Question

The non stoichiometric compound $F{e}_{0.94}O$ is formed when $x$ $\%$ of $F{e}^{2+}$ ions are replaced by as many $\frac{2}{3}F{e}^{3+}$ ions, $x$ is

• A. 18
• B. 12
• C. 15
• D. 6

Answer 1

Answer: (A)

The number of $F{e}^{3+}$ ions replacing $\times F{e}^{2+}$ ions
$=\frac{2x}{3}$ vacancies of cations
$=x-\frac{2x}{3}=x/3$
But $x/3=1-0.94=0.06,x=0.06\times 3$
$=0.18=18\%$