# Why are the densities of europium and ytterbium anomalously low?

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?

• 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. – Jon Custer Jun 20 '16 at 14:42
• Mostly because these two tend to be 2+, while the rest is 3+. – Ivan Neretin Jun 20 '16 at 14:46

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.
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.