Explain CO Using Valence Electrons...

I’ve always been bothered by how CO (Carbon Dioxide) does not seem to fit the rules for covalent bonding. Carbon has four valence electrons, and Oxygen has two valence electrons. So, I can form two bonds between C and O, but what about the two remaining free electrons of the Carbon atom? :dubious:

In short, I’d have something that looks like this:
:C=O (where the colon would be two valence electrons waiting to bond with something)

Maybe someone can tell me if CO can really be explained using valence electrons alone? Or, do I have to know something about orbital hybrids, I think it’s called (IIRC)? Also, bear in mind I never had a course in organic chemistry…where they tell me the rules are different.

Thanks, y’all!

  • Jinx

Oops! I meant Carbon Monoxide!

Carbon Dioxide is O=C=O, and that makes sense balancing the valence electrons!

Seems you’ve figured it out - carbon monoxide does not stay around very long in the presence of oxygen. In fires or engines, CO forms because there is not enough oxygen present to complete the reaction.

One problem you may be having is that oxygen has six valence electrons, not two.

Carbon Monoxide (CO):
Carbon supplies 4 valence electrons; oxygen supplies 6 valence electrons, for a total of 10 valence electrons to work with.

Each atom can form an octet by forming a triple bond between the carbon and the oxygen atom, with a lone pair of electrons attached to each atom.

Carbon Dioxide (CO[sub]2[/sub]):
The bonds in your model above are correct, but should show two lone pairs on the two oxygen atoms. In your model, you have not utilized all available valence electrons, nor do your oxygen atoms have octets.

(You have 16 valence electrons to work with, 4 from the carbon atom, and 6 each from the oxygen atoms. The two double bonds use up 8 of the valence electrons. The remaining 8 valence electrons go on the oxygen atoms in the form of two lone pairs on each atom.)

No difficulties using valence electrons (Lewis structures) for either of these compounds.

Check out this site for the full story. The short version seems to be that our notation system is not adequate to properly describe the CO molecule, which does not satisfy the octet rule and is therefore fairly unstable anyway.

Actually, oxygen has six valence electrons, so the structure is
Here’s a simple explanation, and here’s a more complex explanation.
Hope this helps!

Slight correction to my above post: CO does in fact satisfy the octet rule for oxygen, just not for carbon (if I understand this article correctly).

While your comments are true enough, they have no bearing on constructing Lewis structures for these compounds. According to the (incorrect) Lewis structures presented in the OP, neither compound would be stable.

Reading the article, the first resonance structure for CO is the model I presented in my first post, (and nicely illustrated by jk1245). Indeed the triple bond predicted in this structure agrees quite nicely with the experimentally observed bond length This would be the end of the story, except that the experimentally observed dipole for the molecule does not agree with this simple structure. This brings up the necessity for the other two resonance structures illustrated, which do not satisfy the octet rule for carbon.

Incidentally, I wouldn’t expect to see this come up in any chemistry class taught at the level where Lewis structures were first being introduced. The first resonance structure with all octets satisfied and all valence electrons accounted for would, no doubt, suffice.