From this answer, I came to know that 1,4-diiodobenzene has a dipole moment of $0.19~\rm D$ in benzene at $20\ ^\circ\rm C$. On the basis of symmetry, I'd expect the bond moments due to the difference in electronegativities of iodine and carbon atoms to cancel each other out, similar to other 1,4-dihalobenzenes. However, it turns out to be non-zero. What is the reason for this observation?
I asked the same question in the comments, and I was suggested that this could be due to the high polarizability of the large iodine atoms. I can understand this reason since this is how books explain the increase in boiling point of alkanes with increase in chain length. And in this case, the boiling point of 1,4-diiodobenzne ($285.0\ ^\circ\rm C$) is higher than 1,4-dibromobenzene ($218.5\ ^\circ\rm C$). But this is the first time I'm seeing that dipole moment is altered by dispersion forces and I'm also confused by the issues described below.
We know that dipole moment $\vec{p}$ of a charge distribution is given by:
$$\vec{p} = \sum_i q_i \vec{r_i}$$
where $q_i$ and $\vec{r_i}$ are the charge and position vector of the $i$th charge about a particular origin. If it's the dispersion forces, arising due to the lack of symmetry of the fluctuating electron cloud, how can we ascertain a particular value of dipole moment as both $q_i$ and $\vec{r_i}$ are indeterminate?
