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1 SOLID STATE

INTRODUCTION

Matter exists mainly in three states, viz. solids, liquids and gases. The existence of matter in any
of these three forms depends upon two factors

1. Intermolecular forces of attraction (keeps particle closer)

2. Thermal energy (keeps particles apart)

Some of the common properties of solids, which distinguish them from other two states of matter,
are:


Solids are rigid and have definite shapes.

Solids have definite volume irrespective of the size or shape of the container in which they
are placed.
 Solids are almost incompressible.
 Solids diffuse very slowly as compared to liquids and gases. Constituent particles are very
closely packed in solids permitting very little space for their movement.
 Solids have a much higher density (mass to volume ratio) than that of gases and liquids.
 Most solids become liquids when heated. Some undergo sublimation on heating. The
temperature at which a solid changes into liquid is called the melting point and the process
is called as melting. Due to the varying natures of solids their melting temperatures vary
considerably.
CLASSIFICATION OF SOLIDS

Solids are divided into two classes, namely crystalline and amorphous solids. A solid is said
to be crystalline if the constituents arrange themselves in regular manner throughout the three-
dimensional network. The ordered arrangement of building constituents extends over a large
distance (long range order). On the other hand, in amorphous solids, the arrangement of
building constituents is not regular (short range order).

i) Crystalline solids: A crystalline solid is a homogeneous solid in which the constituent particles
(atoms, ions or molecule) are arranged in a definite repeating pattern.

Example:Diamond, Quartz, NaCl, K2SO4 etc.

General characteristics of Crystalline solids:

i) A crystalline solid is a homogeneous solid in which the constituent particles (i.e. atoms, ions
or molecules) are arranged in a definite repeating pattern in all dimensions.
ii) The total intermolecular force of attraction in crystalline solid is maximum thus imparting
maximum stability, the forces responsible for the stability involves ionic bonds, covalent
bonds, hydrogen bonds and Van der Waal’s forces.
iii) A crystalline solid usually consist of a large number of small tiny crystals called unit cell,
each having a definite characteristic geometrical shape.
iv) Crystalline solid has regular arrangement of particles which repeats periodically over entire
crystal, thus exhibiting short and long range order.
v) Crystalline solids have sharp melting point, thus have definite heat of fusion.

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 SOLID STATE

vi) Crystalline solids are true solids.

vii) Crystalline solid on cutting gives a clean cleavage.
viii) Crystalline solid shows different physical properties in different direction, this type of
behavior is called anisotropy and the substances exhibiting this type of behavior are called
anisotropic. The properties like electrical conductivity, refractive index, thermal expansion
etc. have different value in different direction.
ix) Two or more crystalline substances having same crystal structure are said to be
isomorphous. Isomorphous substance contains constituent atom of same atomic ratio.
Example: a) NaF and MgO (Ratio is 1:1)

b) NaNO3 and CaCO3 (Ratio is 1:1:3)

c) K2SO4 and K2SeO4 (Ratio is 2:1:4)
d) Cr2O3 and Fe2O3 (Ratio is 2 : 3)

Exceptional: NaCl and KCl have all properties identical [same atomic ratio, similar
molecular formula or similar chemical properties] but are not isomorphous.

x) A single substance that crystallises in two or more forms under different conditions is called
polymorphous. (allotropic forms)
. Example: a) Carbon has two allotropes graphite and diamond.

b) Sulphur has two polymorphic forms monoclinic and rhombic.

c) CaCO3 and SiO2 have two allotropic forms.

ii) Amorphous solids:Substances that appear like solids but do not have perfectly
orderedcrystalline structure and no regular arrangement of constituentparticles in structure is
called amorphous solids.

Example: Tar, glass, plastic, rubber, butter etc.

General characteristics of Amorphous solids:

i) Amorphous substances appear like solids but they do not have perfectly ordered crystalline
structure, hence they are not real solids.
ii) An amorphous solid does not have regular arrangement of constituent particles.
iii) The arrangement of constituent particles like atoms or molecules has only short range
order hence periodically repeating regular pattern is only over a short distance.
iv) Regular patterns are scattered and hence the arrangement is disordered.
v) Amorphous solids are called supercooled liquids of very high viscosity or pseudo solids.
vi) Physical properties do not change with change in directions hence amorphous solids are
isotropic in nature.
vii) Amorphous solids behave like fluids and very slowly float under gravity.
viii) Amorphous solids do not have sharp melting points.
ix) When cut, they split into pieces with irregular and rough surfaces.

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 SOLID STATE

B
D

A B A B A B

B A B A B A

A B A B A B

B B A B A
A

A B A B A B

C
A

Anisotropic behaviourof crystal

Uses of amorphous solids:

i) The most widely used amorphous solid are the inorganic glasses viz. construction, house-
ware, laboratory ware, etc.
ii) Used as rubber in making tyres, shoe soles, etc.
iii) Used in plastics.
iv) Amorphous silica used for converting sunlight into electricity (in photovoltaic cell).
Anisotropy:The ability of crystalline solids to change their physical properties when measured in
different directions is called anisotropy.

Explanation: This property is due to different arrangement of constituents in different directions.

Different types of particles fall on the way of measurements in different directions. Hence, the
composition of crystalline solids changes with directions changing their physical properties.

Isotropy: The ability of amorphous solids to have same physical properties when measured in
different directions is called isotropy.

Explanation:This property is due to no regular arrangement of particles in any direction. Hence

the properties like electrical conductivity, thermal expansion are identical in all the direction.

Difference between Crystalline and Amorphous Solid:

Crystalline solids Amorphous solids

1) They have definite and regular They do not have any pattern of arrangement
geometry due to definite and orderly
arrangement of atoms, ions or molecules of atoms, ions or molecules and, thus do not
in three-dimensional space. have any definite geometrical shape.

2) They have sharp melting points and Amorphous solids do not have sharp melting
change abruptly into liquids
points and do not change abruptly into liquids.

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 SOLID STATE

3) Crystalline solids are anisotropic. Amorphous solids are isotropic. Their physical
Some of theirphysical properties are
different in different directions properties are same in all direction

4) These are considered as true solids These are considered pseudo solids or

super cooled liquid.

5) Crystalline solids are rigid and their Amorphous solids are not very rigid. These
shape isnot distorted by mild distorting
forces force can be distorted by bending or compressing

6) Crystals are bound by plane faces. Amorphous solids not have well defined planes.
The angle between any two faces is When an amorphous solid is broken, the
called interfacial angle. For a given surfacesof thebroken pieces are generally and
crystalline solid, it is a definite angle and not flatand intersect at random angles.
remains always constant no matter how (Amorphoussolids do not have any symmetry)
the faces develop.When a crystalline
solid is hammered, it breaks up into
smaller crystals of the same geometrical
shape.

7) Ex.: NaCl, KCl, Sugar, Quartz etc. Ex.: Plastic, Glass, Rubber etc.

Types of crystalline solids:Crystalline solids are classified into four main types as follows:

i) Molecular solids: They are further classified into three types:

a. Polar molecular solids.
b. Non-polar molecular solids.
c. Hydrogen bonded molecular solids.
ii) Ionic solids.
iii) Metallic solids.
iv) Covalent solids.
Molecular solids:

i) In the crystalline molecular solids, the constituent particles aremolecules of same compound.
ii) Depending upon the type of molecules involved in crystal formation and the nature of
intermolecular force of attraction between the neighboring molecules, they are further sub-
divided as :
a) Non-polar molecular solids:

1. These are those crystalline solid in which the constituent particles are either atoms [Noble
gases] or non-polar molecules [H2, Cl2, I2, CH4, etc.] or weakly polar molecules like CO or
other hydrocarbons.
2. They are formed at relatively lower temperature and are in usually gaseous state at normal
temperature.

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 SOLID STATE

3. In these atoms or non-polar molecules are held by weak dispersion forces or London forces.
4. These solids are generally soft, having low melting and boiling point and are non-conductor
of electricity.
5. As polar molecules exist in gaseous state, polar molecular solids is obtained by subjecting
the gas high pressure and low temperature.
b) Polar-molecular solids:

1. These are the crystalline solid in which the constituent particles are polar molecules [HCl,
SO2, etc.]
2. In polar molecule there is separation of charges, in which the opposite charges of
neighboring molecules are brought closer.
3. The forces holding these molecules are dipole-dipole forces of attraction, this force of
attractionis stronger than London forces.
4. These solids show following characteristic:
i) They are soft.

ii) Their melting point and boiling point are comparatively higher than non-polar molecular

solids but lowerthan ionic and metallic.

iii) They also exist as liquid or gases at room temperature.

iv) They are non-conductor of electricity.

v) They possess permanent dipole moment.

c) Hydrogen bonded molecular solids:

1. In these solids, the constituent particles are such molecules which contain hydrogen atom l
inked to a highly electronegative atom [O, N or F]. Example: solid ice, NH3, etc.
2. In these, molecules are held by hydrogen bond in which the H atom of one molecule is
bonded to electronegative atom of another molecule.
3. This intermolecular force of attraction existing among the molecules is strong hydrogen
bonds.
4. Characteristic of hydrogen bonded molecular solid :
i) They exist as volatile liquid or soft solids at room temperature.

ii) They are non-conductors of electricity.

iii) Their melting point and boiling are usually higher than non-polar molecular solids and

polar molecular solids.

5. They solidify on cooling.

Ionic solids:

i) In these crystalline solids, the constituent particles are positive and negative ions i.e. cations
and anions. e.g. Na+ and Cl– ions in case of NaCl.
ii) These ions in the solid are held in their lattice points by strong electrostatic force of
attraction resulting into well ordered three dimensional arrangement of ions.

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 SOLID STATE

iii) All salts are crystalline in nature and are called Ionic solids.
iv) In Ionic solids, the charges on the ions and the arrangement of ions are in such a manner that
they balance each other and hence the molecule is electrically neutral.
v) The arrangement of ions in the solid depends upon :
a) Size of cation and anion
b) charges on the ion
c) ease with which anion is polarized (i.e. polarizability of anions).
vi) Ionic solids are hard and brittle and have high melting point and boiling point.
vii) They are electrical insulators in solid state because their ions are not free to move.
viii) In aqueous solution or in molten state as the ions become free they are good conductor of
electricity.
ix) They are soluble in polar solvent but insoluble in non-polar solvent.
x) On application of shearing force, ionic crystals undergo distortion and fracture in crystal
structure.
Metallic solids:
i) In metallic solids the constituent particles are positively charged metal ion and free electron.
ii) Due to low ionization energy of metal atom the metal atom loses their valence electron and
becomes positively charged ions.
iii) Thus electrons lost are delocalized over the crystal space and flows throughout the crystal like
water in sea, hence also called sea of free electrons.
iv) The force of attraction between positively charged metallic ion and negatively charged sea of
delocalized electron is called metallic bond.
v) If energy is supplied, the valence electron from sea of electrons move from one place to
another, this presence of mobile electrons makes all the metal good conductor of heat and
electricity.
vi) On application of shearing force, the layers slide on one another and hence the structure is not
fractured imparting the properties of malleability and ductility.
vii) They have high melting point, boiling point and density.
viii) The mixtures of metals can be fused together to form alloys, which exhibit all properties of
metals.
ix) They possess lusture and colour. [Gold metal exhibit yellow lusture and copper has reddish
lusture].
x) Metallic bonds are stronger than ionic and covalent bond.
Covalent solids or Network solids:
i) These are crystalline solids in which the constituent particles are non-metal atoms linked to
the adjacent atoms by covalent bonds throughout the crystal forming a giant three-dimensional
structure.
ii) Hence covalent solids are called giant solids and the constituting molecules are called giant
molecules.
iii) Since covalent bonds are strong and directional, atoms are held strongly at their lattice
positions.
iv) They are hard or brittle depending on the event of bonding.
v) They have high melting points.
vi) They act as good conductors of electricity or insulators depending upon the availability of free
electrons.
The examples of covalent solids are diamond, graphite, Silicon carbide (carborundum), fullerene,
boron nitride, etc.

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Figure. A covalent crystal contains a three-dimensional network of covalent bonds, as illustrated by the structures
of diamond, silicon dioxide, silicon carbide, and graphite. Graphite is an exceptional example, composed of planar
sheets of covalent crystals that are held together in layers by noncovalent forces. Unlike typical covalent solids,
graphite is very soft and electrically conductive.

Type of Solid Constituent Bonding/ Examples Physical Electrical Melting

Particles Attractive Nature Conductivity Point
Forces
1. Molecular
solids
(i) Non-polar Molecules Dispersion Ar, CCl4, H2, I2, Soft Insulator
or London CO2
forces
(ii) Polar Dipole- HCl, SO2 Soft Insulator
dipole
interactions
(iii) Hydrogen Hydrogen H2O (ice) Hard Insulator
bonded bonding
2. Ionic solids Ions Coulombic NaCl, MgO, ZnS, Hard but Insulators in High
or CaF2 brittle solid state but
electrostatic conductors in
molten state

solutions
3. Metallic solids Positive Metallic Fe, Cu, Ag, Mg Hard but Conductors Fairly high
ions in bonding malleable in solid state
a sea of and ductile
delocalised molten state
electrons
4. Covalent or Atoms Covalent SiO2 Hard Insulators Very high
bonding C (diamond), AlN,
C (graphite) Soft Conductor

Comparison of different types of crystalline solids:

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Figure. Diamond is extremely hard because of the strong bonding between carbon atoms in all directions. Graphite
(in pencil lead) rubs off onto paper due to the weak attractions between the carbon layers. An image of a graphite
surface shows the distance between the centers of adjacent carbon atoms. (credit left: modification of work by
Steve Jurvetson; credit middle: modification of work by United States Geological Survey)

TERMS AND CONCEPTS

Space lattice or crystal lattice:

Itmay be defined as a regular three-dimensional arrangement of constituent particles of a solid

substancein space.

The positions which are occupied by atoms, ions or molecules in the crystal lattice are called
lattice points or lattice sites.

A portion of a three dimensionalcubic lattice and its unit cell

Characteristics of Crystal lattice:

1. Lattice points or lattice sites:The crystal lattice of a substance is represented by showing the
position of the particle in space. These positions are represented by points and are referred to
as Lattice points.

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 SOLID STATE

2. Each point in a crystal lattice represents one constituent particle which may be an atom, a
molecule (group of atoms) or an ion.
3. Lattice points are joined by straight lines to bring out the geometry of the molecule.

Unit Cell: It is the smallest portion of a crystal lattice which, when repeated in three
dimensions produces crystal lattice.

Characteristics of Unit cell: A unit cell is characterized by following parameters,

1. Edges or edge length:The intersection of two faces of crystal lattice is called as edge. The
three edges denoted by a, b and c represent the dimensions (lengths) of the unit cell along
three axes. These edges may or may not be mutually perpendicular.
2. Angles between the edges (or planes):There are three angles between the edges of the unit
cell represented as α , β and γ . The crystal is defined with the help of these parameters of its
unit cell.
a) The angle α is between edges b and c.

b) The angle β is between edges a and c. c

b
a
c) The angle γ is between edges a and b.
Illustration ofparameters of a unit cell

A space lattice can be sub-divided into a number of small cells known as unit cells. It can be
defined asthe smallest block from which entire crystal can be built up by its translational
repetition in three dimensions

1) Total number of edges in a cube= 12

2) Total number of faces in a cube= 6

3) Total number of corners in a cube= 8

4) Total number of body centre in a cube= 1

5) Total number of face diagonals in a cube= 6 × 2 = 12

6) Total number of body diagonals in a cube= 4

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TYPES OF LATTICES AND TYPES OF UNIT CELL

Unit Cells are of two types: Primitive & Non- primitive.

(i) Primitive or simple Unit Cells: In a primitive unit cell, the same type of particles is present at
all the corners of the unit cell.
 
 

 
 
simple cubic

(ii) Non-primitive or centered unit cells:

There are three types of non-primitive unit cells as follows:

(a) Face Centered:Whenatomsarepresentinall 8-corners and six face centersin a cubic unit cell
then this arrangement is known as FCC.
 

 

 

 

 
face centred cubic (fcc)

(b) Body Centered(BCC):When atoms are present at 8 corners as well as in the body centre in a
cubic unit cell then this arrangement is known as BCC.
 
 

 
 
Body centred cubic (bcc)
(c) End-Centered:Inadditiontoparticlesatthe corners, there are particles at the centers of two
oppositefaces.
 

 

 

 
end centred cubic (fcc)
SEVEN CRYSTAL SYSTEMS

By considering the symmetry of the axial distance and the axialangles between the edges, the
various crystals can be divided into 7 systems. Although each system is expected to have 4

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 SOLID STATE

different unit cell, but actually all of them cannot exist in each case and only 14 different type
of lattices called Brava is lattice had been established.

There are 14 possible three-dimensional lattices. These are called Bravais Lattices.

Serial Crystal System Possible Edge Axial Examples

No. variations lengths angles
Primitive NaCl
Body – ZnS
α =β = γ =
1 Cubic Centered a=b=c
90o
Face – Cu
Centered
Primitive SnO2
α =β = γ =
2 Tetragonal Body – a=b≠c TiO2
90o
centered
Primitive Rhombic
Sulphur
Body – KNO3
α =β = γ =
3 Orthorhombic Centered a≠b≠c
90o
Face – BaSO4
Centred
End – Centred MgSO4.7H2O
α =β =
4 Hexagonal Primitive a=b≠c 90o Graphite, ZnO
γ = 120o
Rhombohedral or α =β = γ Calcite (CaCO3)
5 Primitive a=b=c
Trigonal ≠90o Cinnabar (HgS)
Primitive α= γ= Monoclinic
6 Monoclinic a≠b≠c 90o sulphur
End – Centred β ≠90o Na2SO4.10H2O
Primitive α ≠β ≠ γ K2Cr2O7, H3BO3
7 Triclinic a≠b≠c
≠90

a
a

Primitive (or simple) Body-centred

Face Centerted
The three cubic lattices: all sides of samelength, angles between faces all 90°

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Primitive End-centred Body-centred Face-centred

The four orthorhombic
lattices: unequal sides,angles
between faces all 90°

The two tetragonal: one side different in

length from the other,two angles between faces
all 90°

More than
900

Less than
900

Primitive End-centred
The two monoclinic lattices: unequal sides,two faces have angles other than 90°

a 600 Less than 900

a
a a
Hexagonal lattice: Rhombohedral
one side different lattice: all sides of
in length to the equal length,
other two, the angles on two
marked angles on faces are less than
two faces are 60°
90°
and the remaining
angle is 120°.

a Triclinic lattice:
b unequal sides a, b, c,
A A, B, C are unequal
B C angles with none
equal to 90°
c

Unit Cells of 14 Types of Bravais Lattices

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CALCULATION OF NUMBER OF PARTICLES IN A UNIT CELL (Z)

In a crystal, atom located at the corner and face center of a unit cell are shared by other cells
and only a portion of such an atom actually lies within a given unit cell.

