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Written according to the New Text book (2012-2013) published by the Maharashtra State Board of
Secondary and Higher Secondary Education, Pune.
Std. XII Sci.
Perfect Chemistry - I
Prof. Santosh B. Yadav
(M. Sc., SET, NET)
Department of Chemistry
R. Jhunjunwala College, Ghatkopar
Prof. Anil Thomas
(M.Sc., Chemistry)
Salient Features:
Exhaustive coverage of syllabus in Question Answer Format.
Covers answers to all Textual and Intext Questions.
Textual Questions are represented by * mark.
Intext Questions are represented by # mark.
Covers relevant NCERT Questions.
Charts for quick reference.
Simple and Lucid language.
Target PUBLICATIONS PVT. LTD.
Mumbai, Maharashtra
Tel: 022 – 6551 6551
Website: www.targetpublications.in
www.targetpublications.org
email : mail@targetpublications.in

Std. XII Sci.
Perfect Chemistry - I
© Target Publications Pvt Ltd.
Fourth Edition : May 2012
Price : ` 220/-
Printed at:
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PREFACE
In the case of good books, the point is not how many of them you can get through, but rather
how many can get through to you.
Chemistry is a science concerned with the composition, structure and porperties of matter as
well as the changes they undergo under different conditions. From the making of medicines
to the manufacture of steel, to the thunder of clouds, everything is encapsulated in the study
of physical chemistry. A science that calls for such deep knowledge of concepts and an
intrinsic study of all aspects must obviously, not be easy to conquer. For this we bring to you
“Std. XII : PERFECT CHEMISTRY - I” a complete and thorough book which analyses
and extensively boost confidence of the student.
Topic wise classified “question and answer” format of this book helps the student to
understand each and every concept thoroughly. Important definitions, statements and laws
are specified with italic representation. Solved problems are provided to understand the
application of different concepts and formulae. Practice problems and multiple choice
question help the students, to test their range of preparation and the amount of knowledge of
every topic.
And lastly, we would like to thank our publishers for helping us to take this exclusive guide
to all students. There is always room for improvement and hence we welcome all suggestions
and regret any errors that may have occurred in the making of this book.
Your’s faithfully
Publisher
A book affects eternity; one can never tell where its influence stops.
Best of luck to all the aspirants!

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Paper Pattern
• There will be one written paper of 70 Marks in Chemistry.
• Duration of the paper will be 3 hours.
• Chemistry paper will have two parts viz: Part I of 35 marks and Part II of 35 marks
• Same Answer Sheet will be used for both the parts.
• In the question paper for each part there will be 4 Questions.
• Students have freedom to decide the sequence of answers.
• The paper pattern as per the marking scheme for Part I and Part II will be as follows:
Question 1: (7 Marks)
There will be 7 multiple choice Questions (MCQs), each carrying 1 mark.
Total marks = 7
Question 2: (12 Marks)
There will be 8 Questions out of which 6 Questions to be answered, each carrying 2 marks.
Total marks = 12
Question 3: (9 Marks)
There will be 4 Questions out of which 3 Questions to be answered, each carrying 3 marks.
Total marks = 9
(There will be 3 Questions based on numericals from part I)
Question 4: (7 Marks)
There will be 2 Question out of which 1 Question has to be answered.
It will carry 7 marks.
Total Marks = 7
(There will be 2/3 marks Questions based on numericals from Part I)
Distribution of Marks According to Type of Questions
Type of Questions Marks Marks with option Percentage (%)
Objectives 14 14 20
Short Answers 42 56 60
Brief Answers 14 28 20
Total 70 98 100

No. Topic Name
Contents
Page
No.
Marks
Without
Option
Marks
With
Option
1 Solid State 1 04 06
2 Solutions and Colligative Properties 51 05 07
3 Chemical Thermodynamics and Energetics 112 06 08
4 Electrochemistry 182 05 07
5 Chemical Kinetics 256 04 06
6
General Principles and Processes of
Isolation of Elements
330 03 05
7 p−Block Elements 366 08 10
Note: All the Textual questions are represented by * mark
All the Intext questions are represented by # mark

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01 Solid state
1.0 Introduction
Prominent Scientists:
Std. XII Sci.: Perfect Chemistry -I
Scientists Contribution
W.H.Bragg Derived the Bragg equation nλ = 2d sinθ to explain why cleavage of crystal reflect x-ray
beam.
Determined crystal structure of NaCl, ZnS and Diamond.
William Lipscomb Determined structure of boron hydride using technique of x-ray crystallography.
Isabella Karle Determined molecular structure by x-ray diffraction, worked on crystallography of
materials in the field of Organic chemistry and Biochemistry.
Q.1. Under what conditions does a substance exist in solid state?
Ans: i. Matter can exist in three states namely, solid, liquid and gas.
ii. Under a given set of conditions of temperature and pressure, the most stable state of a substance
depends upon the net effect of two opposing forces: intermolecular forces and thermal energy.
iii. Intermolecular forces tends to keep the constituent particles (atoms, ions or molecules) closer,
whereas thermal energy tend to keep them apart by making them move faster.
iv. The competition between molecular interaction energy due to intermolecular forces and thermal
energy determines whether a given substance under a given set of condition is a gas, a liquid or a solid.
v. At sufficiently low temperature, the thermal energy is low and molecular forces are very strong. As a
result, the intermolecular forces keep the constituents so close that they cling to one another and
occupy fixed positions and the substance exists in solid state.
Q.2. How does the solid state differ from other two states?
Ans: i. Solids differ from liquids and gases in the fact that gases and liquids possess fluidity, i.e. they can
flow and hence they are called fluids, whereas solids do not possess fluidity; instead, they possess
rigidity.
ii. The reason for fluidity of gases and liquids lies in the fact that their constituent particles are free to
move about, whereas the rigidity of solids is due to the fact that their constituent particles have fixed
positions and can only oscillate about their mean positions.
iii. Further, because of rigidity possessed by solids, they have definite shape and definite volume.
Q.3 Define a solid.
Ans: A solid is defined as that form of matter which possesses rigidity and hence possesses a definite shape and a
definite volume.
*Q.4. Give characteristics of solid state.
Ans: Characteristics properties of solids:
i. They have definite mass, volume, shape and density.
ii. Intermolecular distances are short.
i.e. constituent particles are very closely packed.
iii. Intermolecular forces are strong.
iv. Their constituent particles (atoms, molecules or ions) have fixed positions and can only oscillate
about their mean positions.
v. They are incompressible and rigid.
vi. Their density is greater than that of liquids and gases.
vii. Depending on the intermolecular forces of attraction melting points of the different solids ranges from
absolute zero (helium) to a few thousand Kelvin (diamond).
Solid State
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Std. XII Sci.: Perfect Chemistry - I
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Q.5. Why are solids rigid? (NCERT)
Ans: In solids, the particles are closely packed and the empty spaces between the particles are very small.
Therefore, they are incompressible and maintain their own shape when subjected to outside force. Hence,
they are rigid.
Q.6. Why do solids have a definite volume? (NCERT)
Ans: The intermolecular forces between the particles in the solid state are very strong. Therefore, they are
strongly held at fixed positions and particles cannot separate from one another. Hence, solids have a definite
volume.
1.1 Classification of Solids
*Q.7. Give the classification of solids.
