So because that $ceSO2$, here"s the solution when finding for the oxidation variety of sulfur.

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

$x$ below would it is in the oxidation variety of sulfur, for this reason the compound would now be $overset+4ce S overset-2ceO2$.

Now if i tried making use of oxygen as the $x$ it would 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 around this because we just obtained into this today and haven"t checked numerous examples. Over there is probably something I"m absent in the equipment or a rule. Deserve to someone define to me the exactly one? And additionally can"t sulfur"s and 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
request Feb 21 "17 at 10:36 EthanEthan
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If you have actually learned the electronegativity rule yet, right here is a good place to use. If girlfriend haven"t, greater electronegativity basically means that the facet has more powerful driving pressure to reaching an octet; you can look up an electronegativity table if girlfriend like. The an ext electronegative aspect takes the electrons first. So, in your example, $ceSO2$ has $ce2O^2-$, due to the fact that of the group oxygen is in. Then, due to the fact that the $ceSO2$ is neutral, there should be a $ceS^4+$.
But, in a compound such as $ceNa2S$, you can see that S is closer to getting to an octet 보다 Na. So, this becomes $ceS^2-$ while each sodium come to be $ceNa+$.
Try detect the oxidation claims (numbers) for $cePF5, ceWO3, ceSOCl2$ because that understanding. Services in the spoiler below.
$cePF5~is~P^5+,5F-$ $ceWO3~is~W^6+,3O^2-$ $ceSOCl2~is~ S^4+, O^2-, 2Cl^-$