Question

Formula of iron oxide with metal deficiency defect in its crystal is $F{e}_{0.94}O$. The crystal contains $F{e}^{2+}\text{and}F{e}^{3+}$. The mole percentage of $Fe$ existing as $F{e}^{3+}$ ion in the crystal is:
[0.77 Mark]

• A. $18.2\%$
• B. $87.2\%$
• C. $12.8\%$
• D. $81.8\%$

Answer 1

The correct option is C $12.8\%$

Let, the number of $F{e}^{2+}$ ions be $x$
So, the number of $F{e}^{3+}$ ions be $0.94-x$
So, $2x+(0.94-x)\times 3=2(\because$ the net charge of the molecule is zero)
$\therefore 2x+2.82-3x=2$
$\Rightarrow x=0.82$
$\therefore$ Number of $F{e}^{2+}$ ions $=0.82$ and Number of $F{e}^{3+}$ ions $\Rightarrow 0.94-0.82=0.12$
$\text{Hence mole percentage of}F{e}^{3+}$ ion = $\frac{0.12\times N_A}{0.94\times N_A}\times 100=12.76\%12.8\%$