The majority of metals are known for appearing grey to our eyes, cf. e.g. Why are most metals gray/silver? But the main exceptions to this are the "famous" group eleven metals, where their distinctive colours have been adequately explained before due to electron bands and the jump from d to s shells (along with the relativistic effects having a large effect on Au) and whatnot, but where does the "blue-ish" tint of Osmium originate from relative to the rest of the heavy d-block elements? Is it also from relativistic origin or is it sill a mystery?

Information about Osmium's colour is nearly elusive on the internet and so I'd appreciate an answer, even if it is a glorified "I don't know" presently.

  • $\begingroup$ Personally, I never saw Osmium. Only through the internet. I'm a bit sad for that now, specially because I thought the blue-ish color of the Osmium in the video I saw with it was just a contaminant of sorts :/ . $\endgroup$
    – Vendetta
    Commented Jun 5, 2018 at 16:49

2 Answers 2


I found reflection spectra of osmium here, and I've graphed them so you can see what they look like:

Full scale

(the article doesn't make clear what $R_1$ and $R_2$ - they are presumably reflection coefficients of some form)

Basically osmium isn't coloured. Compared to gold and copper the spectrum is boringly flat. It's only if we zoom in on the $y$ axis that we can see what's causing the colour:

Zoom in

This shows the reflectivity has a peak for green light (500 - 550nm) and the reflectivity actually falls for blue light (the left side of the graph). The $R_2$ reflectivity holds up at the red end (the right side of the graph) though $R_1$ falls and presumably it's the reduced reflectivity at the red end that causes the blue tint (it would help if the article explained exactly what $R_1$ and $R_2$ are). However this is a very small effect. I suspect it's only the sensitivity of human colour perception that makes the blue tint detectable at all.

Ultimately this is a glorified "I don't know" as you suggested in your question. My point is that there aren't really any absorption features as you get with gold and copper so the slight tint is unlikely to be attributable to a specific electronic feature.

A footnote:

According to this page on the site I linked above the reflectivities $R_1$ and $R_2$ are used for pleochroic minerals i.e. minerals where the reflectivity of plane polarised light depends on the orientation of the light relative to the mineral. Given the difference between $R_1$ and $R_2$ for red light that suggests the blue tint would be orientation dependent if osmium is viewed in polarised light.

  • $\begingroup$ "The $R_2$ reflectivity holds up at the blue end though $R_1$ falls." That's the red end, though, isn't it? $\endgroup$
    – user300
    Commented May 29, 2018 at 6:55
  • $\begingroup$ @Rahul oops, doh, yes you're right. I'm used to diagrams showing frequency not wavelength. I'll edit my answer. $\endgroup$ Commented May 29, 2018 at 7:15
  • $\begingroup$ @JohnRennie - It seems that $R_1$ and $R_2$ are dichroic results from rotating the sample under plane-polarized light. This seems to be a result of Osmium's hexagonal structure, which is quite interesting in itself. $\endgroup$
    – ShroomZed
    Commented May 29, 2018 at 11:33
  • $\begingroup$ @ShroomZed judging from the spectra that would imply that the blue tint is orientation dependent if you're using polarised light. $\endgroup$ Commented May 29, 2018 at 11:36
  • $\begingroup$ According to this page, @ShroomZed is right: the two are different values of reflectance given for pleochroic materials. $\endgroup$
    – Ruslan
    Commented May 29, 2018 at 11:40

The answer is here.

In the article "The structure of the energy bands and optical absorption in osmium" (Sov. Phys. JETP 63, 115 (1986)), Nemoshkalenko et al. report measurements of the complex refractive index of osmium (which can be extracted from reflectivity and related to complex conductivity). Unlike cubic metals such as gold, osmium has a hexagonal crystal structure, which means that its optical properties are not isotropic. Light with electric field in the plane of the hexagon ($\mathbf{E}\perp \mathbf{c}$) has different reflectivity than light with electric field perpendicular to the hexagon ($\mathbf{E}\parallel\mathbf{c}$), where $\mathbf{c}$ is the lattice vector normal to the hexagons.

Check out Figure 1(b), which is the measured reflectivity (the visible range corresponds to ~1.75-3 eV). As @JohnRennie pointed out, there is a reflectivity dip in the red (what the authors call absorption band B), especially for $\mathbf{E}\parallel \mathbf{c}$, which leads to a bluer color.

The authors explain this behavior by computing the band structure of osmuim. They find that the theory predicts the absorption band B to occur due to a couple of electronic transitions (their bands $7\to 8$ and $8\to 9$), which they describe as $d\leftrightarrow p$ type transitions.

As is usually the case with metals, the color is described by interband absorption. Essentially, you have a crystal with a mess of energy bands, the details of which are resulting from the type of lattice, the various electron orbitals in each atom, the couplings between them, and other interactions like spin-orbit coupling. For low energies, the conductivity is typically dominated by free-electron (Drude-like) behavior. When the photon energy matches the energy difference between an occupied and an unoccupied band, you get interband absorption. This is, for example, why copper and gold have their colors, but platinum and silver appear colorless (Pt and Ag don't have interband transitions in the visible range or lower). For osmium, apparently a band with $d$-orbital character is full of electrons, and with photons in the 1-1.5 eV range (with $\mathbf{E}\parallel \mathbf{c}$ to make the matrix elements work out) you can promote those electrons to another band with $p$-orbital character. What's a little interesting about osmium is that there are a number of lower-energy (infrared) transitions too, which distinguishes it from Pt, Ag, Au, Cu, etc.

  • $\begingroup$ A good find! :-) $\endgroup$ Commented May 29, 2018 at 16:55
  • $\begingroup$ @JohnRennie I happened to have been recently thinking about this topic for other metals, so I knew the right search terms. $\endgroup$
    – Gilbert
    Commented May 29, 2018 at 17:08

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