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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 $\ce{CO}$?

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  • $\begingroup$ 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
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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. $\endgroup$ Jan 16, 2016 at 19:23
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    $\begingroup$ > 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
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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, 2016 at 19:38

4 Answers 4

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$\ce{CO2}$ has more electrons than $\ce{CO}$. This means that it has a much larger electron cloud as compared to $\ce{CO}$, so its more easily polarised and thus, the ease of forming instataneous dipole-induced dipole bonds increases. Even though $\ce{CO}$ 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 $17 \space year \space old$ A Level Chem student here, I might be wrong

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    $\begingroup$ 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
  • $\begingroup$ Just want to add that CO is not very polar, with dipole moment ~0.112D compared to e.g. water 1.85D. The reason is that effect of dative covalent bond C←O slightly more than cancelled the polarization of C=O, leaving a very small +ve charge on O and -ve charge on C. $\endgroup$
    – R. Wang
    May 2 at 5:24
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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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Apologies for resurrecting an old post but this came up in Chemistry class today. While CO2 is larger and hence has stronger dispersion forces, you would still expect dipole-dipole bonds to be significantly stronger in relative terms.

I suspect that the dipole-dipole bonds are not as strong as we usually expect from a C-O bond. In carbon monoxide, the carbon and oxygen are triple bonded and both carbon and oxygen atoms have a lone pair of electrons. As such, I suspect the electron cloud density is more evenly distributed than we normally expect in C-O bonds and hence the dipole-dipole bonds would be quite weak. Carbon Monoxide Lewis Diagram

I would be quite happy to have someone more authoritative comment on this but that is the reasoning to my guess.

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As the above answers have suggested, the $\ce{CO2}$ molecule has a larger electron density (than the $\ce{CO}$ molecule) surrounding it, which results in higher dipole-induced-dipole interactions (and that is one of the reasons why moist $\ce{CO2}$ is acidic (it forms carbonic acid in water))

$\ce{CO}$ is a neutral molecule, which suggests that weak dipole-dipole, and Vander Waal's interactions are present in the molecule. Further, the molecule is "internally stable". The molecule has Bond order 3, which is stable within itself, and interactions are reduced to a bare minimum.
To explain this, one may consider the Intra molecular Hydrogen bonding versus Intermolecular Hydrogen bonding analogy. To site one example, ortho-nitrophenol has a boiling point of around 214-215(°C), while para-nitrophenol has a boiling point of around 279-281(°C) (These values may vary slightly depending on the specific source and conditions). This is due to the presence of intramolecular H-bonding in O-nitrophenol, which limits the interaction forces between the molecules, thereby reducing its boiling point.

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