You are watching: What is the oxidation state of sulfur in so2−4?
$$x + 2(-2) = 0 \x - 4 = 0 \ x = +4$$
$x$ 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
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request Feb 21 "17 at 10:36
EthanEthan
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$endgroup$
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$egingroup$
Your teacher used an interesting example that doesn"t follow the rules cleanly. Sulfur can have plenty of oxidation says in this method. In general, larger facets can have actually oxidation states various than suggested by the group they room in.
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.
$cePF5~is~P^5+,5F-$ $ceWO3~is~W^6+,3O^2-$ $ceSOCl2~is~ S^4+, O^2-, 2Cl^-$