Question

oxidation number of s in (NH4)S2O8​

Answer 1

Oxidation state of sulphur is +7.

Explanation:

Oxidation state is generally a combining capacity of the element.

Let the oxidation state of Sulphur in given molecule is xx

We know that the oxidation state of ammonium ion NH_{4}NH

4

IS +1.

We also know the oxidation state of oxygen is -2.

(NH_{4} )_{2}S_{2}O_{8} }

2\times 1+2x-2\times8=02×1+2x−2×8=0

-14+2x=0−14+2x=0

x = \dfrac{14}{2} = 7x=

2

14

=7

x=7x=7

So, oxidation state of sulphur is +7.