## Chapter 4

## Q. 4.16

What is the oxidation number of sulfur in (a) SO_{2}, (b) Na_{2}S, and (c) CaSO_{4}?

## Step-by-Step

## Verified Solution

**Collect, Organize, and Analyze** To assign the oxidation number of sulfur in three of its compounds, we apply the rules for determining the oxidation numbers of the elements in each compound. SO_{2} is the molecule sulfur dioxide. Na_{2}S is an ionic compound, and the oxidation number of an ion in an ionic compound equals its charge. Sulfur in CaSO_{4} is part of the sulfate ion, which means its oxidation number added to those of the four O atoms must add up to the charge of the ion; we determine the oxidation number for calcium from rule 1.

**Solve**

a. From rule 5, O.N. = -2 for the O in SO_{2} . Rule 1 says that the sum of the oxidation numbers for S and the two O in the neutral molecule must be zero. Letting O.N. for S be x:

x + 2(-2) = 0

x = +4 O.N. for S in SO_{2}= +4

b. Sodium forms only one ion, Na^{+}, which means that, according to rule 3, O.N. =+1. To balance the O.N. values in Na_{2}S (rule 1), we let y stand for the O.N. of sulfur:

2(+1) + y = 0

y = -2 O.N. for S in Na_{2}S =-2

c. The charge on the calcium ion is always 2+. This means that the charge on the sulfate ion must be 2-. Assigning O.N. = -2 for oxygen (rule 5) and z for sulfur:

z + 4(-2) = -2

z = +6 O.N. for S in CaSO_{4} = +6

**Think About It** We found oxidation numbers for sulfur ranging from -2 to +6. Many other elements have a range of oxidation numbers in their compounds. For example, the oxidation numbers of iodine range from -1 in KI to +7 in KIO_{4}.