Ans: Solids are classified as crystalline and amorphous on the basis of the nature of order present in the
arrangement of their constituent particles (atoms, ions or molecules).
i. Crystalline solids: A crystalline solid is a homogeneous solid in which the constituent particles
(atoms, ions or molecules) are arranged in a definite repeating pattern.
Crystalline solids are further classified as:
a. Isomorphous form: Two or more substances having the same crystal structure are said to be
isomorphous (iso-same, morphous-form).
Eg. i. NaF and MgO
ii. K2SO4 and K2SeO4
b. Polymorphous / Allotropic form: A single substance that crystallises in two or more forms
under different conditions is called polymorphous (many forms).
Eg. i. Graphite and diamond
ii. Rhombic and Monoclinic sulphur
ii. Amorphous / Psuedo solids / Super cooled solids: The substances that appear like solids but do not
have well developed perfectly ordered crystalline structure are called amorphous (no form) solids.
Eg. Jar, glass, plastic, rubber, butter, etc.
Q.8. Describe the term ‘amorphous’. Give a few examples of amorphous solids. (NCERT)
Ans: Refer Q.7. ii.
*Q.9. Explain crystalline solids and amorphous solids.
Ans: i. Crystalline solids:
a. A crystalline solid usually consists of a large number of small crystals, each of them having a
definite characteristic geometrical shape.
b. In a crystal, the arrangement of constituent particles (atoms, molecules or ions) is ordered.
c. It has long range order which means that there is a regular pattern of arrangement of particles
which repeats itself periodically over the entire crystal.
Eg. Sodium chloride and quartz
ii. Amorphous solids:
a. The arrangement of constituent particles (atoms, molecules or ions) in such a solid has only
short range order.
b. In such an arrangement, a regular and periodically repeating pattern is observed over short
distances only.
c. Such portions are scattered and in between the arrangement is disordered.
Eg. Quartz glass, rubber, plastics and amorphous silicon.
2
Quartz
Quartz glass
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Solid State
Std. XII Sci.: Perfect Chemistry -I
#Q.10. Explain the term isomorphism, polymorphism, anisotropy and unit cell.
Ans: i. Isomorphism:
a. Isomorphous substances contain constituent atoms of the substance in the same atomic ratio.
b. Some of the pairs of isomorphous substances are given below:
i. NaF and MgO (atoms in the ratio 1:1)
ii. NaNO3 and CaCO3 (atoms in the ratio 1:1:3)
iii. K2SO4 and K2SeO4 (atoms in the ratio 2:1:4)
iv. Cr2O3 and Fe2O3 (atoms in the ratio 2:3)
c.
The same atomic ratio, similar molecular formula or similar chemical properties of two or more
solids are not enough to claim that the substances are isomorphous.
Sodium chloride NaCl and potassium chloride KCl have almost all the properties identical, but
their crystalline structures are different.
d. Sodium chloride and potassium chloride are not isomorphous.
ii. Polymorphism:
a. A single substance that crystallises in two or more forms under different conditions is called
polymorphous (many forms).
b. Carbon has two polymorphic forms graphite and diamond.
c. Sulphur also exists in two polymorphic forms rhombic and monoclinic.
d. Calcium carbonate and silicon dioxide also exist in nature in two polymorphic forms.
e. Polymorphic forms are also called allotropic forms.
iii. Anisotropy:
a. All pure crystalline solids exhibit sharp melting points and melt completely at a constant
temperature.
b. The sharp melting points of the crystalline solids are due to the presence of uniform orderly
arrangement of its constituent particles.
c. The physical properties like refractive index, electrical conductance, dielectric constant, etc. of
crystalline solids change with the change in direction.
d. The ability of crystalline solids to change values of physical properties when measured in
different directions is called anisotropy.
e. The anisotropy in crystalline solid arises because the composition of solid changes with
direction. From the figure, it is evident that the composition of the medium changes with the
change of directions AB, CD, EF etc.
B D
E A C
Particle 1
Particle 2
Anisotropy in crystals, different arrangements of
constituent particles about different directions,
AB, CD and EF.
iv. Unit cell:
a. Crystalline solids are aggregates of many small, tiny crystals.
b. These tiny crystals are called unit cells.
c. A unit cell is a basic repeating structural unit of a crystalline solid.
Q.11. Classify the following as amorphous or crystalline solids: Polyurethane, naphthalene, benzoic acid,
teflon, potassium nitrate, cellophane, polyvinly chloride, fibre glass, copper.(NCERT)
Ans: Amorphous solids: Polyurethane, teflon, cellophane, polyvinyl chloride, fibre glass.
Crystalline solids: Benzoic acid, potassium nitrate, copper, naphthalene.
F
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Q.12. Refractive index of a solid is observed to have the same value along all directions. Comment on the
nature of this solid. Would it show cleavage property. (NCERT)
Ans: As the solid has same value of refractive index along all directions, it is isotropic and hence amorphous.
Being an amorphous solid, it would not show a clean cleavage when cut with a knife. Instead, it would
break into pieces with irregular surfaces.
*Q.13. What is a glass? Distinguish between crystalline solids and amorphous solids. Give examples.
Ans: i. Glass is an optically transparent material produced by fusing together silicon oxide with sodium
oxide, boron oxide and a trace amount of transition metal oxide is added to impart colour to the glass.
ii. By changing its composition, almost eight hundred different types of glasses are manufactured.
Quartz glass is obtained from only silicon dioxide.
iii. Pyrex glass is obtained by fusing together 60 to 80% SiO2, 10 to 25% B2O3 and remaining amount of
Al2O3.
iv. Soda lime glass is produced by fusing 75% SiO2, 15% Na2O and CaO.
v. Red glass contains trace amount of gold and copper.
vi. Yellow glass contains UO2.
vii. Blue glass contains CoO or CuO.
viii. Green glass contains Fe2O3 or CuO.
Property Crystalline solids Amorphous solids
1. Shape They have definite characteristic geometrical They have irregular shape.
shape.
2. Melting point They have sharp and characteristic melting
point.
3. Cleavage
property
When cut with a sharp edged tool, they split
into two pieces and the newly generated
surfaces are plain and smooth.
They have a definite and characteristic heat
4. Heat of
fusions of fusion.
5. Anisotropy They are anisotropic, i.e. have different
physical properties in different direction.
They do not have sharp melting
point. They gradually soften over a
range of temperature.
When cut with a sharp edged tool,
they cut into two pieces with
irregular surfaces.
They do not have definite heat of
fusion.
They are isotropic, i.e. have same
physical properties in all directions.
6. Nature They are true solids. They are pseudo solids or super
cooled liquids.
7. Order in They have long range order. They have only short range order.
8.
arrangement
of constituent
Eg. Copper silver, iron, zinc sulphide, common
salt, potassium nitrate, etc.
Glass, rubber, plastics, etc.
Q.14.What makes a glass different from a solid such as quartz? Under what conditions could quartz be
converted into glass? (NCERT)
Ans: Glass is an amorphous solid in which the constituent particles (SiO4 tetrahedra) have only a short range
order. In quartz, the constituent particles (SiO4 tetrahedra) have short range as well as long range order.
Quartz can be converted into glass by melting the quartz and then cooling it rapidly.
1.2 Classification of Crystalline Solids
Q.15. Give the classification of solids on the basis of different forces present in them.
