# Why are the dispersion forces in CS2 stronger than the dipole-dipole forces in COS?

London dispersion forces supposedly have the least strength out of all the intermolecular forces. But $$\ce{CS2}$$, which has only dispersion forces, has a higher boiling point (and thus stronger intermolecular forces) than $$\ce{COS}$$, which has dipole-dipole attraction in addition to dispersion forces. Why is this?

I suppose that it has something to do with $$\ce{CS2}$$ having a thicker/more inducible electron shell, but then a new question arises: how would you know if the dispersion forces in one molecule are stronger that the dipole-dipole forces in another?

(Theoretically, without using boiling points or other experimental data. Also, this is based on question 4a from the 2018 AP chemistry free response.)

• COS is not that much of a dipole. To the point of your question: generally speaking, you don't know it until you try it. Chemistry is an experimental science, after all. – Ivan Neretin May 9 '19 at 8:38

## 1 Answer

Although individual dispersion forces are weak, they are cumulative, and increase with molar mass. As a general rule, boiling point increases with molar mass.

Polar molecules will have higher boiling points when compared to molecules with similar molar masses. For example, ethanol($$\ce{CH3CH2OH}$$) has a higher boiling point than dimethyl ether ($$\ce{CH3OCH3}$$).

$$\ce{CS2}$$ is ~16 g/mol heavier than COS.