Question

In mineral wurtzite, iron oxide is usually non-stoichiometric with approximate formula $F{e_{0.91}}O$. What percentage of cationic sites are occupied by $Fe(III)$?
$(A)32.45\%$
$(B)19.78\%$
$(C)18.71\%$
$(D)12.67\%$

Answer 1

Iron oxide is non stoichiometric because it reflect the ease of oxidation of $F{e^{2 + }}$ to $F{e^{3 + }}$ effectively replacing a small portion of $F{e^{2 + }}$ with two thirds their number of $F{e^{3 + }}$ . thus, for every three missing $F{e^{2 + }}$ ions, the crystal contains two $F{e^{3 + }}$ ions to balance the charge.

Complete answer:
In the formula $F{e_{0.91}}O$, the total concentration of $Fe(II)$ and $Fe(III)$ Is $0.91$.
let $Fe(II)$ in $F{e_{0.91}}O$=$x$
And, $Fe(III)$ in F{e_{0.91}}O$$$ = (0.91 - x)$ The positive charge comes from ferrous and ferric ions which should be balanced by the negative charge from the oxygen ion that is - 2so, for electrical neutrality. $2x + 3(0.91 - x) = 2$ $2x + 2.73 - 3x = 2$ $x = 0.73$ As we know,Fe(III) = (0.91 - x)$ Putting value of $$x$$ we get$$Fe(III) = (0.91 - 0.73) = 0.18 Therefore, percentage of $$Fe(III)$$ is = \dfrac{{F{e^{3 + }}}}{{F{e^{2 + }} + F{e^{3 + }}}} \times 100$$ = \dfrac{{0.18}}{{0.91}} \times 100$$ = 18.35% \approx 18.71$
Therefore, correct answer is option C.

Note:
In this question we were need to calculate the percentage of $Fe(III)$ ions that’s why er are using the formula $\dfrac{{F{e^{3 + }}}}{{F{e^{2 + }} + F{e^{3 + }}}} \times 100$, but if we need to calculate the percentage of $Fe(II)$ ions in the compound than we will be using $\dfrac{{F{e^{2 + }}}}{{F{e^{2 + }} + F{e^{3 + }}}} \times 100$.