Why are most metals gray/silver? Why do most metals (iron, tin, aluminum, lead, zinc, tungsten, nickel, etc.) appear silver or gray?
What makes copper and gold have different colors?
What atomic characteristics determine the color?
 A: Let's start with what "that thingy is X in colour" fundamentally means:

You notice silver is not one of the colours here. However, silver is a LOT like white, as we will see in a second.
There is another factor involved called specular vs diffuse reflection.

White reflects all wavelengths diffusely (the reflected rays go every which way). Silver (e.g., a mirror) reflects all wavelengths specularly (the reflected rays bounce off nicely).
Now, metals do not necessarily always look like mirrors - they are often bumpier than that, so their reflection is a little bit diffuse as opposed to totally specular.
Anyway, the point is that "silver colour" means "reflects all wavelengths specularly (more or less)".
Why do these metals reflect most visible light? Because they have lots of free electrons (that also happens to be why they're good conductors). When light (electromagnetic radiation) hits the surface of a metal, it gets absorbed by electrons orbiting the metal atoms, and re-emitted as the electrons fall back to a more stable configuration. The size of the band gaps determines which frequencies get absorbed and emitted.
A coloured metal like gold has most of these properties, but it absorbs just a little bit of radiation in the green-blue-violet area. So whatever it reflects out has a bit of green-blue light removed and the result looks (by subtraction) yellowish red.
A metal like lead also has most of these properties, but it absorbs a little more of the entire spectrum, so it looks grey.
PS This answer is provided by "Ian Pollock, Sci/Phil dilettante" at quora.com.
A: Metallic band structure allows absorption and re-emission of light as depicted on this  site. 

Metals are colored because the absorption and re-emission of light are dependent on wavelength. Gold and copper have low reflectivity at short wavelengths, and yellow and red are preferentially reflected. Silver has good reflectivity that does not vary with wavelength, and therefore appears very close to white.

A: D electrons in metals allow optical transitions in the visible regime. Visible light can be absorbed by elements having unbound valence electrons in the d shell. So
Chemistry: optical d->s$^2$ transition

*

*Iron [Ar] 3d$^6$ 4s$^2$

*Tin [Kr] 4d$^{10}$ 5s$^2$ 5p$^2$ (full d shell)

*Aluminium [Ne] 3s$^2$ 3p$^1$ (is a special case: no d valence electrons, but  Aluminium reflectivity. I have no other explanation than the calculation of Fresnel equations. However I can't grasp the reason for this distinction.)

*Lead [Xe] 4f$^{14}$ 5d$^{10}$ 6s$^2$ 6p$^2$ (full d shell)

*Zinc [Ar] 3d$^{10}$ 4s$^2$ (full d shell)

*Tungsten [Xe] 4f$^{14}$ 5d$^4$ 6s$^2$

*Nickel [Ar] 4s$^2$ 3d$^8$ or 4s$^2$ 3d$^9$

*Copper [Ar] 3$d^{10}$ 4s$^1$ (one s and full d shell)

*Gold [Xe] 4f$^{14}$ 5d$^{10}$ 6s$^1$ (one s and full d shell)

The shiny metals, except aluminium, have d electrons. A single s electron and a full d shell hint at an important d to s$^2$ orbital transition in the visible spectrum. A full s shell is energetically preferred. There seems to be no explanation for the colored appearance of gold and copper, other than a distinctive electron configuration - at least chemistry does not provide an answer.

