[alancalverd (OP)]

Metals - arrays of atoms with a shared delocalised electron cloud that accounts for high electrical and thermal conductivity. 

The electrical conductivity of a metal at a given temperature is independent of the voltage gradient - no band gap or any evidence of quantum behavior: the conduction electrons are free and indistinguishable.

So why do different metal surfaces have distinct optical absorption spectra? Gold is gold, sodium is white, silver is silver, and titanium is grey. Surely they should all be equally and totally reflective (white) across the entire spectrum,or totally absorbent (black)?

[Eternal Student]

Hi.

   I can answer some of this but I'd only be using stuff suitable for school "A" level chemistry.   
A lot of your question sounds like exactly the sort of thing @chiralSPO  was working on.  Is @chiralSPO still around and using the forum?   A little while ago (maybe a year), Chiral was working on models to predict colours and properties of the elements based on quantum mechanical and relativistic considerations of just how the protons, neutrons and electrons must behave.   I believe the models did very well but there was one issue with predicting the colour of one metal (.. hmm...  which might have been gold).
    Anyway, it's got too late for me to write anything sensible and it will be a lot better if you get a response from Chiral.

Best Wishes.

[paul cotter]

Different crystal arrays I would guess. Caesium has a hint of "gold" in it's appearance and is the only other metal I know of with that colour but there may be others.

[alancalverd (OP)]

Happy to await ES's "A" level chemistry response - my degree in solid state physics was a long time ago, and IIRC didn't address the question!

The oddity just occurred to me because the conductivity of gold is so high. Taking Ag= 100 we have Cu = 95 and Au = 72, way above the next one Al = 60 and any of the other "fully reflective" metals. You'd think that if the matrix structure generates photon absorption bands, it would somehow constrict the notional continuum of conduction electron transport.   

[Eternal Student]

Hi.

Happy to await ES's "A" level chemistry response

    I wouldn't be happy with it.   Presumably you'd like a definitive response instead of some partially correct and very simplified stuff.
    Since the forum is better off with either some reliable information from main stream science  OR  some reasonable discussion instead of being filled with spurious thoughts off the top of someones head,  let's try and make this more of a discussion.

Your idea:
   The delocalised sea of electrons should matter.  It should influence the colour of the metal and more generally the absorption spectra.  Most metals should be pretty much the same colour because they all have a similar sea of electrons.

My idea:
    It doesn't influence the visible spectrum much.   The free electrons aren't as free to move as you'd think.   It does influence the absorption spectrum but not in the visible spectrum.
Evidence:
    The absorption in the radio frequency range is obvious.  Pass a radio wave over a metal rod (an antenna) and you generate an alternating electrical current.  This is how and why pocket radios and Mobile phones work.  We model the situation as if the electrons are completely free to move and so they do move up and down with the e-m wave and we get an alternating electrical current.
   Shine a light on an antenna and nothing much hapens.   There is no discernible electric current produced.   It seems the "free" electrons aren't as free to move as you'd like - they can't oscillate at the frequencies of visible light.

Your idea:
   Electrons, their movement and ability to interact with photons of a certain wavelength is what causes colours to be observed.

My idea:
    Much the same,   except that there isn't much interaction with the sea of electrons,  visible light can just pass through a lot of that.   Most of the electrons in a metal aren't de-localised,  most of them are just in the usual orbitals around a nucleus.
Evidence:
   I don't know what to put here.  It's the standard approximation that is always made from school level chemistry through to degree.   When you have a molecule of some kind,  most of what is there is still recognisable as the isolated atoms,  there are only a few orbitals that are modified and participate in some kind of bonding.   It's only an approximation, of course, but it is quite a good one.  In reality the proximity of other nuclei and electrons changes the potential function and you'd have to re-determine the solutions to the Schrodinger wave equations and all the orbitals.  However, in practice it does seem that most of the atom(s) and their electron orbitals are left much the same.
   If you perform an analysis of the absorption or emission spectrums from a molecule, then you'd still expect to observe some of the strong emission and absorption lines from the individual atoms of that molecule.

