The valency of sulphur in SO2 and SO3 is 4,6. Pls explain how?

Rules which are followed to find oxidation number of sulphur in SO2 and SO3 are:
1) The sum of all oxidation numbers in a neutral compound is zero.
2) The oxidation number of oxygen in a compound is usually –2. If, however, the oxygen is in a class of compounds called peroxides (for example, hydrogen peroxide), then the oxygen has an oxidation number of –1. If the oxygen is bonded to fluorine, the number is +1.

Now in SO2
Let oxidation state of sulphur be 'x'
Oxidation state of oxygen = -2 {Using rule-2}
Total charge = 0 {Using rule-1}
The equation is:  x + 2(-2) = 0
x - 4 = 0
x = 4
Therefore oxidation state of Sulphur in SO2 is +4.

Now in SO3
Let oxidation state of sulphur be 'y'
Oxidation state of oxygen = -2 {Using rule-2}
Total charge = 0 {Using rule-1}
The equation is:  y + 3(-2) = 0
y - 6 = 0
y = 6
Therefore oxidation state of sulphur in SO3 is +6

 

  • 20
What are you looking for?