9
$\begingroup$

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?

Improve this question
$\endgroup$
8
  • $\begingroup$ Related: Does 1-bromo-4-chlorobenzene have a permanent dipole moment? $\endgroup$
    – Vishnu
    Commented Jun 18, 2020 at 13:21
  • 5
    $\begingroup$ I guess "in benzene" is crucial here. You're treating these molecules as separate, while aromatic compounds have pi-stacking and even CT-complexes. $\endgroup$
    – Mithoron
    Commented Jun 18, 2020 at 13:24
  • $\begingroup$ @Mithoron: If I understood your comment properly, you mean, the bond moments are not only due to the bonding electron but also due to the delocalised pi-electrons from the benzene ring. This is understandable for me. But I think even this case is symmetrical and by logic the net dipole moment must be zero. This is why I thought of those dispersion forces. $\endgroup$
    – Vishnu
    Commented Jun 18, 2020 at 13:31
  • 1
    $\begingroup$ No, I mean that molecules bind to each other a that may be source of polarisation. $\endgroup$
    – Mithoron
    Commented Jun 18, 2020 at 13:32
  • 7
    $\begingroup$ pi stacking is a good suggestion, and so the dipole moment would then be different, i.e. zero, in other solvents. Of course a small amount of other diiodo impurity could cause this result. This compound may be hard to keep pure. It is strange that there is some uncertainty in such a simple measurement. $\endgroup$
    – porphyrin
    Commented Jun 18, 2020 at 13:34

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.