Calculate 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 ${\text{FeS}}_2$, Fe is $+2$ and ${\text{S}}_2^{2-}$ means S is $-1$.
2. In products, Fe is $+3$ in ${\text{Fe}}_2{\text{O}}_3$ and S is $+4$ in ${\text{SO}}_2$.
3. Change for Fe: $2\to 3$ is $+1$.
4. Change for S: $$\begin{array}{rl}2\times (-1\to +4)& =2\times 5\\ & =10\end{array}$$.
5. Total n-factor = $1 + 10 = 11$.
6. Hence the answer is (C).