My idea:
   Most of the colour you observe for a metal is a result of the absorption of photons of a particular wavelength by the electrons in orbit around the nucleus.
    The group I and II metals   (like the Sodium you mentioned) have an electronic structure that is basically   [Noble gas] + an s orbital    (e.g.    Sodium  is   [Ne] 3s¹  ).   Most of the absorption of photons is due to either:
(i) electrons moving from the lower electron shells to the outer s orbital or further out to a p-orbital  (because everything else is usually full) and that's a large energy jump.  Visible light doesn't offer that.
(ii)  Movement of electrons from the outer s orbitals to a p orbital  and that's still a bit too large a jump, visible light isn't offering that energy.

    So Sodium and most other group I and II  metals and their salts don't absorb much visible light.  That means what is reflected is almost every colour of visible light and a white colour is what you see (because that is the composition of all colours).

    The transition metals or d-block elements are quite a different thing.   They have electronic structures that terminate with d-orbitals.
    This post is already too long, so we'll just speed up...   For school A level chemistry, it's stated that  there are 5 different d-orbitals although they are degenerate and have the same energy when the atom is isolated.  However, those d orbitals will split into slightly different energy levels at the drop of a hat.   Bond some ligands and the d-orbitals split in energy levels slightly.   Now we've got a small energy gap that's just about right for the energy of some visible light and these outer d-orbitals aren't full,  we've got room available for electrons to move from d-orbital to d-orbital.  Electrons do skip between these d orbitals by absorbing light of a particular frequency.   (If you get the transition metal ion into a very high oxidation state then the d-orbitals become full or empty and the colours do disappear).
    Anyway, it's not a great stretch to assume this sort of split is exhibited in the d-orbitals even when you just have the pure metal there and nothing obviously around to be a ligand attached to the metal ion.  E.g. there is some sort of bonding from one atom to another and the other atom is acting like some sort of ligand.   More generally when you look at a metal you're usually seeing a thin layer of the metal oxide on it's surface and not really the metal in it's pure elemental state anyway.
   So transition metals are much more likely to be coloured in appearance.
Evidence
    Some A level chemistry texts of your choice (the ones I'm finding with Google all seem to be strongly linked with advertising for private Chemistry tutoring services and I don't really want to list that here).   This discussion seems to summarise the points rapidly:   https://chemistry.stackexchange.com/questions/4667/why-do-transition-elements-make-colored-compounds.

Best Wishes.

[alancalverd (OP)]

The free electrons are certainly very free, which is why conductivity is high (particularly for Au and Cu!) and independent of potential gradient. Resistivity is somewhat a matter of phonon interference with the cloud motion, hence the temperature dependence of conductivity of all metals.

But what you say about d orbitals makes sense, I think. Many thanks for a plausible hypothesis!   

I think this could be a fun interview question. Would you care to consider my question on shoe polish?

[Petrochemicals]

Ive had a quick Google and Gold absorbs blue end light, that's why it's yellow, pretty much like most other material colurs. The shine of the metallic surface is similar to other metals. It is particularly good at thin film reflectivity of heat radiation which means it is excellent at heat shielding in space despite its weight and cost.

[alancalverd (OP)]

Why wouldn't aluminum, which doesn't absorb blue light, be cheaper, lighter and better? The absorbed light ends up as heat!

[Petrochemicals]

Why wouldn't aluminum, which doesn't absorb blue light, be cheaper, lighter and better? The absorbed light ends up as heat!

In space I would think it is down to the extreme environment, with gold being so much more ductile and non reactive. The thing that flags on gold is the photo electric effect with it able to absorb blue end light, maybe its that.

[evan_au]

The thing that flags on gold is the photo electric effect with it able to absorb blue end light

The photoelectric Work Function of Gold (5.1eV) is not that much different from Aluminium (4.08eV)
- But they are in the "wrong" order for this hypothesis - if gold absorbed blue light by the photoelectric effect, then Aluminium should do it even more strongly.
See: http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/photoelec.html

For visible light, the range of photon energies is 1.63eV (red) to 3.26 eV (violet), which is less than the work function of gold or aluminium, so no electrons will be ejected. That is why the photoelectric effect is measured with UV light.
- Interestingly, Calcium, with a work function of 2.9 eV would see a photoelectric effect with visible light at the violet end of the spectrum
- But we usually observe Calcium in oil (no photoelectric effect) or air (coated with an oxide).
- Measurements of the photoelectric effect have to be conducted on bare metal, in a vacuum.

