Can we understand the strong reflectivity of metals from band theory? I know that solids, including metals, have electronic bands and bandgaps. If we consider some typical metal such as copper, we know that it strongly reflects visible light. From the point of view of bands, this means that the bandgap energy must be large in comparison to the energy of the visible light and therefore, it cannot be absorbed. Is it so? I am not sure.
 A: 
From the point of view of bands, this means that the bandgap energy
  must be large in comparison to the energy of the visible light and
  therefore, it cannot be absorbed.

I don't think so. When there is a large gap and photons can not be absorbed, the material is transparent to visible light. It is the case of some isolators as quartz. 
In order to absorb a photon, it is necessary that the energy gap between adjacent electronic bands allows a transition of one of the electrons. Normally, the momentum of the electron ($\hbar k$) is much higher than the moment of the photon, and moment conservation requires that there is an available state in a higher energy band with (almost) the same $k$.
After being absorbed, the electron returns to the lower energy band, scattering the incoming light, what explains the bright reflective surface of metals.
Metals don't have large band gaps as insulators, and electrons can find available states at high energy band.
A: J. Murray is correct and yes you can: because most metallic bands are continuous, the promotion of electrons from valence to conduction will occur at a wide range of energies. The relaxation of these electrons then emits a photon resulting in reflection. Metal colours come from differing electron densities at points in their valence bands (DoS).This is also why most metallic materials are opaque and why there is so much interest in transparent materials that are electrically conductive.