(i)A face-centered point is shared by two unit cells and only one half of it is present in given unit
cell, hence the contribution of the particle per unit cell is 1/2.

(ii) A point along an edge is shared by four-unit cells and only one-fourth of it lies within one
cell,hence the contribution of the particle per unit cell is 1/4.

(iii)A point that lies at the corner of a unit cell is shared among eight-unit cells and therefore, only
one eighth of each such point lies within the given unit cell, hence the contribution of the
particle per unit cell is 1/8.

(iv) A body-centered point lies entirely within the unit cell and contributes one complete point to
the cell.

Type of Lattice point Contribution to one unit cell

Face-center 1/2
Edge 1/4
Corner 1/8
Body Center 1

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i. Calculation of number of particles per unit cell in a primitive cubic unit cell:In this
type of unit cell, there are eight particles at the corners of the unit cell.
 Number of particles per unit cell = No. of particles in unit cell x share of particles per unit
1
cell = 8x
8

=1 particle.

ii. Calculation of number of particles per unit cell in a Body- centred cubic unit cell:

This type of unit cell has 8 particles at the corners and 1 particle at the
centre of the unit cell.
1
 Number of particles per unit cell =8 x +1x 1
8

= 2 particles.

iii. Calculation of number of particles per unit cell in a Face- centred cubic unit cell:

This type of unit cell has 8 particles at the corners and 6 particles at centre of 6 faces of the
unit cell.
1 1
 Number of particles unit cell =8 x + 6 x
8 2

=1 +3

=4 particles.

Calculation of number of particles in a unit cell

Type of unit Lattice points Lattice points at Lattice points at Z = no. of lattice points
cell at corners face-centered body centered per unit cell

SC 8 0 0 1
8 × = 1
8
BCC 8 0 1 1
8 × +1×1 = 2
8
FCC 8 6 0 1 1
8× +6 × = 4
8 2

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CALCULATION OF NEAREST NEIGHBOURS

Note: In a simple cube, the atoms at the corners touch each other. But, the atoms at the corners of
face centered cube and body centered cube do not touch each other. In fcc, each sphere at the
corner touches the three spheres at the face centres of three adjuring faces similarly, in bcc, all
the atoms at the corners touch the central sphere.

The relationship between the nearest neighbor distance (d) and the radius of atom ® (for
crystals of pure elements) and the edge of unit cell (a) are given below:

1] Simple Cube:

d = AB = a

r = a/2

2] Face Centered Cube:

AC
d = 2 in right angled ΔABC AC2 = AB2 + BC2

AC2 = a2 + a2 = 2a2

AC = √2. a

AC √2. a a d a
d= = = ∵r= =
2 2 √2 2 2√2
AD
3] Body Centered Cube:d = 2

In right angled ΔABC, AC2 = AB2 + BC2 = 2a2

AC = √2. a

In right angled ΔADC

AD2 = AC2 + DC2 = (√2.a) 2 + a2

√3. a d √3
AD2 = 3a2  AD = √3. a  d = r= = a
2 2 4
(i) Close packing in two dimensions:

Initiallythe spheres arrange themselves in a row to form an edge of the crystal.

There are two ways to build a crystal plane

(a) Sphere are packed in a such a way that the rows have a horizontal as well as vertical
alignment. In this arrangement, the spheres are found to form square. This type of packing is
also called square close packing.

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The number of spheres which are touching a given sphere is called co-ordination number.
Thus, the coordination number of each sphere in square close packing is four.

(b) The sphere is packed in such a way that the spheres in the second row are placed in the
depressions between the spheres of the first row and so on. This gives rise to hexagonal close
packing of spheres and the coordination number of each sphere is six.

(ii) Close packing in three dimensions:

It is clear from the figure (X) that there are two types of voids or hollows in the first layer.
These are marked as b and c. All the hollows are equivalent but the sphere of second layer
may be placed either on hollows which are marked b or on the other set of hollows marked c.
The second layer is indicated as dotted circles in figure (Y).

When a third layer is to be added, again there two types of hollows available. One type of
hollows marked ‘c’ are unoccupied hollows of the first layer. The other type of hollows are
hollows in the second layer (marked a). Thus, there are two alternatives to build the third
layer.

(i) When the third layer is placed over the second layer so as to cover the tetrahedral or ‘a’ voids,
a three-dimensional closest packing is obtained where the spheres in every third layer are
vertically aligned to the first layer. This arrangement is called ABAB.,…pattern or
hexagonal (HCP) close packing(calling first layer as A and second layer B).

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Hexagonal close packing (hcp)

(a) For HCP geometry Coordination number = 12

(b) For HCP geometry no. of atoms per unit cell

1 1
= 12 (corners) × + 2(face centres) × + 3 (inside the body) × 1 = 6
6 2
(c) For HCP geometry packing efficiency = 74 %

(ii) When the third layer is placed over the second layer such that the spheres cover the octahedral
or ‘a’ voids, a layer different from A and B is formed. This pattern is called
ABCABC……pattern or cubic close packing (CCP).

The ABC ABC....... packing has cubic symmetry and is known as cubic close packing (ccp).
The cubic close packing has face centered cubic (fcc) unit cell.

Cubic close packing (ccp)

(i) For CCP geometry coordination number = 12

(ii)For CCP geometry no. of atoms per unit cell = 4(as calculated before)

(iii)For CCP geometry packing efficiency = 74 %

VOIDS

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In the close packing of spheres, certain hollows are left vacant. These holes or voids in the
crystals are called interstitial sites or interstitial voids.

(i) Triangular (ii) tetrahedral

(iii) Octahedral (iv) cubical void
(i) Triangular:The vacant space (void) formed by touching three spheres.

rvoid
= 0.155
rsphere

(ii) Tetrahedral:The vacant space among four spheres having tetrahedral arrangement is called
r
tetrahedral site or tetrahedral hole. For tetrahedral void r void = 0.225.
sphere

(iii) Octahedral:This type of site is formed at the centre of six sphere. The void formed by two
equilateral triangles with apices in opposite direction is called octahedral site or octahedral
hole. For octahedral void
rvoid
= 0.414
rsphere

iv) Cubical void:The vacant space (void) formed by touching eight spheres
rvoid
= 0.732
rsphere

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Note:If a close packing (array) is made up of n number of atoms or ions then it has n no.
ofoctahedral voids and 2n no. of tetrahedral voids.

DENSITY OF UNIT CELL

 The length of edge of the cell= a cm

 Volume of unit cell = a3 cm3
Mass of unit cell
Density =
Volume of unit cell

 Mass of unit cell = number ofatoms in a unit cell  mass of each atom = Z.m
Number of atom in unit cell×mass of one atom
 Density =
Volume of unit cell
Gram Atomic Weight M
 Mass of one atom (m) = =N
Avogadro′ snumber A
Z×M
 Density = Z → Number of atoms per unit cell.
a3 ×NA

PACKING EFFICIENCY

Packing efficiency is the % of total space occupied by particles. Both types of close packing
(hcp and ccp) are equally efficient and occupy 74% of available value. In bcc, the efficiency
is 68% while is simple cubic structure, it is 52.4%
Vol.occupied by all atoms in unit cell v
Packing efficiency = =V
Total vol.of unit cell

Let a be the cube edge length and r the radius of atom.

4
V =volume of unit cell = a3; Volume of sphere,v = 3 πr 3
a. Packing efficiency in CCP or FCCarrangement:
Both type of close packing (hcp and ccp) are equally efficient. Let us calculate the efficiency
of packing in ccp structure.

In ccp, the unit cell is face centred. In face centred cubic unit cell,
there are four spheres per unit cell. Let ‘r’ is the radius of sphere and
‘a’ be the edge length of the cube.

4 3
Volume of the sphere = πr
3

4 3
Since there are four spheres per unit cell of fcc, volume of it = 4 x πr
3
In a face centred cubic unit cell, the spheres at corners are in contact with the sphere at the
centre of the face and the particles at the corners are not in touch with each other.
Therefore we can find the radius of the sphere as follows:

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 SOLID STATE

From Δle ABC,

AC2 = AB2 + BC2
 b2 = a2 + a2 C a B

 b2 = 2a2
b
 b= 2a
2

A
Since b = 4r, we have

 4r = 2a 2

2a 2
r = 4
AC = face diagonal = b = 4r
2a 4r
r = or a = Edge length of the cube = a
4 2

2x2xr 2x 2 x 2 xr
 a= 2 or a= 2

 a=2 2r

Volume of space occuped by atoms in one unit cell

Packing Efficiency = 100
Volume of one cubic unit cell

4
4x x π x r3
3 x 100
 
3
2 2r
=

4 x 4 x π x r3
3
x 100
= 3 x 8 x 2 x 2 r

π
x 100
= 3 2

Packing Efficiency = 74.04% 74%

b) HCP arrangement:

Volume of unit cell = base area height = 6√342 × 4r √2/3 = 24√2r 3

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 SOLID STATE

1 1 4
no. of atoms in hcp = 12 × + 2 × + 3 × 1 = 6 v = 6 × 4πr 3 = 8πr 3
6 2 3
8πr 3 π
Packing efficiency = × 100 = × 100 = 74%
24√2r 3 3√2

c) BCC arrangement:

InBody centred cubic unit cell, atoms are located at the corners of the cube and 1 particle at the
centre of the cube.

Number of particles per unit cell of BCC structure = 2

4 8
 Volume occupied by spheres = 2× πr3 = πr3
3 3

Where r  radius of the spheres. a

A B
Edge length of cube = a b
C
In a body centred cubic unit cell, the spheres at the
corners are not touching each other but are in contact
with the sphere at the centre of the cube. 4r
c=
From the above figure, we can find D

Face-diagonal as, from Δle ABC,

AC2 = AB 2 + BC2 = a2 +a2

AC = face diagonal = b
2 2
b = 2a
CD = body diagonal = c = 4r
To find body diagonal consider the Δ ACD,
le
Edge length of the cube = a

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 SOLID STATE

Body diagonal CD can be calculated as

CD2 = AC2 + AD2

c2 = b2 + a2 = 2a2 +a2

 CD = c = 3a 2 = 3 a

CD = 3 a = 4r

 4r = 3a

3 4
 r= a or a = r
4 √3

4 3
Volume of unit cell = a3 = [ 𝑟]
√3

 a3 = 64
𝑟3
3√3

Volume of space occuped by atoms in one unit cell

Packing Efficiency = × 100
Volume of one cubic unit cell

8 3
πr 8πr3 3√3
Packing Efficiency = 3
64 3 x 100 = x 100
𝑟 64 x 3 r3
3√3

Packing efficiency = 67.98 % ≃ 68 %

d) Simple cubic arrangement: In simple cubic unit cell, atoms arelocated only at the corners of
the cube. The particles touch one another along the edge.
If edge length of the cube= a, and radius of each particle is r; then a is
related to r as a = 2r

The volume of the cubic unit cell =a3

= 2r3 = 8r3

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 SOLID STATE

Since a simple cubic unit cell contains only 1 atom

4 3 a
The volume of the occupied space = πr
3
Volume of space occuped by atom
Packing Efficiency= × 100% 2r
Volume of cubic unit cell

4/3πr3
= ×100
8r3

4πr3
= ×100
3×8r3

𝜋
= ×100
6

= 52.36%

RADIUS RATIO RULES

In ionic crystals, the coordination numbers as well as the geometrical shapes of the crystals
depend mainly on the relative sizes of the ions. The ration of the radii of the positive and
negative ions is called radius ratio.

Radius of postive ion (cation) rc+

Radius ratio = =
Radius of negative ion (anion) ra−

Common coordination numbers are 3, 4, 6 and 8.

𝐫
Limiting 𝐫𝐜 radius ratio Co-ord. No. Shape Example
𝐚

i) < 0.155 2 Linear BeF2

ii) 0.155 – 0.225 3 Trigonal planar B2O3

iii) 0.225 – 0.414 4 Tetrahedral ZnS

iv) 0.414 − 0.732 6 octahedral NaCl

v) 0.732 − 0.999 8 B.C.C. CsCl

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 SOLID STATE

STRUCTURE OF IONIC COMPOUNDS

(A) Ionic Compounds of AB type:

These compounds can have following three type of

structures.

1] Rock salt structure (NaCl):

+
(a)Cl¯ is forming a FCC unit cell in which Na is in the
+
octahedral voids. The co-ordination number of Na is 6
and that of Cl¯ would also be 6.
rNa+
(b) Ratio of ionic radii = ( = 0.525)
rCl−

1
(c) No. of sodium ions = 12 (At edge centre) × + 1 (At body centre) × 1 = 4
4
1 1
No. of Cl− ions = 8 (At corners) × + 6 (At face centres) × = 4
8 2
(Thus formula is Na4Cl4 i.e. NaCl)

(d)Most of the halides of alkali metals and oxides of alkaline-earth metal have this type of
structure.e.g. NaI, KCl, RbI and RbF. FeO also has rock-salt structure in which oxide ions are
2+
arranged in ccp and Fe ions occupy octahedral voids.

2] Caesium chloride structure (CsCl):

(a)CsCl has body-centered cubic (bcc) arrangement. This structure has 8: 8 co-ordinations, i.e.,
each Cs+ ion is touching eight Cl– ions and each Cl– ions in touching eight Cs+ ions. (bcc).
rCs+
(b) = 0.933
rCl−

1
(c)No. of Cl− ions = 8 (At corners) × = 1
8
No. of Cs+ ions = 1 (At the body centre) × 1 = 1

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 SOLID STATE

No. of CsCl unit per unit cell = 1

(d) Compounds having this type of structure are CsBr, CsI, TlCl, and TlBr.

3] Zinc blende structure or Sphalerite structure (ZnS):

(a)Sulphide ions are face centered and zinc is present in alternate tetrahedral voids.
rZn2+
(b) = 0.40
rS2−

– 1 1
(c)No. of S2 ions = 8 at corners × + 6 at face centres × = 4
8 2

2+
No. of Zn ions = 4 (within the body) × 1 = 4

No. of ZnS units per unit cell = 4

(Formula is Zn4S4, i.e. ZnS)

(d) Ionic solids having zinc blende structure are CuCl, CuBr, CuI & AgI

(B) Ionic Compounds of AB2 type:

Fluorite structure(CaF2):

(a) The cations are arranged in cubic close packing (ccp) while the anion occupy all the
tetrahedral voids. Calcium fluoride has 8: 4 co-ordination. (ccp)

(b) In unit cell no. of calcium ions

1 1
= 8 at corners × 8 + 6 at face centres × 2 = 4

No. of fluoride ions = 4 (within the body) × 1 = 4

No. of CaF2 units per unit cell = 4

(c) Compounds having fluorite structure are SrF2, BaCl2, BaF2, PbF2 and CdF2.

Structure of calcium fluoride

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 SOLID STATE

(C) Ionic Compounds of A2B Type:

Antifluorite Structure:

In Antifluorite structure e.g., (Na2O)

(a) The anions are arranged in cubic close packing (ccp) while the captions occupy allthe
tetrahedral voids.

(b) Na2O has 4: 8 co-ordinations

(c) Compounds having ant fluorite structures are: Li2O, K2O, Rb2O and Rb2S

Sr. Type of crystal Ions occupying voids ions forming close Coord no.
packing (cation: anion)
1. Rock salt (AB) (NaCl) Na+ ions occupy Cl- 6: 6
octahedral voids
2. Cesium chloride (AB) Cs+ ion occupy cubic Cl- 8: 8
(CsCl) hde
3. Sphalerite (AB) (ZnS) Zn2+ occupy tetrahedral S2- 4: 4
voids alternatively
4. Fluorite (AB2) (CaF2) F¯occupy tetrahedral Ca2+ 8: 4
voids
5. Antifluorite (A2B) Cations occupy Anions 4: 8
(Na2O) telrahedrd voids

IMPERFECTIONS OR DEFECTS IN SOLIDS

At absolute zero, crystals tend to have a perfectly ordered arrangement. This arrangement
corresponds to state of lowest energy. As the temperature increases, the crystals start
deviating from the perfectly ordered arrangement. Any deviation from the perfectly ordered
arrangement constitutes a defect or imperfection. These defects are sometimes called
thermodynamic defects because the number of these defects depends on the temperature.
Crystals may also possess addition defects due to the presence of impurities. Many properties
of crystalline solids such as electrical conductivity and mechanical strength can be explained
in terms of imperfections. Imperfections not only modify the properties of solids but also give

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 SOLID STATE

rise to new properties. The defect which arises due to the irregularity in the arrangement of
atoms or ions are called atomic imperfections.

These imperfections in the crystalline solid are called defects in crystalline solid. The defects in
crystalline solids are of two types viz.,

a. Point defect
b. Line defect
Point Defect: This defect is produced because of the faulty arrangement of a point i.e.
constituent practice like atom, ion or molecules in a crystalline solid.

The point defects are classified into three types:

Types of point Defects:

i) Stoichiometric defects.These are those defects in which the stoichiometry of the solid is not
disturbed as a result of the defect.

ii) Non-stoichiometry defects. These are those in which the stoichiometry is disturbed due to
the defect.

iii) Impurity defect. These arise when some foreign material is added into the crystal.

Types of Stoichiometric defect

(i) Vacancy defect is a stoichiometric defect generated

Vacancy
during crystallization, in which some of the places of the
constituent particle remain unoccupied. This defect results
in the decrease of density of thecrystal than expected. This
defect possibly occurs when a substance is heated. Vacancy defect
(ii) Interstitial defect:Interstitial defect is a type of
stoichiometric defect, in which the constituent particles
(atoms or molecules) occupy an interstitial site (occupies
the space between the lattice site). This defect results in the
increase of density of the crystal than expected.

In case of ionic solids, always electrical neutrality must be Particle at the

Interstitial defect interstitial site
maintained. The above two defects in case of ionic solids
can be explained as follows:

a) Schottky defect:

It is due to equal number of cations and anions missing from their lattice sties. As a result,
density decreases. This type of defect is shown byionic compounds which have high
coordination number and small difference in the size of cations and anions, e.g., NaCl, KCl,
KBr, AgBr and CsCl.

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 SOLID STATE

b) Frenkel defect:
It arises when cations are missing from their lattice sites and occupy interstitial sites. As a
result, density remains unchanged. This defect is also called dislocation defect as smaller ions
(usually cations) are dislocated from normal sites to interstitial sites.

