Indian Forest Service (Mains), Chemistry optional · 2023 · Paper I
Q1(c). Ions per unit cell of FeO
Question
Iron(II) oxide, FeO, crystal has a cubic structure and each edge of the unit cell is 5.0 Å. Taking density of oxide as 4.0 g cm⁻³, calculate the number of Fe²⁺ and O²⁻ ions present in each unit cell.
Answer in brief
The supplied rounded data give Z ≃ 4.19 FeO formula units; the intended integral result is four Fe²⁺ ions and four oxide ions per conventional cell.
Key points
- Convert the cell edge: a = 5.0 × 10⁻⁸ cm, so a³ = 1.25 × 10⁻²² cm³.
- The mass of one cell is ρa³ = (4.0)(1.25 × 10⁻²²) = 5.0 × 10⁻²² g.
- Using M(FeO) = 71.85 g/mol, Z = ρa³NA/M = (5.0 × 10⁻²²)(6.023 × 10²³)/71.85 = 4.19 ≃ 4.
- Each formula unit contains one ion of each type, so the nearest integral cell content is 4Fe²⁺ + 4O²⁻, consistent with the rock-salt structure.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.
Coversunit cell densityFeOrock-salt structurenumber of formula units Zcrystal density calculation
All parts of Question 1
- 1(a) Quantum numbers of oxygen electrons8 marks
- 1(b) Formation enthalpy of lithium fluoride8 marks
- 1(c) Ions per unit cell of FeO8 marks
- 1(d) Derivatives of thermodynamic potentials8 marks
- 1(e) Ionic atmosphere and electrode currents8 marks