The oxidation state of $F e$ in ${F e}_{3} O_{4}$ is

+ 3

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+ \frac{8}{3}

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+ 6

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+ 2

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Solution

The correct option is D $+ \frac{8}{3}$
Let the oxidation state of $F e$ in ${F e}_{3} O_{4} = x$
$∴ 3 x + 4 \times (- 2) = 0$
or $3 x - 8 = 0$
$∴ x = + \frac{8}{3}$

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