate trigonal bipyramid forms with T (the ligand trans to X), X and Y in the trigonal plane, then X leaves. The k₁ term is the same path with solvent as the attacking group.SN
• Evidence: rate depends on Y (nucleophilicity order); large negative ΔS‡ and ΔV‡; the geometry is retained (cis gives cis, trans gives trans) IFoS 2022 Q7(b)(i) 10m.SN