Test Date: 28.04.2016 (Thursday)
Test Time: 10:00 AM to 11:30 AM
Test Venue:
Lajpat Bhawan, Madhya Marg,
Sector 15-B, Chandigarh
Dr. Sangeeta Khanna Ph.D
1
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Dr. Sangeeta Khanna Ph.D
Test Date: Level – 1
READ INSTRUCTIONS CAREFULLY
1.
2.
3.
4.
5.
The test is of 3 hour duration.
The maximum marks are 400
This test consists of 98 questions.
For each question you will be awarded 4 marks if you have darkened only the bubble
corresponding to the correct answer and zero mark if no bubbles are darkened. Minus one
(-1) mark will be awarded for wrong answer
Keep your mobiles switched off during Test in the Halls.
SECTION – A (Single Answer)
This Section contains 50 multiple choice questions. Each question has four choices A), B), C) and
D) out of which ONLY ONE is correct.
50 × 4 = 200 Marks
1.
Dipole-dipole interaction energy between stationary polar molecules (as in solids) is proportional to
....A... and between the rotating polar molecules is proportional to .....B.... . Here, A and B refer to
a. A 
c. A 
1
6
r
1
r3
,B 
,B 
1
b. A 
3
r
1
d. A 
r4
1
3
r
1
r4
,B 
,B 
1
r6
1
r3
B
Sol. Dipole-dipole interaction energy between stationery polar molecules (as in solids) is proportional to
1
1
and that between rotating polar molecules is proportional to 6 , where r is the distance between
3
r
r
the polar molecules. Besides dipole-dipole interaction, polar molecules can interact by London forces
also.
2. An ionic solid A+B– crystallizes as a fcc structure of NaCl. If the edge length of cell is 508 pm and
radius of anion is 144 pm, the radius of cation is:
a. 110 pm
b. 364 pm
c. 220 pm
d. 288 pm
A
a
Sol. r+ + r– = for fcc
2
3. A molecule AB2 (mol. Wt. = 166.4) occupies orthorhombic lattice with a = 5 Å, b = 8 Å and c = 4 Å. If
density of AB2 is 5.2 g cm–3, the number of molecules present in one unit cell is:
a. 2
b. 3
c. 4
d. 5
B
Sol. Volume of unit cell = a × b × c = 5 × 10–8 × 8 × 10-8 = 1.6 × 10-22 cm3
Mass of unit cell = 1.6 × 10-22 × 5.2 = 8.32 × 10-22 g
8.32  10 22  6  10 23
Number of molecules in one unit cell =
=3
166.4
4. Na and Mg crystallize in bcc and fcc type crystals respectively, then the number of atoms of Na and
Mg present in the unit cell of their respectively crystal is:
a. 4 and 2
b. 9 and 14
D
Sol. No. of unit cell in bcc is 2 and fcc is 4.
Dr. Sangeeta Khanna Ph.D
c. 14 and 9
2
d. 2 and 4
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5.
The volume of atoms present in a face-centred cubic unit cell of a metal (r is atomic radius) is:
a.
20 3
r
3
b.
24 3
r
3
c.
12 3
r
3
d.
16 3
r
3
D
4 3
(n = 4 for fcc)
r  n
3
4
16 3
= r 3  4 
r
3
3
In spinel structure, oxide ions are cubic closest packed, whereas 1/8th of tetrahedral holes are
occupied by cations A2+ and ½ of octahedral holes are occupied by cations B3+ ions. The general
formula of the compound having spinel structure is:
Sol. Volume of atom in a cell =
6.
7.
8.
9.
a. AB2O4
b. A2B2O4
c. A2B6O
d. A4B3O
A
The crystal system of a compound with unit cell dimensions a = 0.387, b = 0.387 and c = 0.504 nm
and  =β= 900 and γ = 1200 is:
a. Hexagonal
b. Cubic
c. Rhombohedral
d. Orthorhombic
A
In a compound PQ2O4, oxide ions are arranged in CCP and cations P are present in octahedral voids.
