What is the $n$-factor of ${\text{FeS}}_2$ in the reaction ${\text{FeS}}_2+{\text{O}}_2\to {\text{Fe}}_2{\text{O}}_3+{\text{SO}}_2$?

Solution

1. In ${\text{FeS}}_2$, the oxidation states are ${\text{Fe}}^{2+}$ and ${\text{S}}_2^{2-}$, so $\text{S}$ is $-1$.
2. In the products ${\text{Fe}}_2{\text{O}}_3$ and ${\text{SO}}_2$, $\text{Fe}$ is $+3$ and $\text{S}$ is $+4$.
3. Change for $\text{Fe}$ is $(+3-+2)=+1$.
4. Change for two $\text{S}$ atoms is $$\begin{array}{rl}2\times (+4-(-1))& =2\times 5\\ & =+10\end{array}$$.
5. Total $n$-factor is $1 + 10 = 11$. Hence the answer is (C).