Ans: Depending upon the nature of intermolecular forces present in the constituent particles, solids are classified
into the following four classes:
i. Molecular solids:
Molecular solids are those solids in which the constituent particles are molecules.
These are further subdivided into the following categories:
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Solid State
Std. XII Sci.: Perfect Chemistry -I
a. Polar molecular solids:
In these solids, molecules of the substance are formed by polar covalent bonds which are held
together by relatively stronger dipole-dipole interactions. On cooling and subjecting to high
pressures the gases first liquefy and then solidify to yield polar molecular solids.
Characteristics:
i. They are soft.
ii. They are non-conductors of electricity.
iii. Their melting points are higher than those of non polar molecular solids. Yet most of
these are usually gases or liquids under room temperature and pressure.
Eg. Solid SO2, solid NH3 and solid HCl.
b. Non polar molecular solids:
They comprise of either atoms or molecules formed by non polar covalent bonds. In these
solids, the atoms or molecules are held by weak dispersion forces or London forces.
Characteristics:
i. They are soft.
ii. They are non-conductors of electricity.
iii. They have low melting points and are usually in liquid or gaseous state at room
temperature and pressure.
Eg. Argon, helium, iodine, solid carbondioxide, etc.
c. Hydrogen bonded molecular solids:
The molecules of these solids contain polar covalent bonds between hydrogen and highly
electronegative atoms of small size such as F, O or N. Strong hydrogen bonding binds
molecules of such solids. The additional bond formed is called hydrogen bond.
Characteristics:
i. They are non-conductors of electricity.
ii. They are volatile liquids or soft solids under room temperature and pressure.
Eg. H2O, NH3.
ii. Ionic Solids:
All salts are crystalline in nature and are called ionic solids.
These solids are formed by the three dimensional arrangement of cations and anions bounded by
strong coulombic (electrostatic) forces. Ionic solids are formed by the force of attraction between ions
of opposite charges and force of repulsion between ions of same charges.
“These two opposite forces result into well ordered three dimensional arrangement of ions in ionic
solids”.
The actual arrangement of ions depends on three factors.
a. Sizes of the cation and anion.
b. The charges on the ions.
c. The ease with which the anion can be polarised.
Characteristics:
i. These solids are hard and brittle in nature.
ii. They have high melting and boiling points.
iii. Since the ions are not free to move about, they are electrical insulators in the solid state.
iv. However, in the molten state or when dissolved in water, the ions become free to move about
and they conduct electricity.
iii. Metallic solids:
a. Metallic solids are the crystalline solids formed by atoms of the same metallic element.
b. Metals are orderly collection of positive ions (called kernels) surrounded by and held together
by a sea of free electrons.
c. These electrons are mobile and are evenly spread throughout the crystal.
d. Each metal atom contributes one or more electrons towards this sea of mobile electrons.
e. The force of attraction between positively charged metallic ion and negatively charged sea of
delocalised electrons is called metallic bond.
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Std. XII Sci.: Perfect Chemistry - I
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Characteristics:
i. They possess high electrical and thermal conductivity. This is because on applying electric
current, electrons start flowing towards positively charged electrode. Similarly, when heat is
supplied to one portion of a metal, the thermal energy is uniformly spread throughout by free
electrons.
ii. They possess lustre and colour in some cases.
iii. They are highly malleable and ductile.
iv. They possess high melting points and high densities.
+ e−
+ e−
+ e−
+ e−
Metallic solid − Positive ions in a sea
of mobile electrons
iv. Covalent or Network Solids:
a. Covalent solids are those in which the constituent
particles are non-metal atoms linked to the adjacent
atoms by covalent bonds throughout the crystal.
b. As a result, a network of covalent bonds is formed
as shown in the figure.
c. They are also called giant molecules.
Characteristics:
i. As covalent bonds are strong and directional in
nature, these solids are very hard and brittle.
ii. They have extremely high melting points and may even decompose before melting.
iii. Depending upon availability of mobile electrons, they are either good conductors of electricity
or act as insulators.
Eg. Diamond and silicon carbide.
Note:
Molecular solid, ionic solid, metallic solid and covalent or network solid are the types of crystalline solids.
*Q.16. Explain:
i. Molecular solids ii. Hydrogen bonded molecular solids iii. Ionic solids
iv. Metallic solids v. Covalent solids
Ans: i. Molecular solids: Refer Q.15. i.
ii. Hydrogen bonded molecular solids: Refer Q.15. i. c.
iii. Ionic solids: Refer Q.15. ii.
iv. Metallic solids: Refer Q.15.iii.
v. Covalent solids: Refer Q.15. iv.
+ e−
+ e−
+ e−
+ e−
+ e−
+ e−
+ e−
+ e−
+ e−
*Q.17. Write a note on
i. Diamond ii. Graphite iii. Fullerene
Ans: i. Diamond:
a. It is the most precious crystal used in jewellery and it is an
allotropic form of carbon.
b. All carbon atoms in diamond are sp 3 hybridised carbon atoms
and the bonding continues in all direction to form big giant
network of covalent solid.
c. The strong covalent bonds between sp 3 hybridised carbon atoms
make diamond very strong and hard solid.
d. It is the hardest material with very high melting point of 3550°C.
154 pm
Network structure
of diamond
154 pm
Network structure
of diamond
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ii. Graphite:
a. Graphite is soft and a good conductor of electricity, due to its
typical structure as shown in the figure.
b. All carbon atoms in graphite are sp 2 hybridised. Each carbon
atom is covalently bonded to three other sp 2 hybridised carbon
atoms and form interlinked six membered rings of carbon atoms.
c. The remaining half filled unhybridised 2pz orbital is used for π
bonding so that layers of carbon atoms i.e. graphite are formed.
d. Because of the presence of these free eletrons in different layers,
graphite becomes a good conductor of electricity.
e. Further, the distance between the adjacent layers is greater than
carbon-carbon bond length. Hence, these layers are not bonded
to each other and can easily slip over each other. The adjacent
layer of carbon atoms are held together by weak van der Waal’s
forces of attraction.
f. Thus, graphite is soft and acts as a good solid lubricant.
iii. Fullerene:
a. A new allotrope of carbon was discovered when a high power laser
was focused on carbon.
b. The product formed was found to have formula C60 and had a shape
of soccer ball i.e. hollow sphere.
c. On this sphere there are sixty equidistant places occupied by carbon
atoms.
d. Like graphite all carbon atoms are sp 2 hybridized.
e. The delocalized molecular orbitals are spread over the complete
structure of the fullerene.
f. The structure of fullerene is similar to soccer-ball structure formed
by arranging carbon hexagons and carbon pentagon to form hollow
spheres to accommodate sixty carbon atoms present at the corners of
hexagons and pentagons.
g. Fullerenes are observed in carbon soot.
Std. XII Sci.: Perfect Chemistry -I
141.5 pm
Structure of graphite
h. Fullerene reacts with potassium to form a compound K35C60 which acts as super conductor of
electricity at 18 K.
i. Fullerene reacts with transition metal to form a catalyst.
j. Tubes made from fullerene and graphite are called nanotubes. These are used as high strength
materials, electric conductors, molecular sensors and semi conductors.
