Is the triiodide ion polar? Three professors argue it is non-polar.

My professor argues that it is a monopole, like most ions. 

The structure of the triiodide ion places a negative formal charge on the central iodine atom. The molecular geometry is also linear (at least according to VSEPR), and its electronic geometry is trigonal bipyramidal. Given the high amount of symmetry, shouldn't triiodide be non-polar? Sure, it might have a charged central atom, but so does carbon dioxide. 

What's the answer?
The structure of the triiodide ion places a negative formal charge on the central iodine atom.


No it doesn’t. The two resonance structures that describe the four-electron three-centre bond put the negative formal charge on the outer iodines ($\ce{1/2-}$ each).

That said, polarity is usually defined as having a non-zero dipole moment. The dipole moment’s vector must display the same symmetry as the entire molecule. Since the molecule is linear and both $\ce{I\bond{...}I}$ distances are equal, its point group is $D_{\infty \mathrm{h}}$ which includes $i$ which means the overall dipole moment must be zero.

The charge it carries does not matter. Any single-atom ion also has zero dipole moment and would thus be called non-polar. ‘Non-polar’ does not mean ‘free of electrostatic interaction’.