Cations Q are equally distributed among octahedral and tetrahedral voids. The fraction of the
octahedral voids occupied is:
a. 1/4
b. 1/2
c. 1/3
d. 2/3
B
A solid has 3 types of atoms namely X, Y and Z; X form a fcc lattice with Y atoms occupying all the
tetrahedral voids and Z atoms occupying half of the octahedral voids. The formula of the solid is:
a. XYZ
b. X4Y4Z
C
Sol. It X1Y2 Z 1 So simplest formula X2Y4Z
c. X2Y4Z
d. X4YZ2
2
10. Iridium crystallizes in a face-centred cubic unit cell that has an edge length at 3.832 Å. The atom in
the center of the face is in contact with the corner atoms, as shown in figure. Find the atomic radius of
an iridium atom:
a. 2.71 Å
D
b. 0.675 Å
c. 1.916 Å
d. 1.355 Å
Sol. 4r = 2a
11. In a CCP lattice of A and B, A atoms are present at the corners while B atoms are at face centres.
Then the formula of the compound would be if one of the A atoms from a corner is replaced by C
atoms (also monovalent)?
a. A7B24C8
b. A7B24C
c. A24BC
d. AB24C
B
12. A solid compound contains X, Y and Z atoms in a cubic lattice with X atoms occupying the corners, Y
atoms in the body centered positions and Z atoms at the centers of faces of the unit cell. What is the
empirical formula of the compound?
a. XYZ
B
b. XYZ3
Dr. Sangeeta Khanna Ph.D
c. XY2Z3
3
d. X2Y2Z3
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13. First two nearest neighbour distance for Body centered cubic unit cell will be: (edge length of unit cell
= )
a.
3
, 
2
b.
3 ,
2
c. ,
2
d. ,
2
, 2
2
A
14. Which of the following expression is correct for packing fraction of NaCl if the ions along with one
face are diagonally removed?
13 3 16 3
r 
r
3
a. 3
8(r  r )3
A
13 3 4 3
r  r
3
b. 3
8(r  r )3
16 3 13 3
r 
r
3
c. 3
8(r  r )3
4 3 13 3
r 
r
3
d. 3
8(r  r )3
1 1
13
Sol. Effective number of remaining Cl– = 4  2    
 8 2 4
Effective number of Na+ = 4
Occupied volume
Packing fraction =
Total volume
13 4 3
4
 r  4  r3
3
= 4 3
8(r  r )3
15. If the radii of A+ and B– in the crystalline solid AB are 96 pm and 200 pm respectively. Then expected
structure of void will be:
a. trigonal
b. octahedral
c. hexagonal
B
r
96
Sol.  
 0.48
r 200
16. Which of the following relation is correct for the unit cell of NaCl?
a. rNa  rCl 
a
2
b. rNa  rCl  4a
c. rNa  rCl 
d. cubic
a
2
d. rNa  rCl 
3
a
4
A
Sol. 2(rNa  rCl )  a
17. In which of the following systems interfacial angles  =  = 90° but   90°?
a. Monoclinic
b. Rhombohedral
c. Triclinic
d. Orthorhombic
A
18. Radii of Cs+ and Cl– ions are 1.69 Å and 1.81 Å; what is the edge length of CsCl unit cell?
a. 4.50 Å
b. 4.04 Å
B
a 3
Sol.
 rCs  rCl
2
= 1.69 + 1.81
a = 4.04 Å
19. What is true about solids?
c. 3.50 Å
d. 3.80 Å
a. interparticle distance is maximum
b. interparticle attraction is large
c. K.E. of particles is high
d. all solids have sharp m.pt. and b.pt.
B
Sol. Only statement B is correct. The other statements are against the facts
Dr. Sangeeta Khanna Ph.D
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20. Which statement among the following applies to graphite?
I.
III.
It is a covalent network solid
It can be used as lubricant
II. It has a layered structure
IV. C atoms are sp3 hybridized
a. II, III
b. I, II, IV
c. I, III, IV
d. I, II, III
D
Sol. Graphite has a layered structure and is also used as lubricant
21. The coordination number of h.c.p. or c.c.p. structure is 12, that is, one atom touches 12 other atoms.
The 12 atoms lie as
a. 6 atoms are on the same plane, 3 above and 3 below the plane
b. 8 atoms are on the same plane, 2 above and 2 below the plane
c. 4 atoms are on the same plane, 4 above and 4 below the plane
d. 2 atoms are on the same plane, 5 above and 5 below the plane
A
22. In a simple cubic crystal, the corner atom touches
a. 6 other atoms
b. 4 other atoms
c. 3 other atoms
d. 8 other atoms
A
23. An ionic compounds has a unit cell constituting of A ions at the corners of a cube and B ions on the
centres of the faces of the cube. The empirical formula for this compound will be:
a. AB
b. A2B
D
Sol. Number of A ions per unit cell = 1
Number of B ions per unit cell = 3
 Empirical formula of compound = A1B3
c. A3B
d. AB3
or AB3
24. In a crystal, with simplest formula AB2C2, lattice is made up of ‘C’ atoms ‘A’ atoms are occupied in
tetrahedral void & B atoms in octahedral void. What percentage of octahedral voids are unoccupied.
a. 100%
b. 25%
c. zero
d. 50%
C
Sol. Total octahedral void = 2; all are occupied by B.
