Find the oxidation number of `Fe` in `Fe_(3)O_(4)` and in `Fe(III)_(4)[Fe(II)(CN)_(6)]_(3)`.

Answer

A, C, D

Explanation

In `Fe_(3)O_(4)`, each `O` has oxidation number `= -2` and therefore we have:
`4(-2)+3 ("oxidation number" of Fe)=0`
or the oxidation number of `Fr= 8//3`
Actually, `Fe_(3)O_(4)`, known as magnetic oxide of iron, is a stoichiometric mixture of ferrous `(FeO)` and ferric `(Fe_(2)O_(3))` oxides and the value `8//3` for the oxidation number is the average of all the `Fe` atoms in the molecule.
In the ferriferrocyanide molecule, `Fe_(4)[Fe(CN)_(6)]`, we have `4` iron atoms with oxidation number `+3` and `3` iron atoms with oxidation number `+2` and the oxidation number of iron in the molecuels is the average of all these atosm equal to `18//7`.
`7Fe= 4xx(+3)+3xx(+2)`
`7Fe= 18`
`Fe=(18)/(7)`