I have understood that the more polar is a compound, the higher is its boiling temperature. But I would like to know what is the justification for this answer.

I am new to chemistry so I would appreciate a simple answer.

  • 1
    $\begingroup$ Do you mean perhaps "polarity" rather than polarizability? They are related but not the same. $\endgroup$ – Buck Thorn Jul 16 '20 at 21:47

In non-polar liquids, the intermolecular forces are known as London dispersion forces (or dipole induced dipole interaction). This is the weakest interaction that exists between molecules.

The strength of this interaction depends on polarisability of non-polar molecule. As an example, let's take the following examples:

$$ \begin{array}{lr|lr|lr|lr} \hline \text{Compound} & \mathrm{bp}/\pu{°C} & \text{Compound} & \mathrm{bp}/\pu{°C} & \text{Compound} & \mathrm{bp}/\pu{°C} & \text{Compound} & \mathrm{bp}/\pu{°C} \\ \hline \ce{CO2} & -78.5 & \ce{CCl4} & 76.7 & \ce{CBr4} & 189.5 & \ce{CI4} & 329.2 \\ \ce{CS2} & 46.3 & \ce{CBr4} & 189.5 & \ce{SiCl4} & 57.7 & \ce{SiBr4} & 153.0\\ \ce{CSe2} & 125.5 & \ce{CI4} & 329.2 & & & \\ \hline \end{array} $$

In these molecules, polarisability depends on polarisability of anions $(\alpha),$ which depends as

$$\alpha \propto q \times r,$$

where $q$ is the charge and $r$ is the ionic radius.

As charge on each anion is same, hence we can conclude that, greater the radius, greater is the polarisability, and thus greater is London dispersion forces.

Therefore, greater boiling point.


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