Q.18. Classify the following solids in different categories based on nature of intermolecular forces operating
in them: (NCERT)
Potassium sulphate, tin, benzene, urea, ammonia, water, zinc sulphide, graphite, rubidium, argon,
silicon carbide
Ans: Ionic solids : Potassium sulphate, zinc sulphide
Molecular solid : Benzene, argon
(non-polar)
Molecular solids : Urea
(Polar)
Molecular solids : Ammonia, water
(Hydrogen bonded)
Covalent or Network solids : Graphite, Silicon carbide
Metallic solids : Tin, rubidium
Q.19. Solid A is very hard, electrical insulator in solid as well as in molten state and melts at extremely high
temperature. What type of solid is it? (NCERT)
Ans: Covalent network solid like SiO2 (quartz) or SiC or C (diamond).
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Solid State
340 pm
The structure of C60,
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Std. XII Sci.: Perfect Chemistry - I
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Q.20. Ionic solids conduct electricity in molten state but not in solid state. Explain. (NCERT)
Ans: In solid state, the ions are present in fixed positions in the crystal lattice and cannot move when an electric
field is applied. However, when melted, the well ordered arrangement of the ions in the crystal is destroyed
and the ions are in a position to move about when an electric current is applied. Hence, ionic solids conduct
electricity in molten state.
Q.21. What type of solids are electrical conductors, malleable and ductile? (NCERT)
Ans: Metallic solids are electrical conductors, malleable and ductile.
Q.22. Classify the following solids into different types
*i. Plastic *ii. P4 molecule *iii. S8 molecule
*iv. Iodine molecule *v. Tetra phosphorous decoxide *vi. Ammonium phosphate
*vii. Brass *viii. Rubidium *ix. Graphite *x. Diamond
*xi. NaCl *xii. Silicon. xiii. SiC (NCERT) xiv. LiBr (NCERT)
Ans: i. Amorphous solid − Plastic
ii. Molecular Solid − P4 molecule, S8 molecule, Iodine molecule, Tetra phosphorous decoxides,
iii. Ionic solid − Ammonium phosphate, NaCl, LiBr
iv. Metallic solid − Brass, Rubidium
v. Network covalent − Graphite, Diamond, Silicon, SiC.
Q.23. ‘Stability of a crystal is reflected in the magnitude of its melting points’. Comment. Looking at the
melting points of solid water, ethyl alcohol, diethyl ether and methane, what can you say about the
intermolecular forces between these molecules? (NCERT)
Ans: i. The melting point of a crystal depends upon the magnitude of forces holding the constituent particles
together, which determine the stability. Higher the melting point, greater are the forces holding the
constituent particles together and hence greater is the stability.
Eg. Ionic crystals such as NaCl, KNO3, etc. have very high melting points and are stable. On the other
hand, molecular solids such as naphthalene, iodine, etc. are less stable because they have low values
of melting points.
The melting points of some compounds are:
Water = 273 K, ethyl alcohol = 155.8 K, diethyl ether = 156.8 K and methane = 90.5 K.
ii. The intermolecular forces in water and ethyl alcohol are mainly hydrogen bonding. The higher
melting point of water as compared to ethyl alcohol indicate that the hydrogen bonding in water is
stronger than in ethyl alcohol. Diethyl ether is a polar molecule and, therefore, the intermolecular
forces in diethyl ether are dipole-dipole interactions. On the other hand, methane is a non-polar
molecule and the only forces present in them are the weak van der Waals forces (London / dispersion
forces). These are weaker than dipole-dipole interactions and hence methane has very low melting
point as compared to diethyl ether.
Q.24. Explain:
i. The basis of similarities and difference between metallic and ionic crystals. (NCERT)
*ii. Ionic solids are hard and brittle. (NCERT)
*iii. Solid ice is lighter than water.
Ans: i. Metallic and ionic crystals:
a. Similarities:
Both ionic and metallic crystals have electrostatic forces of attraction. In ionic crystals, these
are between the oppositely charged ions. In metals, these are among the valence electrons and
the kernels. This is why both have high melting point. In both cases, the bond is nondirectional.
b. Differences:
In ionic crystals, the ions are not free to move. Hence, they cannot conduct electricity in the
solid state. They can do so only in the molten state or in aqueous solution. In metals, the
valence electrons are free to flow. Hence, they can conduct electricity in the solid state.
Ionic bond is strong due to electrostatic forces of attraction. Metallic bond may be weak or
strong depending upon the number of valence electrons and the size of the kernels.
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Std. XII Sci.: Perfect Chemistry -I
ii. Ionic crystals are hard because there are strong electrostatic forces of attraction among the oppositely
charged ions. They are brittle because ionic bond is non-directional.
iii. Solid ice is lighter than water:
a. X-Ray studies of liquid water shows that structure of liquid water and solid ice are almost
identical.
b. However on melting of ice some of the hydrogen bonds are broken and some of the empty
spaces are occupied by water molecules.
c. Hence, the density of liquid water becomes more than solid water (ice). So, solid ice is lighter
than water.
1.3 Unit cell and two and three dimensional lattices
*Q.25. What is a unit cell? Explain Bravais lattices.
Ans: A unit cell is the smallest repeating structural unit of a crystalline solid.
i. When unit cells of the same crystalline substance are repeated in space in all directions, a crystalline
solid is formed.
ii. For convenience, unit cell is drawn on paper by drawing lines connecting centres of constituent
particles.
iii. Each point at the intersection of the lines in the unit cell represents constituent particles, i.e. an ion or
an atom or molecule of the crystalline solid.
iv. Any point at the intersection of lines is called a lattice point.
v. The particles in a crystal are arranged in a definite repeating pattern. Therefore, any one lattice point
in the crystal is identical to the number of other lattice points in the crystal that have identical
environments.
vi. The collection of all the points in the crystal having similar environment is called space lattice.
vii. Lattice means a structure made of strips which cross each other diagonally.
a b
c
β
γ
α
A portion of a three dimensional cubic lattice and its unit
Bravais Lattices:
The crystalline solids have definite orderly arrangement of their constituent particles in three dimensions.
The positions of these particles in a crystal, relative to one another, are usually shown by points. The
arrangement of an infinite set of these points is called space lattice. Thus, a crystal lattice is a regular
arrangement of the constituent particles (atoms, ions or molecules) of a crystalline solid in three
dimensional space.
There are only 14 possible three dimensional lattices. These are called Bravais Lattices.
i. Each point in a lattice is called lattice point or lattice site.
ii. Each point in a crystal lattice represents one constituent particle which may be an atom, a molecule
(group of atoms) or an ion.
iii. The three dimensional arrangement of lattice points represents a crystal lattice.
iv. Lattice points are joined by straight lines to bring out the geometry of the lattice.
v. Shape of any crystal lattice depends upon the shape of the unit cell which in turn depends upon
following two factors
a. Its dimensions along the three edges are a, b and c. These edges may or may not be mutually
perpendicular.
b. The angle between the edges are α (between b and c), β (between a and c) and γ (between a and
b). Thus, a unit cell is characterised by six parameters a, b, c, ∝, β and γ. This is shown in the
figure. The complete crystal lattice can be obtained by extending the unit cell in all three
directions.
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Q.26. How are unit cells classified?
Ans: Unit cells can be broadly classified into two categories:
i. Primitive unit cells:
When constituent particles are present only on the corner positions of a unit cell, it is called as
primitive unit cells.
These are also called simple unit cells.
In all, there are seven types of primitive unit cells.
ii. Centred unit cells:
When a unit cell contains one or more constituent particles present at positions other than corners in
addition to those at corners, it is called a centred unit cell.
