IIRC, Carbon Monoxide has a triple bond between the carbon © and oxygen (O). But, “C” needs four electrons to be satisifed (as in CH4) and “O” can only give two electrons (as in H2O). So, how do chemists explain this? Is this one reason why they say organic chem doesn’t follow the rules? - Jinx
a) First “O” is accepting two, not giving two
b) Actually in both examples, the electrons are shared in covalent bonds as opposed to traded as-in an ionic bond.
Just clarifying, but the question remains that the carbon wants to share four, but the oxygen only wants to share two.
- Jinx
It is a dative bond, where both the electrons in a pair come from one of the atoms. In this case, a pair of electrons orbiting the O is “donated” to form a bond. Another example is acids. Acids occur when a dative bond forms between an H[sub]2[/sub]O and an H[sup]+[/sup] ion to form a H[sub]3[/sub]O[sup]+[/sup] ion. A lone pair on the O forms a dative bond with the H[sup]+[/sup].
http://www.webchem.net/notes/chemical_bonding/dative_bonding.htm
CO should be drawn with a “-” on the C and a “+” on the O. The C is surrounded by the six triple-bond electrons and a lone pair, while the O sees these six plus a single lone pair (rather than the usual two, much like the O in H3O+.
Jinx, didn’t you ask the same question last year? Did you not like the replies you got then? 
Anyway, just to reiterate, an oxygen atom has six valence electrons that are free to form bonds, not two.
Josh, I believe you have these charges reversed. Oxygen is more electronegative than carbon. In the CO molecule, this causes the oxygen end of the molecule to have a partial negative charge, and the carbon end of the molecule to have a partial positive charge.
True.
Wrong. In CH4, the carbon has 8 electrons, two in each C-H bond. 4 come from the carbon, and the other four come from the four hydrogens. In H2O, the oxygen has 8 electrons; 2 in each O-H bond and 4 in two lone pairs.
In CO, there are ten total electrons, 4 from C and 6 from O. 6 of these go into the triple bond, shared between the atoms. The other 4 electrons go into 2 lone pairs, one on each atom. So, each atom sees 8 electrons (fulfilling the octet rule).
Does anyone else think it’d be so much simpler if CO would just be a carbene?
This is exactly what Jinx was told the last time he asked this question (although you need to replace the term “electrons” with “valence electrons”):
Jake4, be aware that there are more than ten total electrons present in the CO molecule. (There are actually 14 total electrons.) There are, however, only ten valence electrons present.
Nope. It’s not a question of partial charges, but of formal charges. These have nothing to do with electronegativities; they follow from the electron configuration in the Lewis structure (the electronegativities of course are relevant to which resonance structures make what contributions). If you’re going to draw CO with a triple bond, then the C implicitly has one lone pair, and the O has one lone pair too. Each “owns” five valence electrons (lone pair plus one from each of the covalent bonds) in addition to its two inner electrons, giving a net -1 charge for the C (with its six protons) and a +1 for the O (with its eight protons). In general, atoms with an “extra” covalent bond and a complete octet have a +1 formal charge (such as the N of a protonated amino group or the O of hydronium ion), while those “missing” a covalent bond have a -1 charge (for example hydroxide ion or a carbanion).
There are other ways to draw CO. It can be drawn with a double bond and no formal charges. In this case the C lacks a complete octet, much like a carbocation or the resonance structure of a carbonyl group that has a single bond. I imagine that this resonance structure of CO makes a significant contribution, but that’s beside the point. If you’re going to draw it with a triple bond, the charges should be drawn as I describe, and they’re important to understanding the meaning of that structure.
You’re absolutely correct, of course. I didn’t realize you were referring to formal charges–my mistake.
My only (lame) defense is that it’s more obvious what you are referring to if you put the charges in circles to indicate formal charges, and preface the charges with lowercase deltas to indicate partial charges.