Answer:
the oxidation number of sulfur in the S2O8 2- ion is +7.
Step by step explanation:
The oxidation number of sulfur (S) in the S2O8 2- ion can be determined by applying the following rules:
Oxygen (O) has an oxidation number of -2 in most compounds, except in peroxides where it has an oxidation number of -1.
The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to its charge.
In the S2O8 2- ion, there are two sulfur atoms and eight oxygen atoms. Let x be the oxidation number of sulfur.
From the first rule, each oxygen atom contributes -2 to the overall charge of the ion. Therefore, the total contribution from the eight oxygen atoms is -16.
From the second rule, the sum of the oxidation numbers of all the atoms must equal the charge of the ion. Since the ion has a charge of -2, we can write:
2x + (-16) = -2
Simplifying and solving for x, we get:
2x = 14
x = +7
Therefore, the oxidation number of sulfur in the S2O8 2- ion is +7.