Centred unit cells are of three type:
a. Body-centred unit cell:
These unit cells contain one constituent particle (atom, molecule or ion) at its body-centre
besides the ones that are at its corners.
b. Face- centred unit cells:
These unit cells contain one constituent particle present at the centre of each face, besides the
ones that are at its corners.
c. End- centred unit cells:
In such a unit cell, one constituent particle is present at the centre of any two opposite faces
besides the ones present at its corners.
Q.27. Distinguish between face-centred and end-centred unit cells. (NCERT)
Ans: A face-centred unit cell has one constituent particle present at the centre of each face in addition to the
particles present at the corners. It has four atoms per unit cell.
An end-centred unit cell has one constituent particle each at the centre of any two opposite faces in addition
to the particles present at the corners. It has two atoms per unit cell.
*Q.28. Explain with the help of diagram
i. Seven types of unit cells. ii. Three types of cubic cells
iii. Two types of tetragonal unit cells. iv. Four types of orthorhombic unit cells.
v. Two types of monoclinic unit cells. vi. Triclinic unit cell
vii. Primitive hexagonal unit cell.
Ans: i. There are seven types of simple or primitive unit cells among crystals. These unit cells are
characterised by the lengths a, b and c and the angles α, β, γ.
These are as follows:
a. Cubic:
All the three axes are of equal length and are at right angles to each other (a = b = c, all angles =
90°).
b. Tetragonal:
The three axes are at right angles to each other but only two axes are equal (a = b ≠ c, all angles
= 90°).
c. Orthorhombic:
It has three unequal axes which are at right angles to each other (a ≠ b ≠ c, all angles = 90°).
d. Monoclinic:
The three axes are of unequal length and two angles are of 90° (a ≠ b ≠ c, two angles = 90° and
one angle ≠ 90°).
e. Hexagonal:
It has two edges of equal length and two angles of 90° and one angle of 120° (a = b ≠ c, two
angles = 90° and one angle = 120° ).
f. Rhombohedral:
The three axes are of equal length which are inclined at the same angle but the angle is not
equal to 90° (a = b = c, all three angles equal but none equal to 90°).
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Solid State
Std. XII Sci.: Perfect Chemistry -I
g. Triclinic:
The three axes are of unequal length, and all angles are different but none is perpendicular to
any of the others (a ≠ b ≠ c, all angles different and none equal to 90°).
CUBIC
a = b = c
α = β = γ = 90°
HEXAGONAL
a = b ≠ c
α = β = 90°, γ ≠ 120°
ii. Three types of cubic cells:
There are three types of cubic cells
a. Simple cubic
b. Body centred cubic
c. Face centred cubic
Primitive
(or simple)
TETRAGONAL
a = b ≠ c
α = β = γ = 90°
The seven crystal systems
iii. Two types of tetragonal unit cells:
There are two types of tetragonal unit cells
a. Primitive or simple tetragonal
b. Body centred tetragonal
ORTHORHOMBIC
a ≠ b ≠ c
α = β = γ = 90°
RHOMBOHEDRAL
a = b = c
α = β = γ ≠ 90°
Body-centred
Primitive Body-centred
MONOCLINIC
a ≠ b ≠ c
α = γ = 90°, β ≠ 90°
TRICLINIC
a ≠ b ≠ c
α ≠ β ≠ γ ≠ 90°
Face-centred
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iv. Four types of orthorhombic unit cells:
There are four types of orthorhombic unit cells
a. Primitive or Simple orthorhombic
b. Body centred orthorhombic
c. End centred orthorhombic
d. Face centred orthorhombic
TARGET Publications
Primitive
Body-centred End- centred Face- centred
v. Two types of monoclinic unit cells:
There are two types of monoclinic unit cells
a. Primitive or Simple monoclinic
b. End centred monoclinic
Primitive
vi. Triclinic unit cell:
Triclinic unit cell exists in only one type
a. Primitive
β
γ
α
vii. Primitive hexagonal unit cell:
β
γ
α
HEXAGONAL
a = b ≠ c
α = β =90°, γ = 120°
More than
90°
Less than
90°
End- centred
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Q.29. Name the parameters that characterise a unit cell. (NCERT)
Ans: Refer Q.26 and 28. i.
Q.30. Distinguish between hexagonal and monoclinic unit cells. (NCERT)
Ans: For hexagonal unit cell, a = b ≠ c, α = β = 90°, γ = 120°
Solid State
For monoclinic unit cell, a ≠ b ≠ c, α = γ = 90°, β ≠ 90°
Std. XII Sci.: Perfect Chemistry -I
Q.31. Give the significance of a lattice point. (NCERT)
Ans: Each lattice point represents one constituent particle of the solid, which may be an atom, a molecule or an
ion.
*Q.32. Explain coordination number. OR
What is meant by the term ‘coordination number’? (NCERT)
Ans: The coordination number of constituent particle of the crystal lattice is the number of particles surrounding
a single particle in the crystal lattice.
In case of ionic crystals, coordination number of an ion in the crystal is the number of oppositely charged
ions surrounding that ion.
Q.33. Explain the following:
i. Open structure
ii. Space filling structure
Ans: i. Open structure:
Open structures are those structures where each small sphere represents only the centre of the
particle occupying that position and not the actual size.
In this structure, the arrangement of particles is easier to follow. This is shown in the figure below.
Simple
Face-centred
Open structures
Body-centred
ii. Space filling structure:
Space filling structures are those structures which show how the particles are pack within the solids.
(a)
(b)
(c)
Space filling structures of (a) Simple cubic (b) Close cubic or face centred
(c) Body centred arrangements.
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Q.34. Calculate the number of atoms present per unit cell in:
i. Simple or primitive cubic lattice
ii. Body-centred cubic lattice
iii. Face-centred cubic lattice
Ans: i. Simple or primitive cubic lattice:
In this unit cell, the points (atoms, ions or molecules) are present at all corners of a cube. This is
shown in figure (a). It is clear from figure (c) that atom present at each corner contributes 1/8 to each
cube because it is shared by 8 cubes. Now, there are 8 atoms at the corners.
1
8
(a) Open structure (b) Space filling structure (c) Actual portion of atoms belonging to
one unit cell
Fig. Simple cubic arrangement and number of spheres per unit cell.
atom
∴ The number of atoms present in each unit cell = 8 corner atoms × 1
atom per unit cell = 1 atom
8
Thus, simple cubic lattice has one atom per unit cell.
ii. Body-centred cubic lattice (bcc):
Body centred unit cell has an atom at each of its corners and also one atom at its body centre as
shown in the figure.
1
8 atom
(a) Open structure
(b) Space filling structure
Body-centred cubic unit cell
It is clear from the above figures that there are eight atoms at the corners and each is shared by 8 unit
cells so that the contribution of each atom of corner is 1/8.
In addition, there is one atom in the body of the cube as shown in figure (c) which is not shared by
any other cube.
Thus, the number of atoms present at the corners per unit cell
= 8 corner atoms × 1
atom per unit cell = 1 atom
8
∴ The number of atoms present at the centre of the cube = 1 × 1 = 1 atom
∴ The number of atoms present in the cell = 1 + 1 = 2 atom
Thus, body centred cubic has 2 atoms per unit cell.
