The oxidation state of $F e$ in ${F e}_{3} O_{8}$ is :

\frac{3}{2}

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\frac{4}{5}

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\frac{5}{4}

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\frac{16}{3}

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Solution

The correct option is D $\frac{16}{3}$
Let $x$ be the oxidation state of $F e$ in $F e_{3} O_{8}$ .
Since, the overall charge on the complex is $0$ , the sum of oxidation states of all elements in it should be equal to $0$ .
Therefore,
$3 x + 8 (- 2) = 0$
or, $x = + \frac{16}{3}$ .
Hence, the oxidation state of $F e$ in $F e_{3} O_{8}$ is $+ \frac{16}{3}$ .

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