# Unit VI: Solid State - Online Test

Q1. Graphite and diamond are
Answer : Option C
Explaination / Solution:
No Explaination.

Q2. An ionic compound AxBy crystallizes in fcc type crystal structure with B ions at the centre of each face and A ion occupying entre of the cube. the correct formula of AxBy is
Answer : Option B
Explaination / Solution:

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

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

Q3. The ratio of close packed atoms to tetrahedral hole in cubic packing is
Answer : Option B
Explaination / Solution:

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

Q4. Solid CO2 is an example of
Answer : Option C
Explaination / Solution:

lattice points are occupied by CO2 molecules

Q5.

Assertion : monoclinic sulphur is an example of monoclinic crystal system

Reason: for a monoclinic system, a≠b≠c and α = γ = 90 0 , β ≠ 900
Answer : Option A
Explaination / Solution:
No Explaination.

Q6. In calcium fluoride, having the flurite structure the coordination number of Ca2+ ion and F- Ion are
Answer : Option C
Explaination / Solution:

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

Q7. The number of unit cells in 8 gm of an element X ( atomic mass 40) which crystallizes in bcc pattern is (NA is the Avogadro number)
Answer : Option B
Explaination / Solution:

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

Q8. The number of carbon atoms per unit cell of diamond is
Answer : Option A
Explaination / Solution:

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

Q9. In a solid atom M occupies ccp lattice and (1/3) of tetrahedral voids are occupied by atom N. find the formula of solid formed by M and N.
Answer : Option D
Explaination / Solution:

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

Q10. The composition of a sample of wurtzite is Fe0.93 O1.00 what % of Iron present in the form of Fe3+?
Answer : Option B
Explaination / Solution:

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%