Question

Find the n-factor of the $\left(FeS{O}_4\right)$ in the following chemical changes.
$(FeS{O}_4\to F{e}_2{O}_3)$

• A. 0.0
• B. 1
• C. -1
• D. $\frac{1}{2}$

Answer 1

The correct option is B 1

Steps for finding valency factor/n-factor:
1. Find the oxidation state (O.S.) of metal in reactant side.
2. Find the oxidation state of metal in product side.
3. Take the magnitude of difference of two oxidation state $|O.S{.}_{Product}-O.S{.}_{Reactant}|\times \text{number of atom}$ for the valency factor
$(FeS{O}_4\to F{e}_2{O}_3$
O.S. of Fe in product side
$=2x+3\times (-2)=0$
$x=+3$
O.S. of Fe in reactant side
$=x-2=0$
$x=+2$
$\text{valency factor}=\text{number of Fe atom}\times |O.S{.}_{Product}-O.S{.}_{Reactant}|$
$\text{valency factor}=1\times |+3-2|$
$\text{valency factor}=1$
In this reaction, ferrous ions get oxidized to ferric ions.