25. In a crystal, lattice is made up of ‘A’ atoms & ‘B’ atoms are occupied in half of the tetrahedral voids &
‘C’ atoms gets occupied in all octahedral voids. What is the simplest formula of crystal.
a. AB2C
b. ABC
c. ABC2
B
26. What will be the number of neighbours in FCC unit cell.
d. A2BC
a. 8
b. 12
c. 6
d. 4
B
27. In a cubic type unit cell, atoms of X are at the corners as well as at the centre of a cube. Atoms of Y
are at one half faces of the cube. Drive the formula of the compound.
a. X4Y3
A
Sol.
b. X2Y3
No. of atoms of X = 8 corners 
c. X2Y5
d. X3Y4
1
atom per unit cell = 1
8
No. of atoms of X at centre = 1.
Hence, total no. of atoms of X = 1+1=2
1
1
3
No. of atoms of Y = (6 faces)
atom per unit cell =
2
2
2
Y
Formula of compound = X2 3 or X4 Y3
2
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28. If three elements P, Q and R crystallize in a cubic lattice with P atoms at the corners, Q atoms at the
cube centre and R atoms at the centre of edges of the cube, then write the formula of the compound.
a. P2Q2R3
D
b. PQR
c. PQR2
Sol. No. of P atoms per unit cell = 8 corners 
d. PQR3
1
atom per unit cell = 1
8
No. of Q atoms per unit cell = 1
No. of R atoms per unit cell = 12 ×
1
=3
4
Formula of compound = PQR3
29. In face centred cubic arrangement of A and B atoms, whose A atoms are at the corners of the unit
cell and B atoms at the face centres and A atoms are missing from two corners in each unit cell. What
is the formula of the compound.
a. AB4
b. AB3
A
Sol. No. of A atoms per unit cell
c. A2B
d. AB
1
6 3
atom per unit cell = 
8
8 4
1
No. of B atoms per unit cell = 6 faces  atom per unit cell = 3
2
Hence, the formula of the compound = A3/4B3 or A3B12 i.e., AB4 Ans.
30. Which of the following pairs contain molecular solids but different nature of intermolecular forces?
= 6(8 – 2 = 6) corners 
a. Ice, NaCl
b. Ice, dry ice
c. SiO2, Diamond
d. AgCl, NaCl.
B
Sol. Ice (H2O) and dry ice (Solid CO2) are both molecular solids. Intermolecular forces in H2O are H-bonds
but in dry ice, these are van der Waal forces.
31. Fe0.95O can be due to presence of iron in +2 and +3 oxidation number. The iron present in +3
oxidation state will be
a. 15%
b. 13.5%
c. 10.5%
d. 8.85%
C
32. If AgCl is doped with 10-4 mol% of CdCl2, then the concentration of cation vacancies per mol will be
a. 6.02 × 1016
b. 6.02 × 1017
c. 6.02 × 1018
d. 6.02 × 1014
B
33. Substance whose physical properties are independent of direction is called
a. isotropic
b. anisotropic
c. isotopic
d. allotrope
A
34. A given crystal has lattice parameters a = 4.94 Å, b = 7.94 Å and c = 5.72 Å. What is the structure of
the unit cell?
a. Cubic
b. Orthorhombic
c. Hexagonal
d. Rhombohedral
B
35. If the three interaxial angles defining the unit cell of a crystal are all equal in magnitude, the crystal
cannot belong to the
a. rhombohedral system
c. hexagonal system
C
Dr. Sangeeta Khanna Ph.D
b. cubic system
d. tetragonal system
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36. The number of molecules per unit cell for a compound that crystallizes in the form of an orthorhombic
end-centred lattice with a molecule at each lattice site is
a. 1
b. 2
c. 4
B
Sol. Contribution of two end faces = 1
corner = 1
37. In which of the following pairs does each member have two Bravais lattices?
d. 8
a. Cubic and tetragonal
b. Tetragonal and monoclinic
c. Triclinic and tetragonal
d. Trigonal and monoclinic
B
38. In A BCC unit cell made up of A+ & B– ions A+ ions are present at corners & B– ion at centre. What will
be the distance between cation and anion if edge lengths of BCC unit cell is ‘a’.
a.
3a
C
Sol. 2r+ + 2r – =
b.
2a
c.
