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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?

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3 Answers 3

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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.

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Chloroform numerically has a relatively strong dipole moment, but the negative charge is distributed over three large chlorine atoms.

The positively charged hydrogen atom finds nothing it could really be attracted to.

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    $\begingroup$ That's a little incomplete. Chloroform has a stronger dipole and that does matter. But it also has weaker london forces than carbon tetrachloride and those outweigh the strength of the dipole forces (it isn't that the dipole forces don't exist). $\endgroup$
    – matt_black
    Feb 22 at 16:42
  • $\begingroup$ @matt_black Ja, see other answer. I should probably just have commented there. $\endgroup$
    – Karl
    Feb 22 at 20:14
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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

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    $\begingroup$ What about Polarity of molecules? Why are you not considering them as well ?? $\endgroup$
    – Soumik Das
    May 16, 2018 at 12:37

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