I notice that Gd has a second ionization potential which is significantly higher than would be expected from the general trend in the lanthanides (see this paper p. 945 for a graph). What is the reason for this effect ?

The transition corresponds to $[Xe] 4f^{7} 5d^{1} 6s^{2} => [Xe] 4f^{7} 5d^{1}$. I understand that the 5d is occupied to prevent inter-electron repulsion between antiparallel f electrons, and this can be used to rationalize why the correspnding Tb transition requires less energy, but I'm not sure why the potential should be significantly higer than that for Eu.


1 Answer 1


The transition corresponds to $[Xe] 4f^{7} 5d^{1} 6s^{2} \rightarrow [Xe] 4f^{7} 5d^{1}$.

Well actually, the second ionization energy being plotted in the graph is only the energy to lose the second electron so the transition is

$[Xe] 4f^{7} 5d^{1} 6s^1 \rightarrow [Xe] 4f^{7} 5d^{1}$.

For all the other elements Pr through Yb, there is no 5d electron, and instead an additional 4f electron. In all cases, it is the 6s electron that is being lost in the second ionization.

The more the nuclear charge is screened, the easier it is for the 6s electron to be lost.

The 5d electron is less effective than a 4f electron at screening the 6s electron from the nuclear charge, because 4f electron density is concentrated much closer to the nucleus (see Fig. 2 of below reference). Therefore it takes more energy to loss the 6s electron for gadolinium.

For more information on nuclear charge screening see Atomic Screening Constants from SCF Functions. II. Atoms with 37 to 86 Electrons J. Chem. Phys. 47, 1300 (1967).


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