C M Y K

OBJECTIVE

CHEMISTRYCHEMISTRY

MH –CET

MBD House, Gulab Bhawan 6, B.S.Z. Marg, New Delhi-110002

An ISO 9001:2008 Certified Company

ByDr. S. P. Jauhar

(Author of Modern's abc + of Chemistry Series)Formerly Professor of Chemistry

Department of ChemistryPanjab University, Chandigarh

NEW EDITION 2016-17

Dr. Sheenu JauharB.Sc. (Gold Medalist), M.Sc. (Gold Medalist),

Ph.D.(Panjab University)

Price : 460.00`

Maharashtra State Engineering & Pharmacy,BAMS, BHMS, Paramedical Entrance Examinations

MBD House, Railway Road, Jalandhar City.MODERN PUBLISHERS

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MODERN'S abc SERIES OFOBJECTIVE BOOKS

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Modern's abc JEE-Mainof Objective Chemistry for

Modern's abc JEE-Mainof Objective Mathematics for

Modern's abc NEETof Objective Biology for

Modern's abc NEETof Objective Physics for

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Modern's

Modern's

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of Biology

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MODERN'S abc + SERIES OFSCIENCE TEXTBOOKS

FOR CLASS XI & XII

C M Y K

Modern's abc MH-CETof Objective Physics for

Modern's abc MH-CETof Objective Chemistry for

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We feel pleasure in presenting the book 'Modern's abc of Objective Chemistry' for students appearing in MH-CETexaminations . The book has been thoroughly written in accordance with the latest syllabus and changing trends ofexaminations.

This provides thorough discussion of the chapters emphasising all the important formulae, facts and terms. This canserve as quick revision of the chapter before the examination. This part has been enriched with new materials, Keypoints, Learning Plus, Facts to Memorise, In Focus etc.

These include a variety of questions in the form of multiple choice questions. These questions are graded according todegree of difficulty.These are:

Level I (Basic Conceptual Questions),Level II (Comprehensive Questions) andRecent Examination Questions (Questions from AIEEE/JEE , AIPM/NEET and MH- CET examinations).

The book provides answers to all questions. This is an important feature of the book. Some of the difficult questionshave been fully solved whereas hints to many other questions are given.

This is yet another important feature of the book. Revision tests are given at the end of each chapter. These will helpstudents to check their performance after they have gone through the chapters. Four unit evaluation tests are given atthe end of each unit. Five mock test papers covering the complete syllabus are also given at the end of the book. All thesewill provide sufficient materials to students for practice during their preparation.

We express my sincere thanks to my friends, colleagues and dear students who have been helping us in different waysduring the revision of our book. We acknowledge with thanks the untiring efforts of our publisher, Mr. BalwantSharma National Sales Head, Mr. Manik Juneja National Head–Content Operations, Modern Publishers and hisefficient staff in bringing out this book. We are also indebted to Mr. S.K. Sikka, Mr. Ravinder Pathania and Mr. B.S.Rawat for their valuable help.

We hope the book will be well-received by all aspirants and will induce confidence in them and prepare them for thechallenges ahead.

Suggestions for the improvement of the book will be gratefully acknowledged.

REVISION NOTES

QUESTIONS

ANSWERS

REVISION TESTS, SELF-EVALUATION TESTS AND MODEL PAPERS

ACKNOWLEDGEMENT

– Dr. S. P. Jauhar– Dr. Sheenu Jauhar

C M Y K

We feel pleasure in presenting the book for students

appearing in MH-CET examinations . The book has been thoroughly written in accordance

with the latest syllabus and changing trends of examinations.

'Modern's abc of Objective Physics'

MODERN'SOF

We feel pleasure in presenting the book for students

appearing in MH-CET examinations . The book has been thoroughly written in accordance

with the latest syllabus and changing trends of examinations.

'Modern's abc of Objective Chemistry'

ABOUT THE BOOK

MODERN'SOF

ABOUT THE BOOK

OBJECTIVE CHEMISTRY

OBJECTIVE PHYSICS

SPECIAL FEATURES

SPECIAL FEATURES

C M Y K

MH–CETFor

�

�

�

�

�

••

�

Comprehensive, precise and exclusively written for

Includes previous years'

Review Section emphasizes important definitions, terms, facts, formulae,

principles, theories, etc.

Large variety and number of Multiple Choice Questions covering all aspects

of each chapter/concept

Questions arranged in increasing order of difficulty –

Level 1 (basic conceptual questions)

Level 2 (application-based questions)

Five mock tests based on recent years' competitive examination questions

Maharashtra

MH-CET questions

�

�

�

�

�

••

�

Comprehensive, precise and exclusively written for

Includes previous years'

Review Section emphasizes important definitions, terms, facts, formulae,

principles, theories, etc.

