Unit VI: Solid State - Online Test

number of A ions = (N_c/8) = (8/8)=1

number of B ions = (N_f/2) = (6/2) =3

if number of close packed atoms =N; then,

The number of Tetrahedral holes formed = 2N

number of Octahedral holes formed = N

therefore N:2N = 1:2

CaF₂ has cubical close packed arrangement

Ca²⁺ ions are in face centered cubic arrangement, each Ca²⁺ ions is  surrounded by 8 F⁻ ions and each F⁻ ion is surrounded by 4 Ca²⁺ ions.

Therefore coordination number of Ca²⁺ is 8 and of F⁻  is 4

in bcc unit cell,

2 atoms ≡ 1 unit cell

Number of atoms in 8g of element is ,

Number of moles = 8g / 40 g mol⁻¹ = 0.2 mol

1 mole contains 6.023 ×10²³ atoms

0.2 mole contains 0.2 × 6.023 ×10²³ atoms

( 1unit cell / 2 atoms ) × 0.2 × 6.023 × 10²³

6.023 ×10²² unit cells

in diamond carbon forming fcc. Carbon occupies corners and face centres and also occupying half of the tetrahedral voids.

(N_c/8) + (N_f/2) + 4 C atomos in Td voids

(8/8) + (6/2) + 4 = 8

if the total number of M atoms is n, then the number of tetrahedral voids

 =2n

given that (1/3)^rd of tetrahedral voids are occupied i.e., (1/3)x2n are occupied i.e., by N atoms

∴ M:N ⇒ n : (2/3) n

1 : (2/3)

3 : 2 ⇒ M₃ N₂

let

the number of Fe²⁺ ions in the crystal be x

the number of Fe³⁺ ions in the crystal be y

total number of Fe²⁺ and Fe³⁺ ions is

x + y

given that x + y = 0.93

the total charge =0

 x (2+) + (0.93 –x) (+ 3) −2 = 0

2x + 2.97 −3x −2 = 0

x = 0.79

Percentage of Fe³⁺

= [ ( 0.93 −0.79) / (0.93) ] 100 = 15.05%