Oxidation state of $F e$ in $F e_{3} O_{4}$ is:

\frac{5}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{4}{5}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{8}{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{3}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\frac{8}{3}$
Let, the oxidation state of $F e$ in $F e_{3} O_{4}$ be $x$ :
$3 x + 4 (- 2) = 0 x = + \frac{8}{3}$

Suggest Corrections

Similar questions

F e S O_{4} (N H_{4})_{2} S O_{4} .6 H_{2} O

F e_{3} - -- - O_{4}

(F e_{3} O_{4})

8 g

F e = 56

F e + H_{2} O \to F e_{3} O_{4} + 4 H_{2}

F e

{F e}_{3} O_{8}

Join BYJU'S Learning Program