# Oxidation states for covalent compounds [duplicate]

Yesterday, I came across $$\ce{SO2}$$ and was supposed to write the reduction and oxidation half reactions. However, I do not understand the reason why S is assigned the oxidation number of 4+. Oxygen shares 4 electrons with S in total, and oxygen is more electronegative than S. I don't get how we somehow "assigned" +4 to S. Shouldn't S also have a negative charge? I am so confused and need help. What is the difference between the oxidation charge and ionic charge?

## marked as duplicate by Mithoron, Jon Custer, Tyberius, A.K., Todd MinehardtNov 28 '18 at 23:02

• What would you suggest that we assign it instead? – Ivan Neretin Nov 27 '18 at 5:35
• We do it for our own convenience, there is no deeper reason. – Ivan Neretin Nov 27 '18 at 7:06
• electronegativity? @IvanNeretin – ten1o Nov 27 '18 at 7:07
• Yes, electronegativity does play a role, but why? That's a matter of convenience. We imagine that the more electronegative atom "owns" the shared electrons, and the other one kinda "borrows" them. – Ivan Neretin Nov 27 '18 at 7:10
• Maybe you forgot an important property/definition for a while, that the sum of all oxidation numbers gives a total charge, i.e. 0 for neutral molecules. – mykhal Nov 27 '18 at 15:55