Question

Oxidation number of iron in $F{e}_{0.94}O$ is :

• A. $2\times 0.94$
• B. $\frac{200}{94}$
• C. $\frac{94}{200}$
• D. $0.94$

Answer 1

The correct option is B $\frac{200}{94}$

Let $x$ be the oxidation number of iron in $F{e}_{0.94}O$.
The oxidation number of oxygen is $-2$.
The sum of all the oxidation numbers is $0$ as the compound is neutral.
Hence, $0.94x+1(-2)=0$ or, $x=\frac{200}{94}$
Hence, $\frac{200}{94}$ is the oxidation state of iron.