Question

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

• A. $18$
• B. $15$
• C. $12$
• D. $6$
• E. $24$

Answer 1

Answer: (A)

$18$

Let $100$ O atoms and $94$ Fe atoms are present.

Out of $94$ Fe atoms, $y$ atoms are in $+3$ oxidation state and $94-y$ atoms are in $+2$ oxidation state.

The charges will be balanced when the following equation is satisfied. $-100(-2)=(94-y)(+2)+y(+3)$

or, $200=188+y$
or, $y=12$

Hence, the number of iron atoms in $+2$ oxidation state is $94-y=94-12=82$

Hence, the percentage of $F{e}^{2+}$ atoms replaced $=\frac{100-82}{100}\times 100=18$%

Hence, the correct option is $A$