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Electronic Configuration and Oxidation States

Consider the ...

Question

Consider the reaction: $F e_{2} O_{3} ⟶ F e_{3} O_{4}$ .
$n$ -factor for $F e_{3} O_{4}$ is :

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Solution

The reaction is
$F e_{2} O_{3} ⟶ F e_{3} O_{4}$
The oxidation number of Fe in $F e_{2} O_{3}$ is +3.
The oxidation number of Fe in $F e_{3} O_{4}$ is +2.67.
During the reaction, the change in the oxidation number for one Fe atom is
$3 - 2.67 = 0.334$ .
The change in the oxidation number for three Fe atom is
$3 \times 0.334 = 1$ .
The n factor for $F e_{3} O_{4}$ is 1.

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