(c) Actual portion of atoms belonging to
one unit cell
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Std. XII Sci.: Perfect Chemistry -I
iii. Face-centred cubic lattice (fcc):
It is also called cubic close packed unit cell. It has atoms at all the corners as well as at the centre of
each of the six faces. In this arrangement, there is one atom at each of the eight corners. It is clear
from figure (c) that atoms present at each corner contribute 1/8 to each cube because it is shared by 8
cubes. In addition, there are six atoms at the faces of the cube and each is shared by two unit cells.
Therefore, the contribution of each atom at the face per unit cell is 1/2 as shown in figure (d).
(b) Space filling structure
(a) Open structure (c) Actual portion of atoms
belonging to one unit cell
Thus, the number of atoms present at corners per unit cell = 8 corner atoms × 1
atom per unit cell = 1
8
The number of atoms present at faces per unit cell = 6 atoms at the faces × 1
Cubic close packed or face centred cubic arrangement and share
of each atom per unit cell.
atom per unit cell = 3
2
∴ Total number of atoms per unit cell = 1+3 = 4 atom
Thus, a face centred cubic unit cell has 4 atoms per unit cell.
Note:
Number of atoms per unit cell is represented by letter ‘z’
Q.35. What is a lattice? Explain a unit cell in two dimensional lattices.
Ans: The repetition of unit cells in a crystal results in the regular spacing of the atoms called the space lattice.
Two dimensional lattices:
i. A two dimensional lattice is a regular arrangement of points in the plane of paper.
ii. Unit cell is obtained by choosing any four points and connecting them by lines. The points represent
the constituent particles, i.e. atoms, ions or molecules.
iii. In some cases, the unit cell may be centred unit cell or primitive unit cell.
90° 90°
(a) (b) (c) (d) (e)
Five two dimensional lattices with their unit cells as (a) Sqauare, (b) Rectangular,
(c) Parallelogram, (d) Rectangular with interior point and (e) Rhombus
iv. There are five types of two dimensional lattices with their unit cell as follows.
No. Lattice Unit cell
i. Square lattice Square
ii. Rectangular lattice Rectangle
iii. Parallelogram lattice Parallelogram
iv. Rhombic lattice Rectangular with interior point
v. Hexagonal lattice Rhombus with an angle of 6Dc
(d) An atom at face centre
of unit cell is shared
between 2 unit cells
90°
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v. The complete lattice can be generated by repeatedly moving the unit cell in the direction of its edges
by a distance equal to the cell edge as shown in the figure.
Generating complete two dimensional lattice by
repeatedly moving the unit cell in the direction of the cell
edges by distance equal to cell edge
vi. Such two dimensional lattices are often seen in the wallpapers and tiled floors.
*Q.36. Explain how to deduce coordination number of cation.
Ans. i. The coordination number of constituent particle of the crystal lattice is the number of particles
surrounding a single particle in the crystal lattice.
ii. If particular sphere is shared, then its contribution to a given cation is accounted accordingly.
Q.37. Explain how much portion of an atom at (i) corner, (ii) body-centre of a cubic unit cell is part of its
neighbouring unit cell. (NCERT)
Ans: i. An atom at the corner is shared by eight adjacent unit cells. Hence, portion of the atom at the corner
that belongs to one unit cell = 1
8 .
ii. The atom at the body centre of a cubic unit cell is not shared by any other neighbouring unit cell.
*Q.38. Give the number of lattice points in one unit cell of the crystal structures.
i. Simple cubic ii. Face-centred cubic
iii. Body centred cubic
OR
iv. Face-centred tetragonal
How many lattice points are there in one unit cell of each of the following lattice? (NCERT)
i. Face-centred cubic ii. Face-centred tetragonal
iii. Body centred cubic iv. Simple cubic
Ans: i. Lattice points in face centred cubic
= 8 (at corners) + 6 (at face centred) = 14.
Lattice points per unit cell = 8 × 1 1
+ 6 × = 4.
8 2
ii. In face centred tetragonal, number of lattice points are:
= 8 (at corners) + 6 (at face centred)
Lattice points per unit cell = 8 × 1 1
+ 6 × = 4
iii. In body centred cubic arrangement number of lattice points are:
= 8 (at corners) + 1 (at body centred)
Lattice points per unit cell = 8 × 1
+ 1 = 2
8
iv. Lattice points in simple cubic = 8 (at corners) = 8
Lattice points per unit cell = 8 × 1
= 1.
8
8
2
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1.4 Packing in solids
Solid State
Std. XII Sci.: Perfect Chemistry -I
Q.39. Explain one dimensional close packing in solids.
Ans: In solids, the constituent particles are close packed, leaving the minimum vacant space. Consider the
constituent particles as identical hard spheres.
Close packing in one dimension:
i. There is only one way of arranging spheres in a one dimensional close packed structure, that is to
arrange them in a row and touching each other as shown in the figure.
ii.
Close packing of spheres in one dimension
In this arrangement, each sphere is in contact with two of its neighbours.
iii. Thus, in one dimensional close packed arrangement, the coordination number is 2.
Q.40. Explain two dimensional close packing in solid.
Ans: Two dimensional close packed structures can be generated by stacking (placing) the rows of close packed
spheres. This can be done in two different ways.
i. Square close packing:
a. The second row may be placed in contact with the first
one such that the spheres of the second row are exactly
above those of the first row.
b. The spheres of the two rows are aligned horizontally
as well as vertically.
c. If we call the first row as ‘A’ type row, the second row
being exactly the same as the first one, is also of ‘A’
type.
d. Similarly, we may place more rows to obtain AAA type of arrangement as shown in figure (a).
e. In this arrangement, each sphere is in contact with four of its neighbours. Thus, the two
dimensional coordination number is four.
f. Also if the centre of these 4 immediate neighbouring spheres are joined, a square is formed.
Hence, this packing is called square close packing in two dimension.
ii. Hexagonal close packing:
a. The second row may be placed above the first one in
staggered manner such that its spheres fit in the
depressions of the first row.
b. If the arrangement of spheres in the first row is called
‘A’ type, the one in the second row is different and
may be called ‘B’ type.
c. When the third row is placed adjacent to the second in
staggered manner, its spheres are aligned with those of
the first layer. Hence this layer is also of ‘A’ type.
(a) Square close packing
d. The spheres of similarly placed fourth row will be aligned with those of the second row
(‘B’ type).
e. Hence, this arrangement is of ABAB type. In this arrangement there is less free space and this
packing is more efficient than the square close packing.
f. Each sphere is in contact with six of its neighbours and the two dimensional coordination
number is six.
g. The centres of these six spheres are at the corners of a regular hexagon as shown in the figure
(b). Hence this packing is called two dimensional hexagonal close packing.
Q.41. What is two dimensional coordination number of a molecule in square close packed layer? (NCERT)
Ans: In the two-dimensional square close packed layer, the atom touches 4 nearest neighbouring atoms. Hence,
its coordination number = 4.
B
B
(b) Hexagonal close packing
of spheres in two dimensions
A
A
A
A
A
A
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Std. XII Sci.: Perfect Chemistry - I
Q.42. What is the coordination number of atoms: (NCERT)
i. in a cubic close-packed structure?
ii. in a body-centred cubic structure?
Ans: i. The coordination number of atoms in a cubic close-packed structure is 12
ii. The coordination number of atoms in a body-centred cubic structure is 8.