3a
2
d. a
3 a
2d = 3 a
39. A crystal is made up of atoms of an element ‘A’ with FCC unit cell. Edge length is 200 pm. What will
be the distance between neighbour.
a. 173 pm
D
Sol. 4r = 2 a;
b. 346 pm
2d =
c. 200 pm
d. 141 pm
2 a;
2
 200 = 141 pm
2
40. Which of the following is not a covalent solid?
d
a. silicon
b. graphite
c. SiC
D
41. Choose the correct matching sequence from the possibilities given
(a)
(b)
(c)
(d)
(e)
a.
c.
B
42. I.
II.
Column – I
Crystal defect
hcp
CsCl
Diamond
NaCl
(A)
(r)
(r)
(B)
(p)
(t)
(p)
(q)
(r)
(s)
(t)
(C)
(q)
(p)
d. Rhombic sulphur
Column – I
AB AB AB type crystal
Covalent crystal
Frenkel
Face centered in cube
Body centered in cube
(D)
(t)
(q)
(E)
(s)
(s)
b.
d.
(A)
(r)
(t)
(B)
(p)
(p)
(C)
(t)
(s)
(D)
(q)
(q)
(E)
(s)
(p)
Crystalline solids have definite heat of fusion whereas amorphous solids lack it.
Crystalline solids have definite geometrical shape whereas amorphous solids have no definite
shape.
III. Crystalline solids are anisotropic whereas amorphous solids are isotropic.
IV. Crystalline solid do not have sharp melting point whereas amorphous solids have sharp melting
point.
The correct statement(s) is/are
a. I, II, IV
B
b. I, II, III
Dr. Sangeeta Khanna Ph.D
c. II, III, IV
7
d. I, II, III, IV
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43. Match the following Columns
Column – I
(a) Non-polar molecular
(b) Metallic
(c) Network
(d) Ionic
(1)
(2)
(3)
(4)
Column – I
CCl4
ZnS
SiO2
Mg
a. A  (1); B  (2); C  (3); D  (4)
b. A  (1); B  (4); C  (3); D  (2)
c. A  (3); B  (4); C  (1); D  (2)
d. A  (3); B  (2); C  (1); D  (4)
B
44. A metal crystallises in a bcc lattice. Its unit cell edge length is about 300 pm and its molar mass about
50 g mol–1. What would be the density of the metal (in g cm–3)?
a. 3.1
b. 6.2
c. 9.3
d. 12.4
B
Sol. Given, Molar mass, M = 50 g/mol
NA = 6.02 × 1023
Z = 2 (for bcc crystal)
Edge length a = 300 pm = 3 × 10-8 cm
ZM
2  50

 6.15  6.2
d
23
3
6.02  10  (3  10  8 )3
NA  a
45. Give the correct order of initial T (true) or F (false) for following statements:
I. In an anti-flourite structure, anions form FCC and cations occupy all the tetrahedral voids.
II. If the radius of cation and anion is 20 and 95 pm, then the coordination number of cation in the
crystal is 4.
III. An atom or ion is transferred from a lattice site to an interstitial position in Frenkel defect.
IV. Density of crystal always increases due to substitutional impurity defect.
a. TTTT
b. FFFF
c. FFTT
d. TFTF
D
46. The theoretical density of ZnS is d g/cm3. If the crystal has 4% Frenkel defect, then the actual density
of ZnS should be
a. d g/cm3
b. 0.04d g/cm3
c. 0.96d g/cm3
d. 1.04d g/cm3
A
Sol. No change in density in Frenkel defect.
47. A ferromagnetic substance becomes a permanent magnet when it is placed in a magnetic field
because
a. all the domains get oriented in the direction of magnetic field
b. all the domains get oriented in the direction opposite to the direction of magnetic field
c. domains get oriented randomly
d. domains are not affected by magnetic field
A
48. Which of following oxide shows electrical properties like metals?
a. SiO2
b. MgO
c. SO2(s)
d. CrO2
D
49. A mineral having the formula AB2 crystallizes in the CCP lattices with A atoms at lattice points. What
is the coordination number of the B atoms?
a. 4
b. 6
c. 8
d. 12
A
50. A metallic element exists as cubic lattice. Each edge of the unit cell is 4.0 Å. The density of the metal
is 6.25 g/cm3. How many unit cells will be present in 100 g of the metal?
a. 1.0 ×1022
B
Sol. No. of unit cell 
b. 2.5 × 1023
c. 5.0 × 1023
d. 2.0 × 1023
Mass
Mass of one unit cell
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LEVEL – II
SECTION – B (ASSERTION & REASON) Negative Marking [-1]
This Section contains 16 multiple choice questions. Each question has four choices A), B), C) and
D) out of which ONLY ONE is correct.
