No Explaination.

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

lattice points are occupied by CO

Assertion : monoclinic
sulphur is an example of monoclinic crystal system

No Explaination.

CaF_{2} has
cubical close packed arrangement

Ca^{2+} ions are
in face centered cubic arrangement, each Ca^{2+} ions is surrounded by 8 F^{−} ions and each F^{−}
ion is surrounded by 4 Ca^{2+} ions.

Therefore coordination
number of Ca^{2+} 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 ×10^{23} atoms

0.2 mole contains 0.2 × 6.023 ×10^{23} atoms

( 1unit cell / 2 atoms ) × 0.2 × 6.023 × 10^{23}

6.023 ×10^{22} 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 ⇒ M_{3} N_{2}

let

the number of Fe^{2+} ions
in the crystal be *x*

the number of Fe^{3+} ions
in the crystal be *y*

total number of Fe^{2+} and
Fe^{3+} 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 Fe^{3+}

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