| 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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