Types of Non-stoichiometricDefects:

These are of two types

i) Metal excess:These defects arise when thereare excess metal ions in the crystal. these defects
arises in two ways:

a) By anion vacancies:It arises due to a negative ion missing from the lattice site leaving a hole
which is occupied by electron thereby maintaining electrical neutrality. The lattice sites
occupied by electrons are called F-centres as they are responsibleforcolour of Crystal:

Ex: When NaCl is heated in the atmosphere of sodium vapors, sodium atoms get deposited on the
surface of the crystal. The Cl¯ ions from the crystal lattice leaves their sites and diffuse to
thesurface. These Cl¯ ions combine with the sodium atoms and these Na atoms ionizes and
the electrons that are released are trapped by the anion vacancies.These are called f centers

(a) Metal excess defect due to anion vacancies

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 SOLID STATE

b) By presence of extra cations in the interstitial sites:

Excess metal ions are entrapped into the vacant interstitials sites and electrons in the
neighbouring interstitial sites to maintain electrical neutrality.

For example, when ZnO is heated, it loses O2 and turns yellow due to the following reaction:

1
ZnO → Zn2+ + O2 + 2e−
2
The Zn2+ ions occupy certain interstitial siteswhereas electrons released will occupy the
neighboring sites in the lattice. On heating the crystal turns yellow, due to transition of
freeelectrons froma lower energy state.

(b) Metal excess defect due to interstitial cation

ii) Metal Deficiency due to cations vacancies:

This defect occurs when the metal shows variable vacancy, i.e., transition metals, e.g., in FeO,
2+ 3+
FeS, NiO etc. This is because in FeO, 3 Fe ions may be replaced by 2 Fe ions to maintain
electrical neutrality. That is why we never have the idea composition FeO. Instead, we have
FexO with x = 0.93 to 0.96.

Metal deficiency due to cation vacancy

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 SOLID STATE

Types of Impuritydefects:

Impurities are added to change the properties of the crystals. The process is called doping.

i) In ionic solids:For example, SrCl2 may be added to NaCl. Two Na+ ions will be replaced by
2+
one Sr ion thereby creating a whole (a cation vacancy) and imparting conductivity.

ii) In covalent solids:

a) Doping with electron rich impurities: For example, Group 14 elements like Si or Ge
(having 4 valence electros) may be doped with Group 15 elements like P oras (having 5
valence electrons). For every atom of Group 14 replaced by that of Group 15 one extra
electron is present making it n-type semiconductor

b) Doping with electron deficit impurities: For example, when Group 14 elements is doped
with Group 13 elements like B, Al or Ga (having 3 valence electrons), for every atom of
Group 14 replaced by that of Group 13, a hole is created. On applying electric field, electrons
move to occupy these holes. Thus, holes move towards the negative plate as if they carry
positive charge. Hence, we get p-type semiconductors.

PROPERTIES OF SOLIDS

(i) Electrical Properties: Solids can be broadly classified into three types, on the basis of
electrical conductivity.

(a) Metals (conductors)

(b) Insulators

(c) Semi-conductors
6 8 –1 –1
Electrical conductivity of metals is very high and is of the order of 10 –10 ohm cm whereas
–12 –1 –1
for insulators, it is of the order of 10 ohm cm . Semi-conductors have intermediate
2 –9 –1 –1
conductivity in the range of 10 –10 ohm cm . Electrical conductivity of solids may arise
through the motion of electrons and holes (positive) or through the motion of ions. The
conduction through electrons is called n-type conduction and through (positive) holes is called
p-type conduction. Pure ionic solids where conduction can take place only through movement

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 SOLID STATE

of ions are insulators. The presence of defects in the crystal structure increases their
conductivity.

The conductivity of semi-conductors and insulators is mainly due to the presence of

interstitial electrons and positive holes in the solids due to imperfections. The conductivity of
semi-conductors and insulators increases with increase in temperature while that of metals
decrease.

(ii) Magnetic Properties:

Diamagnetic Materials: Those materials which are weakly repelled by the magnetic field are
called diamagnetic materials. e.g. Cu+, TiO , NaCl and benzene. They do not have unpaired
2
electrons.

Paramagnetic Materials:The materials which are weakly attracted by magnetic field are
called paramagnetic materials. These materials have permanent magnetic dipoles due to
2+ 2+
presence of atoms, ions or molecules with unpaired electron. e.g. O2, Cu , Fe etc. But these
materials lose their magnetism in the absence of magnetic field.

Ferromagnetic Materials: The materials which show permanent magnetism even in the
absence of magnetic field are called ferromagnetic materials. These materials are strongly
attracted by the magnetic field. e.g. Fe, Co, Ni and CrO2. Ferromagnetism arises due to
spontaneous alignment of magnetic moments of ions or atoms in the same direction.

Alignment of magnetic moments in opposite directions in a compensatory manner and

resulting in zero magnetic moment gives rise to anti-ferromagnetism.

Alignment of magnetic moments Alignment of magnetic moments

Ferromagnetic substances Antiferromagnetic substances

Alignment of magnetic moments

Ferrimagnetic substances

(a ) Ferromagnetic ( ) Antiferromagnetic ( ) Ferrimagnetic

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 SOLID STATE

for example,MnO,Mn2O3 andMnO2.

Alignment of magnetic moments in opposite directions resulting in a net magnetic moment

due to unequal number of parallel and anti-parallel magnetic dipoles give rise to ferri-
magnetism e.g. Fe3O4.

Ferromagnetic and ferrimagnetic substances change into paramagnetic substances at higher

temperature due to randomization of spins. Fe3O4, is ferrimagnetic at room temperature and
becomes paramagnetic at 850 K.

(iii) Dielectric Properties:

The electrons in insulators are closely bound to the individual atoms or ions and thus they do
not generally migrate under the applied electric field. However, due to shift in charges,
dipoles are createdwhichresultsin polarisation. The alignments of these dipoles in different
ways i.e. compensatory way (zero dipole) or non-compensatory way (net dipole) impart
certain characteristic properties to solids.

If the dipoles align in such a way that there is net dipole moment in the crystals, these crystals
are said to exhibit piezoelectricity or piezoelectric effect i.e. when such crystals are subjected
to pressure or mechanical stress, electricity is produced. Conversely, if an electric field is
applied to such a crystal, the crystal gets deformed due to generation of mechanical strain.
This is called inverse piezoelectric effect.

Some crystals which on heating, acquire electric charges on opposite faces, are said to exhibit
pyroelectric effect.

The solids, in which dipoles are spontaneously aligned in a particular direction, even in the
absence of electric field are called ferroelectric substances and the phenomenon is known as
Ferroelectricity. If the alternate dipoles are in opposite direction, then the net dipole moment
will be zero and the crystal is called anti-ferroelectric.

Ferroelectric solids – Bariumtitanate (BaTiO3), sodium potassium tartrate (Rochelle salt) and
potassium hydrogen phosphate (KH2PO4). Anti-ferroelectric – Lead Zircon ate (PbZrO3).

SuperConducting Materials:The materials which offer no resistance to the passage of electricity

is called superconductor or super conducting material. In this state, the materials become
diamagnetic and are repelled by the magnets. Most of the metals become super conducting at
low temperatures (2 – 5K). Highest temperature at which super conductivity is known is 23K
in alloys of niobium (e.g. Nb3Ge). Many complex metal oxides have been found to possess
super-conductivity at somewhat higher temperatures.]

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 SOLID STATE

Material Temperature

Nb3Ge 23 K

Bi2Ca2Sr2Cu3O10 105 K

Ti2Ca2Ba2Cu3O10 125 K

Band theory of solids:

1. According to band Theory, the atomic orbitals of atom in the crystal combine to form
molecular orbital which spreads over the complete crystal structure.
2. As the number of atoms in crystal increases, the number of molecular orbital containing
electrons increased. As the number of molecular orbitals increases the energy difference
between the adjacent orbitals decreases.
3. Until finally the energy gap becomes very small and molecular discrete energy levels merge
into one another to form continuous band of molecular orbitals which extend over the entire
length of crystal.
4. All the molecular orbitals are very close to each other and are collectively called a band.
5. There are two types of bands of molecular orbitals as follows :
1.Valence band:The atomic orbitals with filled electrons from
the inner shells form valence bands, where there are no free Conduction
band
mobile electrons since they are involved in bonding.
2.Conduction band: Atomic orbitals which are partially filled Forbidden zone
Energy

or empty on overlapping form closely placed molecular (large energy gap)

orbitals giving conduction bands where electrons are Valence

delocalized and can conduct, heat and electricity. band
6. In metallic crystals, the valence bands and conduction bands are
very close to each other and a very little energy is required to Insulator
excite electrons from valence bond in to the conduction band. In
conduction band the electron are delocalized and are free to move
from one end to the other end of the metal piece, this migration of
Conduction
electron makes the metal good conductor of heat and electricity. band
Energy

7. In substance which are bad conductor of heat and electricity, the

spacing between the valence band and conduction band is
Valence
relatively more so that more energy is required to promote
band
electrons from valence band to conduction bond, hence electrons
remains in valence band and thus cannot move freely thus do not
conduct heat and electricity and act as insulators. Metal

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 SOLID STATE

8. If the energy difference between valence band and conduction

band is moderate, then the substance in ordinary condition is non- Conduction
conductor, but if heated it becomes conductor due to transition of band

Energy
electrons into conduction band. Such conductors are less energy gap
semiconductors.
Valence
Conduction of electricity in Semiconductors:The substance which band

have poor electrical conductance at low temperature but increases

with increase in temperature is called semiconductor. A substance Semiconductor
containing filled band with electrons and a completely empty band
behaves as a semiconductor.

Example: Si, Ge, etc.

In excitation, empty conduction bands contain electrons to conduct electricity. However electrons
can be added to conduction band by adding impurity (like Arsenic with extra electrons to silicon
conduction band) and hence can become conductor of electricity.

Electron rich impurity: n-type semiconductor:If the impurity from Group 15 i.e. (Arsenic) is
added to group 14 (i.e. silicon), some of the sites of silicon in the crystal are occupied by arsenic
atoms each with one extra electron in the conduction band and will be available for transport of
electricity. Such type of semiconductor with impurity having extra negative charge due to extra
electron of impurity atom is called n-type semiconductor.

Electron deficient impurity: p-type semiconductor: If the impurity from Group 13 i.e. (Boron)
added to Group 14 (i.e. silicon) then some atoms of born will occupy some of the sites of silicon
atoms. At all sites of boron atoms one valence electron will be shorter as compared to silicon
atoms and there will be a positive hole in the lattice. Hence an electron from neighbouring silicon
atoms jumps into the electron hole and continues till the electron hole is transferred to the edge of
the crystal lattice and movement of electron takes place. This type of semiconductor is called p-
type semiconductor.

Examples for n and p –type semiconductors:

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 SOLID STATE

Semiconductor Type
1 B doped with Si p-type
2 As doped with Si n-type
3 P doped with Si n-type
4 Ge doped with In p-type

Packing in Solids:

In solids the constituent particles are closely packed due to stronger Intermolecular forces of
attraction so as to attain a state of maximum possible density and stability.

Considering the constituent particles as identical hard spheres, we shall build up the three
dimensional structures in 3 steps:

1. Close-packing in one-dimension:
i. In one dimensional close packing there is only one way of arranging
spheres. In this arrangement the spheres are arranged in a row touching one 1 2
another.
ii. In this arrangement each spheres is in contact with two spheres. The number of nearest
neighbours of a particle is called its coordination number. The coordination Number in one-
dimension close packing is 2.
2. Close packing in two dimension:

Two- dimension close packing is also called as planer arrangement. The arrangement of
identical spheres can by of two types:
A
A
i. AAA type of arrangement: The spheres of identical rows (1-
A
dimension) are placed exactly one above the other to form
A
multiple rows so that each sphere in a row is placed over another
A
sphere of another row and so on to forma planner structure. If we A
cell 1st row as A, then the upper rows is also A, A,A……….so on.
Hence this arrangement is called as AAA type of Arrangement. Each sphere in this
arrangement is in touch with 4 spheres; hence the coordination number is 4. If the centres of
these 4 nearest neighbours are joined, it forms a square. Hence type of packing is also called
as square closed packing in two-dimensions.
ii. ABAB type of arrangement: This type of arrangement is made by arranging successive rows
in such a way that the spheres of adjacent rows are placed in the
depressions of lower layer and so on. In this type of arrangement B
A
the unoccupied space is considerably reduced and the
B
arrangement is closed and firm. If we name first row arrangement A
as A and second row arrangement as B, then third row will be A B
and the fourth row will be B and so on. Hence this type of A
arrangement is called as ABAB type of arrangement.Each sphere
in this arrangement is in touch with 6 spheres; hence the coordination number is 6. If the

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 SOLID STATE

centres of these nearest neighbors are joined, it forms a hexagon; hence this type of packing is
also called as hexagonal close packing in two dimensions.

Voids (Vacant space or holes) in two dimensional arrangements:

1. In AAA type of arrangement:

i. Shape of the Voids is square. Square shaped vacant
space (void)
ii. Each sphere is surrounded by 4 square
shaped voids.
iii. Size of void is comparable with the size of the sphere.
2. In ABAB type of arrangement:
i. Shape of the voids is triangular &
hence it is called as triangular voids.
ii. Each sphere is surrounded by six Triangle shaped vacant
triangular voids. space (void)
iii. Size of the triangular void is much
smaller as compared to the size of the sphere.
iv. In one row the apex of triangle is facing upwards and in another row the apex of the
triangle is facing downwards. (The Number of triangle voids are doubled as the apex of
triangular holes is alternatively pointing upwards and downwards).
3. Close-packing in three dimensions:Three dimension close-packing structures can be obtained
by placing two dimension structures on above the other. They are of two types.
i. Three- dimensional close-packing from two dimensional square close-packed
layers:AAAA type, simple cubic structure:
a) In this arrangement, AAAA type of two dimensional
crystal layers are placed one over the other such that all
the spheres of the successive layers are exactly above the
spheres of the lower layers.
b) All the spheres of different layers of the crystal structure
are perfectlyaligned horizontally and also vertically such
that any unit cell hassimple cubic structure.
c) Since, all the layers are identical and if each layer is
labeled aslayer 4, then the whole dimensional crystal
lattice will be of AAAAtype.
d) In simple cubic structure, each sphere is surrounded by
Simple cubic lattice formed by
six spheres,4 in the layer of the sphere (same plane) and
AAA... 3D- arrangement
1 in layer above and 1below the layer. Hence the
coordination number of sphere participatingin simple
structure is six.
e) In such type of structure there is more empty space.

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 SOLID STATE

ii. Three- dimensional close-packing from two dimensional hexagonal close-packed layers:
ABAB type, Hexagonal Close Packed structure:
a. Formation of second layer:

Layer A

Layer B

Tetrahedral void Tetrahedron geometry

 Take two dimensional hexagonal close packed layer ‘A’ and pace a similar layer above it
such that the spheres of the second layer are placed in the depressions of the first layer.
 The second layer is not assigned in the same way as the first layer; let the second layer
be called as ‘B’.
 When second layer is placed over the first layer, some of the triangular voids of the first
layer are occupied by second layer spheres and some of them are not occupied.
 When a triangular void of the first layer is occupied by second layer or vice versa a
tetrahedral void is formed, since when the centres of these 4 spheres are joined it forms
tetrahedron.
b. Hexagonal close packing (HCP) :

Layer 1
Layer 3
Layer 3

tetrahedral void

Hexagonal close packed

structure in three dimension

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 SOLID STATE

 If the packing of third layer is done in such a way that third layer is placed on
thetetrahedral voids formed in the second layer are covered.
 Three dimensional closest packing structure is obtained in which the spheres of third
layer lie directly above the spheres of first layer, i.e., first and third layers are identical.
Following same placing of layers, fourth layer will be identical to second layer. If the
first layer is labeled A and second layerB, then the arrangement of pacing will be of
ABAB type. This is also called hexagonal close packing (hcp).
 In this, packing efficiency is 74 %. The coordination number of each sphere is 12. The
metals Be, Mg, Zn, Cd crystallize in HCP crystalline structure.
iii. Three- dimensional close-packing from two dimensional hexagonal close-packed layers:
ABCABC type, Cubic Close Packed structure:
a. Formation of second layer:

Layer A

Layer B

octahedron geometry

octahedral void
 Take two dimensional hexagonal close packed layer ‘A’ and pace a similar layer above it
such that the spheres of the second layer are placed in the depressions of the first layer.
 The second layer is not assigned in the same way as the first layer; let the second layer be
called as ‘B’.
 When second layer is placed over the first layer, some of the triangular voids of the first
layer are occupied by second layer spheres and some of them are not occupied.
 When a triangular void of the first layer is not occupied by second layer or vice versa an
octahedral void is formed, since when the centres of these 6 spheres are joined it forms
octahedron.
Cubic close packing (CCP) :

Layer 1

Layer 3
Layer 2

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 SOLID STATE

Geometry of packing
 If the packing of third layer is done in such a way that third layer is placed on octahedral
voids formed in the second layer.
 Three dimensional closest packing structure is obtained in which the spheres of third
layer are not aligned above the spheres of first layer, i.e., first and third layers are not
identical. Fourth layer is identical with first layer. Following same placing of layers, fifth
layer will be identical to second layer and so on. If the first layer is labeled A and second
layerB, and third layer as C, then the arrangement of packing will be of ABCABC type.
This is also called cubic close packing (ccp).
 In this, packing efficiency is 74 %. The coordination number of each sphere is 12. The
metals like Fe, Ni, Cu, Ag have ccp crystal line structure.

Figure. The diffraction of X-rays scattered by the atoms within a crystal permits the determination of the distance
between the atoms. The top image depicts constructive interference between two scattered waves and a
resultant diffracted wave of high intensity. The bottom image depicts destructive interference and a low intensity
diffracted wave.

