Iron(II) oxide has a cubic structure and each unit cell has a size of 5. If density of this oxide is 4 g cm-3, calculate the number of Fe2+ and O- ions present in each unit cell.
(Atomic mass of Fe = 56, O = 16, NA = 6.0231023 and 1 = 10-8 cm)
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Given : = 4 g cm-3
a = 5A = 510-8 cm M = 72 g/mol, Z = ?
Using the formula for cubic crystals
= Z =
Z = = 4.18 4
There are four formula units of FeO present per unit cell. Hence it has face-centred cubic lattice where each Fe+ and O2- are four in number.