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So for $\ce{SO2}$, here's the solution when finding for the oxidation number of sulfur.

$$x + 2(-2) = 0 \\ x - 4 = 0 \\ x = +4$$

$x$ here would be the oxidation number of sulfur, so the compound would now be $\overset{+4}{\ce S} \overset{-2}{\ce{O2}}$.

Now if I tried using oxygen as the $x$ it would become: $$1(-2) + 2x = 0 \\ -2 + 2x = 0 \\ 2x = 2 \\ \frac{2x}{2} = \frac 22 \\ x=1$$

So, in this case the compound would actually be $\overset{-2}{\ce S} \overset{+1}{\ce{O2}}$.

I'm pretty confused about this since we just got into this today and haven't checked many examples. There is probably something I'm missing in the solution or a rule. Can someone explain to me the correct one? And also can't sulfur's and oxygen's oxidation number both be -2 as they're in group VI(A)?

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    $\begingroup$ Your second example seems to have a sign flipped somewhere. It assumes Sulphur exist as S(-2) which would lead to O(+1)*2, but you then describe SO2 as S(+2)O(-1)*2. $\endgroup$
    – MSalters
    Feb 21, 2017 at 15:28
  • $\begingroup$ @MSalters Agreed; edited to reflect. $\endgroup$
    – hBy2Py
    Feb 21, 2017 at 15:52

2 Answers 2

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Your teacher used an interesting example that doesn't obey the rules cleanly. Sulfur can have many oxidation states in this method. In general, larger elements can have oxidation states different than suggested by the group they are in.

If you have learned the electronegativity rules yet, here is a good place to use. If you haven't, higher electronegativity basically means that the element has stronger driving force to reaching an octet; you can look up an electronegativity table if you like. The more electronegative element takes the electrons first. So, in your example, $\ce{SO2}$ has $\ce{2O^2-}$, because of the group oxygen is in. Then, because the $\ce{SO2}$ is neutral, there must be a $\ce{S^4+}$.

But, in a compound such as $\ce{Na2S}$, you can see that S is closer to reaching an octet than Na. So, this becomes $\ce{S^2-}$ while each sodium become $\ce{Na+}$.

Try finding the oxidation states (numbers) for $\ce{PF5}, \ce{WO3}, \ce{SOCl2}$ for understanding. Solutions in the spoiler below.

$\ce{PF5~is~P^5+,5F-}$
$\ce{WO3~is~W^6+,3O^2-}$
$\ce{SOCl2~is~ S^4+, O^2-, 2Cl^-}$

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This is a straight forward application of rules that should be in your text. These rules are based, in general, on preference given to more electronegative elements. From Zumdahl, and in order of application:

  1. raw element = 0
  2. monatomic ion = same as charge
  3. fluorine = -1
  4. oxygen = -2 except peroxides and when combined with fluorine.

So, oxygen is a bully until fluorine shows up.

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