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I understand that polarity is electronegativity difference and that the more the electronegativity difference, the more polar the bond.

However, I have read that O=C=O is nonpolar. This doesn't make sense to me.

In $\ce{C=O}$, the carbonyl carbon is partially positive and thus carbonyls are polar. $\ce{O=C=O}$ is like 2 of these bonded together so wouldn't it be that you would have an even more positive carbon and 2 partially negative oxygens causing $\ce{O=C=O}$ to be polar?

I mean this would explain why more of this dissolves in water than $\ce{O2}$ and makes water more acidic than $\ce{O2}$ does.

So why is $\ce{O=C=O}$ nonpolar when 2 oxygens are pulling electron density from THE SAME CARBON?

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which direction do you think should be more positive? –  Grady Player Jul 26 at 18:47
towards the carbon. and towards the oxygen more negative. –  caters Jul 26 at 22:26
which oxygen is more negative? –  Grady Player Jul 27 at 2:04
they are equally negative –  caters Jul 27 at 12:41

6 Answers 6

up vote 10 down vote accepted

so wouldn't it be that you would have an even more positive carbon and 2 partially negative oxygens

Yes, your analysis is correct to this point. A chemist would say that the bonds in $\ce{CO2}$ are polar (or polarized) and therefor each $\ce{C=O}$ bond has a bond dipole moment. However the molecule itself is linear and the two bond dipole moments are oriented 180 degrees with respect to one another and cancel each other out, so overall the molecule does not have a dipole moment and is non-polar.

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EDIT: There are a couple of reasons why $\ce{CO2}$ is more soluble in water than $\ce{O2}$. Because the two $\ce{C=O}$ bonds in $\ce{CO2}$ are polarized (whereas in $\ce{O2}$ the bond is not polarized) it makes it easier for the polar water molecule to solvate it and to form hydrogen bonds. Both of these factors will stabilize a $\ce{CO2}$ molecule more than an $\ce{O2}$ molecule in water; stabilization translates into greater solubility. Another factor enhancing the solubility of $\ce{CO2}$ in water is the fact that $\ce{CO2}$ reacts with water to set up an equilibrium with carbonic acid. $$\ce{CO2(aq) + H2O <=> H2CO3(aq)}$$ This reaction will also enhance $\ce{CO2}$'s solubility in water compared to oxygen which does not react with water.

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but this does not explain why more of this dissolves in water than how much O2 dissolves in water. –  caters Jul 26 at 14:32
@caters You should probably ask about that in another question. –  ntoskrnl Jul 26 at 14:34
I accept this answer as valid and correct. However, I really have problems with this definition of polarity solely based on on the dipole moment. I would consider toluene to be an unpolar solvent, yet the molecule itself has a dipole of 0.36D, it is small, but it is there. Does that mean toluol is a polar molecule? On the other hand $\ce{CO2}$ has no net dipole moment, but its reactivity towards nucleophiles/ electrophiles is similar to polar molecules. Therefore I'd rather consider it polar. This is in no way a critique on your answer, I'd just value your opinion on the matter. –  Martin Jul 28 at 10:59
@Martin The dipole moment or polarity of a molecule never mattered that much to me. I was usually interested in the reactivity of a molecule, which usually translates into the reactivity of a bond. So the polarity of a bond is what I would consider. If the hybridization in the bond wasn't symmetrical, then the bond was polarized; the asymmetry of the hybridization told me how polarized. –  ron Jul 28 at 15:18

In understanding molecular polarity you need to take the whole structure into account.

Your reasoning is correct as far as the parts of the molecule are concerned. The individual bonds are polar.

But, a molecule can only be polar if it has a net dipole moment (that is, the charges don't balance out in direction across the whole molecule). So CO is polar as the polarity of the carbon-oxygen bond is unbalanced and the distribution of partial charge on the bond affects the whole molecule (giving the whole molecule a dipole moment). But CO2 is a linear molecule and the partial dipoles of the two bonds are in exactly opposite directions. Hence they exactly balance out giving a molecule which is non-polar.

If CO2 were not linear (like SO2 which is bent) it would be a polar molecule.

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This may be the source of the OP's confusion, yes. Otherwise what would the difference between O=C=O and, say, H-O-H be? Maybe you could explain why O=C=O is linear and H-O-H is "bent". –  Mr Lister Jul 27 at 9:34

The formula for the net dipole moment $\vec{\mu_{net}}$ of an overall neutral system of $n$ charged point particles is given by:

$$ \vec{\mu_{net}} = \sum\limits_{i}^{n} q_i\vec{r_i} $$

where $q_i$ is the charge of the $i$th point particle and $\vec{r_i}$ is the position vector for said particle; each individual dipole moment points from a negative charge to a positive charge. Carbon dioxide is a linear and symmetrical molecule, meaning that in the ground state the bonds between each respective oxygen atom and the carbon atom have the same lengths, and each oxygen atom bears an identical (partial) negative charge. All of this ultimately means that the two individual dipole moments between the carbon and oxygens one and two, respectively, perfectly oppose each other geometrically and have the same magnitude, i.e., $q\vec{r_1} = -q\vec{r_2}$, hence the vector sum $q\vec{r_1} + q\vec{r_2} = \mathbf{0}$.

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It's like they are at tug-of-war with each other, and they both have the same strength; it's not going anywhere. Since the oxygen is more electronegative than carbon, it has a partial negative charge and the carbon has a partial positive charge (twice of the oxygen since there are 2 oxygen).

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The lone pairs on both oxygen atoms cancel each others effect, as simple as that. And the structure is symmetrical from all aspects, which again leads to zero dipole moment. Plus there is no net negative charge or lone pair of Carbon as well, therefore the compound cannot be polar. REMEMBER polarity is a vector.

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Dipole moments determine it. Polarity is determined by charge, electronegativity and geometricity.

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This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. –  Greg E. Jul 28 at 19:03

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