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)?

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edited Mar 2 "18 in ~ 3:23 Avyansh Katiyar
asked Feb 21 "17 in ~ 10:36 EthanEthan
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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.
$cePF5~is~P^5+,5F-$ $ceWO3~is~W^6+,3O^2-$ $ceSOCl2~is~ S^4+, O^2-, 2Cl^-$