NEET Chemistry The Solid State Questions Solved

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Iron(II) oxide has a cubic structure and each unit cell has a size of 5A. 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 = 6.023×1023 and  1A = 10-8 cm) 

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Concept Questions :-

Density/Formula/packing fraction/ Semiconductors

Given : ρ = 4 g cm-3

a = 5A = 5×10-8 cm M = 72 g/mol, Z = ?

Using the formula for cubic crystals

ρ = Z×Ma3×N0  Z = ρ×a3×N0M

 Z = 4g cm-3×(5×10-8cm)3×6.022×102372

 Z = 3011×10-172 = 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.

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