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Q.43. Define interstitial voids.
Ans: 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.
These are of two types.
i. Tetrahedral voids
ii. Octahedral voids
Q.44. Explain the following terms.
a. Tetrahedral voids
b. Octahedral voids
Ans: a. Tetrahedral voids:
A sphere in the second layer is placed above three spheres touching one another in the first layer.
This is shown in the figure. The centres of these spheres lie at the apices of a tetrahedron. It may be
noted that the shape of the void is not tetrahedral, but the arrangement around this void is tetrahedral.
Thus, the vacant space among four spheres having tetrahedral arrangement is called tetrahedral site
or tetrahedral void.
b. Octahedral void:
Octahedral void is formed at the centre of six spheres as shown in the figure. It is clear that each
octahedral site is produced by two sets of equilateral triangles which point in opposite directions.
Thus, the void formed by two equilateral triangles with apices in opposite directions is called
octahedral site or octahedral void. This site is, therefore, surrounded by six spheres lying at the
vertices of a regular octahedron.
Octahedral void
Tetrahedral void
Octahedral void
Tetrahedral void
Note:
The number of these two types of voids depends upon the number of close packed spheres.
Octahedral void
Q.45. How will you distinguish between tetrahedral void and octahedral void? (NCERT)
Ans: A void surrounded by four spheres is called a tetrahedral void while a void surrounded by six spheres is
called an octahedral void.
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Solid State
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*Q.46. Explain with the help of neat diagram AAAA and ABAB and ABCABC type of three dimensional
packings.
Ans: i. AAAA type of three dimensional packing:
a. For placing the second square close-packed layers above the first row, follow the same
procedure that was followed when one row was placed adjacent to the other row.
b. The second layer is placed over the first layer such that these spheres of the second layers are
exactly above those of the first layer.
c. In this arrangement, spheres of both layers are perfectly aligned horizontally as well as
vertically. Similarly, more layers can be placed one above the other. This is shown in figure (a).
d. If the arrangement of spheres in the first layer is called ‘A’ type, the other layers will also be
‘A’ type because they have same arrangement.
e. This type of packing is referred to as AAAA type. This type of packing is also called simple
cubic packing and its unit cell is the primitive unit cell as shown in figure (b).
(a) (b)
ii.
Simple cubic lattice formed by AAAA…. Arrangement.
ABAB type of three dimensional packing:
a. This packing can be obtained by stacking two dimensional layers one above the other.
b. Mark the spheres in the first row as A. It is clear from figure (a) that in the first layer there are
some empty spaces or hollows called voids. These are triangular in shape.
c. These triangular voids or hollows are of two types marked as ‘a’ and ‘b’.
d. All the hollows are equivalent but the spheres of second layer may be placed either on hollows
which are marked ‘a’ or on other set of hollows marked ‘b’.
e. It may be noted that it is not possible to place spheres on both types of hollows.
f. Let us place the spheres on hollows marked ‘b’ to make the second layer which may be labelled
as B layer. Obviously the holes marked ‘a’ remain unoccupied while building the second layer.
The second layer is indicated as dotted circles in figure (b).
a
b
a a
b b
a
a
b b
a
b
a a
b b
c c
a a
b b
(a)
(b)
Packing of second layer (b) on first layer (a)
g. Whenever a sphere of second layer is placed above the void of first layer, a tetrahedral void is
formed.
h. The voids ‘a’ are double triangular voids. The triangular void in the second layer are above the
triangular voids in the first layer and the triangular shapes of these voids do not overlap. Such
voids are surrounded by six spheres and are called octahedral voids.
i. When a third layer is to be added, again there are two types of hollows available. One type of
hollow marked ‘a’ are unoccupied hollows of the first layer. The other type of hollows are
hollows in the second layer (marked c).
j. Tetrahedral voids of the second layer may be covered by the spheres of the third layer.
k. In this case, the spheres of the third layer are exactly aligned with those of the first layer.
B
A
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l. Thus, the pattern of spheres is repeated in alternate layers. This pattern is often written as
ABAB pattern.
m. This type of packing is also known as hexagonal close packing. It is abbreviated as hcp. For
simplicity, hcp arrangement can be drawn as shown in figure (b).
n. This sort of arrangement of atoms is found in many metals like magnesium and zinc.
A
A
B
B
Hexagonal
packing
(a)
Hexagonal
(b)
B
(a) Hexagonal close-packing exploded view showing stacking of layers of spheres
(b) four layers stacked in each case and (c) geometry of packing
iii. ABCABC type of three dimensional packing:
The third layer may be placed above the second layer in a manner such that its spheres cover the
octahedral voids. When placed in this manner, the spheres of the third layer are not aligned with
those of either the first or the second layer. This arrangement is called ‘C’ type. Only when fourth
layer is placed, its spheres are aligned with those of the first layer as shown in figure (a) and figure
(b). This pattern of layers is often written as ABCABC. This structure is called cubic close packed
(ccp) or face-centred cubic (fcc) structure. Metals such as copper and silver crystallise in this
structure.
A
C
B
A
(b)
cubic
B
A
C
A
B
Cubic close
(d)
packing
(a)
fcc
(c)
(a) Cubic close-packing exploded view showing stacking of layers of sphere(b) Stacking of
four layers (c) geometry of packing (d) ABCABC arrangement of layers when octahedral
void is covered (e) fragment of structure formed by this arrangement resulting in cubic
closed packed (ccp) or face centred cubic (fcc)
A
A
Both these types of close packing are highly efficient and 74% space in the crystal is filled. In either
of them, each sphere is in contact with twelve spheres. Thus, the coordination number is 12 in
either of these two structure.
A
C
B
A
hcp
(c)
(e)
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Solid State
Std. XII Sci.: Perfect Chemistry -I
Octahedral voids are bigger is size and are as many in number as is the number of close packed spheres.
Formula of a compound can be deduced from the information about the nature and fraction of voids
occupied in its structure.
Imperfections or defects in crystals are deviations from the ideal structures.
Defect in crystals are broadly of two types-point defects (irregularities around a point) and line or plane
defects (irregularities present in entire row or layer of the particles).
Point defects are of three types-stoichiometric defects, impurity defects and non-stoichiometric defects.
Stoichiometric defects do not disturb the stoichiometry of the substance. They are basically of two types,
vacancy defect (absence of a particle at a lattice point) and interstitial defect (presence of a particle at
interstitial position).
Impurity defects in ionic solids create voids in the crystal structure and increase their conductivity.
Non-stoichiometric defects are of two types-metal excess defects and metal deficiency defects.
On the basis of conductivity, substances can be classified as conductors, insulators and semiconductors.
Electrical conductivity of semiconductors can be increased by doping them with electron-rich or electrondeficit
impurities.
On the basis of magnetic properties, substances can be classified as paramagnetic, diamagnetic,
ferromagnetic, antiferromagnetic and ferrimagnetic.
Formula
Isomorphous Form
Eg.
i. NaF and MgO
ii. K2SO4 and K2SeO4
1. Density of unit cell:
z.M
d = 3
a.N A
where, a is edge of unit cell
NA = Avogadro number (6.023 × 10 23 )
M = Molar mass
z = number of atoms per unit cell
For fcc, z = 4
for bcc, z = 2
for simple cubic, z= 1
Quick Review
Solids
Crystalline Amorphous/Pseudo
solids/ Super cooled
solids
Polymorphous/Allotroplic
Form
Eg.
i. Graphite and
diamond
ii. Rhombic and
Monoclinic sulphur
Eg. Jar, glass, plastic,
rubber, butter, etc.
