# Topic: Unit VI: Solid State (Test 1)

Topic: Unit VI: Solid State
Q.1
Graphite and diamond are
A. Covalent and molecular crystals
B. ionic and covalent crystals
C. both covalent crystals
D. both molecular crystals
Explaination / Solution:
No Explaination.

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Q.2
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
A. AB
B. AB3
C. A3B
D. A8B6
Explaination / Solution:

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

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

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Q.3
The ratio of close packed atoms to tetrahedral hole in cubic packing is
A. 1:1
B. 1:2
C. 2:1
D. 1:4
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

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Q.4
Solid CO2 is an example of
A. Covalent solid
B. metallic solid
C. molecular solid
D. ionic solid
Explaination / Solution:

lattice points are occupied by CO2 molecules

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Q.5

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
A. Both assertion and reason are true and reason is the correct explanation of assertion.
B. Both assertion and reason are true but reason is not the correct explanation of assertion.
C. Assertion is true but reason is false.
D. Both assertion and reason are false.
Explaination / Solution:
No Explaination.

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Q.6
In calcium fluoride, having the flurite structure the coordination number of Ca2+ ion and F- Ion are
A. 4 and 2
B. 6 and 6
C. 8 and 4
D. 4 and 8
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

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Q.7
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)
A. 6.023 X 1023
B. 6.023 X 1022
C. 60.23 X 1023
D. ( [6.023 x 1023 ] / [8 x 40] )
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

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Q.8
The number of carbon atoms per unit cell of diamond is
A. 8
B. 6
C. 1
D. 4
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

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Q.9
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.
A. MN
B. M3N
C. MN3
D. M3N2
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

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Q.10
The composition of a sample of wurtzite is Fe0.93 O1.00 what % of Iron present in the form of Fe3+?
A. 16.05%
B. 15.05%
C. 18.05%
D. 17.05%
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%

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