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 SOLID STATE

IMPORTANT PRACTICE QUESTION SERIES

1. Schottky defect generally appears in

a) NaCl b) KCl c) CsCl d) All of these
2. Which arrangement of electrons leads ferromagnetism?
a) ↑ ↑ ↑ ↑ b) ↑ ↓ ↑ ↓ c) ↑ ↑ ↑ ↓ ↓ d) None of these
3. The crystal are bounded by plane faces (𝑓), straight edges (𝑒) and interfacial angel (𝑐). The
relationship between these is :
a) 𝑓 + 𝑐 = 𝑒 + 2 b) 𝑓 + 𝑒 = 𝑐 + 2 c) 𝑐 + 𝑒 = 𝑓 + 2 d) None of these
4. The melting point of RbBr is 682C, while that of NaF is 988C. The principle reason that melting
point of NaF is much higher than that of RbBr is that :
a) The two crystals are not isomorphous
b) The molar mass of NaF is smaller than that of RbBr
c) The internuclear distance 𝑟c + 𝑟a is greater for RbBr than for NaF
d) The bond is RbBr has more covalent character than the bond in NaF.
5. If a crystal lattice of a compound, each corner of a cube is enjoyed by sodium, each edge of a
cube has oxygen and centre of a cube is enjoyed by tungsten (W), then give its formula
a) Na2 WO4 b) NaWO3 c) Na3 WO3 d) Na2 WO3
6. In antifluorite structure, the negative ions:
a) Occupy tetrahedral voids
b) Occupy octahedral voids
c) Are arranged in ccp
d) Are arranged in hcp
7. An insulator oxide is :
a) CuO b) CO O c) Fe2 O3 d) All of these
8. A solid with high electrical and thermal conductivity from the following is :
a) Si b) Li c) NaCl d) ice
9. 𝑟
The radius ratio ( +) of an ionic solid (𝐴+ 𝐵− ) is 0.69. What is the coordination number of 𝐵− ?
𝑟−
a) 6 b) 8 c) 2 d) 10
10. The axial angles in triclinic crystal system are
a) 𝛼 = β = 𝛾 = 90° b) 𝛼 = 𝛾 = 90°, β ≠ 90° c) 𝛼 ≠ β ≠ 𝛾 ≠ 90° d) 𝛼 = β = 𝛾 ≠ 90°
11. In NaCl crystal each Cl ion is surrounded by
−

a) 4 Na+ ions b) 6 Na+ ions c) 1 Na+ ion d) 2 Na+ ions

12. For an ionic crystal of the general formula 𝐴+ 𝐵− and co-ordination number 6, the radius ration
will be :
a) Greater than 0.73
b) Between 0.73 and 0.41
c) Between 0.41 and 0.22
d) Less than 0.22
13. The ratio of cations to anion in a octahedral close packing is :
a) 0.414 b) 0.225 c) 0.02 d) None of these
14. Electrons in a paramagnetic compound are
a) Shared b) Unpaired c) Donated d) Paired
15. Crystals which are good conductor of electricity and heat are known as :
a) Ionic crystals b) Covalent crystals c) Metallic crystals d) Molecular crystal
16. An element has bcc structure having unit cells 12.08 × 10 . The number of atoms in these cells
23

is :

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 SOLID STATE

a) 12.08 × 1023 b) 24.16 × 1023 c) 48.38 × 1023 d) 12.08 × 1022

17. Among the following types of voids, which one is the largest void?
a) Triangular b) Cubic c) Tetrahedral d) Octahedral
18. The crystalline structure of NaCl is
a) Hexagonal close packing b) Face centred cubic
c) Square planar d) Body centred cubic
19. Metals have conductivity of the order of (ohm−1 cm−1 ) :
a) 1012 b) 108 c) 102 d) 10−6
20. Of the elements Sr, Zr, Mo, Cd and Sb, all of which are in V period, the paramagnetics are:
a) Se, Cd and Sb b) Zr, Mo and Cd c) Sr, Zr and Cd d) Zr, Mo and Sb
21. The radius ratio of CsCl is 0.93. The expected lattice structure is
a) Tetrahedral b) Square planar c) Octahedral d) Body centred cubic
22. Which one of the following defects in the crystals lowers its density?
a) Frenkel defect b) Schottky defect c) F-centres d) Interstitial defect
23. The yellow colour of ZnO and conducting nature produced in heating is due to:
a) Metal excess defects due to interstitial cation
b) Extra positive ions present in an interstitial site
c) Trapped electrons
d) All of the above
24. A metal has bcc structure and the edge length of its unit cell is 3.04 Å. The volume of the unit cell
in cm3 will be
a) 1.6 × 10−21 cm3 b) 2.81 × 10−23 cm3 c) 6.02 × 10−23 cm3 d) 6.6 × 10−24 cm3
25. The edge length of a face centred cubic cell of an ionic substance is 508 pm. If the radius of the
cation is 110 pm, the radius of the anions is
a) 288 pm b) 398 pm c) 618 pm d) 144 pm
26. An ionic compound is expected to have tetrahedral structure if 𝑟+ /𝑟− lies in the range of
a) 0.414 to 0.732 b) 0.225 to 0.414 c) 0.155 to 0.225 d) 0.732 to 1
27. The interparticle forces in solid hydrogen are :
a) Hydrogen bonds b) Covalent bonds c) Co-ordinate bonds d) Van der Waals’ forces
28. If Z is the number of atoms in the unit cell that represents the closest packing sequence
−𝐴𝐵𝐶 𝐴𝐵𝐶 −, the number of tetrahedral voids in the unit cell is equal to :
Z Z
a) Z b) 2Z c) d)
2 4
29. Quartz is an example of :
a) Chain silicate b) Infinite sheet silicate c) Framework silicate d) Cyclic silicate
30. For 𝐴𝑋 ionic crystal to exist in bcc structure, the ratio of radii ( 𝑟cation ) should be
𝑟anions
a) Between 0.41 and 0.73 b) Greater then 0.73
c) Less than 0.41 d) Equal to 1.0
31. Which crystal is expected to be soft and have low melting point?
a) Covalent b) Metallic c) Molecular d) Ionic
32. The elements commonly used for making transistors are
a) C and Si b) Ga and In c) P and As d) Si and Ge
33. Silver (atomic weight = 108 g mol ) has a density of 10.5 g cm . The number of silver atoms
−1 −3

on a surface of area 10−12 m2 can be expressed in scientific notation as 𝑦 × 10𝑥 . The value of 𝑥
is
a) 3 b) 5 c) 7 d) 9

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 SOLID STATE

34. The first order reflection (𝑛 = 1) from a crystal of the X-ray from a copper anode tube
(𝜆 = 1.54 Å)occurs at an angle of 45°. What is the distance between the set of plane causing the
diffraction?
a) 0.1089 nm b) 0.1089 m c) 0.905 Å d) 1.089 × 10−9 m
35. What is the number of tetrahedral voids per atom in a crystal?
a) 1 b) 2 c) 6 d) 8
36. Iodine is a
a) Electrovalent solid b) Atomic solid c) Molecular solid d) Covalent solid
37. In CsCl type structure the coordination number of Cs +and Cl− are
a) 6, 6 b) 6, 8 c) 8, 8 d) 8, 6
38. Structure of a mixed oxide is cubic close-packed (c.c.p). The cubic unit cell of mixed oxide is
composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal 𝐴 and
the octahedral voids are occupied by a monovalent metal 𝐵. The formula of the oxide is :
a) 𝐴𝐵 𝑂2 b) 𝐴2 𝐵𝑂2 c) 𝐴2 𝐵3 𝑂4 d) 𝐴𝐵2 𝑂2
39. The example of orthosilicate is :
a) MgCaSi2 O6 b) Mg 2 SiO4 c) Fe2 O3 SiO2 d) Ba3 Al2 Si6 O8
40. A compound CuCl has face centred cubic structure. Its density is 3.4 g cm . The length of unit cell
−3

is :
a) 5.783Å b) 6.783Å c) 7.783Å d) 8.783Å
41. The orthorhombic, the value of 𝑎, 𝑏and 𝑐are respectively 4.2 Å, 6.8𝐴 Å and 8.3 Å. Given the
molecular mass of the solute is 155 g mol−1 and that of density is 3.3g/cc, the number of formula
units per unit cell is
a) 2 b) 3 c) 4 d) 6
42. Which one of the following is a covalent crystal?
a) Rock salt b) Ice c) Quartz d) Dry ice
43. LiF is a/an :
a) Ionic crystal b) Metallic crystal c) Covalent crystal d) Molecular crystals
44. A binary solid (𝐴 𝐵 ) has a rock salt structure. If the edge length is 400 pm and radius of cation
+ −

is 75 pm the radius of anion is :

a) 100 pm b) 125 pm c) 250 pm d) 325 pm
45. The limiting radius ratio for tetrahedral shape is
a) 0 to 0.155 b) 0.255 to 0.414 c) 0.155 to 0.225 d) 0.414 to 0.732
46. A metallic element has a cubic lattice. Each edge of the unit of cell is 2Å. The density of the metal
is 2.5 g cm−3 . The unit cells in 200 g of metal are
a) 1 × 1024 b) 1 × 1020 c) 1 × 1022 d) 1 × 1025
47. Potassium has a bcc structure with nearest neighbour distance 4.52 Å. Its atomic weight is 39. Its
density will be :
a) 454 kg m−3 b) 804 kg m−3 c) 852 kg m−3 d) 910 kg m−3
48. Lithium forms body centred cube structure. The length of the side of its unit cell is 351 pm. Atomic
radius of the lithium will be :
a) 300 pm b) 240 pm c) 152 pm d) 75 pm
49. Bragg’s equation is :
a) 𝑛𝜆 = 2𝜃 sin θ b) 𝑛𝜆 = 2𝑑 sin θ c) 2𝑛𝜆 = 𝑑 sin θ d) 𝜆 = (2𝑑/𝑛) sin θ
50. The intermetallic compound LiAg has a cubic crystalline structure in which each Li atom has 8
nearest neighbor silver atoms and 𝑣𝑖𝑐𝑒 − 𝑣𝑒𝑟𝑠𝑎. What is the type of unit cell?
a) Body centred cubic
b) Face centred cubic

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 SOLID STATE

c) Simple cubic for either Li atoms alone or Ag atoms alone

d) None of the above
51. In the face centred cubic lattice, atom A occupies the corner positions and atom B occupies the
face centre positions. If one atom of B is missing from one of the face centred points, the formula
of the compound is
a) 𝐴2 𝐵 b) 𝐴𝐵2 c) 𝐴2 𝐵2 d) 𝐴2 𝐵5
52. Which compound has highest lattice energy?
a) LiBr b) LiCl c) LiI d) LiF
53. In a face centred cubic cell, an atom at the face centre is shared by :
a) 4 unit cells b) 2 unit cells c) 1 unit cell d) 6 unit cells
54. Extremely pure samples of Ge and Si are non-conductors, but their conductivity increases
suddenly on introducing ….in their crystal lattice.
a) As b) B c) Both (a) and (b) d) None of these
55. Iodine crystals are :
a) Metallic solid b) Ionic solid c) Molecular solid d) Covalent solid
56. Which of the following statements about amorphous solids is incorrect?
a) They melt over a range of temperature b) They are anisotropic
c) There is no orderly arrangement of particles d) They are rigid and incompressible
57. The number of atoms present in a simple cubic unit cell are :
a) 4 b) 3 c) 2 d) 1
58. An 𝐴𝐵2 type structure is found in :
a) NaCl b) CaF2 c) Al2 O3 d) N2 O
59. A cubic crystal possesses in all ……elements of symmetry.
a) 9 b) 13 c) 1 d) 23
60. A solid compound contains 𝑋, Y and Z atoms in a cubic lattice with X atom occupying the
corners. Y atoms in the body centred positions and Z atoms at the centres of faces of the unit
cell. What is the empirical formula of the compound?
a) 𝑋𝑌2 𝑍3 b) 𝑋𝑌𝑍3 c) 𝑋2 𝑌2 𝑍3 d) 𝑋8 𝑌𝑍6
61. The oxide which shows transition from metal to insulation, i.e., semiconductors are :
a) V2 O3 b) VO2 c) Ti2 O3 d) All of these
62. Edge length of a cube is 400 pm. Its body diagonal would be :
a) 600 pm b) 566 pm c) 693 pm d) 500 pm
63. Crystals can be classified into ……. Basic crystal habits.
a) 7 b) 4 c) 14 d) 3
64. The unit cell with crystallographic dimensions 𝑎 = 𝑏 ≠ 𝑐; 𝛼 = 𝛽 = 𝛾 = 900 is :
a) Cubic b) Tetragonal c) Monoclinic d) Hexagonal
65. The number of octahedral void(s) per atom present in a cubic close-packed structure is :
a) 2 b) 4 c) 1 d) 3
66. The hardness of metals increases with increase in number of ……involved in metallic bonding.
a) Atoms b) Molecules c) Electrons d) All of these
67. The substance which possesses zero resistance as 0 K :
a) Conductor b) Super conductor c) Insulator d) Semiconductor
68. Sodium metal crystallises at room temperature in a body centred cubic lattice with a cell edge
𝑎 = 4.29 Å. The radius of sodium atom is
a) 1.40 b) 2.65 c) 1.85 d) 2.15
69. The oxide which shows metallic conduction:
a) ReO3 b) VO c) CrO2 d) All of these
70. The number of hexagonal faces that are present in a truncated octahedron is

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 SOLID STATE

a) 2 b) 4 c) 6 d) 8
71. Which of the following statement is true?
a) Some complex metal oxides behave as b) Zinc oxide can act as superconductor
superconductor
c) An impurity of tetravalent germanium in d) A Frenkel defect is formed when an ion is
trivalent gallium creates electron deficiency displaced from its lattice site to an interstitial
site
72. Schottky defect defines imperfection in the lattice structure of a :
a) Solid b) Gas c) Liquid d) Plasma
73. When electrons are trapped into the crystal in anion vacancy, the defect is known as :
a) Schottky defect b) Frenkel defect c) Stoichiometric defect d) F-centres
74. Which of the following has highest value of energy gap?
a) Aluminum b) Silver c) Germanium d) Diamond
75. If ‘a’ stands for the edge length of the cubic systems : simple cubic, body-centred cubic and face-
centered, then the ratio of radii of the spheres in these systems will be respectively,
1 1 1 1 1 √3 1
a) 𝑎: √3 𝑎: 𝑎 b) 𝑎: √3 𝑎: √2 𝑎 c) 𝑎 ∶ √3𝑎 ∶ √2𝑎 d) 𝑎: 𝑎: 𝑎
2 √2 2 2 2 2 2 4 2√2
76. In a face centred cubic lattice the number of nearest neighbours for a given lattice point are :
a) 6 b) 8 c) 12 d) 14
77. Percentage of free space in cubic close packed structure and in body centred packed structure
are respectively
a) 30% and 26% b) 26% and 32% c) 32% and 48% d) 48% and 26%
78. Lithium borohydride crystallizes in an orthorhombic system with 4 molecule per unit cell. The
unit cell dimensions are 𝑎 = 6.8 Å, 𝑏 = 4.4 Å and 𝑐 = 7.2 Å. If the molar mass is 21.76, then the
density of crystals is :
a) 0.6708 g cm−3 b) 1.6708 g cm−3 c) 2.6708 g cm−3 d) None of these
79. Total volume of atoms present in a face centred cubic unit cell of a metal is
(r=atomic radius )
20 24 12 16
a) 𝜋𝑟 3 b) 𝜋𝑟 3 c) 𝜋𝑟 3 d) 𝜋𝑟 3
3 3 3 3
80. Which has no rotation of symmetry?
a) Hexagonal b) Orthorhombic c) Cubic d) Triclinic
81. The unit cell with dimensions 𝛼 = 𝛽 = 𝛾 = 90°, 𝑎 = 𝑏 ≠ 𝑐 is
a) Cubic b) Triclinic c) Hexagonal d) Tetragonal
82. A fcc element (atomic mass = 60) has a cell edge of 400 pm. Its density is :
a) 6.23 g cm−3 b) 6.43 g cm−3 c) 6.53 g cm−3 d) 6.63 g cm−3
83. For a crystal system 𝑎 = 𝑏 = 𝑐 and 𝛼 = 𝛽 = 𝛾 ≠ 90°
a) Tetragonal b) Hexagonal c) Rhombohedral d) Monoclinic
84. The number of atoms (𝑛) contained within a cubic cell is :
a) 1 b) 2 c) 3 d) 4
85. All the substances becomes diamagnetic at :
a) 4 K b) 10 K c) 20 K d) 25 K
86. The co-ordination number of Ca ion in fluorite crystal is :
2+

a) 2 b) 8 c) 6 d) 4
87. What is the structure of NaCl?
a) BCC b) FCC c) Interpenetrating fcc d) None of these
88. Which of the following statements is not correct?
a) Molecular solids are generally volatile

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 SOLID STATE

b) The number of carbon atoms in an unit cell of diamond is 4

c) The number of Bravais lattices in which a crystal can be categorized is 14
d) The fraction of the total volume occupied by the atoms in a primitive cell is 0.48.
89. Which is the wrong statement regarding a crystal containing Schottky defect?
a) Electrical neutrality of the crystal is maintained
b) Entropy of the crystal increases
c) The density of the overall crystal remains the same
d) The density of the overall crystal reduces
90. How many ‘nearest’ and ‘next nearest’ neighbours respectively potassium have in bcc lattice?
a) 8, 8 b) 8, 6 c) 6, 8 d) 8, 2
91. Ferrimagnetic is converted into paramagnetic at :
a) 300 K b) 400 K c) 600 K d) 850 K
92. A match box exhibits :
a) Cubic geometry
b) Monoclinic geometry
c) Orthorhombic geometry
d) Tetragonal geometry
93. The oxide that possesses electrical conductivity :
a) V2 O5 b) CrO2 c) NiO d) MnO
94. The arrangement 𝐴𝐵𝐶 𝐴𝐵𝐶……is referred to as,
a) Octahedral close packing
b) Hexagonal close packing
c) Tetrahedral close packing
d) Cubic close packing
95. The lattice points of a crystal of hydrogen iodide are occupied by
a) HI molecules b) H atoms and I atoms
c) H cations and I anions
+ −
d) H2 molecules and I2 molecules
96. 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
97. The radius of the Na+ is 95 pm and that of Cl− ion is 181 pm. Predict the co-ordination number of
Na+ :
a) 4 b) 6 c) 8 d) Unpredictable
98. How many unit cells are present in a cube shaped ideal crystal of NaCl of mass 1.00g?
[Atomic masses ∶ Na = 23, Cl = 35.5]
a) 2.57 × 1021 b) 5.14 × 1021 c) 1.28 × 1021 d) 1.71 × 1021
99. For a covalent solid, the units which occupy lattice points are :
a) Atoms b) Ions c) Molecules d) Electrons
100.The metal surfaces are excellent reflectors because of absorption and re-emission of light by :
a) Protons in atom b) Electrons in atom c) Neutrons in atom d) None of these
101.The fraction of total volume occupies by the atoms present in a simple cube is :
𝜋 𝜋 𝜋 𝜋
a) b) c) d)
3√2 4√2 4 6
102.If we mix a pentavalent impurity in a crystal lattice of germanium, what type of semiconductor
formation will occur?
a) 𝑝 −type b) 𝑛 −type c) Both (a) and (b) d) None of the two
103.A metal crystallizes with a face-centered cubic lattice. The edge of the unit cell is 408 pm. The
diameter of the metal atom is :

