3
$\begingroup$

Why is there only little difference between molecular dipole moments in $\ce{CHX3}$ compounds ($\ce{X}$ = $\ce{Cl}$, $\ce{Br}$, $\ce{I}$)?


As we can see there is a huge difference in dipole moment of methyl halides $\ce{F}$, $\ce{Cl}$, $\ce{Br}$, $\ce{I}$ excluding $\ce{CHX3}$.

enter image description here

Please explain the reason why is it so only in $\ce{CHX3}$ but not in other methyl halides?

$\endgroup$
  • 2
    $\begingroup$ Monohalides: electronegative element just on one corner. dihalides: electronegative elements on one side. haloforms: electronegative elements on three of four sides, and they have to divide the electrons of carbon and just one hydrogen between them. Doesn't leave much for each. $\endgroup$ – Karl Jul 21 at 7:53
  • $\begingroup$ Plus: Cl, Br and I are so bulky that the molecule is no longer nearly tetragonal. I don't find a web source for the actual bond angles now. The gist is that there are several effects that give haloforms a low dipole moment. That the resulting number is practically identical for three of them is an "accident". And you notice that except for CH2F2, already the dihalides have a significantly lower dipole moment, esp. methylene iodide. $\endgroup$ – Karl Jul 21 at 12:48
  • $\begingroup$ Data here: According to the Computational Chemistry Comparison and Benchmark DataBase, the experimental H-C-X angles are 110.4 deg for chloroform, and 107.2 deg for bromoform Refs: 1. doi.org/10.1063/1.1743153 2. Landolt-Bornstein: Group II: Atomic and Molecular Physics Volume 15: Structure Data of Free Polyatomic Molecules $\endgroup$ – The_Vinz Jul 21 at 17:18
  • 1
    $\begingroup$ If I had to guess, I would look into the definition of the dipole moment. The dipole moment for two point charges depends both on the charge difference and on the distance. If I had to give a rationalization for the results, I would rely more on the VDW radius rather than the bond angle: Cl, Br and I have decreasing electronegativity, but increasing VDW radius. Probably, in haloforms the two effects (luckily) "balance" themselves, and this results in a small difference in dipole moments. $\endgroup$ – The_Vinz Jul 21 at 17:37
  • $\begingroup$ Every sum of vectors does. It is why falling is faster than sliding. Even treating electronegativity as not influenced by what goes on the rest of the bonds. That beyond a legitimate but rudimentary analysis I guess it is not true. $\endgroup$ – Alchimista Jul 22 at 10:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.