[chiralSPO]

Multiple effects can be blamed for the colors of metals--sometimes it is from the electronic structure of the metal itself (as in the case of gold, freshly polished copper, bronze, etc. see below), and sometimes it is due to the electronic structure of what is on the surface (often dull colors due to "tarnish" "rust" "patina" etc., like red iron, green copper/bronze, dark silver), and sometimes it is due to the thickness of an otherwise transparent oxide layer on the surface, creating colors due to interference, just like with soap bubbles or petroleum residues on the surface of water (creating the rainbow colors observed on bismuth crystals, or can be done intentionally, as when anodizing aluminum to achieve certain desired colors). There can be even more unusual phenomena involved when the metal is in very small particles (nanoparticles), where the electronic structure can become non-metallic (at least at "defect sites", which become more densely present in smaller and smaller particles, and then there's plasmonic resonance, which is how gold or silver particles in stained glass can adopt so many colors)

So, with that out of the way, I will now just focus on answering why gold is gold:

Gold is metallic because it has a partially-filled "conduction band" that spans the entire lattice. The electrons in this band can be considered delocalized across the whole "piece" of gold, and are very much free to be anywhere, or "moving" anywhere given a potential. There are ALSO bands that are completely filled (lower in energy than the conduction band) AND bands that are completely empty (higher in energy than the conduction band). Photons of suitable energies (frequencies, wavelengths) will be absorbed, promoting an electron from one band to another. And in the case of gold there are a few absorbtions in the blue part of the visible spectrum, so we perceive it as yellow.

(being able to calculate the electronic structure of gold is no small feat--it requires QM principles, modified by general relativity, because the electrons close to the gold nucleus experience both relativistic time dilation and mass dilation, compared to electrons far from the nucleus!)

[Eternal Student]

Hi @chiralSPO  and thanks for replying.

   Your reply looks good.   
First paragraph:   "There exists complications, many of them".
    I didn't know there were electronic defects in nanoparticles of metals.   I always assumed it was just due to diffraction from the shape of the nanostructures.

Example:
  The morpho butterfly is not blue and certainly not iridescent.   This is revealed by viewing the wings in a medium with a different refractive index, it just has structures of nanometre size that cause diffraction and hence interference phenomena, eliminating certain wavelengths and reinforcing others.
   
 
Second paragraph:    "Conduction bands and photon absorption".
   Are the outer d-orbitals of Gold all used up and just become all or part of  "the conduction band"?   Specifically, the absorption of blue light,  that's caused by an electron jumping from where to where?

No rush to reply.   I'm already grateful for your answer, it's enough to keep me busy for a while.

Best Wishes.

[William Hardy]

Gold is gold because of the atomic structure within it. In fact gold and other elements have the same electron, nucleon, and proton. It is just the difference in number and configuration. Like gold has 79 protons 118 neutrons and 79 electron. If any other element has this configuration then it becomes gold. If any confusion do let me know.

[wollastoni]

William : my question may be stupid, but cannot we imagine a metal with the same number of electron, nucleon and proton but with a different organization... which would not have a "golden" color ?

[Bored chemist]

Surely they should all be equally and totally reflective (white) across the entire spectrum,or totally absorbent (black)?

They are totally absorbent- which is why they reflect so well.
The conduction electrons are practically free and that means they are free to emit as well as being free to absorb.
For low energy photons there's "nowhere for the energy to go" so it gets re-emitted and the requirements for the wave function and the EM fields to be continuous etc can only be met if the reflection is symmetrical- so you get specular reflection.

But, if the energy is high enough, you can promote an electron from the conduction band into an excited state, and that gives rise to an absorption spectrum- in particular, gold has a sufficiently low lying state that blue light can kick electrons into it.