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The Solid State

Question
CBSEENCH12005500

Iron (II) oxide has a cubic structure and each unit cell has side 5A. If the density of the oxide is 4 g cm–3, calculate the number of Fe2+ and O2– ions present in each unit cell. (Molar mass of FeO = 72 g mol–1, NA = 6.02 x 1023 mol–1).

Solution

Solution:
We  have given
volume of the unit cell = (5 x 10-8cm)3 = 1.25 x 10-22cm3

Density of FeO = 4g cm-3

Density,

ρ = z×Ma3×NA4= z×72(5×10-8)3×6.02×1023z= 4×1.25×10-22 ×6.02×102372 = 4.18 4

Each unit cell has four units of FeO. So it has four Fe2+ and four O2– ions.