number of A ions = (Nc/8) = (8/8)=1
number of B ions = (Nf/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
Assertion : monoclinic
sulphur is an example of monoclinic crystal system
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.
(Nc/8) + (Nf/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%