Is molecular vibration just phonon modes for a single molecule? I'm reading about Raman Scattering, of which a big part is measuring the energy lost to/gained from Molecular Vibrations. I wasn't totally clear on exactly what is "vibrating" in vibrational modes (is it the electron around the nucleus? Or the whole atom, with respect to its center of mass? Or the individual atoms that make up the molucule?), but it seems like it's the atoms (and I guess with them, their respective electrons), from the wiki page:

The coordinate of a normal vibration is a combination of changes in the positions of atoms in the molecule.

And then they have some handy gifs showing it.
However, (collective) normal modes of nuclei sound exactly like phonons to me. So are vibrational modes very similar to phonons?
thanks!
 A: To kick things off: you're right. Molecular vibrational excitations are exactly the same as phonon modes. We don't use that language very much because the system is too small (so we can't have things like travelling waves which have momentum, and we need to work only with stationary waves) but the analogy is indeed exact.
Now, as to what exactly is vibrating, the quick answer is the nuclei, but there are some subtleties involved through the Born-Oppenheimer approximation. The degrees of freedom that are involved are exactly the spatial coordinates of all the $N$ nuclei in the molecule. These nuclei sit at the bottom of the (mean) electronic energy well and oscillate about its minimum. The vibrational modes are the normal modes of these oscillations, obtained by diagonalizing the potential in the corresponding $3N$-dimensional configuration space. 
Similarly, the oscillator masses at stake are just the nuclear masses. Any contribution from the kinetic energy of the electrons would get subsumed into the electronic potential energy surface for the nuclei. However, this is not incorporated in the Born-Oppenheimer approximation, and to include the kinetic energy of the electrons as they accompany the nuclei in their motion you need to use complete adiabatic wavefunctions to appropriately describe the electron current. For more details see these lecture slides.
If the nuclei are oscillating, the electron motion is rather hard to describe. The core electrons follow the nuclei quite closely, but the valence electrons generally roam over large chunks of the molecule, so they generally won't follow any individual atom. As the nuclei move, the molecular orbitals will shift and stretch (by not much), and this is the only meaningful sense in which the electrons can be said to move during the oscillation. 
However, as mentioned above, within the Born-Oppenheimer approximation none of this is really included in the dynamics of the molecular vibration beyond the mean electronic energy at each position. This is really a very good approximation, because the electrons are much lighter than the nuclei. In turn, that means that (i) the electrons' kinetic energy is a negligible contribution to the kinetic energy of the molecular vibration, and more importantly that (ii) the electronic motion is much faster than the nuclear oscillations. As far as the electrons are concerned, the nuclei are frozen in place, and as far as the nuclear motion is concerned the electrons are a blur which you can average over. As a consequence, the mean electronic kinetic energy is much more important for the nuclear motion than the slight average motion as the electrons follow (to some extent) the nuclear motion.
A: *

*In this context, vibration usually refers to the relative motion of the nuclei.

*In a periodic solid (crystal), vibrational modes are called phonons. We can say, for example, that a normal vibrational mode of a branch $s$ with wavevector $\mathbf{k}$ is in its $n$th excited state, or equivalently, that there are $n_{\mathbf{k}s}$ phonons of branch $s$ with wavevector $\mathbf{k}$. In molecules these are called just vibrations.

