Chemistry : Solid State: Choose the best answer with Answers, Solution and Explanation

**Choose
the best answer:**

1. Graphite and diamond
are

a) Covalent and
molecular crystals

b) ionic and covalent
crystals

**c) both covalent crystals **

d) both molecular
crystals

2. An ionic compound A_{x}B_{y}
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) AB _{3 }**

c) A_{3}B

d) A_{8}B_{6}

**Solution**

number of A ions = (N_{c}/8) = (8/8)=1

number of B ions = (N_{f}/2) = (6/2) =3

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

**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

4. Solid CO_{2} is an example of

a) Covalent solid

b) metallic solid

**c) molecular solid **

d) ionic solid

**Solution**

lattice points are
occupied by CO_{2} molecules

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 ≠ 90^{0}

**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.

c) Both assertion and reason are false.

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

**Solution**

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

7. The number of unit
cells in 8 gm of an element X ( atomic mass 40) which crystallizes in bcc
pattern is (N_{A} is the Avogadro number)

a) 6.023 X 10^{23}

**b) 6.023 X 10 ^{22} **

c) 60.23 X 10^{23}

d) ( [6.023 x 10^{23}
] / [8 x 40] )

**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 ×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

8. The number of carbon atoms per unit cell of diamond is

**a) 8 **

b) 6

c) 1

d) 4

**Solution**

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

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) M_{3}N

c) MN_{3}

**d) M _{3}N_{2}**

**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 ⇒ M_{3} N_{2}

10. The composition of a sample of wurtzite is Fe_{0.93} O_{1.00}
what % of Iron present in the form of Fe^{3+}?

a) 16.05%

**b) 15.05% **

c) 18.05%

d) 17.05%

**Solution**

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%

11. The ionic radii of A^{+}
and B^{−} are 0.98 × 10^{−}^{10} m and 1.81 × 10^{−}^{10} m . the coordination
number of each ion in AB is

a) 8

b) 2

**c) 6 **

d) 4

**Solution**

r_{C+}/ r_{A-}
= 0.98 ×10^{−10} / 1.81 ×10^{−10} = 0.54

it is in the range of 0.
414 - 0.732 , hence the coordination number of each ion is 6

12. CsCl has bcc
arrangement, its unit cell edge length is 400pm, its inter atomic distance is

a) 400pm

b) 800pm

c) √3 x100pm

**d) (√3 / 2) x 400pm**

**Solution**

√3 a = r_{Cs+} +
2r_{Cl-} + r_{Cs+}

(√3/2) a = ( r_{Cl-}
+ r_{Cs+})

(√3/2) x 400 = inter
ionic distance

13. A solid compound XY
has NaCl structure. if the radius of the cation is 100pm , the radius of the
anion will be

**a) (100/0.414)**

b) (0.732/100)

c) 100x0.414

d) (0.414 / 100)

**Solution**

for an fcc structure r_{x+} / r_{y−} = 0.414

given that r_{X+} = 100 *pm*

r _{y−} = 100 *pm*
/ 0.414

14. The vacant space in bcc lattice unit cell is

a) 48%

b) 23%

**c) 32% **

d) 26%

**Solution**

packing efficiency = 68%

therefore empty space
percentage = (100-68) = 32%

15. The radius of an
atom is 300pm, if it crystallizes in a face centered cubic lattice, the length
oif the edge of the unit cell is

a) 488.5pm

**b) 848.5pm **

c) 884.5pm

d) 484.5pm

**Solution**

let edge length = a

√2a = 4r

a = [4 × 300] / √2

a = 600 ×1.414

a = 848.4 pm

16. The fraction of
total volume occupied by the atoms in a simple cubic is

a) π / 4√2

**b) π / 6**

a) π / 4

a) π / 3√2

**Solution**

17. The yellow colour in NaCl crystal is due to

**a) excitation of
electrons in F centers**

b) reflection of light from Cl- ion on the surface

c) refraction of light from Na+ ion

d) all of the above

18. if ‘a’ stands for
the edge length of the cubic system ; sc , bcc, and fcc. Then the ratio of
radii of spheres in these systems will be respectively.

**Ans:
c) (1/2 a : √3/4 a : 1/2√2 a)**

**Solution**

19. If ‘a’ is the length
of the side of the cube, the distance between the body centered atom and one
corner atom in the cube will be

**Ans: d) (****√3/2)a**

**Solution**

if a is the length of the side, then the length of the leading
diagonal passing through the body centered atom is √3a

Required distance (√3/2) a

20. Potassium has a bcc structure with nearest neighbor distance
4.52 Aº . its atomic weight is 39. its density will be

**a) 915 kg m ^{-3} **

b) 2142 kg m^{-3}

c) 452 kg m^{-3}

d) 390 kg m-3

**Solution**

ρ = n × M / a^{3}N_{A}

for bcc

n = 2

M=39

nearest distance 2r =
4.52

21. Schottky defect in a
crystal is observed when

a) unequal number of
anions and anions are missing from the lattice

**b) equal number of anions and anions are missing from the lattice**

c) an ion leaves its
normal site and occupies an interstitial site

d) no ion is missing
from its lattice.

22. The cation leaves
its normal position in the crystal and moves to some interstitial position, the
defect in the crystal is known as

a) Schottky defect

b) F center

**c) Frenkel defect **

d) non-stoichiometric
defect

23. Assertion: due to
Frenkel defect, density of the crystalline solid decreases.

Reason: in Frenkel
defect cation and anion leaves the crystal.

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**

24. The crystal with a
metal deficiency defect is

a) NaCl

**b) FeO**

c) ZnO

d) KCl

25. A two dimensional
solid pattern formed by two different atoms X and Y is shown below. The black
and white squares represent atoms X and Y respectively. the simplest formula
for the compound based on the unit cell from the pattern is

**a) XY _{8 }**

b) X_{4}Y_{9 }

c) XY_{2 }

d) XY_{4 }

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