In the compound $\ce{CO}$ I know that the carbon has an oxidation number of $+2$ and the oxygen has an oxidation number of $-2$. However up until now I have thought of oxidation number and charge being the same, and it is apparent that the $\ce{CO}$ molecule is most definitely covalent and has no ionic character hence the molecule is neutral and each atom should not have an actual "charge" or we would call the molecule ionic. So are oxidation number and charge different or do $\ce{C}$ and $\ce{O}$ have charges in carbon monoxide.


1 Answer 1


Oxidation State and formal charge are most definitely not the same thing; in fact they make opposite assumptions to exist.

The molecule CO is, as you correctly identified, covalent, and you’re likely to only see oxidation states used when it comes to ionic molecules, so focusing on covalent molecules:

Formal charge totally neglects the existence of electro negativity in covalent bonds. It makes the assumption that it any covalent bond, shared electrons belong equally split between the two atoms that share them.

Hence the formal charge of an atom can be defined as :

Number of valence electrons in uncharged atom - number of electrons in lone pairs in the atom in the molecule - (1/2 * the number of electrons shared in covalent bonds by the atom in the molecule).

Oxidation states, however, make the assumption that electronegativity is an all-powerful effect, such that in a covalent bond with an electro negativity difference, shared electrons belong entirely to the more electronegative element.

Hence the oxidation state of an atom in a covalent bond can be defined as:

Number of valence electrons in uncharged atom of element - number of electrons in atom in the molecule (assigning electrons to more electronegative atoms)

The ONLY real similarity between the two concepts is that in any molecule/ion, the sum of formal charges is equal to the charge of the molecule, and so is the sum of the formal charges. But the individual charges on atoms will often differ from the individual oxidation states on atoms.

Considering the molecule CO, where there is a triple bond between oxygen and carbon , and each atom has one lone pair:

Oxygen is more electronegative than carbon and oxygen normally has 6 VE and carbon has 4VE. So for oxidation states, all 6 electrons in the triple bond as considered to belong to the oxygen; for formal charge they are split 3 each. Hence:

Formal charges:

Carbon: 4 - 2 - (0.5*6) = -1

Oxygen: 6 - 2 - (0.5*6) = 1+

Oxidation states:

Carbon: 4 - 2 = 2+

Oxygen: 6 - 2 - 6 = -2

Note that although these are different, both pairs of values sum to 0 as your molecule is neutral.

Finally, you’re right that the molecule is neutral and does not have an overall charge, although it does obviously have a dipole towards the oxygen nonetheless.

But the truth is that these formal charges and oxidation states are both just numerical but artificial concepts. They’re mainly taught as they can be extremely useful in other areas of chemistry, such as organic synthesis in the case of formal charge or knowing when acidic conditions are needed in the case of oxidation states.

  • $\begingroup$ The dipole in CO is actually with the negative end towards carbon, basically formal charge beats electronegativity difference. $\endgroup$ Dec 8, 2022 at 15:50

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