Question: 86: What fraction of Fe exists as Fe (III) in \mathrm{Fe}_{0.96} \mathrm{O} ?
(Consider \mathrm{Fe}_{0.96} to be made up of Fe (II) and Fe (III) only)
(1) \frac{1}{12}
(2) 0.08
(3) \frac{1}{16}
(4) \frac{1}{20}
Answer: Option (1)
Explanation:
In normal ferrous oxide, the formula is \mathrm{FeO},
where iron exists as \mathrm{Fe^{2+}} and oxygen as \mathrm{O^{2-}}.
In the given compound \mathrm{Fe}_{0.96}\mathrm{O}, there is a deficiency of iron,
which means some \mathrm{Fe^{2+}} ions are oxidized to \mathrm{Fe^{3+}} to maintain electrical neutrality.
Let the fraction of iron present as \mathrm{Fe^{3+}} be x.
Then fraction of iron present as \mathrm{Fe^{2+}} is 0.96 - x.
Total positive charge contributed by iron ions is:
3x + 2(0.96 - x) = 3x + 1.92 - 2x = x + 1.92Total negative charge contributed by oxygen is:
2 \times 1 = 2For electrical neutrality:
x + 1.92 = 2 x = 0.08This means 0.08 moles of iron per mole of compound are present as
\mathrm{Fe^{3+}}.
The fraction of iron existing as \mathrm{Fe^{3+}} is:
\frac{0.08}{0.96} = \frac{1}{12}Hence, the correct answer is Option (1).