How can I calculate the oxidation number for${\mathrm{FeSO}}_4$?

Calculating the oxidation number of${\mathrm{FeSO}}_4$:

• The oxidation number is the number of electrons an atom gains or loses to form a chemical bond which also helps in deciding whether an element gets reduced or oxidized.
• For the compound${\mathrm{FeSO}}_4$ , the oxidation number of each element present in the compound is represented as;

Oxidation number of $\mathrm{S}=+6.$

Oxidation number of $\mathrm{Fe}=+2.$

Oxidation number of $\mathrm{O}=(-2✕4)=–8$

• But the compound ${\mathrm{FeSO}}_4$is a neutral compound, for which the oxidation number can be calculated by adding the oxidation number of the elements in it.
• Thus, the oxidation number of ${\mathrm{FeSO}}_4$can be calculated as:
• = $(+2)+(+6)+(-8)=0$

=$(+8)–8=0.$

Hence, the oxidation number of ${\mathrm{FeSO}}_4$ is calculated as $0$, as it is a neutral compound.