(16 × 4 = 64 Marks)
a. Both A and R are correct; R is the correct explanation of A
b. Both A and R are correct; R is not correct explanation of A
c. A is correct; R is incorrect
d. R is correct; A is incorrect
1.
Assertion: London forces are applicable for short distance (~ 500 pm).
Reason: The magnitude of London forces depends on the polarisiability of the particle.
a. (a)
b. (b)
c. (c)
d. (d)
B
Sol. London forces are applicable for short distance (~500 pm) and its magnitude depends on the
polarisability of the particle.
Forces of attraction acts between two temporary dipoles is known as London force. Another name for
this force is dispersion force. This force of attraction was first proposed by the German physical Fritz
London. These forces are always attractive and interaction energy is inversely proportional to the
1
sixth power of the distance between two interacting particles (i.e.,
where, r is the distance
r6
between two particles). These forces are important only at short distance (~500 pm) and their
magnitude depends on the polarisability of the particle.
2. Assertion: Three states of matter are the result of balance between intermolecular forces and
thermal energy of the molecules.
Reason: Intermolecular forces tend to keep the molecules together but thermal energy of molecules
tends to keep them apart.
a. (a)
b. (b)
c. (c)
d. (d)
A
Sol. Intermolecular forces tend to keep the molecules together but thermal energy tends to keep them a
part. Three states of matter are the result of balance between intermolecular forces and the thermal
energy of the molecules.
Gases do not liquefy on compression only, although molecules come very close to each other and
intermolecular forces operate to the maximum. However, when the thermal energy of molecules is
reduced by lowering of temperature, the gases can be very easily liquefied.
Gas  liquid  Solid
Predominance of intermolecular interaction
Gas  liquid  Solid
Predominance of thermal energy
3.
Assertion: In ccp arrangement, a tetrahedral void is surrounded by four spheres whereas an
octahedral void is surrounded by six spheres.
Reason: Size of tetrahedral void is Bigger than that of octahedral void.
a. (a)
C
b. (b)
Dr. Sangeeta Khanna Ph.D
c. (c)
9
d. (d)
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4.
Assertion: Na2O has antifluorite structure. O2– ions occupy ccp arrangement while Na+ ions occupy
100% of the tetrahedral voids.
Reason: If ‘a’ is the edge length of the unit cell then the nearest neighbour distance in the antifluorite
3a
structure is
.
4
a. (a)
B
b. (b)
c. (c)
d. (d)
1
of Body diagonal
4
Assertion: When potassium chloride is heated in the vapour of potassium then its colour becomes
light blue-violet.
Reason: Anionic vacancy having trapped electrons is called F-centre. F-centre is responsible for
colour of ionic solid.
Sol. Distance of corner & centre of tetrahedral voiod is
5.
6.
7.
a. (a)
b. (b)
c. (c)
d. (d)
A
Assertion: For atomic crystalline solids, the packing efficiency lies in the sequence:
Face centred cubic < body centre cubic < simple cubic unit cell.
4
Z  r 3
3
Reason: Packing efficiency =
 100
a3
Z = Number of atoms per unit cell
r = Radius of atom
a = Edge length of unit cell
a. (a)
b. (b)
D
Assertion: Old glass panes turns milky
Reason: It is due to environmental pollution
c. (c)
d. (d)
8.
a. (a)
b. (b)
c. (c)
d. (d)
C
Assertion: Ionic solids are brittle, even though they have strong columbic force of attraction
Reason: On applying force same charge comes very close & repulsion between them is responsible
for their brittle ness.
9.
a. (a)
b. (b)
c. (c)
d. (d)
A
Assertion: Transition metals normally have high melting point.
Reason: They have additional covalent bonding due to overlapping of unpaired electron in ‘d’ orbitals
and strong columbic force between metal ion & electrons.
a. (a)
b. (b)
c. (c)
A
10. Assertion: Calcite and aragonite are Isomorphs.
Reason: These two forms have different crystal arrangements.
d. (d)
a. (a)
b. (b)
c. (c)
d. (d)
D
Sol. There are polymorphs
11. Assertion: Band gap in germanium is small.
Reason: The energy spread of each germanium at atomic energy level is infinitesimally small.
a. (a)
C
b. (b)
Dr. Sangeeta Khanna Ph.D
c. (c)
d. (d)
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Sol. A semiconductor has only small difference in energy between the filled valence gap between valence
and conduction band for germanium is 68 kJ mol–1 at room temperature. In order to increase its
conduction doping is done.
