Why does symmetry reduce heat capacity? For monatomic gasses, the heat capacity approaches $3R/2$ because there are three degrees of freedom (DOF) of translational motion. For diatomic gasses it approaches $7R/2$ because the symmetry about its axis cancels out one rotational DOF, but it also has two vibrational modes. And for complex molecules you get up the $(3 + DOF_{Rotational} + 2DOF_{Vibrational})R/2$
What I don't understand is why rotational symmetry means heat cannot be "stored" in that degree of freedom. Just because a disc is symmetrical doesn't mean we can't use it as a flywheel. Just because a football is symmetrical doesn't mean that its rotational energy won't get used up as its path through the air is bent.
My first thought was maybe the answer is "Because quantum", but that's ridiculous because one of the first operators you learn about is the orbital angular momentum operator and you apply that to atoms, so you can definitely have angular momentum being stored in an atom. It also reminded me that a lot of atoms aren't even spherically symmetrical because the electron shells extend out in bulbs and things.
Why can't molecules store heat in their axes of rotational symmetry?
 A: Your understanding is incorrect. The symmetry about the axis for diatomic molecules doesn't cancel that rotational degree of freedom. The system has translation symmetries, and if symmetry canceled degrees of freedoms out, you'd have zero heat capacity.
To understand what's going on, you have to look at the energy written in terms of the quantum mechanically relevant variables:
$$ E = \frac{P^2}{2M} + \frac{1}{2}\vec{L} \cdot I^{-1} \vec{L},$$
where $P$ is the momentum of the center of mass, $M$ is the total mass, $\vec{L}$ is the angular momentum about the center of mass, and $I$ is the moment of inertia tensor. For a diatomic molecule the moment of inertia for two of the rotations is quite large - about $M\ell^2 / 8$, where $M$ is the total mass of the molecule and $\ell$ is the total distance from nucleus to nucleus. The third moment of inertia is quite small, though, of order $M$ times the size of the nucleus squared. Because you divide by the moment of inertia tensor, that means that the energy needed to excite rotations along that axis is very high.
For comparison, rotations around the long axes take of order $1\,\mathrm{meV}$ of energy. Rotations around the small axis take $10\,\mathrm{MeV}$ (i.e., 10 billion times as much energy)! Every atom I'm familiar with would fully ionize long before this degree of freedom unfroze.
To calculate this I assumed the nucleus has a mass of one proton, the molecule is about 1 ångström, and the nucleus is about 1 femtometer. You then just have to assume that the minimum non-zero angular momentum is $\hbar$, and you can calculate the answer yourself.