Physics: sign change of $\epsilon(\lambda)$ near blue
If the absorbed light is reemitted (in fact reflected) for the whole visible spectrum, the metal appears shiny as a mirror. In fact, our bathroom mirrors are made of an aluminum backside coated glass.
Here physics has to explain more than just "is there a d valence electron". A second more physical reason doesn't describe its origin: Reflectivity, out of the Fresnel equations using
$$n=\sqrt{\epsilon_r\cdot \mu_r}\qquad\text{with}\qquad
\epsilon_r=1-\frac{n_e e^2}{\epsilon_0m\omega^2}\qquad\text{with a sign change at}\qquad \omega=\omega_p
$$
out of the Drude free electron gas model for electrons (and density of electrons $n_e$), is high through the whole visible spectrum for these metals. This sign change at $\omega=\omega_p$, plasma frequency is the reason for a changing $\epsilon_r$, therefore a changing refractive index $n$, due to the Fresnel equations, a changing reflectivity. If this change happens to be in the visible spectrum, then there are colored reflections like gold.
Blue absorption of gold happens, because special relativity has to be taken into account for this heavy element. See top answer. Copper and Gold don't have a high reflectivity for blue ($\approx 475\,$nm).
A: Taken from http://www.webexhibits.org/causesofcolor/9.html
"The color of metals can be explained by band theory, which assumes that overlapping energy levels form bands.
In metallic substances, empty conduction bands can overlap with valence bands containing electrons. The electrons of a particular atoms are able to move to  a higher-level state, with little or no additional energy. The outer electrons are said to be "free," and ready to move in the presence of an electric field.
The highest energy level occupied by electrons is called the Fermi energy, Fermi level, or Fermi surface. 
Above the Fermi level, energy levels are empty (empty at absolute zero), and can accept excited electrons. The surface of a metal can absorb all wavelengths of incident light, and excited electrons jump to a higher unoccupied energy level. These electrons can just as easily fall to the original energy level (after a short time) and emit a photon of light of the same wavelength. 
So, most of the incident light is immediately re-emitted at the surface, creating the metallic luster we see in gold, silver, copper, and other metals. This is why most metals are white or silver, and a smooth surface will be highly reflective, since it does not allow light to penetrate deeply. 
If the efficiency of absorption and re-emission is approximately equal at all optical energies, then all the different colors in white light will be reflected equally well. This leads to the silver color of polished iron and silver surfaces.
For most metals, a single continuous band extends from valence energies to 'free' energies. The available electrons fill the band structure to the level of the Fermi surface.
If the efficiency decreases with increasing energy, as is the case for gold and copper, the reduced reflectivity at the blue end of the spectrum produces yellow and reddish colors.
Silver, gold and copper have similar electron configurations, but we perceive them as having quite distinct colors. 
Gold fulfills all the requirements for an intense absorption of light with energy of 2.3 eV (from the 3d band to above the Fermi level). The color we see is yellow, as the corresponding wavelengths are re-emitted. 
Copper has a strong absorption at a slightly lower energy, with orange being most strongly absorbed and re-emitted. 
Silver. The absorption peak lies in the ultraviolet region, at about 4 eV. As a result, silver maintains high reflectivity evenly across the visible spectrum, and we see it as a pure white. The lower energies corresponding to the entire visible spectrum of color are equally absorbed and re-emitted making silver a good choice for mirror surfaces.
A: Why do most metals appear silver in color, with gold being an exception? 
It is hardly surprising that the answer to this question relies heavily on quantum theory, but most people will be surprised to hear that the full answer brings relativistic considerations into the picture. So we are talking quantum relativistic effects.
The quantum bit of the story tells us that the colour of metals such as silver and gold is a direct consequence of the absorption of photons by d electrons. This photon absorption results in d electrons jumping to s orbitals. Typically, and certainly for silver, the 4d→5s transition has a large energy separation requiring ultraviolet photons to enable the transition. Therefore, photons with frequencies in the visible band have insufficient energy to be absorbed. With all visible frequencies reflected, silver has no colour of its own: it's reflective, an appearance we refer to as 'silvery'. 
Now the relativistic bit. It is important to realize that electrons in the s orbitals have a much higher likelihood of being in the neighborhood of the nucleus. Classically speaking, being close to the nucleus means higher velocities (cf speed of inner planets in solar system with that of the outer planets). 
For gold (with atomic number 79 and hence a highly charged nucleus) this classical picture translates into relativistic speeds for electrons in s orbitals. As a result, a relativistic contraction applies to the s orbitals of gold, which causes their energy levels to shift closer to those of the d orbitals (which are localized away from the nucleus and classically speaking have lower speeds and therefore less affected by relativity). This shifts the light absorption (for gold primarily due to the 5d→6s transition) from the ultraviolet down to the lower frequency blue range. So gold tends to absorb blue light while it reflects the rest of the visible spectrum. This causes the yellowish hue we call 'golden'.

Reflectivity as function of wavelength. Purple/blue light corresponds to 400 - 500 nm, the red end of the visible spectrum to about 700 nm.
See: the color of gold, relativistic quantum chemistry.
A: This question has another interesting aspect which has more to do with neuroscience than physics: why do we perceive metals with a neutral colour (such as silver) as grey, even why they are shiny and therefore simply reflect the colours of their surroundings?
One answer is that such metals always have some roughness and therefore scatter light from a range of angles, and these rays typically have a range of wavelengths. The mixing of these wavelengths tends to desaturate the perceived colour, and moves it towards a neutral tone. However, some simple experiments suggest there's more to it than this. Even when the surface is reflecting one dominant colour our perception of the surface colour is grey.
The reason for this is connected with the way the brain processes colour information. Colour constancy ensures that our perception adjusts for colour bias in the ambient light conditions: we tend to perceive an object's intrinsic colour rather than the colour of the light reflected from it. The apparent greyness of metallic surfaces (both shiny and matt) seems to be an interesting variant of this phenomenon.
A: An electron can get excited to a higher energy level either by absorbing a photon, or through the vibrational kinetic energy of the atom itself, or by resonance energy transfer from an adjacent excited atom. An electron in an excited state can return to its ground state by the reverse of either of the above processes.
In metals, photons can get scattered after collision with electrons or get absorbed by them, kicking them to higher energy levels. In metals with white metallic luster, the photons in the visible spectrum get absorbed by electrons in the conduction band and get emitted immediately. However, in metals like gold, the blue photons (which is a bad way of labeling photons, but bear with me here) just have enough energy to also allow transition of electrons from the d-band to the conduction band. And some of these electrons return to their ground state without emitting the blue photons back. Thus, the reflected light will miss some of the incident blue photons, resulting in gold's yellowish hue. In metals like silver, the d-s transition can't be brought about by any of the visible photons, but only by ultraviolet photons, whose absence can't be perceived by our eyes.
