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 '16 at 18:40
  • $\begingroup$ I don't think it answers my question. $\endgroup$ – Lyndt Jan 16 '16 at 18:53
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    $\begingroup$ 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. $\endgroup$ – Ivan Neretin Jan 16 '16 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 '16 at 11:20
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    $\begingroup$ 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 '16 at 19:38

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

  • $\begingroup$ I believe you are on the right track here. Additionally to your thought, CO2 has a well developed quadrupole moment. $\endgroup$ – Martin - マーチン Dec 15 '16 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.)


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