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 SOLID STATE

a) 144 pm b) 204 pm c) 288 pm d) 408 pm

104.Metallic crystalline solids :
a) Have low melting point and boiling point
b) Are bad conductors
c) Are good conductors of heat and electricity
d) Only conduct heat
105.Most crystals show good cleavage because their atoms, ions and molecules are:
a) Weakly bonded together
b) Strongly bonded together
c) Spherically symmetrical
d) Arranged in planes
106.The structure of MgO is similar to NaCl. The co-ordination number of Mg is :
a) 2 b) 6 c) 4 d) 8
107.If NaCl is dopped with 10−4 mole % of SrCl2 the concentration of cation vacancies will be:
a) 6.02 x 1016 mol−1 b) 6.02 x 1017 mol−1 c) 6.02 x 1014 mol−1 d) 6.02 x 1015 mol−1
108.What type of crystal defect is indicated in the diagram below?

a) Frenkel defect b) Schottky defect

c) Interstitial defect d) Frenkel and Schottky defects
109.An ion leaves its regular site occupy a position in the space between the lattice sites is called
a) Frenkel defect b) Schottky defect c) Impurity defect d) Vacancy defect
110.Schottky defects occurs mainly in electrovalent compounds where
a) Positive ions and negative ions are of different size
b) Positive ions and negative ions are of same size
c) Positive ions are small and negative ions are big
d) Positive ions are big and negative ions are small
111.Sodium metal crystallizes in a body centred cubic lattice with the cell edge 𝑎 = 4.29 Å. The radius
of sodium atom is :
a) 1.8574 Å b) 2.8574 Å c) 3.8574 Å d) None of these
112.The cation-anion bond have the largest amount of covalent character for:
a) NaBr b) SrS c) CdS d) BaO
113.In a cubic close packing of spheres in three dimensions, the co-ordination number of each sphere
is :
a) 6 b) 9 c) 3 d) 12
114.In a cubic structure of diamond which is made from 𝑋and Y, where X atoms are at the corners of
the cube and Y at the face centres of the cube. The molecular formula of the compound is
a) 𝑋2 𝑌 b) 𝑋3 𝑌 c) 𝑋𝑌2 d) 𝑋𝑌3
115.Which of the following statements is not correct?
a) The units of surface tension are dynes Cm−1
b) The units of viscosity coefficient of a liquid are ‘poise ‘
c) CsCl crystallizes in body centred cubic type of lattice
d) The coordination number of S 2− in ZnS is 6

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 SOLID STATE

116.The ability of a given substance to assume two or more crystalline structure is called
a) Amorphism b) Isomorphism c) Polymorphism d) Isomerism
117.With which one of the following element silicon should be doped so as to give 𝑝-type
semiconductor?
a) As b) Se c) B d) Ge
118.If the radius of K and F are 133 pm and 136 pm respectively, the distance between
+ −

K + and F − in KF is
a) 269 pm b) 134.5 pm c) 136 pm d) 3 pm
119.Copper crystallises in fcc with a unit cell length of 361 pm. What is the radius of copper atom?
a) 108 pm b) 127 pm c) 157 pm d) 181 pm
120.Which species is paramagnetic?
a) NO b) Fe3+ c) Fe2+ d) All are correct
121.Density of a crystal remains unchanged as a result of
a) Ionic defect b) Schottky defect c) Frenkel defect d) Crystal defect
122.A metallic element crystallises into lattice containing a sequence of layers of 𝐴𝐵𝐴𝐵𝐴𝐵𝐴𝐵……. Any
packing of spheres leaves out void in the lattice. The empty space in percentage by volume in this
lattice is :
a) 26% b) 32% c) 20% d) 30%
123.For a solid with the following structure, the co-ordination number of the point 𝐵 is :

a) 3 b) 4 c) 5 d) 6
124.The phenomenon in which crystals on subjecting to a pressure or mechanical stress produce
electricity is called :
a) Pyro-electricity b) Piezo-electric effect c) Ferro-electricity d) Ferri-electricity
125.Which arrangement of electron decides ferrimagnetism?
a) ↑ ↑ ↑ ↑ ↑ b) ↑ ↓ ↑ ↓ c) ↑ ↑ ↑ ↓ ↓ d) None of these
126.The 8 : 8 type of packing is present in
a) MgF2 b) CsCl c) KCl d) NaCl
127.Which is not the correct statement for ionic solids in which positive and negative ions are held
by strong electrostatic attractive forces?
a) The radius 𝑟 + /𝑟 − increases as coordination number increases
b) As the difference in size of ions increases, coordination number increases
c) When coordination number is eight, the 𝑟 + /𝑟 − ratio lies between 0.225 to 0.414
In ionic solid of the type 𝐴𝑋(ZnS, Wurtzite), the coordination number of Zn2+ andS 2−
d)
respectively are 4 and 4
128.Which set of characteristics of ZnS crystal is correct?
a) Coordination number (4 : 4): ccp; Zn2+ ion in the alternate tetrahedral voids
b) Coordination number (6 : 6); hcp; Zn2+ ion in all tetrahedral voids
c) Coordination number (6 : 4); hcp; Zn2+ ion in all octahedral voids
d) Coordination number (4 : 4); ccp; Zn2+ ion in all tetrahedral voids
129.NaCl structure consists of :
a) Na and Cl atoms

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 SOLID STATE

b) Na+ and Cl atoms

c) Na atoms and Cl− ions
d) Na+ and Cl− ions
130.A solid metal has ccp or fcc structure. The relation of side of cube (𝑎)and radius of atom (𝑟) will
be
4 3
a) 𝑎 = 2𝑟 b) 𝑎 = 2√2𝑟 c) 𝑎 = 𝑟 d) 𝑎 = √ 𝑟
√3 2
131.𝐴 solid 𝑋 melts slightly above 273 K and is a poor conductor of heat and electricity. To which of
the following categories does it belong?
a) Ionic solid b) Covalent solid c) Metallic d) Molecular
132.Lubricating properties of graphite are diminished in presence of :
a) High pressure b) Low pressure c) Vacuum d) None of these
133.Lithium metal crystallises in a body centred cubic crystal. If the length of the side of the unit cell
of lithium is 351 pm, the atomic radius of the lithium will be :
a) 300.5 pm b) 240.8 pm c) 151.8 pm d) 75.5 pm
134.Close packing is maximum in the crystal lattice of :
a) Simple cubic b) Face centred c) Body centred d) None of these
135.The radii of Na and Cl ions are 95 pm and 181 pm respectively. The edge length of NaCl unit
+ −

cell is
a) 276 pm b) 138 pm c) 552 pm d) 415 pm
136.The ionic radii of Rb and I are 1.46 Å and 2.16 Å. The most probable type of structure
+ −

exhibited by it is
a) CsCl type b) ZnS type c) NaCl type d) CaF2 type
137.Which one is diamagnetic?
a) ClO3 b) Cu2+ c) F − d) Ni2+
138.The statement that “All crystals of the same substance possess the same elements of symmetry”
is known as :
a) Hauy’s law of rationality of indices
b) The law of constancy of interfacial angles
c) The law of constancy of symmetry
d) None of the above
139.A solid 𝐴𝐵 has NaCl type structure with edge length 580.4 pm. The radius of 𝐴+ is 100 pm. What
is the radius of 𝐵− ?
a) 190.2 b) 540.13 c) 525 d) 78.12
140.In a face centred cubic arrangement off 𝐴 and 𝐵 atoms whose 𝐴 atoms are at the corner of the unit
cell and 𝐵 atoms at the face centres. One of the 𝐴 atom is missing from one corner in unit cell. The
simplest formula of compound is :
a) 𝐴7 𝐵3 b) 𝐴𝐵3 c) 𝐴7 𝐵24 d) 𝐴7/8 𝐵3
141.Which one of the following is a covalent crystal?
a) Rock salt b) Ice c) Quartz d) Dry ice
142.The coordination number of Al in the crystalline state of AlCl3 is
a) 2 b) 4 c) 6 d) 8
143.In crystal structure of rock salt (NaCl), the arrangement of Cl ion is :
a) Fcc b) Bcc c) Both (a) and (b) d) None of these
144.In which of the following crystals alternate tetrahedral voids are occupied?
a) NaCl b) Zns c) CaF2 d) Na2 O

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 SOLID STATE

145.A compound of ‘A’ and ‘B’ crystallises in a cubic lattice in which ‘A’ atoms occupy the lattice
points at the corners of the cube. The ‘B’ atoms occupy the centre of each face of the cube. The
probable empirical formula of the compound is
a) 𝐴𝐵2 b) 𝐴3 𝐵 c) 𝐴𝐵 d) 𝐴𝐵3
146.Amorphous solids:
a) Possess sharp melting points
b) Undergo clean cleavage when cut with knife
c) Do not undergo clean cleavage when cut with knife
d) Possess orderly arrangement over long distances
147.For which crystal anion-anion contact is valid?
a) NaF b) NaI c) CsBr d) KCl
148.The crystal system of a compound with unit cell dimensions 𝑎 = 0.387, 𝑏 = 0.387, and 𝑐 =
0.504 nm and 𝛼 = 𝛽 = 90 and 𝛾 = 120 is ∶
a) Cubic b) Hexagonal c) Orthorhombic d) Rhombohedral
149.Possible number of different type of crystal lattice present in all types of crystals, is
a) 23 b) 7 c) 230 d) 14
150.Doping of silicon (Si) with boron (B) leads to
a) 𝑛 −type semiconductor b) 𝑝 −type semiconductor
c) Metal d) Insulator

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 SOLID STATE

1) d 2) a 3) a 4) c
5) b 6) c 7) d 8) b
9) a 10) c 11) b 12) b
13) a 14) b 15) c 16) b
17) d 18) b 19) b 20) d
21) d 22) b 23) d 24) b
25) d 26) b 27) d 28) b
29) c 30) b 31) c 32) d
33) c 34) c 35) b 36) c
37) c 38) d 39) b 40) a
41) c 42) c 43) a 44) b
45) b 46) d 47) d 48) c
49) b 50) a 51) d 52) d
53) b 54) c 55) c 56) b
57) d 58) b 59) d 60) b
61) d 62) c 63) a 64) b
65) c 66) c 67) b 68) c
69) d 70) d 71) d 72) a
73) d 74) d 75) d 76) c
77) b 78) a 79) d 80) d
81) d 82) a 83) c 84) a
85) a 86) b 87) b 88) b
89) c 90) b 91) d 92) c
93) b 94) d 95) a 96) b
97) b 98) a 99) a 100) b
101) d 102) b 103) c 104) c
105) d 106) b 107) b 108) b
109) a 110) b 111) a 112) c
113) d 114) d 115) d 116) c
117) c 118) a 119) b 120) d
121) c 122) b 123) d 124) b
125) c 126) b 127) c 128) a
129) d 130) b 131) d 132) c
133) c 134) b 135) c 136) c
137) c 138) c 139) a 140) c
141) b 142) c 143) a 144) b
145) d 146) c 147) a 148) b
149) d 150) b

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 SOLID STATE

1 (d)
Schottky defect arises when equal number of a cations and anions are missing from their
sites. This defect is generally found in ionic compounds like NaCl, KCl, CsCl, etc.
2 (a)
Ferromagnetism is due to spontaneous alignment of the magnetic dipoles in same direction.
3 (a)
𝑓 + 𝑠𝑐 = 𝑒 + 2; where 𝑓 is plane faces, 𝑐 is interfacial angle and 𝑒 is straight edges.
4 (c)
This leads to stronger coulombic forces of attractions in NaF.
5 (b)
1
No. of Na atoms present at each corner = 8 × = 1
8
1
No. of O atoms present at the centre of edges = 12 × 4 = 3
No. of W atoms present at the centre of cube = 1
Formula of the compound = NaWO3
6 (c)
In antifluorite crystal (Na2 O) the anions are arranged in cubic close packing while the
cations occupy all the tetrahedral voids.
7 (d)
All are insulator
8 (b)
In the given choices lithium has high thermal and electrical conductance.
9 (a)
Relation between radius ratio and coordination number
𝒓𝒄
Coordination
𝒓𝒂
number
0.155 − 0.225 3
0.225 − 0.414 4
0.414 − 0.732 6
0.732 − 1 8
10 (c)
The axial angles in triclinic crystal system are different and none is perpendicular to any of
the others 𝑖. 𝑒., 𝛼 ≠ β ≠ 𝛾 ≠ 90°.
11 (b)
In NaCl crystal, Cl− ions adopt cubic close packed arrangement and Na+ ions occupy all the
octahedral sites. Therefore, Na and Cl have 1 : 1 stoichiometry. In other words, each Na+ ion
is surrounded by six Cl− ions which are disposed towards the corners of a regular
octahedron. Similarly, each Cl− ion is surrounded by six Na+ ions.
12 (b)
The radius ratio for co-ordination and has 4, 6, and 8 lies in between the ranges
[0.225 − 0.414], [0.414 − 0.732] and [0.732 − 1] respectively.
13 (a)
𝑟+ 𝑟+
𝑟−
for octahedral void = 0.414; 𝑟− for cubic = 0.732 − 1

15 (c)
Metallic crystals are good conductor of heat and current due to free electrons on them.
16 (b)
One unit cell of bcc has atoms = 2. Hence 12.08 × 1023 unit cells will have atoms

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 SOLID STATE

= 2 × 12.08 × 1023
= 24.16 × 1023
17 (d)
The vacant spaces between the spheres in closed packed structures is called void. The voids
are of two types, tetrahedral voids and octahedral voids. Also, radius of tetrahedral voids
and octahedral voids are 𝑟void = 0.225 × 𝑟sphere and 𝑟void = 0.414 × 𝑟sphere respectively.
Thus, octahedral void is larger than tetragonal void.
18 (b)
Sodium chloride (NaCl) has face centred cubic structure. It contains 4 Na+ and 4 Cl− in the
unit cell. Each Na+ is surrounded by 6 Cl− ions and 𝑣𝑖𝑐𝑒 − 𝑣𝑒𝑟𝑠𝑎.
19 (b)
The conductance order of metals is 106 to 108 ohm−1 cm−1
20 (d)
Each possess unpaired electrons.
21 (d)
The radius ratio of CsCl is 0.93 hence, its structure is body centred cubic.
22 (b)
Schottky defects - This defect is due to vacancy at a cation site accompanied by vacancy at
an anion site so that the electrical neutrality of the system is maintained. Due to this defect,
density decreases.
23 (d)
These are characteristics of metal excess defects due to interstitial cation.
24 (b)
Edge length 𝛼 = 3.04 Å
= 3.04 × 10−8 cm
Volume of bcc (cubic) cell = 𝑎3
= (3.04 × 10−8 )3
= 2.81 × 10−23 cm3
25 (d)
For fcc arrangement
2(𝑟 + + 𝑟 − ) = edge length
2(110 + 𝑟 − ) = 508
So, 𝑟 − = 114 pm
26 (b)
Radius ratio(𝒓+ / Structure
𝒓− )
< 0.155 linear
0.155 − 0.225 planar
triangular
0.225 − 0.414 tetrahedral
0.414 − 0732 octahedral
0.732 − 1 bcc
27 (d)
Solid hydrogen involves van der Waals’ forces.
28 (b)
In ccp or fcc and hcp, number of tetrahedral voids is double the number of atoms forming
the main lattice.
29 (c)

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 SOLID STATE

Quartz is a covalent crystal having a framework of silicates, i.e., a three dimensional

network when all the four oxygen atoms of each of SiO4 tetrahedron are shared.
30 (b)
For body centred cubic (bcc) structure, the ratio of radii (𝑟+ /𝑟−) lies in between
0.732−1.00.
∴ The ratio of radii for bcc is greater than 0.73.
31 (c)
Follow characteristics of molecular solids.
32 (d)
Si and Ge are used for making transistors.
33 (c)
108
Volume of one mole of silver atoms = 10.5 cm3 /mol
108 1
Volume of one silver atom = 10.5 × 6.022 ×1023 cm3
4 108 1
So, 𝜋𝑟 3 = × = 1.708 × 10−23
3 10.5 6.022 ×1023
𝑟 3 = 0.407 × 10 cm = 0.407 × 10−29 m3
−23 3

Area of each silver atom,

𝜋𝑟 2 = 𝜋(0.407 × 10−29 m3 )2/3
So, number of silver atoms in given area
10−12 108
= =
(0.407 × 10−29 m3 )2/3 𝜋 × 2
= 1.6 × 107 = 𝑦 × 10𝑥
So, 𝑥 = 7
34 (c)
𝑛𝜆 = 2𝑑 sin θ
1 × 1.54 = 2𝑑 sin 45°
1 × 1.54 = 2𝑑 × 0.850
1.54
2𝑑 = = 0.905 Å
0.850
35 (b)
In the close packing of ‘𝑛’ atoms, the number of tetrahedral voids are ‘2𝑛’. Hence, their
number per atom is 2.
37 (c)
The coordination number is 8 : 8 in Cs + ∶ Cl−
The coordination number is 6 : 6 in Na+ ∶ Cl−
38 (d)
In a cubic close packing, the number of octahedral voids is equal to number of atoms and
number of tetrahedral voids is equal to the twice the number of atoms
Number of atoms is a ccp array = 1
∴ 𝐴2+ 𝐵+ 𝑂2−
1
1×2× 4 1 1
1
1 1
2
𝑜𝑟 1 2 2
𝐴𝐵2 𝑂2
39 (b)
In orthosilicate SiO2−
4 ion exist as discrete unit.
40 (a)

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 SOLID STATE

Molecular mass of CuCl = 99

𝑛 = 4 for face centred cubic cell
𝑛 × mol. wt.
∵ 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 =
𝑉 × av. no.
4 × 99
= 3
𝑎 × 6.023 × 1023
4 × 99
Or 3.4 = 3 23
𝑎 × 6.023 × 10
−8
∴ 𝑎 = 5.783 × 10 cm
= 5.783Å
41 (c)
𝑉×𝑁×𝑑
𝑍=
𝑚
4.2 × 8.6 × 8.3 × 10−21 × 6.023 × 1023 × 3.3
=
155
= 3.14
≈4
42 (c)
Quartz (SiO2 ) is a covalent crystal.
43 (a)
LiF is an ionic crystal. An ionic solid has ions as constituent units at lattice points held by
oppositely charged ions.
44 (b)
Edge = 2𝑟 + + 2𝑟 −
∴ 400 = 2 × 75 + 2𝑟 −
∴ 𝑟 − = 125 pm
45 (b)
For tetrahedral shape, limiting radius ratio is 0.225 − 0.414.
46 (d)
mass of metal
Number of unit cells = mass of one unit cell
Given, edge length of unit cell = 2Å = 2 × 10−8 cm
Mass of metal = 200 g
Density of metal = 2.5 g cm−3
Volume of unit cell = (edge length)3 = (2 × 10−8 )3
= 8 × 10−24 cm3
Mass of one unit cell = volume × density
= 8 × 10−24 × 2.5
= 20 × 10−24
mass of metal
∴ No. of unit cells in 200 g metal = mass of one unit cell
200
=
20 × 10−24
= 10 × 1024 = 1.0 × 1025
47 (d)
√3
For bcc, 𝑟 = 2
=𝑎
2𝑟 2 × 4.52
Or 𝑎 = =
√3 1.732
= 5.219 𝐴̊ = 522 pm.