12. Assertion: Glass is an amorphous material, whereas quartz is a crystalline substance.
Reason: Glass has only short – range order, whereas quartz has long – range order.
a. (a)
b. (b)
c. (c)
d. (d)
B,A
13. Assertion: The stability of a crystal gets reflected in its melting point.
Reason: The stability of a crystal depends upon the strength of the inter particle force. The melting
point of a solid depends on the strength of the attractive force acting between the constituent
particles.
a. (a)
b. (b)
c. (c)
A
14. Assertion: HCl, SO2 are the examples of polar molecular solids.
Reason: These are good conductors of electricity.
d. (d)
a. (a)
b. (b)
c. (c)
d. (d)
C
Sol. HCl, SO2 are polar molecular solids. These are non-conductors of electricity
15. Assertion: CrO2 shows metallic as well as ferromagnetic property
Reason: It is conductor like metal and have permanent magnetism
a. (a)
b. (b)
c. (c)
d. (d)
A
16. Assertion: Increasing temperature increases the density of point defects.
Reason: The process of formation of point defects in solids is endothermic and has S > 0
a. (a)
A
b. (b)
c. (c)
d. (d)
SECTION – C (Paragraph Type) Negative Marking [-1]
This Section contains 3 paragraphs. Each of these questions has four choices A), B), C) and D) out
of which ONLY ONE is correct.
10 × 4 = 40 Marks
Comprehension – 1
Answer the following on the basis of given diagram:
A
B
D
C
1.
The shaded plane represents diagonal plane of symmetry. How many such planes are possible in the
cubic system?
a. 1
b. 2
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c. 4
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2.
D
The line joining the points A and C is called:
3.
a. face diagonal and three fold axis of symmetry
b. body diagonal and three fold axis of symmetry
c. four fold axis of symmetry
d. only two fold axis of symmetry
B
How many body diagonal like AC are possible in the cubic system?
a. 2
B
b. 4
c. 6
d. 8
Paragraph – 2
The mineral perovskite has a structure that is the prototype of many ABX 3 solids; particularly oxides.
The perovskite structure is cubic with A atoms surrounded by 12X atoms and the B atoms surrounded
by 6X atoms. The sum of charges on the cations A and B must be +6; it can be achieved in several
ways (A2+ B4+ and A3+ B3+ among them).
The perovskite structure is closely related to the materials that show interesting electrical properties,
such as piezoelectricity, ferroelectricity, ferroelectricity and high temperature superconductivity.
Perovskite is the name of a mineral containing calcium, oxygen and titanium having following unit cell
structure:
Titanium
Oxygen
Calcium
4.
Molecular formula of perovskite is:
d. CaTiO4
5.
a. CaTiO3
b. CaTiO
c. CaTiO2
A
Total number of atoms in the unit cell of perovskite is:
a. 2
D
d. 5
Sol. Number of Ca2+ ions = 8 ×
b. 3
c. 4
1
1
8
Number of Ti4+ = 1
1
=3
2
Total Number of atoms = 5
Oxidation state and co-ordination number of titanium in CaTiO3 (Perovskite) is:
Number of O2– ions = 6 ×
6.
a. + 2, 6
C
Sol. CaTiO3
+2 + x – 6 = 0
x=+4
b. + 3, 8
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c. + 4, 6
12
d. + 4, 8
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7.
A non stoichiometric compound Cu1.8S is formed due to incorporation of Cu2+ ions in the lattice of
cuprous sulphide. What percentage of Cu2+ ion in the total copper content is present in the
compound?
a. 88.88
b. 11.11
c. 99.8
d. 89.8
B
Sol. Let x Cu2+ and (1.8 – x)Cu+ ions are present in the compound Cu1.8S. Compound is electrically
neutral.

+2x + (1.8 – x) = 2
2x + 1.8 – x = 2
x = 0.2
0 .2
% Cu2+ =
× 100 = 11.11
1 .8
Paragraph-3
In a primitive cubic unit cell, all the eight corners of the cube are occupied by the same atoms/ions
and not found anywhere else in the cube. The number of atoms within a unit cell is called the rank of
a unit cell. For primitive cubic unit cell, the rank (z) is 1. In a bcc, the same atoms/ion is also present
at the center of the cube. These atoms/ions are not present any – where else in the cube. The rank of
bcc is 2. In an fcc, the same atoms/ions are present at all the corners of the cube and are also
present at the center of each square face. These atoms/ions also present at the center of each
square face. These atoms/ions are not present anywhere else in the unit cell. The rank of an fcc is 4.
8.
