I've already posted this question in Physics Stack Exchange, but the answer that I received (actinide contraction similar to lanthanide contraction) is not convincing for me, or at least is not detailed enough to explain the huge difference.

According to Villars and Daams [Journal of Alloys and Compounds, 197 (1993) 177] the atomic volume of U is 2.073×10−2nm3, whereas that of thorium is much larger, 3.295×10−2nm3.

This is confirmed by comparing bond distances in pure elements. Even if Th and U crystallize with different structures, they have the same coordination number (12). In U the U-U bond distance is 0.275 nm, whereas in Th the Th-Th bond distance is much larger 0.360 nm. Why this difference? This huge difference is not observed for Ce and Pr (where 0.333 and 0.363 bond distances are measured).

The external electronic configurations of Th and U atoms are:

Th: [Rn]6d27s2

U: [Rn]5f36d17s2

What's the origin of this strong difference?

Why, then, their ionic sizes (for example the sizes of Th4+ and U4+) are quite similar?

  • 1
    $\begingroup$ Due different relations of 4f vs 5d and 5f vs 6d orbital energies, the first actinoids, in contrary to the first lanthanoids, behaving more like d than f elements. So, in a sense, you are like comparing $\ce{Hf}$ and $\ce{W}$. OTOH, when ionized enough, this deviation more or less disappears. $\endgroup$
    – Poutnik
    Apr 13 at 6:55
  • $\begingroup$ Are you sure the coordination number of U in the crystalline metal is 12? Greenwood and Earnshaw state "For U, Np and Pu are rather irregular so that the coordination number is not a precise concept". webelements.com/uranium/crystal_structure.html gives the structure as orthorhomobic with space group Cmcm, and references C. S. Barrett, M. H. Mueller, and R. L. Hitterman, Phys. Rev., 1963, 29, 625 $\endgroup$
    – Ian Bush
    Apr 13 at 9:53
  • 1
    $\begingroup$ 12 (anti-cuboctahedron) is the coordination number reported in the Pearson's crystal structure database for the Cmcm structure of U. $\endgroup$
    – gryphys
    Apr 13 at 9:58

1 Answer 1


Uranium has more valence electrons capable of populating bonding molecular orbitals (or bonding electronic bands) when compared to thorium - basically, uranium is closer to the middle of its block, compared to thorium. More bonding interactions will pack the atoms tighter. Indeed, uranium has an atomic weight ~2.5% greater than thorium, but metallic uranium is 63% denser than metallic thorium (I am assuming that these densities were measured for metallic phases with the same packing for both elements).

Now, if the way you define "atomic size" is by using distance measurements performed on the solid metallic elements (e.g. from x-ray diffraction data), you will pick up on this extra bonding effect, and it will look like uranium atoms are much smaller than thorium. Indeed, the ratio of the atomic volumes you quote is 1.59, which is suspiciously close to the (inverse) ratio of 1.63 for the densities of the pure metals, so I'm guessing this is how these values were obtained.

However, you could in principle make a measurement of atomic size for isolated atoms in the gas phase. In this situation, there are no bonding interactions, so uranium and thorium are likely to have much more similar sizes. So as it almost always happens in these discussions, it comes down to what do you mean by size?

The reason there is no such difference between cerium and praseodymium is that the 4f orbitals lack radial nodes and are subject to a considerable effective nuclear charge, so they behave almost as core electrons in the lanthanides, where they should be valence electrons. Therefore, placing or removing electrons in the lanthanide 4f orbitals as you move along the block has comparatively little effect on the bonding interactions for these elements.

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    $\begingroup$ I have since realized the atomic volume ratio of 1.59 is reproduced exactly by the ratio of the densities of the pure metals (1.63) multiplied by the ratio of the atomic masses (232/238 = 0.975), so the atomic volume data quote in the question indeed almost certainly is based on the pure metals. $\endgroup$ Sep 28 at 10:31

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