Question

why are Co2+ complexes labile with respect to ligand exchange while Co3+ complexes are relatively inert?

Answer 1

For an inert complex, it is a large activation energy which prevents ligand substitution. Inert complexes are therefore kinetically stable compounds. example,

[Co(NH₃)₆]³⁺(aq) + 6H₃O⁺(aq) ------> [Co(H₂O)₆]³⁺(aq) + 6NH₄⁺(aq),

. The large equilibrium constant suggests that the complex ion [Co(NH₃)₆]³⁺ is thermodynamically unstable. This reaction is highly thermodynamically favored yet the inert [Co(NH₃)₆]³⁺ complex ion lasts for weeks in acidic solutions because the rate of the reaction is very slow. Thus, the large activation energy acts as an efficient barrier for ligand substitution rendering the [Co(NH₃)₆]³⁺ ion thermodynamically metastable.

example of a labile complex:

[Co(NH₃)₆]²⁺(aq) + 6H₃O⁺(aq) ------> [Co(H₂O)₆]²⁺(aq) + 6NH₄⁺(aq).

This reaction is virtually complete in a few seconds. The [Co(NH₃)₆]²⁺ complex is thermodynamically unstable and also labile. Notice that the only difference between the two ammine cobalt complexes shown in the examples above is the oxidation number of the cobalt atom. The inert complex has Co(III) while the labile one has Co(II). Both ammine complexes are octahedral and in the case of Co(III), a d⁶species, the t_2g levels are filled. Co(II), on the other hand, has partially filled e_g orbitals. It is straightforward to demonstrate lability since changes will occur and ligand substitution in complexes is normally accompanied by a color change. Inertness, however, is somewhat dull since nothing happens