Iron (II) oxide crystallise in cubic structure with unit cell edge of

1.	Iron (II) oxide crystallise in cubic structure with unit cell edge of 5.0 Å If the density of the oxide is 3.8 g cm-3 . Calculate the no.of Fe2+ and O2- present in each unit cell, [atomic mass of Fe = 56, O = 16]
Answer» Cell edge a = 5.0Å = 5.0 × 10 m = 5 × 10^-8 cm Density = 3.8 g/cm³ Molecular mass of FeO = 56 + 16 = 72 u N_A = 6.022 × 10²³ d = \(\frac{ZM}{N_Aa^3}\) Z = \(\frac{dN_Aa^3}{M}\) = \(\frac{3.8\times6.022\times10^{23}}{72}\times\)(5 × 10^-8)³ = 3.97 = 4 Each cell contain 4 FeO molecule every FeO molecule contain one Fe²⁺ and one O^2- ion. So no. of Fe²⁺ ion =4 no.of O^2- ion = 4.

Iron (II) oxide crystallise in cubic structure with unit cell edge of 5.0 Å If the density of the oxide is 3.8 g cm-3 . Calculate the no.of Fe2+ and O2- present in each unit cell, [atomic mass of Fe = 56, O = 16]

Answer»

Cell edge a = 5.0Å = 5.0 × 10 m = 5 × 10^-8 cm

Density = 3.8 g/cm³

Molecular mass of FeO = 56 + 16 = 72 u

N_A = 6.022 × 10²³

d = \(\frac{ZM}{N_Aa^3}\)

Z = \(\frac{dN_Aa^3}{M}\)

= \(\frac{3.8\times6.022\times10^{23}}{72}\times\)(5 × 10^-8)³

= 3.97 = 4

Each cell contain 4 FeO molecule every FeO molecule contain one Fe²⁺ and one O^2- ion.

So no. of Fe²⁺ ion =4 no.of O^2- ion = 4.