Finding oxidation states
 
What are the oxidation states in a.)S8, b.)SCl2, c.)SO4-2, and d.)Na2SO3?
 
a.)S8
Since sulfur is in its elemental form, Rule 1 tells us that its oxidation state will be 0.
 
b.) SCl2
Cl is the more electronegative species in this compound. For this reason, we should give it an oxidation number equal to its ionic charge; this is -1. The sum of the oxidation numbers has to equal 0.
-1(2 Cl) + ?= 0
-2 + ?= 0
0 + 2= +2
Sulfur's oxidation number is +2 in the example.
 
c.) SO4-2
O is the more electronegative species in this example, so it must get an oxidation number equal to its ionic charge; this is -2. The sum of the oxidation numbers is -2 because it is the overall charge.
-2(4 O) + ?= -2
-8 + ?= -2
8 + -2= +6
The oxidation state of sulfur in this example is +6.
  
d.) Na2SO3
Oxygen will take on it's -2 oxidation state, while sodium will take its usual +1 oxidation state. We need to find the oxidation state of sulfur.
+1(2 Na) + ? + -2(3)= 0
+2 + ? -6= 0
+2 + -6= -4
-4 + ?= 0
-4 + +4= 0
The oxidation state of sulfur in this problem is +4.

 

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