[Sshirazi]

Hi all
I am trying to understand the oxidation state of carbon in different molecules. Carbon has 4 valence electrons so oxidation state should be +- 4 right? However in acetic acid the two carbons are +3 and -3 respectively. I think that I do not understand the concept of redox properly. I would appreciate any *delete me*
Many thanks

[DevaDevil]

look at the structure of the molecules. Also, remind yourself that organic molecules have covalent, not ionic bonds, so oxidation numbers are fine, but do not see it as a full electron transfer as in salts.

acetic acid:

CH₃-COOH

the C-C bond no charge transfer as both atoms are equally electronegative.
the C-H bond: C more electronegative than H, oxidation state of H = +1, hence in CH₃, the oxidation state of C is -3 (overall neutrality has to be retained)

then look at the other part:
C=O: O is more electronegative and has oxidation state -2.
C-O-H: O again most electronegative, oxidation state -2, H has again oxidation state +1.
This leaves C with an oxidation state of -( 1 -2 -2 ) = +3

again, do not think this means the bonds are ionic, as they are covalent!

[opti384]

A little tip about oxidation state (a.k.a. oxidation number) is that it's an imaginary electrical charge of atoms that bond with covalent bonds. So, if same atoms form a covalent bond, their oxidation state will be 0 because we assume that they equally share the electrons. Hope this helped.

[vmelkon]

In acetic acid, the 2 carbons  has covalent bonds. The important thing to know here is that carbon has 4 valence electrons and they pair up with another 4 electrons. Carbon makes 4 bonds.
The first carbon has 3 bonds with each H. Another bond with the C. Total = 4 bonds.
The second C has 1 bond with the first C. Double bond with O. Single bond with the other O. Total = 4 bonds.