Ozone is a bent molecule, and therefore, Ozone is a polar molecule. Polar molecules have dipole-dipole intermolecular forces. CO2 is a linear molecule, and therefore, CO2 is a nonpolar molecule. Nonpolar molecules have london dispersion intermolecular forces.

Having dipole-dipole intermolecular forces means you have greater intermolecular forces than London dispersion forces, and therefore means that you have a higher boiling point. Therefore, Ozone should have a higher boiling point than CO2.

However, after looking at multiple sources across the internet, Ozone has a lower boiling point than CO2. Could someone provide an explanation as to why this is true?

  • 1
    $\begingroup$ CO2 does not melt and hence boil at normal pressure, but sublimes instead. A better comparison would be to find ozone's melting or boiling point at a pressure where CO2 melts or boils. Alternatively consider comparing the triple points. $\endgroup$
    – porphyrin
    Jan 31 '20 at 9:10

The primary thing to realize is that, although the ozone molecule has a bent structure, it is a homonuclear molecule. This means that between the individual oxygens, the bond moments being formed are quite less, as inherently, both the oxygens have the same Pauling electronegativity values. Only a slight difference in the same will occur, due to the asymmetrical bonding scheme in ozone.

It's worth noting that bond moments in carbon dioxide are much more pronounced than those in ozone, even though they end up being canceled vectorially (like you said).

For comparing the boiling points, we need to see the extent of intermolecular attraction as a measure for the same. Due to the difference in electronegativities between $\ce{C}$ and $\ce{O}$, $\ce{CO2}$ ends up having more partial charges on its constituent atoms as compared to ozone. Hence, the boiling point of carbon dioxide is more than ozone


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.