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

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.


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.

  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.

  • 1
    Two questions. 1) Can you correct the 1987 DOI? The given link fails. 2) Have you considered crystal field theory? It seems to fit the pattern quoted here. – Oscar Lanzi Mar 25 '17 at 11:38
  • 2
    1) Thanks, I have fixed it. 2) Yes, CF effects would produce the pattern shown here with a minimum at $\mathrm{f^7}$. However I'm not entirely convinced that CF effects would play a significant role in the chemistry of the lanthanides. I'm working on the premise that the interaction is electrostatic - although I would certainly not mind being proven wrong. – orthocresol Mar 25 '17 at 12:26
  • 1
    I'm also now curious about the transitions at dysprosium: why does malonate ~plateau and, especially, why does DTPA fully turn over?? – hBy2Py Mar 25 '17 at 12:45
  • @hBy2Py I didn't read ref 1 fully (this was originally something I saw in an exam question and I just wanted a reference for the data), but my first guess is that it could be to do with coordination numbers. The authors listed $\Delta H$ and $\Delta S$ values inside there, too, so you may find an explanation inside there. – orthocresol Mar 25 '17 at 18:24
  • 1
    <nod>, ortho, I agree -- DTPA has a seventh ligand site, so it could well be just a steric effect there -- when the cations get small enough, that seventh ligating position can't bind effectively and that frustrated ligand messes with the free energy. Not sure of the malonate, though -- why it'd be level, instead of decreasing, if it were a steric effect there, too. – hBy2Py Mar 25 '17 at 18:55

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.

From $\ce{La}$ to $\ce{Eu}$, the f orbitals are being filled with 1 electron per orbital (all unpaired). Gadolinium has an anomalous configuration, $\ce{[Xe] 4f^7 5d^1 6s^2}$ because it acquires greater stability by promoting an electron to the orbital d, and keeping all the electrons in f unpaired.

However, from $\ce{Gd}$ to $\ce{Lu}$, a new stage begins, the electrons begin to pair in the orbitals f. It is this transition that reflects in the constant formation.

Note that for ions, in the complex they have gained electrons leading the reasoning to the same of the atom.

  • May I clarify that, strictly speaking, the f orbitals are being filled only from Ce to Yb. Thus, La has no f electrons; and the f orbital is complete at Yb, with 14 f electrons. Lu has 14 f electrons and adds one 5d electron. This represents the third appearance of one 5d electron, having been seen for the first time in La, and for the second time in Gd. – Pretty fly for a chem nerd Apr 29 at 6:55

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.