What will be the n-factor $(n_{f})$ for $F e_{2} O_{3}$ in the following chemical change :
$F e_{2} O_{3} \to F e S O_{4}$

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Solution

The correct option is C $2$
For $n_{f}$ :
$| O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} | \times number of atom$
$F e_{2} O_{3} \to F e S O_{4}$
O.S. of Fe in reactant side
$= 2 x + 3 \times (- 2) = 0$
$x = + 3$
O.S. of Fe in product side
$= x - 2 = 0$
$x = + 2$
$n-factor = number of Fe atom \times | O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} |$
$n-factor = 2 \times | + 2 - 3 |$
$n-factor = 2$

Theory :

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 states.
4. Multiply it with no of atoms for the respective compound whose $n_{f}$ is to be found.

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(n_{f})

F e_{2} O_{3}

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F e_{2} O_{3}

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(F e S O_{4})

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