Question

How do you find the oxidation number of $S{O}_4^{2-}$?

Answer 1

Hint: In $S{O}_4^{2-}$, there are sulfur and oxygen atoms, and we know some of the oxidation states of the atoms like hydrogen, oxygen are +1, -2 in almost all the compounds. So, to find the oxidation state of sulfur, add all the oxidation numbers in the compound and keep it equal to -2.

Complete step by step answer:
The oxidation number is the number equal to the charge on the atom in the compound which appears on the atom when it is in a combined state with other atoms or elements.
To find the oxidation state of an element or atom in the compound, we have to add all the oxidation of the atoms in the compound and keep it equal to the overall charge on the compound. If there is no charge on the compound then it must be equal to 0.
In $S{O}_4^{2-}$, there are sulfur and oxygen atoms, and we know some of the oxidation states of the atoms like hydrogen, oxygen are +1, -2 in almost all the compounds.
So, the oxidation state of oxygen in $S{O}_4^{2-}$ will be -2 and the overall charge on the compound is -2. Therefore, to find the oxidation state of sulfur, add all the oxidation numbers in the compound and keep it equal to -2. Taking the oxidation state as x, we get:
$x+4(-2)=-2$
$x-8=-2$
$x=-2+8$
$x=+6$
Hence, the oxidation state or number of sulfur in $S{O}_4^{2-}$is +6.

Note: The oxidation number doesn't need to be positive or negative, it can be zero and even fractional. The zero oxidation is obtained when the element or atom elementary state or in the free state.