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## Q. 4.16

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

## 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 $KIO_{4}$.