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?
Answer
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.
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