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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.)

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    $\begingroup$ 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. $\endgroup$
    – Ivan Neretin
    May 9, 2019 at 8:38

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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.

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