An analysis of iron oxide shows that its formula is \(Fe_{1.9}O_3\). What percentages of iron exists as \(Fe^{3+}\) and \(Fe^{4+}\) ions:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(Fe^{3+}=84.21\%\) and \(Fe^{4+}=15.79\%\)
In a non-stoichiometric oxide the iron is present in two oxidation states, and the total positive charge of the iron ions must balance the negative charge of the oxide ions.
In Fe1.9O3, three \( \ce{O^{2-}} \) ions carry a charge of 3 × (−2) = −6. Let x of the 1.9 iron atoms be \( \ce{Fe^{3+}} \); then (1.9 − x) are \( \ce{Fe^{4+}} \):
3x + 4(1.9 − x) = 6
7.6 − x = 6
x = 1.6 (\( \ce{Fe^{3+}} \)) and 1.9 − 1.6 = 0.3 (\( \ce{Fe^{4+}} \))
% \( \ce{Fe^{3+}} \) = (1.6 / 1.9) × 100 = 84.21%
% \( \ce{Fe^{4+}} \) = (0.3 / 1.9) × 100 = 15.79%
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic is not there in the latest NCERT book (2026-27), and it is also out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Metal Deficiency Defect – “There are many solids which are difficult to prepare in the stoichiometric composition and contain less amount of the metal as compared to the stoichiometric proportion. A typical example of this type is FeO which is mostly found with a composition of Fe0.95O. It may actually range from Fe0.93O to Fe0.96O. In crystals of FeO some Fe2+ cations are missing and the loss of positive charge is made up by the presence of required number of Fe3+ ions.”