What causes the "Gd break" in the trend of lanthanide-EDTA formation constants?

Smith and Martell obtained a series of data for the binding of trivalent lanthanide ions, $\ce{Ln^3+}$, with various carboxylic acid ligands (amongst them the well-known EDTA).1 A graph of the formation constants is attached ($K\_\mathrm{f} = [\ce{Ln(edta)-}]/[\ce{Ln^3+}][\ce{edta^4-}]$):


[![Graph of formation constants for various lanthanide species](https://i.stack.imgur.com/4Lco0.png)](https://i.stack.imgur.com/4Lco0.png)


Due to the contracted nature of the 4f orbitals, the lanthanide ions characteristically do not exhibit covalency in metal-ligand bonding, which is strongly reflected in their chemistry. Therefore, the almost linear increase in $K\_\mathrm{f}$ going from $\ce{La^3+}$ to $\ce{Lu^3+}$ is entirely to be expected. The ionic radius decreases going across the 4f block, and consequently, the interactions with the ligand (primarily electrostatic in nature) become stronger.


However, as the title suggests, I'm interested in the slight blip at gadolinium ($\ce{Gd^3+}$). The value of $K\_\mathrm{f}$ seems to be lower than would be expected, and the astute reader will notice that this applies not just to the EDTA complexes but also for some of the other ligands investigated.


Is there a reason for this anomaly?


There is a review by Moeller *et al.* from 1965,2 which wrote: 



> 
> For all ligands which have been studied, the gadolinium complex is less stable than would be expected from the simple electrostatic model. This behavior has been called the “gadolinium break” and cannot be explained, as was originally attempted, by assuming a steric effect, since it is still apparent in ligands for which there should be no steric interference.
> 
> 
> 


Of course, this is more than half a century ago. I would be grateful if anybody could provide either a sound theoretical explanation for why the Gd formation constants should be smaller or a newer reference that explains this effect.




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**References:**


1. Smith, R. M.; Martell, A.E. Critical stability constants, enthalpies and entropies for the formation of metal complexes of aminopolycarboxylic acids and carboxylic acids. *Sci. Total Environ.* **1987,** *64* (1–2), 125–147. [DOI: 10.1016/0048-9697(87)90127-6](http://doi.org/10.1016/0048-9697(87)90127-6).
2. Moeller, T.; Martin, D. F.; Thompson, L. C.; Ferrús, R.; Feistel, G. R.; Randall, W. J. The Coordination Chemistry of Yttrium and the Rare Earth Metal Ions. *Chem. Rev.* **1965,** *65* (1), 1–50. [DOI: 10.1021/cr60233a001](http://doi.org/10.1021/cr60233a001).




If you consider the solvent extraction of lanthanides by reagents such as H-DEHPA, then to a first approximation the distribution ratio is given by the equation.


$$\ce{DLn}= k[\ce{H-DEHPA}]^3[\ce{H+}]^{-3}$$


If you plot k as a function of atomic number for DEHPA dissolved in most organic solvents (solvent extraction term is diluent) then the value goes up as a function of Z. But the line has bumps on it.


The reason is that the extraction reaction is


$$\ce{Ln(H2O)\_n^3+(aq) + 3H-DEHPA(org) -> [Ln(DEHPA)3](org) + H^+(aq)}$$


As you go from one side of the lanthanide group to the other the binding of the lanthanide to the anionic DEHPA ligands is likely to become stronger, but the value of $n$ changes in several steps. This will give bumps on the graph of $K$ for both the extraction of the lanthanides and also for their binding to ligands such as EDTA.


As the lanthanide shrinks as it becomes smaller then the hydrated lanthanide $$[Ln(H2O)n]^3+$$ will change in stability, if we assume that for a moment we have a ten coordinate (decaaquo) complex which has a metal atom which is slightly too big for the ten waters, what will happen is that as it shrinks it will go through an optimum size and then become too small for the ten waters. We will then have a complex with nine waters.


Consider the chemical reactions of a trivalent f-block metal with a ligand such as a EDTA tetraanion or something neutral such as a pair of BTBPs. We will have


$$EDTA^(4-) + [Ln(H\_2O)\_n]^(3+) --> [Ln(EDTA)(H\_2O)\_q]^- + (n-q)H\_2O$$


When I checked the cambidge database Ce,Pr, Eu, Gd, Dy, Ho in the solid state has q = 3.


For Yb and Er q = 2


So as n and q do not change at the same time going along the lanthanide series the issue of the number of waters will add some bumps to the graph of stability constant as the stability constant relates to the delta G of the complexation reaction.


If we are changing suddenly the number of waters formed by the reaction then the entropy change (Delta S) of the reaction will change suddenly. This will introduce some sudden changes into the graph of binding constant. You might also want to look at a paper by Mike Hudson, Mike Drew, Mark Foreman and Karren Kennedy. In this paper they form the coordiantion complexes of all the non radioactive lanthanides with a ligand (not EDTA) and then you can compare the strucutres. When I last looked I noticed that the La, Ce and Lu complexes were slightly different to those in the middle.


You might help yourself if you were to compare all the crystal strcutures of the lanthanide EDTA complexes, choose only those which are merely hydrated.


M.G.B.Drew, M.R.StJ.Foreman, M.J.Hudson, K.F.Kennedy, Inorganica Chimica Acta, 2004, 357, 4102, DOI: 10.1016/j.ica.2004.06.032


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