How many unit cells are present in 39 g of potassium that crystallizes in bcc structure (atomic mass of
K = 39 u)?
a. 0.5 NA
b. 0.25 NA
c. NA
d. 0.75 NA
Where NA means Avogadro’s constant
A
mass
Sol. No. of unit cell =
mass of unit cell
39  Na

2  39
9. Li crystallizes in bcc. The edge length of unit cell is 351 pm. What would be radius of Li atom?
a. 151.98 pm
b. 273 pm
c. 290 pm
d. 76 pm
A
Sol. 4r = 3 a
10. Sodium crystallizes in bcc lattice. If the length of the edge of the unit cell is 424 pm, density of sodium
is (atomic mass of sodium = 23 u)
a. 10.4 g cm–3
B
Sol. d 
b. 1.002 g cm–3
2  23
( 424 )3  10  30  6.2  10 23
Dr. Sangeeta Khanna Ph.D

c. 50.4 g cm–3
d. 20.4 gm cm-3
1.005  10 7
 1.002 gm/cm 3
10
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SECTION – D (More than One Answer) No Negative Marking
This Section contains 10 multiple choice questions. Each question has four choices A), B), C) and
D) out of which more than one answer is correct.
(10 × 5 = 50 Marks)
1.
Which of the following statements are correct?
a.
b.
Dislocation of ion from lattice site to interstitial site is called Frenkel defect
Missing of +ve and –ve ions from their respective position producing a pair of holes is called
Schottky defect
c. Presence of ions in the vacant interstitial sites along with lattice point is called interstitial defect
d. Non stoichiometric NaCl is yellow solid
A, B, C, D
Sol. Non stoichiometric NaCl has F-centres due to anion vacancy defect.
2. Select the correct statements about three dimensional HCP system:
3.
(a) Number of atoms in HCP unit cell is six
(b) The volume of HCP unit cell is 24 2 r3
(c) The empty space in HC Punit cell is 26%
(d) It is ABAB…… packing
A, B, C, D
Which of the following statements for crystals having Schottky defect is/are correct?
a.
b.
4.
Schottky defects are more common in ionic compounds with high co-ordiantion numbers
Schottky defect arises due to the absence of a cation and anion from the position which it is
expected to occupy
c. The density of the crystals having Schottky defect is larger than that of the perfect crystal.
d. The crystal having Schottky defect is electrically neutral as a whole.
A,B,D
Select the correct statement
a. There is one Pb+4 ion per cation vacancy in Pb0.94O
b. Three unit cells belong to tetragonal crystal system depending on lattice point.
c. In orthorhombic unit cell all interfacial angles are equal
d. In Frankel defect density remain constant
A,C,D
Sol. (a) two Pb+2 are replaced by one Pb+4
(b) only two system
(c) all are 90°
5. Which of the following is a true statement?
6.
a. Solids are rigid due to strong force of attraction
b. Glass is a supercooled liquid
c. Crystalline solids are anisotropic
d. Amorphous solids have zero entropies
A, B, C
Which of the following statement is true for ionic solid.
a. They have high melting point
b. These are water soluble
c. Ionic solid can show conductance in molten state
d. These are insulator in solid state
A, B, D, C
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7.
In an f.c.c. unit cell, atoms of same size are numbered as shown below. The atoms not touching
each other are
1
2
4
3
6
5
8.
a. 1 & 3
b. 1 & 2
c. 4 & 5
d. 4 & 6
A, B, C
In the given b.c.c. unit cell, atoms are numbered as shown below. The atoms touching each other
are
4
5
2
9.
3
1
a. 2 & 4
b. 1 & 5
c. 1 & 2
B, C, D
Which of the following statement is correct for two dimensional packing
d. 1 & 4
a. Square packing is more efficient than hexagonal packing
b. In hexagonal packing atoms of 2nd row are placed in depression of 1st row.
c. In square packing, atoms are aligned horizontally & vertically
d. Hexagonal packing is more efficient than square packing
B, C, D
10. Which of the following is not true for metallic solid.
a. Strength of metallic bond is directly proportional to size
b. They are good conductor of electricity
c. Conductance increases with increase in temperature
d. Localised metal ions are embedded in sea of electrons.
A,C
1
Sol. Strength 
size
Conductance decrease with temperature.
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SECTION – E (Matrix Type) No Negative Marking
This Section contains 2 questions. Each question has four choices (A, B, C and D) given in Column I
and five statements (p, q, r, and s) in Column II. Any given statement in Column I can have correct
matching with one or more statement(s) given in Column II.
8 × 2 = 16 Marks
1.
Match Column – I with Column – II.
(A)
(B)
(C)
Column – I
The stoichiometry defect that lowers the density
(p)
–
The defect in which e are present in voids, left (q)
vacant by the anion.
Extrinsic semi-conductor shows conductance due to (r)
(D) Only n-type semi-conductance is due to
Sol. A  p; B  q; C  r; D  q, r, s
2.
