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. 6
• B. 12
• C. 15
• D. 18

Answer 1

The correct option is D

18

Number of $F{e}^{3+}$ ions replacing x $F{e}^{2+}$ ions $=\frac{2x}{3}$
Vacancies of cations $=x-\frac{2x}{3}=\frac{x}{3}$
But $\frac{x}{3}=1-0.94=0.06$
Or, $x=0.06\times 3=0.18=18\%$