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 SOLID STATE

𝑛×𝑀
Density =
𝑎 3 × 𝑁𝐴 × 10−30
2 × 39
=
(522)3
× (6.02 × 1023 ) × 10−30
= 0.91g/cm3 = 910 kg m−3
48 (c)
For bcc structure
49 (b)
Bragg’s equation is 𝑛𝜆 = 2𝑑 sin θ
50 (a)
The bcc structure has co-ordination no. of eight.
51 (d)
1
Number of atoms (𝐴) per unit cell = 8 × 8 = 1
1 5
Number of atoms (𝐵) per unit cell = (6 − 1) × 2 = 2
(One atom B is missing)

Thus, formula is 𝐴1 𝐵5/2 = 𝐴2 𝐵5

52 (d)
Due to small anion, it possess maximum ionic nature.
53 (b)
The fcc unit cell has 8 atoms at the eight corners and one atom at each of six faces. The atom
at the face is shared by two unit cells.
54 (c)
Doping of elements of group 14 (Ge and Si) with group 15 (As) elements produces excess of
electrons and shows 𝑛 -type conduction, the symbol 𝑛 indicating flow of negative charge in
them. Doping of elements of group 14 (Ge and Si) with group 13 (B) elements products hole
(electron deficiency) in the crystal and shows 𝑝-type conduction, the symbol 𝑝 indicating
flow of positive charge.
55 (c)
Molecular solids are the substances having molecules as constituent units having
interparticle forces such as van der waal’s forces or hydrogen bonds.
57 (d)
The number of atoms present in sc, fcc and bcc unit cell are 1, 4, 2 respectively.
58 (b)
N2 O is gas; CaF2 is 𝐴𝐵2 type crystalline solid.
59 (d)
These are characteristic elements of symmetry of a cubic crystal.
60 (b)
Since atom 𝑋 is present at corner and one corner is shared by eight unit cells,
1
Number of X atoms per unit cell = × 8 = 1
8
Atom 𝑌 is present at body centred position and used by only one unit cell. So, number of Y
atoms per unit cell = 1

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 SOLID STATE

Atom Z is present at the center of each face, so shared by two unit cells,
1
Thus, number of Z atoms per unit cell = 2 × 6 = 3
Hence, the formula of compound = 𝑋𝑌𝑍3
61 (d)
The transition of metal to insulation occurs at a certain temperature due to imperfection.
62 (c)
Body diagonal in bcc = √3 𝑎 = √3 × 400 = 692.8 pm
63 (a)
The seven basic crystal lattice are cubic, tetragonal, orthorhombic, monoclinic, hexagonal,
rhombohedral and triclinic.
64 (b)
The conditions for tetragonal systems.
65 (c)
The number of octahedral voids in cubic close packed = 4
The number of atoms per unit cell in ccp = 4
The number of octahedral voids per atom = 1
66 (c)
An increase in charge of +ve ions also brings in an increase in number of electrons involved
in metallic crystals, and thereby metallic bonding becomes stronger.
67 (b)
Electrical resistance of metals decreases with decrease in temperature and becomes zero at
zero kelvin. Materials in this state are called super conductors and the phenomenon as
super conductivity.
68 (c)
For a body centred cubic lattice radius, (𝑟)
√3
= 𝑎 = 0.433𝑎
4
Therefore, radius of Na+ = 0.433 × 4.29 = 1.8575
69 (d)
All are conductors however shows insulation at a certain temperature.
70 (d)
The truncated octahedron is the 14-faced Archimedean solid, with 14 total faces : 6 squares
and 8 regular hexagons.
The truncated octahedron is formed by removing the six right square pyramids one from
each point of a regular octahedron as :

Truncated Octachedron

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 SOLID STATE

71 (d)
Frenkel defect is formed by displacement of ion from its lattice to interstitial state.
72 (a)
Inperfections are notice in solids.
73 (d)
Trapping of electrons in anion vacancies develop F-centres.
74 (d)
Diamond has the highest value of energy gap as it is a insulator.
75 (d)
a a √3
sc: r = fcc ∶ r = ; bcc ∶ r = a
2 2√2 4
a √3 a
∴ sc; bcc and fcc are , a,
2 4 2√2
76 (c)
Number of sodium ions are 12 at edge centres in fcc structure which are nearest neighbours
for a given lattice point.
77 (b)
𝜋
Packing fraction of ccp = = 0.74 ⇒ 74%
3√2
% free space in ccp = 26%
𝜋√3
Packing fraction of bcc = 8
= 0.68 ⇒ 68%
% free space in bcc = 32%
78 (a)
𝑛 × mol.wt.
Density
𝑉 × av.no.
𝑛 = 4, 𝑀 = 21.76, av. no. = 6.023 × 1023 and
And 𝑉=𝑎 × 𝑏 × 𝑐
∴ 𝑉 = 6.8 × 10−8 × 4.4 × 10−8 × 7.2 × 10−8
= 2.154 × 10−22 × 6.023 × 1023
4 × 21.76
Density = 2.154 × 10−22 × 6.023 × 1023
= 0.6708 g cm−3
79 (d)
4
Volume of an atom = 3 𝜋𝑟 3
In fcc, number of atoms per unit cell = 4
4
∴ Volume of total atoms = 4 × 3 𝜋𝑟 3
16 2
= 𝜋𝑟
3
80 (d)

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 SOLID STATE

In triclinic lattice, the eight lattice points are located, one each at the corners of triclinic
lattice. Also 𝑎 ≠ 𝑏 ≠ 𝑐 and 𝛼 ≠ β ≠ 𝛾. There is no planes and no axes. Thus, triclinic lattice
has no rotation of symmetry.
81 (d)
The unit cell with dimensions 𝑎 = 𝑏 ≠ 𝑐, 𝛼 = β = 𝛾 = 90 is tetragonal.
82 (a)
𝑛 × at.wt. 𝑛 × at.wt.
Density = =
𝑉 × av.no. 𝑎 3 × av.no.
Given, at.wt. = 60
𝑎 = 4 × 102 pm = 4 × 102 × 10−12 m
= 4 × 10−10 × 102 cm = 4 × 108 cm
(∵ 1 pm = 10−12 )
4 × 60
∴ Density = (4 × 10−8 ) × 6.023 × 1023
−3
= 6.23 g cm
83 (c)
Crystal system Axial Axial angle
distances
Tetragonal 𝑎=𝑏 𝛼=β=𝛾
≠𝑐 = 90°
Hexagonal 𝑎=𝑏 𝛼≠β
≠𝑐 = 90°, 𝛾
= 120°
Rhombohedral 𝑎=𝑏 𝛼=β=𝛾
=𝑐 ≠ 90°
Monoclinic 𝑎≠𝑏 𝛼=𝛾
≠𝑐 = 90, β
≠ 90°
84 (a)
The cubic unit cell has 8 atmos at eight corners. Each atom is shared by 8 unit cells.
1
∴ 𝑛=8 × =1
8
85 (a)
Most of the metals have their transition temperature (i.e., the temperature at which a
substance starts to behave as super conductor) in the range of 2-5 K.
86 (b)
CaF2 has fcc structure with 8 : 4 co-ordination and has 4 units of CaF2 per unit cell.
87 (b)
NaCl has fcc arrangement of ions. The coordination number of Cl− as well as Na+ ion is six.
Therefore, it is termed 6 : 6 coordination crystal.
88 (b)
No. of carbon atoms in unit cell of diamond is 8. Also fraction of volume occupied by the
atoms in primitive cell is 52%.
89 (c)
When equal number of cations and anions are missing from their position in a crystal lattice
so that electrical neutrality is maintained, the defect is called Schottky defect. Due to
missing of ions, the overall density of the crystal decreases. Moreover, defect leads to
randomness, thus entropy also increases.
90 (b)
It is a fact for crystal structure (bcc) potassium.
91 (d)
At high temperature randomization of spins changes.

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 SOLID STATE

92 (c)
Orthorhombic geometry has 𝑎 ≠ 𝑏 ≠ 𝑐 and 𝛼 = 𝛽 = 𝛾 = 90. The shape of match box obey
this geometry.
93 (b)
CrO2 is metallic conductor, V2 O5 , NiO and MnO are insulators.
94 (d)
It represents ccp arrangement.
96 (b)
Given,
Molar mass, 𝑀 =50g/mol
𝑁𝐴 = 6.02 × 1023
𝑍 = 2 (for bcc crystal)
Edge length 𝑎 = 300 pm
= 3 × 10−8 cm
𝑍×𝑀
𝑑=
𝑁𝐴 × 𝑎3
2 × 50
=
6.02 × 10 × (3 × 10−8 )3
23

= 6.15
≈ 6.2
97 (b)
𝑟Na+ 95
= = 0.524, i.e., in between 0.414 to 0.732 and thus, co-ordination no.=6
𝑟Cl− 181
98 (a)
Mass of one unit-cell (𝑚)
= volume × density
𝑀𝑍 𝑀𝑍
= 𝑎3 × 𝑑 = 𝑎3 × 3
=
𝑁0 𝑎 𝑁0
58.5 × 4
𝑚= g
6.02 × 1023
1
∴ Number of unit cells in 1 g = 𝑚
6.02 × 1023
=
58.5 × 4
= 2.57 × 1021
99 (a)
In covalent molecules atoms occupy the lattice points.
100 (b)
The presence of free electrons in metals, they are opaque, strongly reflecting and possess
metallic lustre.
101 (d)
Volume of cube = 𝑎3
4
Volume of unit cell = 1 × 3 πr 3
4 a 3 𝜋𝑎3
= 𝜋 ( ) =
3 2 6
𝜋𝑎 3 𝜋
∴ packing density = =
6 × 𝑎3 6
102 (b)

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 SOLID STATE

On adding a pentavalent impurity with germanium, we get 𝑛 −type of semiconductors

because excess of electrons is responsible for conduction.
103 (c)
𝑎
For fcc structure 𝑟 = 2 2
√
𝑎 408
∴ diameter = 2𝑟 = = = 288.5 pm
√2 1.414
104 (c)
It is a fact.
105 (d)
Due to different plane arrangement, cleavage becomes easier at these points.
106 (b)
Na has 6 co-ordination number (fcc structure).
107 (b)
Dopping of SrCl2 to NaCl brings in replacement of two Na+ by each Sr 2+ ion, but Sr 2+
occupies one lattice point. This produces one cation vacancy.
No. of cation vacancies = 10−4
100 mole of NaCl will have cationic vacancy = 10−4
∴ 1 mole of NaCl will have cationic vacancy = 10−4 /100 = 10−6
∴ No. of cationic vacancies = 10−6 × 6.02 × 1023 = 6.02 × 1017
108 (b)
When equal number of cations and anions (such, that charges are equal) are missing
(1 Na+ , 1 Cl− /1 Fe2+ , 2 Cl− ).
It is a case of Schottky defect.
109 (a)
Frenkel defects arises when an ion is missing from its normal position and occupies an
interstitial site between the lattice points.
110 (b)
When equal number of cations or anions are missing from their lattice sites (to maintain
electrical neutrality), then the defect is called Schottky defect. The defect is observed in
highly ionic compounds which have cations and anions of similar size 𝑒. 𝑔., NaCl, KCl etc.
111 (a)
√3
Radius of Na(if bcc lattice) = 4
𝑎
√3 x 4.29
=
4
= 1.8574 Å
112 (c)
More is deformation in anion more is covalent character.
113 (d)
In hexagonal close packing and cubic close packing, the co-ordination number is 12.
114 (d)
1
Number of atoms at corner = 8 × = 1
8
1
Number of atoms at face centres = 6 × 2 = 3
∴ The formula of the compound is 𝑋𝑌3 .
115 (d)
Zinc blende (ZnS) has ccp arrangement of S 2− and Zn2+ in alternative tetrahedral sites. The
coordination number of Zn2+ = 4 and S 2− = 4 in ZnS

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 SOLID STATE

116 (c)
The phenomenon by which a certain crystalline compound exists in two or more different
crystalline forms, is called polymorphism e.g., CaCO3 occurs in two polymorphic forms, 𝑖. 𝑒.,
calcite (rhombohedral) and aragonite (orthorhombic).
117 (c)
Ge and Si are doped with gp 13(boron) element to give p-type conductor.
118 (a)
Distance between K + and F − in KF
= 𝑟K+ + 𝑟F− = 133 + 136 = 269 pm
119 (b)
In fcc unit cell
√2 𝑎
√2𝑎 = 4𝑟 ⇒ 𝑟 =
4
√2 × 361
= = 127 pm
4
120 (d)
Each possess unpaired electrons.
121 (c)
Due to Frenkel defect, density of a crystal remains unchanged.
122 (b)
𝐴𝐵𝐴𝐵𝐴𝐵…..packing has empty space of 28% in sc, 32% in bcc, 26% in hcp and ccp.
123 (d)
It is evident from figure that 𝐵 occupies tetrahedral voids and thus, co-ordination number is
six.
124 (b)
It is the definition of piezo-electric effect or piezo-electricity.
125 (c)
Ferrimagnetism involves magnetic dipoles oriented in parallel and antiparallel direction in
unequal number to give some net dipole moment.
126 (b)
The 8:8 type of packing is present in caesium chloride (CsCl). In this structure each Cs + ion
is surrounded by 8 Cl− ions and each Cl− ion is also surrounded by 8 Cs + ions.
127 (c)
𝑟+
When coordination number is eight, the radius ratio 𝑟 − lies between 0.732 to 1.000.
128 (a)
ZnS has zinc blende type structure (𝑖. 𝑒., ccp structure). The S 2− ions are present at the
corners of the cube and at the centre of each face. Zinc ions occupy half of the tetrahedral
sites. Each zinc ion is surrounded by four sulphide ions which are disposed towards the
corner of regular tetrahedron. Similarly, S 2− ion is surrounded by four Zn2+ ions.
129 (d)
NaCl has Na+ and Cl− ions in solid state.
130 (b)
In case of ccp or fcc structure
4𝑟
4𝑟 = √2𝑎 ⇒ 𝑎 =
√2
∴ 𝑎 = 2√2𝑟
131 (d)

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 SOLID STATE

Molecular solids just melt above 273 and are poor conductor of heat and electricity.
132 (c)
In vacuum, there is no friction.
133 (c)
√3 √3
In bcc 𝑟 = 4
𝑎 = 4
× 351 = 151.98 pm
134 (b)
The maximum packing or the maximum proportion of volume filled by hard spheres in
various arrangements are :
𝜋
1. Simple cubic = 6 = 0.52

𝜋√3
2. bcc = 8
= 0.68

𝜋√2
3. fcc = 6
= 0.74

𝜋√2
4. hcp = = 0.74
6

𝜋√3
5. Diamond = = 0.34
6

135 (c)
NaCl has fcc structure.
In fcc lattice
𝑎
𝑟+ + 𝑟− =
2
Where, 𝑎 = edge length
𝑟 + = 95 pm, 𝑟 − = 181 pm
Edge length = 2𝑟 + + 2𝑟 −
= (2 × 95 + 2 × 181) pm
= 190 + 362 = 552 pm
136 (c)
Radius ratio Coord Examp
inatio le
n no
0.155 − 0.225 3 B2 O3
0.225 – 0.414 4 ZnS
0.414 – 0.732 6 NaCl
0.732 – 1 8 CsCl
In ionic solids the shape of crystal depends upon relative size of ions.
Given, 𝑟𝑐 + (Rb+ ) = 1.46 Å
𝑟𝑎− (I− ) = 2.16 Å
𝑟𝑐 + 1.46
∴ = = 0.676
𝑟𝑎− 2.16
∴ It will have coordination number 6 and structure will be same as of NaCl.
137 (c)
F − has no unpaired electron and thus, diamagnetic. A diamagnetic does not contain any
unpaired electron.
138 (c)
This is the law of constancy of symmetry.

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 SOLID STATE

139 (a)
NaCl has fcc structure and thus,
𝑎
𝑟𝑐 + 𝑟𝑎 =
2
580.4
100 + 𝑟𝑎 =
2
= 290.2
100 + 𝑟𝑎 = 290.2
𝑟𝑎 = 290.2 – 100
= 190.2
140 (c)
1
No. of atoms of 𝐴 from corners of unit cell = 7 × = 7/8
8
1
No. of atoms of 𝐵 from faces of unit cell = 6 × 2 = 3
Thus, 𝐴 ∶ 𝐵 ∶ ∶ 7/8 ∶ 3 or 7 ∶ 24 Thus, formula is 𝐴7 𝐵24
142 (c)
Coordination number of Al in AlCl3 in (solid) crystalline state is 6.
143 (a)
Rock salt has fcc structure.
144 (b)
In ZnS structure, sulphide ions occupy all (fcc) lattice points while Zn2+ ions are present in
alternate tetrahedral sites.
Therefore, there is one Zn2+ ion for every S 2− ion.
145 (d)
A occupies corners, thus number of A atoms per unit cell
1
=8× =1
8
B occupies face centres, thus number of B atoms per unit cell
1
=6× =3
2
∴ The empirical formula of the compound is 𝐴𝐵3 .
146 (c)
Amorphous solids neither have ordered arrangement (i.e., no definite shape) nor have
sharp melting point like crystals, but when heated they become pliable until they assume
the properties usually related to liquids. If is therefore, they are regarded as super cooled
liquids.
147 (a)
Due to smaller size of F.
148 (b)
For hexagonal 𝑎 = 𝑏 ≠ 𝑐 and 𝛼 = 𝛽 = 90 and 𝛾 = 120.
150 (b)
Doping of silicon with boron leads to 𝑝 −type semiconductor.

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 SOLID STATE

DPP : 1
1. Schottky defect generally appears in
a) NaCl b) KCl c) CsCl d) All of these

2. Which arrangement of electrons leads ferromagnetism?

a) ↑ ↑ ↑ ↑ b) ↑ ↓ ↑ ↓ c) ↑ ↑ ↑ ↓ ↓ d) None of these

3. The crystal are bounded by plane faces (𝑓), straight edges (𝑒) and interfacial angel (𝑐). The
relationship between these is :
a) 𝑓 + 𝑐 = 𝑒 + 2 b) 𝑓 + 𝑒 = 𝑐 + 2 c) 𝑐 + 𝑒 = 𝑓 + 2 d) None of these

4. The melting point of RbBr is 682C, while that of NaF is 988C. The principle reason that melting
point of NaF is much higher than that of RbBr is that :
a) The two crystals are not isomorphous
b) The molar mass of NaF is smaller than that of RbBr
c) The internuclear distance 𝑟c + 𝑟a is greater for RbBr than for NaF
d) The bond is RbBr has more covalent character than the bond in NaF.

