Question

Explain the average oxidation state of ${\mathrm{Fe}}_3{\mathrm{O}}_4$.

Answer 1

Step 1: Calculation of Oxidation state of ${\mathbf{Fe}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{4}}$

1. ${\mathrm{Fe}}_3{\mathrm{O}}_4$ is made up of a stoichiometric blend of ferrous ($\mathrm{FeO}$) and ferric (${\mathrm{Fe}}_2{\mathrm{O}}_3$) oxides combined as $\mathrm{FeO}.{\mathrm{Fe}}_2{\mathrm{O}}_3$
2. Oxygen in $\mathrm{FeO}$ and ${\mathrm{Fe}}_2{\mathrm{O}}_3$ is in the $-2$ oxidation state.
3. Let x and y, respectively, be the oxidation states of iron in $\mathrm{FeO}$ and${\mathrm{Fe}}_2{\mathrm{O}}_3$. As we know the sum of all the atoms or ions in a neutral substance is zero.
4. Therefore, the oxidation number of iron in $\mathrm{FeO}$ is $\mathrm{x}+(-2)=0,\mathrm{x}=+2$
5. And the oxidation number of iron in ${\mathrm{Fe}}_2{\mathrm{O}}_3$ is $2\mathrm{y}+(-2)\times 3=0,\mathrm{y}=+3$

Therefore, the average oxidation state of $\mathrm{Fe}$ in ${\mathrm{Fe}}_3{\mathrm{O}}_4$ is$+2$ and$+3$.