Substitution in Octahedral Complexes: Dissociative or Associative

The two limiting mechanisms

Why octahedral is usually dissociative

Six ligands already surround the metal, leaving little room for a seventh to approach. Adding one is sterically costly, whereas losing one relieves crowding. So the dissociative route is generally lower in energy.

Square planar complexes are the opposite case: the metal is exposed above and below the plane, so an incoming ligand can approach easily and the associative route dominates. Contrasting the two geometries on steric grounds is the expected explanation.

Lability and inertness

Some complexes exchange ligands in fractions of a second; others take days. This is kinetic, and must be distinguished from thermodynamic stability.

The pattern follows crystal field activation energy: forming the intermediate changes the geometry, and configurations with large CFSE in the octahedral arrangement lose the most by distorting. That loss is an extra barrier, so they react slowly.

A complex can be thermodynamically unstable yet kinetically inert, persisting for long periods despite being unfavourable. Confusing these two is one of the most common errors in coordination chemistry, and questions often set it up deliberately.

Evidence for the dissociative pathway

• Rates vary enormously with the leaving group but only slightly with the entering group, which is what a rate-determining loss predicts.
• Positive entropies of activation, indicating a more disordered transition state with a ligand partly departed.
• Positive activation volumes, confirming expansion.
• A limiting rate is reached at high entering-ligand concentration, since the dissociation step cannot be accelerated further.

That last point is a good discriminator in a data question: a rate that saturates as the entering ligand concentration rises indicates a dissociative mechanism.

Base hydrolysis and the conjugate base mechanism

Hydrolysis of some complexes in basic solution is far faster than in neutral solution, and the rate depends on hydroxide concentration — which looks associative.

The accepted explanation is different. Hydroxide deprotonates a coordinated amine ligand, and the resulting amide ligand is a much stronger donor, which greatly accelerates loss of the leaving group by stabilising the five-coordinate intermediate. So the mechanism remains dissociative, with hydroxide acting as a base rather than a nucleophile.

The decisive evidence is that the effect requires an N–H bond on a ligand: complexes with no such proton show no rate enhancement. Being able to cite that test is what distinguishes a full answer.

Frequently asked questions

How is lability different from instability?

Lability is kinetic, describing how fast ligands exchange. Instability is thermodynamic, describing where equilibrium lies. The two are independent, and complexes exist in all four combinations.

Why are low-spin d⁶ complexes so inert?

Because their filled t_2g set gives maximum crystal field stabilisation in the octahedral geometry, so distorting toward any intermediate costs a great deal of that stabilisation.

How is activation volume measured?

From the pressure dependence of the rate constant. A rate that falls with increasing pressure indicates a positive activation volume and a dissociative mechanism.

Why does base hydrolysis not prove an associative mechanism?

Because hydroxide acts as a base rather than a nucleophile, deprotonating a ligand to create a stronger donor. The substitution itself remains dissociative, and the requirement for an N–H bond confirms it.