Why is osmium so dense despite there being heavier elements after it in the periodic table?
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2$\begingroup$ Is there any agreement as to whether Iridium is denser or Osmium? They're very close. $\endgroup$– stoic-santiagoApr 26, 2019 at 16:32
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1 Answer
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The density of an element is related to how many atoms can be placed in a given volume and the weight of the nuclei. Therefore, the smaller the atomic radius of an atom and the higher the atomic number of the nucleus, the greater the density of the element. The very small atomic radius of osmium results in a small metal-metal separation. This small atomic separation along with osmium's relatively high atomic number gives rise to osmium's high density.
The small atomic radius can be attributed to the following factors:
f orbitals are very diffuse and therefore result in poor screening of the electrons further out. In the case of osmium ($\ce{[Xe] 4f^{14} 5d^6 6s^2}$) the poor screening by its 4f orbitals leads to a contraction in the n=5 and n=6 orbitals.
Due to osmium's high atomic number, relativistic effects come into play. These relativistic effects have been discussed previously on SE Chem for gold and mercury. A general discussion as well as the mathematics describing the relativistic effect can be found in these earlier posts. Basically, for heavy nuclei the s electrons must circulate at relativistic speeds in order to maintain a stable orbit (not fall into the nucleus). Therefore the electron mass increases and the s orbital radius decreases (the p orbital radius also decreases, but to a lesser degree).
The orbital contraction caused by these two effects leads to a small atomic radius for osmium. Therefore, short metal-metal bonds result. This is reflected in the small unit cell volume for osmium (27.96 cubic angstroms). For comparison the unit cell volume for lead is 121.3 cubic angstroms! As a result, more osmium atoms can pack into a given volume compared to other elements
Osmium's relatively high atomic number along with its small atomic radius, as explained above, gives rise to osmium's high density.
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2$\begingroup$ That, and the fact that it has a close-packed structure. $\endgroup$– GimelistDec 14, 2014 at 5:33
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1$\begingroup$ Do the relativistic effects reduce the OVERALL size if the atom or just the inner s orbitals (and slightly p)? so basically, is this effect more significant it increasing the mass of the atom or decreasing it's size? $\endgroup$– RobChemDec 14, 2014 at 13:34
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1$\begingroup$ @Michael Lead has a face centered cubic crystalline structure while osmium crystallizes in a hexagonal close packed structure. In both structures the lead and osmium atoms occupy around 74% of the volume, yet the densities of lead and osmium are very different (11.34 and 22.61 g/cm^3 respectively). So I don't see how the crystal structure, per se, plays into the answer. I fall back to unit cell size, which reflects osmium's small radius, and atomic weight. If I'm missing something, please elaborate. $\endgroup$– ronDec 14, 2014 at 15:55
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1$\begingroup$ @user3764899 The relativistic effects reduce the size of all orbitals that have significant density at the nucleus. Therefore, all s and, to a lesser degree, p orbitals are contracted. Osmium's 6s orbital (its outermost orbital) is consequently contracted, so overall osmium's size (atomic radius) is reduced. Look here and notice how osmium's radius sticks out like a sore thumb. Its radius is by far the lowest for its atomic number. $\endgroup$– ronDec 14, 2014 at 16:04
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2$\begingroup$ @ron true. I wasn't saying that it's the only reason - but it definitely helps. A small size does not help if the structure is not close-packed. $\endgroup$– GimelistDec 14, 2014 at 16:09