7
$\begingroup$

London dispersion forces (LDF) are present in all molecules, whether polar or non-polar. Molecules also exhibiting dipole-dipole interactions (in addition to the LDF) must have stronger forces of attraction than those molecules which exhibit only LDF.

Then, why does tetrachloromethane (carbon tetrachloride), which is a non-polar molecule exhibiting only London dispersion forces, have a higher boiling point ($\pu{77 ^\circ C}$) than trichloromethane (chloroform) ($\pu{61 ^\circ C}$) which is a polar molecule, exhibiting dipole-dipole interactions?

$\endgroup$
5
$\begingroup$

You also need to account for the difference in dispersion forces between the two molecules. Chlorine is much larger than hydrogen. Therefore tetrachloromethane has a larger molecular surface area which increases the intermolecular interaction strength. In this particular case, it outweighs the weak dipole interactions present in trichloromethane.

$\endgroup$
0
$\begingroup$

tetrachloromethane has a higher mollecular mass than that of trichloromethane i.e tetrachloromethane has a Rmm=154 and trichloromethane has a Rmm=118.5 thus the heavier the mollecule the greater the forces of attraction between the mollecules hence a high boiling point and vice versa

$\endgroup$
  • 1
    $\begingroup$ What about Polarity of molecules? Why are you not considering them as well ?? $\endgroup$ – Soumik Das May 16 '18 at 12:37

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.