What is the reason for deviation of ytterbium from the general trend of lanthanides in terms of hardness, melting point, density etc.?

  • 2
    $\begingroup$ You can find many 'deviations' in the periodic table - nobody said it would be a set of monotonic series, so don't be surprised when things are not simple. One thing to note for the three properties you listed is that Yb is a fcc crystal at room temperature, while the immediate neighbors (Gd to Hf) are all hcp (although many, including Yb, go bcc at higher temperatures). Why fcc? Well, the electronic structure happens to be happier in that crystal structure based on a number of competing factors. $\endgroup$
    – Jon Custer
    Commented Feb 24, 2016 at 15:03

1 Answer 1


To compliment Jon Custer's comment, we can likely trace this deviation to the electron configuration of ytterbium. Note that ytterbium is not the only deviation among the lanthanides - Europium deviate as well:

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Why is that? Note the electron configurations of europium and ytterbium:

  • Europium - $\ce{[Xe]} 4f^7 6s^2$
  • Ytterbium - $\ce{[Xe]} 4f^{14} 6s^2$

In europium, the $4f$ subshell is half-filled, an arrangement that is particularly desirable from a quantum mechanics standpoint. In ytterbium the $4f$ subshell is completely filled, another 'happy' arrangement. These configurations lead to changes in bonding behavior, for example noting that most lanthanides have the hexagonal close pack or the double hexagonal close pack crystal structures. The only ones that are different are samarium, which is rhombohedral, europium which is body-centered cubic, and ytterbium with is face-centered cubic.

The link between electron configuration and physical properties exists elsewhere in the periodic table, but the trends are not as clear. However, transition elements with filled or half-filled $d$ subshells are likely to have lower melting points than their neighbors.


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