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I was reading about why most metals are gray/silvery in colour and it said something about d orbital electrons transitioning to s orbitals and the visible spectrum not having sufficient energy to raise them.

But take Scandium for example, it forms a $Sc^{+3}$ ion, so it has no d electrons and these are now delocalised. How can these delocalised electrons still occupy the orbital?

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    $\begingroup$ @CuriousOne, shouldn't your comment be an answer? $\endgroup$ – Alfred Centauri Sep 5 '15 at 15:24
  • $\begingroup$ @CuriousOne voting to close that comment because it is actually an answer. $\endgroup$ – Asher Sep 5 '15 at 16:22
  • $\begingroup$ I appreciate all the votes on my comment and have posted it as an answer, even though I think that others have done a better job, already. $\endgroup$ – CuriousOne Sep 6 '15 at 2:31
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How can electrons still occupy orbitals in metals if they are delocalised?

Because they're not even localized in say a hydrogen orbital. Check out atomic orbitals on Wikipedia where you can read this: "The electrons do not orbit the nucleus in the sense of a planet orbiting the sun, but instead exist as standing waves". They exist as waves, and waves are delocalised. Even ordinary waves. Take a look at the Wikipedia Wind wave article where you can see this gif:

enter image description here GNUFDL image by Kraaiennest, see Wikipedia

You might think of an ocean wave as something that's a metre-high hump on top of the sea. But look at the red test particles in the animation above. They go round and round, and this motion still occurs deep under the water. It diminishes with depth, but nevertheless the wave takes "many paths" and isn't localized to the metre-high hump. The wave nature of matter means electrons aren't totally unlike this. Don't think of them as little billiard-ball things. Don't think of an electron as some little ball that has a field, instead think the electron's field is what it is. And this field doesn't have an edge. So an electron in an ordinary orbital is not localized anyway. So when it's say a 3s electron in a metal, it's even less localized, but the change isn't so drastic.

I was reading about why most metals are gray/silvery in colour and it said something about d orbital electrons transitioning to s orbitals and the visible spectrum not having sufficient energy to raise them.

IMHO it's worth having a google on electrons shiny. Metals are shiny because of the free electrons.

But take Scandium for example, it forms a $Sc^{+3}$ ion, so it has no d electrons and these are now delocalised. How can these delocalised electrons still occupy the orbital?

Noting Gert's answer re Scandium Chloride, it's because they're in "molecular" orbitals. See this article by Jim Clark which IMHO explains it fairly well: "The electrons can move freely within these molecular orbitals, and so each electron becomes detached from its parent atom. The electrons are said to be delocalised. The metal is held together by the strong forces of attraction between the positive nuclei and the delocalised electrons." IMHO it would be better if the electrons were shown as red circles fading to pink around the metal ion rather than red dots, but nevermind. I am reminded of a barn dance myself, where you swap partners.

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Do not confuse $Sc$ metal with $Sc^{3+}$ ions(*).

Scandium (metal) has the electron configuration $[Ar]3d^1 4s^2$ (with $[Ar]$ the electron configuration of Ar) and these outer $3d^14s^2$ electrons (aka 'valence electrons') can in the metal lattice join the conduction band, causing the conductivity of the metal and its shiny bulk appearance (due to reflection of photons).

But when scandium is oxidised:

$Sc \to Sc^{3+} + 3 e^-$

... these three valence electrons are removed ions are formed, which have no electrons left to contribute to the conduction band. These electrons are not 'delocalised', they have effectively been removed. The scandium ion now has the extremely stable Ar configuration. Its electrons are tightly bound to the nucleus and cannot participate in conduction bands.

Also, in most situations, these cations (positively charged ions) need to combine with anions (negatively charged ions) like for example chloride ($Cl^-$) ions:

$Sc^{3+}+3 Cl^- \to ScCl_3(s)$

This forms an ionic but electrically neutral and crystalline lattice that doesn't conduct electricity, nor reflects photons (in the same way the scandium bulk metal does).

(*) For ion charges we use the $n+$ notation. $+n$ is reserved for the Oxidation Number of an oxidised element.

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Orbitals are a property of atoms and conventional orbital theory is a strong simplification that treats strongly relativistic multi-electron atoms with high Z as if they were hydrogen. When many atoms are combined in solid state lattices the simple hydrogen-like orbital states available to electrons become complex band structures that are delocalized across the entire crystal. Most importantly electrons that occupy conduction band states can move freely even under the influence of small fields and optical properties of metals are therefor governed by these quasi-free electrons.

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