Ferrous oxide has a cubic structure and each edge of the unit cell is 5.0 Å. Assuming density of the oxide as 4.0 g cm–3, then the number of Fe²⁺ and O₂– ions present in each unit cell will be

Edge length = 8 5 A 5 10 x cm Density = 3 4g/cm Density × Volume = Mass 3 8 4 5 10 x x Mass 24 500 10 g x Mass 22 5 10 g x Mass of 1 unit cell Mass of 1 molecule of FeO 23 56 16 72 6.023 10 A g N x No. of molecule in one unit cell Massof unit cell = Massof 1molecule 22 23 5 10 6.023 10 4 72 x >> x x So the no. of +2 Fe ions = 4 = no. of 2 O ions.