5. If a crystal lattice of a compound, each corner of a cube is enjoyed by sodium, each edge of a
cube has oxygen and centre of a cube is enjoyed by tungsten (W), then give its formula
a) Na2 WO4 b) NaWO3 c) Na3 WO3 d) Na2 WO3

6. In antifluorite structure, the negative ions:

a) Occupy tetrahedral voids
b) Occupy octahedral voids
c) Are arranged in ccp
d) Are arranged in hcp

7. An insulator oxide is :
a) CuO b) CO O c) Fe2 O3 d) All of these

8. A solid with high electrical and thermal conductivity from the following is :
a) Si b) Li c) NaCl d) ice
9. 𝑟
The radius ratio (𝑟+) of an ionic solid (𝐴+ 𝐵− ) is 0.69. What is the coordination number of 𝐵− ?
−
a) 6 b) 8 c) 2 d) 10

10. The axial angles in triclinic crystal system are

a) 𝛼 = β = 𝛾 = 90° b) 𝛼 = 𝛾 = 90°, β ≠ 90° c) 𝛼 ≠ β ≠ 𝛾 ≠ 90° d) 𝛼 = β = 𝛾 ≠ 90°

11. In NaCl crystal each Cl− ion is surrounded by

a) 4 Na+ ions b) 6 Na+ ions c) 1 Na+ ion d) 2 Na+ ions

12. For an ionic crystal of the general formula 𝐴+ 𝐵− and co-ordination number 6, the radius ration
will be :
a) Greater than 0.73

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 SOLID STATE

b) Between 0.73 and 0.41

c) Between 0.41 and 0.22
d) Less than 0.22

13. The ratio of cations to anion in a octahedral close packing is :

a) 0.414 b) 0.225 c) 0.02 d) None of these

14. Electrons in a paramagnetic compound are

a) Shared b) Unpaired c) Donated d) Paired

15. Crystals which are good conductor of electricity and heat are known as :
a) Ionic crystals b) Covalent crystals c) Metallic crystals d) Molecular crystal

16. An element has bcc structure having unit cells 12.08 × 1023 . The number of atoms in these cells
is :
a) 12.08 × 1023 b) 24.16 × 1023 c) 48.38 × 1023 d) 12.08 × 1022

17. Among the following types of voids, which one is the largest void?
a) Triangular b) Cubic c) Tetrahedral d) Octahedral

18. The crystalline structure of NaCl is

a) Hexagonal close packing b) Face centred cubic
c) Square planar d) Body centred cubic

19. Metals have conductivity of the order of (ohm−1 cm−1 ) :

a) 1012 b) 108 c) 102 d) 10−6

20. Of the elements Sr, Zr, Mo, Cd and Sb, all of which are in V period, the paramagnetics are:
a) Se, Cd and Sb b) Zr, Mo and Cd c) Sr, Zr and Cd d) Zr, Mo and Sb

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 SOLID STATE

DPP : 1 SOLUTION
1 (d)
Schottky defect arises when equal number of a cations and anions are missing from their
sites. This defect is generally found in ionic compounds like NaCl, KCl, CsCl, etc.
2 (a)
Ferromagnetism is due to spontaneous alignment of the magnetic dipoles in same direction.
3 (a)
𝑓 + 𝑠𝑐 = 𝑒 + 2; where 𝑓 is plane faces, 𝑐 is interfacial angle and 𝑒 is straight edges.
4 (c)
This leads to stronger coulombic forces of attractions in NaF.
5 (b)
1
No. of Na atoms present at each corner = 8 × 8 = 1
1
No. of O atoms present at the centre of edges = 12 × 4 = 3
No. of W atoms present at the centre of cube = 1
Formula of the compound = NaWO3
6 (c)
In antifluorite crystal (Na2 O) the anions are arranged in cubic close packing while the
cations occupy all the tetrahedral voids.
7 (d)
All are insulator
8 (b)
In the given choices lithium has high thermal and electrical conductance.
9 (a)
Relation between radius ratio and coordination number
𝒓𝒄
Coordination
𝒓𝒂
number
0.155 − 0.225 3
0.225 − 0.414 4
0.414 − 0.732 6
0.732 − 1 8
10 (c)
The axial angles in triclinic crystal system are different and none is perpendicular to any of
the others 𝑖. 𝑒., 𝛼 ≠ β ≠ 𝛾 ≠ 90°.
11 (b)
In NaCl crystal, Cl− ions adopt cubic close packed arrangement and Na+ ions occupy all the
octahedral sites. Therefore, Na and Cl have 1 : 1 stoichiometry. In other words, each Na+ ion
is surrounded by six Cl− ions which are disposed towards the corners of a regular
octahedron. Similarly, each Cl− ion is surrounded by six Na+ ions.
12 (b)
The radius ratio for co-ordination and has 4, 6, and 8 lies in between the ranges
[0.225 − 0.414], [0.414 − 0.732] and [0.732 − 1] respectively.
13 (a)
𝑟+ 𝑟+
for octahedral void = 0.414; for cubic = 0.732 − 1
𝑟− 𝑟−

15 (c)

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 SOLID STATE

Metallic crystals are good conductor of heat and current due to free electrons on them.
16 (b)
One unit cell of bcc has atoms = 2. Hence 12.08 × 1023 unit cells will have atoms
= 2 × 12.08 × 1023
= 24.16 × 1023
17 (d)
The vacant spaces between the spheres in closed packed structures is called void. The voids
are of two types, tetrahedral voids and octahedral voids. Also, radius of tetrahedral voids
and octahedral voids are 𝑟void = 0.225 × 𝑟sphere and 𝑟void = 0.414 × 𝑟sphere respectively.
Thus, octahedral void is larger than tetragonal void.
18 (b)
Sodium chloride (NaCl) has face centred cubic structure. It contains 4 Na+ and 4 Cl− in the
unit cell. Each Na+ is surrounded by 6 Cl− ions and 𝑣𝑖𝑐𝑒 − 𝑣𝑒𝑟𝑠𝑎.
19 (b)
The conductance order of metals is 106 to 108 ohm−1 cm−1
20 (d)
Each possess unpaired electrons.

ANSWER-KEY
Q. 1 2 3 4 5 6 7 8 9 10
A. D A A C B C D B A C

Q. 11 12 13 14 15 16 17 18 19 20
A. B B A B C B D B B D

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 SOLID STATE

DPP : 2
1. The radius ratio of CsCl is 0.93. The expected lattice structure is
a) Tetrahedral b) Square planar c) Octahedral d) Body centred cubic
2. Which one of the following defects in the crystals lowers its density?
a) Frenkel defect b) Schottky defect c) F-centres d) Interstitial defect
3. The yellow colour of ZnO and conducting nature produced in heating is due to:
a) Metal excess defects due to interstitial cation
b) Extra positive ions present in an interstitial site
c) Trapped electrons
d) All of the above
4. A metal has bcc structure and the edge length of its unit cell is 3.04 Å. The volume of the unit cell
in cm3 will be
a) 1.6 × 10−21 cm3 b) 2.81 × 10−23 cm3 c) 6.02 × 10−23 cm3 d) 6.6 × 10−24 cm3
5. The edge length of a face centred cubic cell of an ionic substance is 508 pm. If the radius of the
cation is 110 pm, the radius of the anions is
a) 288 pm b) 398 pm c) 618 pm d) 144 pm
6. An ionic compound is expected to have tetrahedral structure if 𝑟+ /𝑟− lies in the range of
a) 0.414 to 0.732 b) 0.225 to 0.414 c) 0.155 to 0.225 d) 0.732 to 1
7. The interparticle forces in solid hydrogen are :
a) Hydrogen bonds b) Covalent bonds c) Co-ordinate bonds d) Van der Waals’ forces
8. If Z is the number of atoms in the unit cell that represents the closest packing sequence
−𝐴𝐵𝐶 𝐴𝐵𝐶 −, the number of tetrahedral voids in the unit cell is equal to :
Z Z
a) Z b) 2Z c) d)
2 4
9. Quartz is an example of :
a) Chain silicate b) Infinite sheet silicate c) Framework silicate d) Cyclic silicate
10. For 𝐴𝑋 ionic crystal to exist in bcc structure, the ratio of radii ( 𝑟cation ) should be
𝑟anions
a) Between 0.41 and 0.73 b) Greater then 0.73
c) Less than 0.41 d) Equal to 1.0
11. Which crystal is expected to be soft and have low melting point?
a) Covalent b) Metallic c) Molecular d) Ionic
12. The elements commonly used for making transistors are
a) C and Si b) Ga and In c) P and As d) Si and Ge
13. −1 −3
Silver (atomic weight = 108 g mol ) has a density of 10.5 g cm . The number of silver atoms
on a surface of area 10−12 m2 can be expressed in scientific notation as 𝑦 × 10𝑥 . The value of 𝑥
is
a) 3 b) 5 c) 7 d) 9
14. The first order reflection (𝑛 = 1) from a crystal of the X-ray from a copper anode tube
(𝜆 = 1.54 Å)occurs at an angle of 45°. What is the distance between the set of plane causing the
diffraction?
a) 0.1089 nm b) 0.1089 m c) 0.905 Å d) 1.089 × 10−9 m
15. What is the number of tetrahedral voids per atom in a crystal?
a) 1 b) 2 c) 6 d) 8
16. Iodine is a

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 SOLID STATE

a) Electrovalent solid b) Atomic solid c) Molecular solid d) Covalent solid

17. In CsCl type structure the coordination number of Cs +and Cl− are
a) 6, 6 b) 6, 8 c) 8, 8 d) 8, 6
18. Structure of a mixed oxide is cubic close-packed (c.c.p). The cubic unit cell of mixed oxide is
composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal 𝐴 and
the octahedral voids are occupied by a monovalent metal 𝐵. The formula of the oxide is :
a) 𝐴𝐵 𝑂2 b) 𝐴2 𝐵𝑂2 c) 𝐴2 𝐵3 𝑂4 d) 𝐴𝐵2 𝑂2
19. The example of orthosilicate is :
a) MgCaSi2 O6 b) Mg 2 SiO4 c) Fe2 O3 SiO2 d) Ba3 Al2 Si6 O8
20. A compound CuCl has face centred cubic structure. Its density is 3.4 g cm−3 . The length of unit cell
is :
a) 5.783Å b) 6.783Å c) 7.783Å d) 8.783Å

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 SOLID STATE

DPP : 2 SOLUTION
1 (d)
The radius ratio of CsCl is 0.93 hence, its structure is body centred cubic.
2 (b)
Schottky defects - This defect is due to vacancy at a cation site accompanied by vacancy at
an anion site so that the electrical neutrality of the system is maintained. Due to this defect,
density decreases.
3 (d)
These are characteristics of metal excess defects due to interstitial cation.
4 (b)
Edge length 𝛼 = 3.04 Å
= 3.04 × 10−8 cm
Volume of bcc (cubic) cell = 𝑎3
= (3.04 × 10−8 )3
= 2.81 × 10−23 cm3
5 (d)
For fcc arrangement
2(𝑟 + + 𝑟 − ) = edge length
2(110 + 𝑟 − ) = 508
So, 𝑟 − = 114 pm
6 (b)
Radius ratio(𝒓+ / Structure
𝒓− )
< 0.155 linear
0.155 − 0.225 planar
triangular
0.225 − 0.414 tetrahedral
0.414 − 0732 octahedral
0.732 − 1 bcc
7 (d)
Solid hydrogen involves van der Waals’ forces.
8 (b)
In ccp or fcc and hcp, number of tetrahedral voids is double the number of atoms forming
the main lattice.
9 (c)
Quartz is a covalent crystal having a framework of silicates, i.e., a three dimensional
network when all the four oxygen atoms of each of SiO4 tetrahedron are shared.
10 (b)
For body centred cubic (bcc) structure, the ratio of radii (𝑟+ /𝑟−) lies in between
0.732−1.00.
∴ The ratio of radii for bcc is greater than 0.73.
11 (c)
Follow characteristics of molecular solids.
12 (d)
Si and Ge are used for making transistors.
13 (c)

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 SOLID STATE

108
Volume of one mole of silver atoms = cm3 /mol
10.5
108 1
Volume of one silver atom = 10.5 × 6.022 ×1023
cm3
4 108 1
So, 𝜋𝑟 3 = × = 1.708 × 10 −23
3 10.5 6.022 ×1023
𝑟 3 = 0.407 × 10 cm = 0.407 × 10−29 m3
−23 3

Area of each silver atom,

𝜋𝑟 2 = 𝜋(0.407 × 10−29 m3 )2/3
So, number of silver atoms in given area
10−12 108
= =
(0.407 × 10−29 m3 )2/3 𝜋 × 2
= 1.6 × 107 = 𝑦 × 10𝑥
So, 𝑥 = 7
14 (c)
𝑛𝜆 = 2𝑑 sin θ
1 × 1.54 = 2𝑑 sin 45°
1 × 1.54 = 2𝑑 × 0.850
1.54
2𝑑 = = 0.905 Å
0.850
15 (b)
In the close packing of ‘𝑛’ atoms, the number of tetrahedral voids are ‘2𝑛’. Hence, their
number per atom is 2.
17 (c)
The coordination number is 8 : 8 in Cs + ∶ Cl−
The coordination number is 6 : 6 in Na+ ∶ Cl−
18 (d)
In a cubic close packing, the number of octahedral voids is equal to number of atoms and
number of tetrahedral voids is equal to the twice the number of atoms
Number of atoms is a ccp array = 1
∴ 𝐴2+ 𝐵+ 𝑂2−
1
1×2× 4 1 1
1
2
1 1
𝑜𝑟 1 2 2
𝐴𝐵2 𝑂2
19 (b)
In orthosilicate SiO2−
4 ion exist as discrete unit.
20 (a)
Molecular mass of CuCl = 99
𝑛 = 4 for face centred cubic cell
𝑛 × mol. wt.
∵ 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 =
𝑉 × av. no.
4 × 99
= 3
𝑎 × 6.023 × 1023
4 × 99
Or 3.4 = 3 23
𝑎 × 6.023 × 10
−8
∴ 𝑎 = 5.783 × 10 cm
= 5.783Å

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 SOLID STATE

ANSWER-KEY
Q. 1 2 3 4 5 6 7 8 9 10
A. D B D B D B D B C B

Q. 11 12 13 14 15 16 17 18 19 20
A. C D C C B C C D B A

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 SOLID STATE

DPP : 3
1. The orthorhombic, the value of 𝑎, 𝑏and 𝑐are respectively 4.2 Å, 6.8𝐴 Å and 8.3 Å. Given the
molecular mass of the solute is 155 g mol−1 and that of density is 3.3g/cc, the number of formula
units per unit cell is
a) 2 b) 3 c) 4 d) 6
2. Which one of the following is a covalent crystal?
a) Rock salt b) Ice c) Quartz d) Dry ice
3. LiF is a/an :
a) Ionic crystal b) Metallic crystal c) Covalent crystal d) Molecular crystals
4. + −
A binary solid (𝐴 𝐵 ) has a rock salt structure. If the edge length is 400 pm and radius of cation
is 75 pm the radius of anion is :
a) 100 pm b) 125 pm c) 250 pm d) 325 pm
5. The limiting radius ratio for tetrahedral shape is
a) 0 to 0.155 b) 0.255 to 0.414 c) 0.155 to 0.225 d) 0.414 to 0.732
6. A metallic element has a cubic lattice. Each edge of the unit of cell is 2Å. The density of the metal
is 2.5 g cm−3 . The unit cells in 200 g of metal are
a) 1 × 1024 b) 1 × 1020 c) 1 × 1022 d) 1 × 1025
7. Potassium has a bcc structure with nearest neighbour distance 4.52 Å. Its atomic weight is 39. Its
density will be :
a) 454 kg m−3 b) 804 kg m−3 c) 852 kg m−3 d) 910 kg m−3
8. Lithium forms body centred cube structure. The length of the side of its unit cell is 351 pm. Atomic
radius of the lithium will be :
a) 300 pm b) 240 pm c) 152 pm d) 75 pm
9. Bragg’s equation is :
a) 𝑛𝜆 = 2𝜃 sin θ b) 𝑛𝜆 = 2𝑑 sin θ c) 2𝑛𝜆 = 𝑑 sin θ d) 𝜆 = (2𝑑/𝑛) sin θ
10. The intermetallic compound LiAg has a cubic crystalline structure in which each Li atom has 8
nearest neighbor silver atoms and 𝑣𝑖𝑐𝑒 − 𝑣𝑒𝑟𝑠𝑎. What is the type of unit cell?
a) Body centred cubic
b) Face centred cubic
c) Simple cubic for either Li atoms alone or Ag atoms alone
d) None of the above
11. In the face centred cubic lattice, atom A occupies the corner positions and atom B occupies the
face centre positions. If one atom of B is missing from one of the face centred points, the formula
of the compound is
a) 𝐴2 𝐵 b) 𝐴𝐵2 c) 𝐴2 𝐵2 d) 𝐴2 𝐵5
12. Which compound has highest lattice energy?
a) LiBr b) LiCl c) LiI d) LiF
13. In a face centred cubic cell, an atom at the face centre is shared by :
a) 4 unit cells b) 2 unit cells c) 1 unit cell d) 6 unit cells
14. Extremely pure samples of Ge and Si are non-conductors, but their conductivity increases
suddenly on introducing ….in their crystal lattice.
a) As b) B c) Both (a) and (b) d) None of these
15. Iodine crystals are :
a) Metallic solid b) Ionic solid c) Molecular solid d) Covalent solid

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 SOLID STATE

16. Which of the following statements about amorphous solids is incorrect?

a) They melt over a range of temperature b) They are anisotropic
c) There is no orderly arrangement of particles d) They are rigid and incompressible
17. The number of atoms present in a simple cubic unit cell are :
a) 4 b) 3 c) 2 d) 1
18. An 𝐴𝐵2 type structure is found in :
a) NaCl b) CaF2 c) Al2 O3 d) N2 O
19. A cubic crystal possesses in all ……elements of symmetry.
a) 9 b) 13 c) 1 d) 23
20. A solid compound contains 𝑋, Y and Z atoms in a cubic lattice with X atom occupying the
corners. Y atoms in the body centred positions and Z atoms at the centres of faces of the unit
cell. What is the empirical formula of the compound?
a) 𝑋𝑌2 𝑍3 b) 𝑋𝑌𝑍3 c) 𝑋2 𝑌2 𝑍3 d) 𝑋8 𝑌𝑍6

