So because that $ceSO2$, here"s the solution as soon as finding because that the oxidation number of sulfur.

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

$x$ right here would be the oxidation number of sulfur, so the compound would currently be $overset+4ce S overset-2ceO2$.

Now if i tried using oxygen together the $x$ it would certainly become: $$1(-2) + 2x = 0 \ -2 + 2x = 0 \2x = 2 \frac2x2 = frac 22 \x=1$$

So, in this case the compound would actually be $overset-2ce S overset+1ceO2$.

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I"m pretty confused about this since we just obtained into this today and haven"t checked many examples. Over there is most likely something I"m lacking in the solution or a rule. Have the right to someone describe to me the exactly one? And likewise can"t sulfur"s and also oxygen"s oxidation number both it is in -2 together they"re in group VI(A)?

redox oxidation-state
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edited Mar 2 "18 in ~ 3:23

Avyansh Katiyar
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Your teacher used an interesting instance that doesn"t follow the rule cleanly. Sulfur deserve to have plenty of oxidation states in this method. In general, larger elements can have actually oxidation states different than argued by the group they are in.

If you have learned the electronegativity rules yet, here is a an excellent place come use. If friend haven"t, higher electronegativity basically method that the element has more powerful driving pressure to reaching an octet; you can look up an electronegativity table if friend like. The an ext electronegative facet takes the electron first. So, in her example, $ceSO2$ has $ce2O^2-$, due to the fact that of the team oxygen is in. Then, due to the fact that the $ceSO2$ is neutral, there need to be a $ceS^4+$.

But, in a link such as $ceNa2S$, you can see that S is closer to getting to an octet than Na. So, this i do not care $ceS^2-$ while each sodium become $ceNa+$.

Try detect the oxidation says (numbers) because that $cePF5, ceWO3, ceSOCl2$ because that understanding. Solutions in the spoiler below.

$cePF5~is~P^5+,5F-$ $ceWO3~is~W^6+,3O^2-$ $ceSOCl2~is~ S^4+, O^2-, 2Cl^-$