Whilst I understand that $\ce{CO2}$ has a linear structure, resulting in a vector sum of the dipoles of 0, I do not understand why the dipoles formed at the two oxygen atoms as a result of their increased electronegativity cannot interact with other molecules via dipole-dipole forces. Do the vectors equaling 0 somehow negate the present dipoles?

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    $\begingroup$ Do realize that there are other attractive forces besides dipole-dipole interactions. So two carbon dioxide molecules could be attracted to each other by a London dispersion force. $\endgroup$ – MaxW Feb 12 '17 at 23:28
  • $\begingroup$ The attractive induced-dipole, induced-dipole interaction exists even when the molecule has no overall dipole. It is proportional to the product of average polarisabilities and mean of ionisation potentials and inversely as the sixth power of separation of centres of mass. ($\ce{CO2}$ has an asymmetric vibration and two bending vibrations that produce transient dipoles.) $\endgroup$ – porphyrin Feb 13 '17 at 14:33
  • $\begingroup$ Of course it depends whether you consider the chemical reactivity of carbon dioxide as something that can be explained by dipole dipole interactions when other molecules are very close. Most people think of the chemical reactivity as something different even though the reactions are explained by the movement of electrons. $\endgroup$ – matt_black Feb 13 '17 at 21:41

As the other answers have indicated, CO2 has no NET dipole moment. However it does have two dipoles pointed in opposite directions (as OP keeps mentioning). This means that CO2 can possibly interact through higher moments, such as the quadrupole moment. The mathematical procedure behind this is known as the multipole expansion. It is important to note however, the higher order multipoles get weaker and weaker by a factor of 1/R approximately, so quadrupole interactions are usually ignored for example.


Carbon dioxide is more than linear. It's symmetric, and the axis of symmetry perpendicular to the bonds also applies to whatever dipole moment it has. The only vector that looks the same after being rotated 180 degrees is the null vector, so the molecule has zero overall dipole moment.

Therefore, dipole-dipole interactions are not possible because carbon dioxide does not have a dipole.


Exactly that.

Dipole-dipole interactions are caused because the positive end of one dipole is attracted to the negative end of another dipole. This macroscopic attraction can only occur if the centre of positive charge and the centre of negative charge do not coincide.

In molecules like carbon dioxide that contain a centre of symmetry, these charge centres do coincide. They fall on top of each other. So macroscopically, a neighbouring molecule cannot see any positively or negatively charged centre and therefore sees no possibility to attach via dipole interactions.

  • $\begingroup$ so basically the charges are distributed equally within the molecule? But how can that be ? Aren't the 2 o atoms suppose to deprive the c atom os electrons and make it positively charged? $\endgroup$ – Pranoy De Feb 12 '17 at 22:59
  • $\begingroup$ Well yes, each of the oxygens is negatively charged and carbon is positively charged. But these microscopic charge centres both fall together in the molecule’s centre of gravity when viewed from a greater distance. That’s the power of symmetry. $\endgroup$ – Jan Feb 12 '17 at 23:05
  • $\begingroup$ so even though the oxygen is negatively charged, it cannot interact with the positively charged C atom of another CO2 molecule $\endgroup$ – Pranoy De Feb 12 '17 at 23:10
  • $\begingroup$ Thank you Jan. So due to the symmetry the overall dipole is 0? Wouldn't the electrons still be more attracted to the oxygen, resulting in dipoles? And wouldn't these dipoles be able to interact with other dipoles? I.e a positive carbon atom $\endgroup$ – Vihanga Dharmasena Feb 13 '17 at 4:32

Consider the CO2 molecule as a superposition of two dipoles, one for each C=O bond. A nearby polar molecule will feel a dipole-dipole interaction with each C=O bond, but since those two dipoles are equal and opposite, those two interactions will almost cancel out. They may not completely cancel since the interaction with the closer of the two bonds will be stronger.

The resulting net interaction will therefore be much weaker than the interaction with a polar molecule such as water.

Note that a superposition of two equal and opposite dipoles in different locations forms a quadrupole. So, instead of electing to treat the interaction of a polar molecule with CO2 as the interaction with two equal and opposite dipoles, we can treat it as the interaction with a zero dipole but nonzero quadrupole. As user157879 points out, quadrupole interactions are much weaker than dipole interactions.


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