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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?

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

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    $\begingroup$ What would you suggest that we assign it instead? $\endgroup$ – Ivan Neretin Nov 27 '18 at 5:35
  • $\begingroup$ We do it for our own convenience, there is no deeper reason. $\endgroup$ – Ivan Neretin Nov 27 '18 at 7:06
  • $\begingroup$ electronegativity? @IvanNeretin $\endgroup$ – ten1o Nov 27 '18 at 7:07
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    $\begingroup$ 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. $\endgroup$ – Ivan Neretin Nov 27 '18 at 7:10
  • $\begingroup$ 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. $\endgroup$ – mykhal Nov 27 '18 at 15:55
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Strictly speaking, oxidation states should only be used with ionic compounds. However, this concept can be interesting in organic or semi-organic molecules as long as you understand that atoms actually don't have such a black and white oxidation state.

The oxidation state is attributed according to the relative electronegativity of 2 atoms (the most electronegative takes a -1 state and the other a +1 state) and they sum up if an atom has more than one neighbors. For example:

In CCl4, there are 4 C-Cl bonds which you need to consider. Every time, the chlorine atom will get a -1 oxidation state, while the central carbon atom will get 4 times a +1 oxidation state, which gives +4 at the end.

For multiple bonds, you need to consider each single bond individually. In an S=O bond, there are 2 single bonds and each of them gives a -1 state to the oxygen (hence a +1 state for the sulphur).

Combining the two examples, you should see why in SO2 sulphur has a +4 oxidation state and both oxygens have a -2 state.

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