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

Hint: 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 \[ - 2\] so, 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 $\dfrac{{F{e^{2 + }}}}{{F{e^{2 + }} + F{e^{3 + }}}} \times 100$.