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 SOLID STATE

DPP : 3 SOLUTION
1 (c)
𝑉×𝑁×𝑑
𝑍=
𝑚
4.2 × 8.6 × 8.3 × 10−21 × 6.023 × 1023 × 3.3
=
155
= 3.14
≈4
2 (c)
Quartz (SiO2 ) is a covalent crystal.
3 (a)
LiF is an ionic crystal. An ionic solid has ions as constituent units at lattice points held by
oppositely charged ions.
4 (b)
Edge = 2𝑟 + + 2𝑟 −
∴ 400 = 2 × 75 + 2𝑟 −
∴ 𝑟 − = 125 pm
5 (b)
For tetrahedral shape, limiting radius ratio is 0.225 − 0.414.
6 (d)
mass of metal
Number of unit cells = mass of one unit cell
Given, edge length of unit cell = 2Å = 2 × 10−8 cm
Mass of metal = 200 g
Density of metal = 2.5 g cm−3
Volume of unit cell = (edge length)3 = (2 × 10−8 )3
= 8 × 10−24 cm3
Mass of one unit cell = volume × density
= 8 × 10−24 × 2.5
= 20 × 10−24
mass of metal
∴ No. of unit cells in 200 g metal = mass of one unit cell
200
=
20 × 10−24
= 10 × 1024 = 1.0 × 1025
7 (d)
√3
For bcc, 𝑟 = 2
=𝑎
2𝑟 2 × 4.52
Or 𝑎 = =
√3 1.732
= 5.219 𝐴̊ = 522 pm.
𝑛×𝑀
Density =
𝑎 3 × 𝑁𝐴 × 10−30
2 × 39
=
(522)3
× (6.02 × 1023 ) × 10−30
= 0.91g/cm3 = 910 kg m−3
8 (c)
For bcc structure

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 SOLID STATE

9 (b)
Bragg’s equation is 𝑛𝜆 = 2𝑑 sin θ
10 (a)
The bcc structure has co-ordination no. of eight.
11 (d)
1
Number of atoms (𝐴) per unit cell = 8 × 8 = 1
1 5
Number of atoms (𝐵) per unit cell = (6 − 1) × 2 = 2
(One atom B is missing)

Thus, formula is 𝐴1 𝐵5/2 = 𝐴2 𝐵5

12 (d)
Due to small anion, it possess maximum ionic nature.
13 (b)
The fcc unit cell has 8 atoms at the eight corners and one atom at each of six faces. The atom
at the face is shared by two unit cells.
14 (c)
Doping of elements of group 14 (Ge and Si) with group 15 (As) elements produces excess of
electrons and shows 𝑛 -type conduction, the symbol 𝑛 indicating flow of negative charge in
them. Doping of elements of group 14 (Ge and Si) with group 13 (B) elements products hole
(electron deficiency) in the crystal and shows 𝑝-type conduction, the symbol 𝑝 indicating
flow of positive charge.
15 (c)
Molecular solids are the substances having molecules as constituent units having
interparticle forces such as van der waal’s forces or hydrogen bonds.
17 (d)
The number of atoms present in sc, fcc and bcc unit cell are 1, 4, 2 respectively.
18 (b)
N2 O is gas; CaF2 is 𝐴𝐵2 type crystalline solid.
19 (d)
These are characteristic elements of symmetry of a cubic crystal.
20 (b)
Since atom 𝑋 is present at corner and one corner is shared by eight unit cells,
1
Number of X atoms per unit cell = 8 × 8 = 1
Atom 𝑌 is present at body centred position and used by only one unit cell. So, number of Y
atoms per unit cell = 1
Atom Z is present at the center of each face, so shared by two unit cells,
1
Thus, number of Z atoms per unit cell = 2 × 6 = 3
Hence, the formula of compound = 𝑋𝑌𝑍3

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 SOLID STATE

ANSWER-KEY
Q. 1 2 3 4 5 6 7 8 9 10
A. C C A B B D D C B A

Q. 11 12 13 14 15 16 17 18 19 20
A. D D B C C B D B D B

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 SOLID STATE

DPP : 4
1. The oxide which shows transition from metal to insulation, i.e., semiconductors are :
a) V2 O3 b) VO2 c) Ti2 O3 d) All of these

2. Edge length of a cube is 400 pm. Its body diagonal would be :

a) 600 pm b) 566 pm c) 693 pm d) 500 pm

3. Crystals can be classified into ……. Basic crystal habits.

a) 7 b) 4 c) 14 d) 3

4. The unit cell with crystallographic dimensions 𝑎 = 𝑏 ≠ 𝑐; 𝛼 = 𝛽 = 𝛾 = 900 is :

a) Cubic b) Tetragonal c) Monoclinic d) Hexagonal

5. The number of octahedral void(s) per atom present in a cubic close-packed structure is :
a) 2 b) 4 c) 1 d) 3

6. The hardness of metals increases with increase in number of ……involved in metallic bonding.
a) Atoms b) Molecules c) Electrons d) All of these

7. The substance which possesses zero resistance as 0 K :

a) Conductor b) Super conductor c) Insulator d) Semiconductor

8. Sodium metal crystallises at room temperature in a body centred cubic lattice with a cell edge
𝑎 = 4.29 Å. The radius of sodium atom is
a) 1.40 b) 2.65 c) 1.85 d) 2.15

9. The oxide which shows metallic conduction:

a) ReO3 b) VO c) CrO2 d) All of these

10. The number of hexagonal faces that are present in a truncated octahedron is
a) 2 b) 4 c) 6 d) 8

11. Which of the following statement is true?

a) Some complex metal oxides behave as b) Zinc oxide can act as superconductor
superconductor
c) An impurity of tetravalent germanium in d) A Frenkel defect is formed when an ion is
trivalent gallium creates electron deficiency displaced from its lattice site to an interstitial
site

12. Schottky defect defines imperfection in the lattice structure of a :

a) Solid b) Gas c) Liquid d) Plasma

13. When electrons are trapped into the crystal in anion vacancy, the defect is known as :
a) Schottky defect b) Frenkel defect c) Stoichiometric defect d) F-centres

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 SOLID STATE

14. Which of the following has highest value of energy gap?

a) Aluminum b) Silver c) Germanium d) Diamond

15. If ‘a’ stands for the edge length of the cubic systems : simple cubic, body-centred cubic and face-
centered, then the ratio of radii of the spheres in these systems will be respectively,
1 1 1 1 1 √3 1
a) 𝑎: √3 𝑎: 𝑎 b) 𝑎: √3 𝑎: √2 𝑎 c) 𝑎 ∶ √3𝑎 ∶ √2𝑎 d) 𝑎: 𝑎: 𝑎
2 √2 2 2 2 2 2 4 2√2

16. In a face centred cubic lattice the number of nearest neighbours for a given lattice point are :
a) 6 b) 8 c) 12 d) 14

17. Percentage of free space in cubic close packed structure and in body centred packed structure
are respectively
a) 30% and 26% b) 26% and 32% c) 32% and 48% d) 48% and 26%

18. Lithium borohydride crystallizes in an orthorhombic system with 4 molecule per unit cell. The
unit cell dimensions are 𝑎 = 6.8 Å, 𝑏 = 4.4 Å and 𝑐 = 7.2 Å. If the molar mass is 21.76, then the
density of crystals is :
a) 0.6708 g cm−3 b) 1.6708 g cm−3 c) 2.6708 g cm−3 d) None of these

19. Total volume of atoms present in a face centred cubic unit cell of a metal is
(r=atomic radius )
20 24 12 16
a) 𝜋𝑟 3 b) 𝜋𝑟 3 c) 𝜋𝑟 3 d) 𝜋𝑟 3
3 3 3 3

20. Which has no rotation of symmetry?

a) Hexagonal b) Orthorhombic c) Cubic d) Triclinic

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 SOLID STATE

DPP : 4 SOLUTION
1 (d)
The transition of metal to insulation occurs at a certain temperature due to imperfection.
2 (c)
Body diagonal in bcc = √3 𝑎 = √3 × 400 = 692.8 pm
3 (a)
The seven basic crystal lattice are cubic, tetragonal, orthorhombic, monoclinic, hexagonal,
rhombohedral and triclinic.
4 (b)
The conditions for tetragonal systems.
5 (c)
The number of octahedral voids in cubic close packed = 4
The number of atoms per unit cell in ccp = 4
The number of octahedral voids per atom = 1
6 (c)
An increase in charge of +ve ions also brings in an increase in number of electrons involved
in metallic crystals, and thereby metallic bonding becomes stronger.
7 (b)
Electrical resistance of metals decreases with decrease in temperature and becomes zero at
zero kelvin. Materials in this state are called super conductors and the phenomenon as
super conductivity.
8 (c)
For a body centred cubic lattice radius, (𝑟)
√3
= 𝑎 = 0.433𝑎
4
Therefore, radius of Na+ = 0.433 × 4.29 = 1.8575
9 (d)
All are conductors however shows insulation at a certain temperature.
10 (d)
The truncated octahedron is the 14-faced Archimedean solid, with 14 total faces : 6 squares
and 8 regular hexagons.
The truncated octahedron is formed by removing the six right square pyramids one from
each point of a regular octahedron as :

Truncated Octachedron

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 SOLID STATE

11 (d)
Frenkel defect is formed by displacement of ion from its lattice to interstitial state.
12 (a)
Inperfections are notice in solids.
13 (d)
Trapping of electrons in anion vacancies develop F-centres.
14 (d)
Diamond has the highest value of energy gap as it is a insulator.
15 (d)
a a √3
sc: r = fcc ∶ r = ; bcc ∶ r = a
2 2√2 4
a √3 a
∴ sc; bcc and fcc are , a,
2 4 2√2
16 (c)
Number of sodium ions are 12 at edge centres in fcc structure which are nearest neighbours
for a given lattice point.
17 (b)
𝜋
Packing fraction of ccp = 3 2 = 0.74 ⇒ 74%
√
% free space in ccp = 26%
𝜋√3
Packing fraction of bcc = = 0.68 ⇒ 68%
8
% free space in bcc = 32%
18 (a)
𝑛 × mol.wt.
Density
𝑉 × av.no.
𝑛 = 4, 𝑀 = 21.76, av. no. = 6.023 × 1023 and
And 𝑉=𝑎 × 𝑏 × 𝑐
∴ 𝑉 = 6.8 × 10−8 × 4.4 × 10−8 × 7.2 × 10−8
= 2.154 × 10−22 × 6.023 × 1023
4 × 21.76
Density = 2.154 × 10−22 × 6.023 × 1023
= 0.6708 g cm−3
19 (d)
4
Volume of an atom = 3 𝜋𝑟 3
In fcc, number of atoms per unit cell = 4
4
∴ Volume of total atoms = 4 × 3 𝜋𝑟 3
16 2
= 𝜋𝑟
3
20 (d)

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 SOLID STATE

In triclinic lattice, the eight lattice points are located, one each at the corners of triclinic
lattice. Also 𝑎 ≠ 𝑏 ≠ 𝑐 and 𝛼 ≠ β ≠ 𝛾. There is no planes and no axes. Thus, triclinic lattice
has no rotation of symmetry.

ANSWER-KEY
Q. 1 2 3 4 5 6 7 8 9 10
A. D C A B C C B C D D

Q. 11 12 13 14 15 16 17 18 19 20
A. D A A D D C B A D D

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 SOLID STATE

DPP : 5
1. The unit cell with dimensions 𝛼 = 𝛽 = 𝛾 = 90°, 𝑎 = 𝑏 ≠ 𝑐 is
a) Cubic b) Triclinic c) Hexagonal d) Tetragonal

2. A fcc element (atomic mass = 60) has a cell edge of 400 pm. Its density is :
a) 6.23 g cm−3 b) 6.43 g cm−3 c) 6.53 g cm−3 d) 6.63 g cm−3

3. For a crystal system 𝑎 = 𝑏 = 𝑐 and 𝛼 = 𝛽 = 𝛾 ≠ 90°

a) Tetragonal b) Hexagonal c) Rhombohedral d) Monoclinic

4. The number of atoms (𝑛) contained within a cubic cell is :

a) 1 b) 2 c) 3 d) 4

5. All the substances becomes diamagnetic at :

a) 4 K b) 10 K c) 20 K d) 25 K

6. The co-ordination number of Ca2+ ion in fluorite crystal is :

a) 2 b) 8 c) 6 d) 4

7. What is the structure of NaCl?

a) BCC b) FCC c) Interpenetrating fcc d) None of these

8. Which of the following statements is not correct?

a) Molecular solids are generally volatile
b) The number of carbon atoms in an unit cell of diamond is 4
c) The number of Bravais lattices in which a crystal can be categorized is 14
d) The fraction of the total volume occupied by the atoms in a primitive cell is 0.48.

9. Which is the wrong statement regarding a crystal containing Schottky defect?

a) Electrical neutrality of the crystal is maintained
b) Entropy of the crystal increases
c) The density of the overall crystal remains the same
d) The density of the overall crystal reduces

10. How many ‘nearest’ and ‘next nearest’ neighbours respectively potassium have in bcc lattice?
a) 8, 8 b) 8, 6 c) 6, 8 d) 8, 2

11. Ferrimagnetic is converted into paramagnetic at :

a) 300 K b) 400 K c) 600 K d) 850 K

12. A match box exhibits :

a) Cubic geometry
b) Monoclinic geometry
c) Orthorhombic geometry
d) Tetragonal geometry

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 SOLID STATE

13. The oxide that possesses electrical conductivity :

a) V2 O5 b) CrO2 c) NiO d) MnO

14. The arrangement 𝐴𝐵𝐶 𝐴𝐵𝐶……is referred to as,

a) Octahedral close packing
b) Hexagonal close packing
c) Tetrahedral close packing
d) Cubic close packing

15. The lattice points of a crystal of hydrogen iodide are occupied by

a) HI molecules b) H atoms and I atoms
c) H cations and I anions
+ −
d) H2 molecules and I2 molecules

16. 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

17. The radius of the Na+ is 95 pm and that of Cl− ion is 181 pm. Predict the co-ordination number of
Na+ :
a) 4 b) 6 c) 8 d) Unpredictable

18. How many unit cells are present in a cube shaped ideal crystal of NaCl of mass 1.00g?
[Atomic masses ∶ Na = 23, Cl = 35.5]
a) 2.57 × 1021 b) 5.14 × 1021 c) 1.28 × 1021 d) 1.71 × 1021

19. For a covalent solid, the units which occupy lattice points are :
a) Atoms b) Ions c) Molecules d) Electrons

20. The metal surfaces are excellent reflectors because of absorption and re-emission of light by :
a) Protons in atom b) Electrons in atom c) Neutrons in atom d) None of these

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 SOLID STATE

DPP : 5 SOLUTION
1 (d)
The unit cell with dimensions 𝑎 = 𝑏 ≠ 𝑐, 𝛼 = β = 𝛾 = 90 is tetragonal.
2 (a)
𝑛 × at.wt. 𝑛 × at.wt.
Density = 𝑉 × av.no. = 𝑎3 × av.no.
Given, at.wt. = 60
𝑎 = 4 × 102 pm = 4 × 102 × 10−12 m
= 4 × 10−10 × 102 cm = 4 × 108 cm
(∵ 1 pm = 10−12 )
4 × 60
∴ Density = (4 × 10−8 ) × 6.023 × 1023
−3
= 6.23 g cm
3 (c)
Crystal system Axial Axial angle
distances
Tetragonal 𝑎=𝑏 𝛼=β=𝛾
≠𝑐 = 90°
Hexagonal 𝑎=𝑏 𝛼≠β
≠𝑐 = 90°, 𝛾
= 120°
Rhombohedral 𝑎=𝑏 𝛼=β=𝛾
=𝑐 ≠ 90°
Monoclinic 𝑎≠𝑏 𝛼=𝛾
≠𝑐 = 90, β
≠ 90°
4 (a)
The cubic unit cell has 8 atmos at eight corners. Each atom is shared by 8 unit cells.
1
∴ 𝑛=8 × =1
8
5 (a)
Most of the metals have their transition temperature (i.e., the temperature at which a
substance starts to behave as super conductor) in the range of 2-5 K.
6 (b)
CaF2 has fcc structure with 8 : 4 co-ordination and has 4 units of CaF2 per unit cell.
7 (b)
NaCl has fcc arrangement of ions. The coordination number of Cl− as well as Na+ ion is six.
Therefore, it is termed 6 : 6 coordination crystal.
8 (b)
No. of carbon atoms in unit cell of diamond is 8. Also fraction of volume occupied by the
atoms in primitive cell is 52%.
9 (c)
When equal number of cations and anions are missing from their position in a crystal lattice
so that electrical neutrality is maintained, the defect is called Schottky defect. Due to
missing of ions, the overall density of the crystal decreases. Moreover, defect leads to
randomness, thus entropy also increases.
10 (b)
It is a fact for crystal structure (bcc) potassium.
11 (d)
At high temperature randomization of spins changes.

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 SOLID STATE

12 (c)
Orthorhombic geometry has 𝑎 ≠ 𝑏 ≠ 𝑐 and 𝛼 = 𝛽 = 𝛾 = 90. The shape of match box obey
this geometry.
13 (b)
CrO2 is metallic conductor, V2 O5 , NiO and MnO are insulators.
14 (d)
It represents ccp arrangement.
16 (b)
Given,
Molar mass, 𝑀 =50g/mol
𝑁𝐴 = 6.02 × 1023
𝑍 = 2 (for bcc crystal)
Edge length 𝑎 = 300 pm
= 3 × 10−8 cm
𝑍×𝑀
𝑑=
𝑁𝐴 × 𝑎3
2 × 50
=
6.02 × 10 × (3 × 10−8 )3
23

= 6.15
≈ 6.2
17 (b)
𝑟Na+ 95
= = 0.524, i.e., in between 0.414 to 0.732 and thus, co-ordination no.=6
𝑟Cl− 181
18 (a)
Mass of one unit-cell (𝑚)
= volume × density
𝑀𝑍 𝑀𝑍
= 𝑎3 × 𝑑 = 𝑎3 × 3
=
𝑁0 𝑎 𝑁0
58.5 × 4
𝑚= g
6.02 × 1023
1
∴ Number of unit cells in 1 g = 𝑚
6.02 × 1023
=
58.5 × 4
= 2.57 × 1021
19 (a)
In covalent molecules atoms occupy the lattice points.
20 (b)
The presence of free electrons in metals, they are opaque, strongly reflecting and possess
metallic lustre.

ANSWER-KEY
Q. 1 2 3 4 5 6 7 8 9 10
A. D A C A A B B B C B

Q. 11 12 13 14 15 16 17 18 19 20
A. D C B D A B B A A B

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