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.
Practically all the elements heavier than lead are created by neutron capture in the
r-process. This requires the explosive conditions of a supernova or neutron star merger.
In terms of some limit, I'm not sure how to answer. Stuff that's heavier than Uranium has a short half life compared to the age of the Earth, so there's not much to be found. They may have been produced in supernovae, but rapidly decayed.
We do know (at least theoretically) that much heavier elements were created indirectly by a supernova, in that they exist in the crust of a neutron star. Calculations show that atomic masses of more than around 300 are possible at densities of a few $10^{16}$ kg/m$^3$, largely due to the suppression of beta decay by degenerate electrons.
Best Answer
Lithium, beryllium and boron aren't produced in (normal) stellar nucleosynthesis - instead, three atoms of helium fuse to form carbon (the triple-alpha process - two helium nuclei fall apart again almost instantly). But the necessary conditions only arise late in the lifetime of a star, when it has stopped burning hydrogen to helium and instead burns helium to heavier elements.
In massive stars there's a catalytic cycle with carbon, nitrogen and oxygen called the CNO cycle, during main sequence evolution - the equilibrium state is very nitrogen rich (which is why massive stars are usually nitrogen overabundent and carbon and oxygen depleted), but in the latter states of the stellar lifetime the balance shifts as helium burning forms carbon - visible in the spectra of carbon-type Wolf-Rayets - while adding a helium nucleus to carbon gives you oxygen. Nitrogen then dips and becomes somewhat underabundent.
Side branches of the CNO cycle are responsible for some other elements, while successively adding helium nuclei to oxygen gives things like silicon, magnesium, calcium (alpha-process elements) and iron. The heavy elements all require neutron capture in red (super)giants.