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As 2
3
= n × 2
3
TARGET Publications
rd of the octahedral voids are occupied by ferric ions, therefore the number of ferric ion present
= 2n
3
Ratio of Fe 3+ : O2 = 2n
: n
3
∴ Hence the formula of ferric oxide is Fe2O3.
Example 20.
Atoms of element B form hcp lattice and those of the element A occupy 2/3rd of tetrahedral voids. What is
the formula of the compound formed by the elements A and B? (NCERT)
Solution:
The number of tetrahedral voids formed is equal to twice the number of atoms of element B and only 2/3 rd of
these are occupied by the atoms of element A. Hence the ratio of the number of atoms of A and B is 2 × (2/3): 1
or 4:3 and the formula of the compound is A4B3.
*Example 21.
Atoms C and D form fcc crystalline structure. Atom C is present at the corners of the cube and D is at the
faces of the cube. What is the formula of the compound?
Solution:
Given: C is fcc present at corners of the cube
D is at the faces of the cube.
To find: Formula of the compound.
Calculation: As C is present at the 8 corners of the cube, therefore, number of atoms of C in the unit cell
= 1
× 8 = 1
8
As D atoms are present at the face centres of the 6 faces of the cube, therefore, the number of
atoms of D in the unit cell = 1
× 6 = 3
2
∴ Ratio of atoms C : D = 1 : 3.
Hence, the formula of the compound is CD3
*Example 22.
An element A and B constitute bcc type crystalline structure. Element A occupies body centre position and
B is at the corner of cube. What is the formula of the compound? What are the coordination numbers of A
and B? (Formula AB, coordination numbers of A = 8 and B = 8)
Solution:
Given: A is bcc.
B is at the corner.
To find: Formula of the compound
Co-ordination numbers.
Calculation: As atoms B are present at the 8 corners of the cube, therefore, number of atoms of B in the unit cell
= 1
× 8 = 1
8
As atoms A are present at the body centre, therefore, number of atoms of A in the unit cell = 1
∴ Ratio of A : B = 1 : 1
Hence, the formula of the compound is AB.
Co-ordination number of each A and B = 8.
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Solid State
Practice problem
1. The edge length of NaCl unit cell is 564 pm.
What is the density of NaCl in g cm 3 . (NA =
6.023 × 10 23 and atomic masses of Na = 23, Cl
= 35.5)
2. An element occurs in bcc structure. It has a
cell edge of 250 pm. Calculate its atomic mass
if its density is 8.0 g cm 3
3. Tungsten has body centred cubic lattice. Each
edge of the unit cell is 316 pm and density of
the metal is 19.35 g cm 3 . How many atoms are
present in 50g of the element.
4. An element density 6.8 g cm −3 occurs in bcc
structure with cell edge of 290 pm. Calculated
the number of atoms present in 200g of the
element.
5. Lead (II) sulphide crystal has NaCl structure.
What is its density ? The edge length of the
unit cell of PbS crystal is 500 pm.
(atomic masses : Pb = 207, S = 32)
6. If three elements P, Q and R crystallizes in a
cubic solid lattice with P atoms at the corners,
Q atoms at the cube centre and R atoms at the
centre of edges, then write the formula of
compound.
7. A solid is made of two elements X and Y.
Atoms X are in fcc arrangement and Y atoms
occupy all the octahedral sites and alternate
tetrahedral sites. What is the formula of the
compound?
Multiple choice Questions
1. A compound alloy of gold and copper
crystallizes in a cube lattice in which the gold
atoms occupy the lattice points at the corners
of cube and copper atoms occupy the centres
of each of the cube faces. The formula of this
compound is
(A) AuCu (B) AuCu
2
(C) AuCu
3
(D) AuCu4
2. Which is not a property of solids?
(A) Solids are always crystalline in nature.
(B) Solids have high density and low
compressibility.
(C) The diffusion of solids is very slow.
(D) Solids have definite volume.
Std. XII Sci.: Perfect Chemistry -I
3. One among the following is an example of
ferroelectric compound
(A) Quartz (B) Lead chromate
(C) Barium titanate (D) Tourmaline
4. Which is covalent solid?
(A) Fe2O3 (B) Diamond
(C) Graphite (D) All of the three
5. Graphite is an example of
(A) Ionic solid
(B) Covalent solid
(C) Van der Waals crystal
(D) Metallic crystal
6. 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
7. Glass is:
(A) microcrystalline solid
(B) super cooled liquid
(C) Gel
(D) Polymeric mixture
8. Which of the following is/are pseudo solids?
i. KCl
ii. Barium chloride dihydrate
iii. Rubber
iv. Solid cake left after distillation of coal
tar
(A) i, iii (B) ii, iii
(C) iii, iv (D) only iii
9. A solid having no definite shape is called
(A) Amorphous solid
(B) Crystalline solid
(C) Anisotropic solid
(D) Allotropic solids
10 Which is/are amorphous solids
(A) Rubber (B) Plastics
(C) Glass (D) All of the three
11. Amorphous substances show:
(i) Short and long range order
(ii) short range order
(iii) long range order
(iv) no sharp m.p.
(A) (i) and (iii) are correct
(B) (ii) and (iii) are correct
(C) (iii) and (iv) are correct
(D) (ii) and (iv) are correct
47

Std. XII Sci.: Perfect Chemistry - I
48. The flame colours of metal ions are due to
(A) Frenkel defect
(B) Schottky defect
(C) Metal deficiency defect
(D) Metal excess defect
49. Certain crystals produce electric signals on
application of pressure. This phenomenon is
called
(A) Pyroelectricity (B) Ferroelectricity
(C) Piezoelectricity (D) Ferrielectricity
50. Some polar crystals produce small electric
current heating. This phenomenon is called:
(A) Piezoelectricity
(B) Pyroelectricity
(C) Ferroelectricity
(D) Antiferroelectricity
51. Crystals where dipoles may align themselves
in an ordered manner so that there is a net
dipole moment, exhibit:
(A) Pyro-electricity
(B) Piezo-electricity
(C) Ferro-electricity
(D) Antiferro-electricity
50
Answers to Practice Problems
1. 2.16 g cm 3
2. 37.64
3. 1.61 × 10 23 atoms.
4. 2.4 × 10 24 atoms.
5. 12.7 g cm −3
6. PQR3
7. XY2
Answers to Multiple Choice Question
1. (C) 2. (A) 3. (C) 4. (D)
5. (B) 6. (C) 7. (B) 8. (C)
9. (A) 10. (D) 11. (D) 12. (D)
13. (D) 14. (B) 15. (D) 16. (D)
17. (A) 18. (C) 19. (D) 20. (B)
21. (A) 21. (B) 23. (C) 24. (A)
25. (A) 26. (D) 27. (B) 28. (A)
29. (B) 30. (D) 31. (B) 32. (D)
33. (C) 34. (C) 35. (A) 36. (C)
37. (D) 38. (B) 39. (D) 40. (A)
41. (C) 42. (B) 43. (B) 44. (D)
45. (D) 46. (D) 47. (C) 48. (D)
49. (C) 50. (B) 51. (B)
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