Column – II
Schottky defect
Presence of F-centre
Presence of impurity of higher
valency
Electronic imperfection
(s)
Match Column – I with Column – II
(A)
(B)
Column – I
Tetragonal and Hexagonal
Cubic and Rhombohedral
(P)
(Q)
(C)
Monoclinic and Triclinic
(R)
(D)
Cubic and Hexagonal
(S)
Sol. A S & P; B  P, Q ; C  P, R; D  P
Column– II
Are two crystal systems (unit cell)
Have these cell parameters
a=b=c
abc
a=bc
SECTION – F (Integer Type)
This Section contains 10 questions. The answer to each question is a single digit integer ranging
from 0 to 9. The correct digit below the question number in the ORS is to be bubbled.
10 × 3 = 30
1.
If No. of cation vacancy in AgCl is 4 moles, due to impurity of AuCl3. What will be the no. of moles of
AuCl3.
Sol. 2 mol
2. How many moles of total tetrahedral & octahedral voids will be found in a ccp arrangement having 3
moles of sphere in lattice.
Sol. 9 mole; 3 octahdral & 6 tetrahedral
z
3. Calculate the value of where
5
z = Co. No. of 2D square close packing + Co. No. of 2D hexagonal packing + No. of tetrahedral void
in ccp (one unit cell) + Co. No. of ABAB (Hexgonal)……. type of packing.
Sol. 6
Z = 4 + 6 + 8 + 12 = 30
4. The total number of ions per unit cell in sphalerite (zinc blend) structure if there is a Schottky defect
per unit cell, is
Sol. 6
Sphalerite structure have 4 Zn+2 & 4S-2 ion i.e. 8 ions. In A pair of schottky defect two ions will be
missing so total ions left = 6
5. How many are correct statements
(a) The conductance through electrons is called p-type conduction
(b) The conductance through positive holes is called p-type conduction
(c) The conductance through electrons is called n-type conduction
(d) The band gap in Germanium is small
(e) Solids with F-centres are paramagnetic
(f) Ferrimagnetic character of Fe3O4 at room temperature changes to paramagnetic character at 850 K
(g) Anti ferromagnetic compounds changes to paramagnetic at high temperature
(h) Non stoichiometric Cu2O is p-type semiconductor
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Sol. 7
B, C, D, E, F, G, H
6. Which of the following statements is/are correct for a molecular solid?
(a) A low melting solid
(b) A compound that conducts electricity when molten
(c) A solid that is a non – conductor of electricity
(d) A solid formed by the combination of non-metallic elements
(e) Molecules or atoms are present at lattice point.
(f) These are water soluble solids
(g) Fullerene is a molecular solid
(h) Iodine is non-polar molecular solid
Sol. 6
A, C, D, E, G, H
7. Which of the following is correct for three dimensional packing?
(a) If atoms of third layer are placed over octahedral void of 2nd layer, it will be ABCABC ….. type
packing
(b) If atoms of 3rd layer form tetrahedral void with 2nd layer, it will be ABAB …. packing
(c) Coordination number of ABAB….. packing is more than ABCABC packing
(d) Unit cell of ABCABC packing is Body centred cubic
(e) Both ABCABC and ABAB …. Packing Co. No. is 12
(f) Unit cell of ABCABC packing is FCC
(g) If Atoms of 3rd layer are present on tetrahedral voids of 2nd layer its arrangement is same as that
of the first layer in HCP.
Sol. 5
A, B, E, F, G
8. How many of the following statements are correct for defects in solids
(a) Schottky defect is a vacancy defect
(b) Solids with smaller cation have tendency of schottky defects
(c) Zinc oxide become non-stoichiometric on heating
(d) In ferrous oxide crystal, Number of cation vacancy is half, of the number of ferric ion as three
ferrous ions are replaced by two ferric ion.
(e) Transition metal salts having metal in lower oxidation state have tendency to have nonstoichiometric defect
(f) F-centres are cation vacancies having e–.
(g) Boron doped silicon crystal is a p-type semiconductor
(h) Density of crystal remains same both in schottky and Frenkel defect.
(i) Compounds with schottky defect donot obey law of constant composition
Sol. 5
a,c,d,e,g
9. How many atoms are considered for one unit cell. If lattice point are given as under
‘A’ atoms
‘B’ atoms
‘C’ atoms
Sol. Corner atoms = 1
Edge centered atom = 8 ×
1
=2
4
Centered atom = 1
Total atom = 4
10. What will be the coordination number of surface atom in H.C.P. three dimensional close packing.
Sol. 9
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