Why do europium and ytterbium have lower densities than expected in comparison to other lanthanides? I know it has something to do with the fact that they have half-full and full 4f subshells in the +2 oxidation state, but how exactly does it relate to density? And how does it affect the melting point?

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  • 3
    $\begingroup$ Why do you think that properties should vary monotonically? Given the wide variety of room temperature crystal structures across the lanthanides, one really should not be all that surprised that things change. $\endgroup$
    – Jon Custer
    Jun 20, 2016 at 14:42
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    $\begingroup$ Mostly because these two tend to be 2+, while the rest is 3+. $\endgroup$ Jun 20, 2016 at 14:46

2 Answers 2


Expanding upon @Ivan's comment: Eu and Yb can access the +2 oxidation state instead of +3, due to the +2 ions having a relatively stable half-filled or filled f subshell.

That includes the embedded ions in the structure of the metal. Eu and Yb have $\ce{M^2+}$ instead of $\ce{M^3+}$ ions in the metal. The extra electron and reduced effective nuclear charge on the outer subshells make those $\ce{M^2+}$ ions larger, therefore these metals have lower densities than the surrounding lanthanides. The lower ion charge also means fewer electrons binding the ions together, making Eu and Yb easier to melt.

The difference in ion charge can also show up in other ways. Eu and Yb, for instance, resemble heavy alkaline earth metals more than other lanthanides by dissolving in liquid ammonia. Europium is also known to appear in some rocks alongside alkaline earth metals rather than alongside other lanthanides (see Wikipedia):

In anaerobic, and particularly geothermal conditions, the divalent form is sufficiently stable that it tends to be incorporated into minerals of calcium and the other alkaline earths. This ion-exchange process is the basis of the "negative europium anomaly", the low europium content in many lanthanide minerals such as monazite, relative to the chondritic abundance.

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    $\begingroup$ "Eu and Yb do favour the +2 oxidation state instead of +3" is a bit strong I think, in most situations the 2+ ions are very good reducing agents, Yb2+ in particular. But yes, at least part of the difference is due to these two lanthanides having a chemically accessible 2+ state. The interesting case is Sm, which is the next most stable 2+ Lanthanide, but doesn't show the above anomalies. $\endgroup$
    – Ian Bush
    Jun 23, 2016 at 19:58

As you likely know, the structure of metals is a lattice of cations surrounded by freely moving electrons. The structure of most lanthanides consists of +3 anions, but the electronic structures of europium ($\mathrm{(4f)^7 (6s)^2}$) and ytterbium ($\mathrm{(4f)^{14} (6s)^2}$) mean that the +2 state is more stable. The lower charge difference gives weaker metallic bonding and therefore lower density.


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