13
$\begingroup$

From an online lecture, I heard that $d$ orbitals cause metals to have a peak in their reflectivity curve at some wavelength. This is generally the case for most metals. However, the peak lies mostly in the ultraviolet region which makes them still appear silver-greyish. But for copper and gold this peak lies in the blue/violet region, which makes them appear orange or yellow. For gold, the shift of the peak is caused by special relativity due to the great mass of the nucleus.
My underlying question is, why mercury (and all following metals in the transisiton metal group) is not also golden, as it also has $d$ orbitals and is heavier than gold and therefore should have a more shifted peak. With almost the same logic: Why aren’t transition metals before copper in the PSE also colored, when they have $d$ orbitals?
I could not find a lot of sources to this topic to compare with, so I might have some errors in my reasoning. Feel free to correct them.

$\endgroup$
3
  • 1
    $\begingroup$ You need to account for relativistic effects in the electronic structure of mercury and gold to account for their unique properties. $\endgroup$ Feb 12 at 0:33
  • $\begingroup$ Gold is not ‘unique’ - remember copper which does not ‘need’ relativistic treatment. And silver is not too far away in plasma frequency either, but it is in the near uv. $\endgroup$
    – Jon Custer
    Feb 12 at 2:09
  • 1
    $\begingroup$ See: physics.stackexchange.com/a/72412/85785 $\endgroup$ Feb 12 at 15:39

1 Answer 1

13
$\begingroup$

Gold metal conduction (and the associated reflectivity) is from hybridized orbitals of the outer shell, which is 6S electrons only. There's no free-to-respond competition from non-S-orbital P, D, or F electrons. The metallic bond in a lump of gold is composed of these mobile S electrons, and it is these which make the metal reflective.

Those S electrons have no orbital angular momentum, so penetrate the high-Z nucleus. Potential energy when IN the nucleus is low so kinetic energy is high (the sum is a constant). As a result, relativistic mass increase makes them "spend more time there" so to speak. The same mass increase makes them less responsive to the electric field of light, thus less reflective than would otherwise be the case.

Cesium is slightly lower-Z but also has one 6S-electron dominating. It, too, has some color.

$\endgroup$
1
  • $\begingroup$ And copper has no 6s electrons... $\endgroup$
    – Jon Custer
    Feb 13 at 16:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.