The intermolecular forces between $\ce{CO2}$ molecules are dispersion forces, while the forces between $\ce{CO}$ molecules are mostly dipole-dipole attraction forces. So, why does $\ce{CO2}$ have a higher boiling point than that of $\ce{CO}$?

  • $\begingroup$ related chemistry.stackexchange.com/questions/42946/… $\endgroup$
    – Mithoron
    Jan 16, 2016 at 18:40
  • $\begingroup$ I don't think it answers my question. $\endgroup$
    – Lyndt
    Jan 16, 2016 at 18:53
  • 4
    $\begingroup$ There are many other points on the scale from dipole-dipole interaction to dispersion forces. $\ce{CO}$ is a very weak dipole; on the other hand, $\ce{CO2}$ is a pretty strong quadrupole. $\endgroup$ Jan 16, 2016 at 19:23
  • $\begingroup$ > the forces between COCO molecules are mostly dipole-dipole attraction forces. || Orientation of CO molecules in solid is random with weak preference of head-tail. This suggests that dispersion interactions dominates in CO solid. $\endgroup$
    – permeakra
    Jan 17, 2016 at 11:20
  • 1
    $\begingroup$ Also the MW of $\ce{CO2}$ is about 57% greater than that of $\ce{CO}$ meaning that it will take considerably more energy to raise the kinetic energy of the heavier molecule to where it has the necessary surface escape velocity. $\endgroup$
    – ron
    Jan 17, 2016 at 19:38

2 Answers 2


CO2 has more electrons than CO. This means that it has a much larger electron cloud as compared to CO, so its more easily polarised and thus, the ease of forming instataneous dipole-induced dipole bonds increases. Even though CO is a polar molecule and it forms permanent dipole-permanent dipole bonds, in this case the id-id bonds are stronger.

P.s. Just a 17 year old A Level Chem student here, I might be wrong

  • 1
    $\begingroup$ I believe you are on the right track here. Additionally to your thought, CO2 has a well developed quadrupole moment. $\endgroup$ Dec 15, 2016 at 1:29

The larger the small covalent molecule, the greater the intermolecular bonds, hence higher boiling / melting point.

$\ce{CO2}$ has 3 atoms involved in the molecule and is therefore larger than $\ce{O2}$ that has 2 atoms. Hence, $\ce{CO2}$ has a higher boiling / melting point compared to $\ce{O2}$. (Exception to this is water molecules.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.