As I am learning about elements physical properties, i notice that the melting points of Sc, Y, La, Lu decreases down the group. I suppose the increase of atomic mass would result in better Van der Waals interaction and therefore bigger melting point. Any thoughts?

  • 3
    $\begingroup$ There's metallic bonding in there, not vdW, and good luck understanding it. $\endgroup$
    – Mithoron
    Sep 25, 2023 at 16:42
  • 4
    $\begingroup$ Scandium is hcp at room temperature, bcc at the melt (1608K for hcp->bcc, 1814K is the melt). Yttrium is similar, but hcp->bcc is at 1751K and bcc->melt at 1795. Lanthanum starts as dhcp, goes dhcp->fcc at 550K, fcc->bcc at 1134K, and bcc->melt at 1193K. Lutetium stays hcp all the way to the melt at 1936K. Clearly lots of stuff is going on (all those allotropes, etc.). Complex metallic bonding is not really amenable to simple intuition. $\endgroup$
    – Jon Custer
    Sep 25, 2023 at 16:59
  • $\begingroup$ See this question and links therein: chemistry.stackexchange.com/questions/171198/… $\endgroup$ Sep 26, 2023 at 1:36

1 Answer 1


I will restrict my reasoning to the path Sc $\rightarrow$ Y $\rightarrow$ La. I have answered a similar question for the $3d$ series : Why does Cr have higher melting point. I have assumed that the melting point is related to the strength of the bond and the cohesive energy of a rectangular band of a bandwidth $W$.

This time I will assume this time that the strength of the bonds is related to the bandwidth and not directly to the electronic charge with a strength $k \propto W $, this relation is very approximate and not generally exact. Without wanting to lose my audience, I will use the translational operator $\hat{T}$ that generates the hamiltonian : $\hat{H} = -t \sum_{i,j,\sigma} c_{j\sigma}^\dagger c_{i\sigma} + h.c$ with the condition of commutation of a weakly correlated system $[\hat{H}, \hat{T}] = 0$ so that $t \propto W > 0 $, the hopping integral $t$ generated bandwidth, intuitively this translational symmetry is enhances by a strong attractive potential, it means that $t(\ce{Sc})>t(\ce{Y})>t(\ce{La})$, so the melting point associated to this translational symmetry drops from the Sc to La.


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