# Why does ligand exchange take place? [closed]

1. Why does ligand exchange happen?
2. Are there specific factors that determine when a ligand can replace another ligand?
3. When ligand exchange takes place, does that mean that the "previous" ligand is not bonded to the complex ion at all, or both ligands are bonded?

## 2 Answers

1. Ligand exchange is in equilibrium, so it is driven by thermodynamics
2. There are many reasons one ligand would replace another, but it often has to do with Lewis basicity, HSAB (see trans effect), or the thermodynamics of a subsequent reaction.
3. Ligand Exchange is a broad term for a set of mechanisms that include various types of ligand association and dissociation, and these all occur for different kinetic reasons. I'll try to provide a brief overview, but the answers to all of these questions are varied depending on specific situations.

In the covalent model, both $\ce{1e-}$ "X" ligands and $\ce{2e-}$ "L" ligands are called "ligands", so it's important to point out that we're only talking about L ligands here.

Generally speaking, there are three types of ligand exchange mechanisms: association, interchange, and dissociation, and each mechanism is further broken into the associative and dissociative subclasses, depending on the exact character of the transition state. If the higher energy TS of a ligand association has the geometric character of the dissociation, it is called a "dissociative association" (the Ad mechanism), for instance, but this is somewhat rare. The equilibrium expressions of the Dissociation, Interchange, and Association mechanisms are as follows, respectively:(1)

\begin{align}\ce{ML5X &<=>[k_1][k_{-1}] ML5 + X}\\ \ce{ML5 + Y &->[k_2] ML5Y}\\ \ce{ML5X + Y &->[k_1]ML5Y + X}\\ \ce{ML5X + Y &<=>[k_1][k_{-1}] ML5XY}\\ \ce{ML5XY &->[k_2] ML5Y + X}\end{align}

The exact nature of a ligand exchange mechanism is something that requires computations, and physical chemistry to determine, and you should look at (1) to read more about this. Really, the mechanisms are part of a continuum:

$$\mathrm{D_d \cdots I_d \cdots I \cdots I_a \cdots A_a}$$

Once the ligand is fully displaced, it is no longer part of the inner sphere of coordination, so the "'previous ligand" is no longer bound to the complex, but there exist many ligand exchange mechanisms that involve transient intermediate in which both the leaving and the incoming ligand are bound to the central atom. All associative mechanisms will have such an intermediate, while dissociative and interchange mechanisms will not. Another important example is the formation of a preassociation complex, in which "rapid equilibrium occurs between the incoming ligand and the 6-coordinate reactant to form an ion pair" (1) (or a neutral preassociation complex with strong dipole-dipole interactions, depending on oxidation states). In the case of a preassociation complex, the outgoing ligand is no longer in the inner sphere of coordination, but it is still "bound" to the complex, albeit through weaker noncovalent interactions. Such a reaction has equilibrium like this:

\begin{align}\ce{ML5X + Y &<=>[k_1][k_{-1}] ML5X . Y}\\ \ce{ML5X . Y &->[k_2] ML5Y + X}\end{align}

Ligand exchange kinetic and thermodynamic arguments also frequently invoke the chelate effect and the trans effect.

Another important concept is that the driving force for individual ligand dissociations and associations involved in catalytic cycles often implicates the driving force of the overall cycle. In catalytic reactions, the rate of ligand dissociation or association is not dependent not on the thermodynamics/kinetics of the individual reaction, but instead on the step in the cycle with the highest activation energy. This step is known as the turnover limiting step (2). In a catalytic cycle, all equilibria have the same rate. "On a circular track, on average the same number of trains must pass each point per unit time." (2) For example; in this catalytic alkene hydrogenation, the turnover limiting step is the dissociation of the phosphine from Wilkinson's catalyst (3), so this is the step that determines the turnover frequency (TOF). The rate of association of the ethylene ligand later on is driven by the cycle's overall TOF, so such a ligand association would be indirectly driven by a different step in the cycle. If the final step of the cycle is irreversible, the substrate release might also be a thermodynamic driving force.

$\hskip1.4in$(3)

These are just some of the many possible answers to your questions.

(1) G. L. Miessler, P. J. Fischer, D. A. Tarr. Inorganic Chemistry (5th Ed.), Pearson, 2014, p. 442-444

(2) R. H. Crabtree, The Organometallic Chemistry of the Transition Metals (3rd Ed.), John Wiley & Sons, 2001 p. 225

(3) J. Luo, A. G. Oliver, J. S. McIndoe, Dalton Trans. 2013, 42, 11312-11318

• Np! I recommend you look at Miessler, or another inorganic book if you want to read more about the specifics of ligand exchange. Feb 11, 2017 at 23:26

Ligand bonds are labile. A ligand isn't welded to the complex. Remember that it is an equilibrium.

To switch gears a bit consider water just having $\ce{^1H2O}$. A drop of $\ce{^2H2O}$ is added. The $\ce{^1H}$ and $\ce{^2H}$ atoms exchange quickly and the solution is mostly $\ce{^1H2O}$, with some $\ce{^1H^2HO}$ and only the faintest trace of $\ce{^2H2O}$.