Large variety and number of Multiple Choice Questions covering all aspects

of each chapter/concept

Questions arranged in increasing order of difficulty –

Level 1 (basic conceptual questions)

Level 2 (application-based questions)

Three mock tests based on recent years' competitive examination questions

Maharashtra

MH-CET questions

For MH–CET

C M Y K

MODERN'SOF

OBJECTIVE MATHEMATICS

We feel pleasure in presenting the book for

students appearing in MH-CET examinations. The book has been thoroughly written in

accordance with the latest syllabus and changing trends of examinations.

'Modern's abc of Objective Mathematics'

ABOUT THE BOOK

ForMH–CET

SPECIAL FEATURES

�

�

�

�

�

••

�

Comprehensive, precise and exclusively written for

Includes previous years' AIEEE/JEE/IIT questions

Review Section emphasizes important definitions, terms, facts, principles,

theories, etc.

Large variety and number of Multiple Choice Questions covering all aspects

of each chapter/concept

Questions arranged in increasing order of difficulty –

Level 1 (basic conceptual questions)

Level 2 (application-based questions)

Two mock tests based on recent years' competitive examination questions

Maharashtra

For Class XI & XII School Textbooks

Modern's abc + of Biology has been written specially for students of XII under 10+2 system of education of CBSE and other Boards following NCERTpattern of Examination. Ever since the publication of the first edition, the book has been receiving the overwhelming response from teachers and taughtalike. Keeping in view the recent edition of the book has been re-written as per latest syllabus. The book has been supplemented with practice problemsand some interesting facts for competitive examinations at the end of each chapter.All the objective type questions and very short answer questions havebeen answered. Special attempt has been made to make the book useful for students preparing for competitive examinations for entrance to variousmedical Colleges. Superfluous details present the text material in most practical and original way.With all these exclusive features, the book is bound to be the first choice of students all over India for Board and various competitive examinations.

The book designed for Higher Secondary class fulfils the student's need for a basic study of the concepts, methods and logic of modern discreteMathematics. Its subject matter is simple, up-to-date and in accordance with the changing trends of different examinations. Solved examples andunsolved problems have been selected very carefully and graded properly. Keeping in view the modern trend, the exercises have been dividedinto three groups viz. " Very ShortAnswer Type Questions", "ShortAnswer Type Questions" and "LongAnswer Type Questions."Almost each unitis followed by "Competition Corner" in order to meet the requirements of those students who are to appear in various competitive examinations foradmission in I.I.T., Roorkee and other Engineering colleges of the country.

The book in your hands is strictly based upon the new syllabus prescribed by C.B.S.E., New Delhi and Educational Boards of Indian states. Thebook has been written according to N.C.E.R.T. pattern and keeping in view the changing trends of different examinations. The book has beennumber ONE among the teachers and the students all over India for its clear presentation, effective approach of solving numerical problems andattractive figures.

A THOROUGH & SINCERE STUDY OF THE UNIQUE & UNMATCHED BOOK WILL BOOST THE STUDENTS TO ACHIEVE THEIR TARGET

In a NUTSHELL the book provides EXCELLENT GUIDANCE to students for Board's examinations as well as forcompetitive examinations for entrance to professional colleges.

FEATURES OF THE BOOKSimple language and easily reproducible diagrams. Large variety of SOLVED NUMERICAL PROBLEMS. Additional numerical

problems under the heading PRACTICE PROBLEMS for self assessment and practice. REVISION EXERCISES in the form : Very ShortAnswer Type, Short Answer Type and Long Answer Type Questions with HINTS and SOLUTIONS to some questions.

CONCEPTUAL QUESTIONS solved at the end of each chapter. COMPETITION FILE covering additional information, graded numericalproblems and objective questions to prepare for COMPETITIVE EXAMINATIONS for entrance to Medical and Engineering colleges.

COMPLETE coverage of previous year questions from all types of Boards' examinations and competitive examinations such as I.I.T., RoorkeeUniversity, C.B.S.E. (PMT) and other State Boards.

The book presents the subject matter in full conformity with the syllabi prescribed by C.B.S.E., New Delhi and Education Boards of other Indianstates. To keep pace with changing trends in education at national level, the whole text has been arranged strictly according to N.C.E.R.T. pattern.The main stress has been laid on SI. The symbols and signs used for various physical quantities are also in keeping with the recommendations atnational and international levels.The book provides a result-oriented training to young students. The whole text of the book is embedded with short notes in the form of(introducing apparently a new physical term with a proper definition), Key point (highlighting an important point in the text) and (bringingout the difference between the physical and apparent meaning of a physical term). Further, the text has been studded with simple

so as to provide an insight and a proper grip over the topic, as one learns it. The article work in each chapter of a unit is coupled withwell graded and carefully selected for easy comprehension of the beginners. So that the students can prepare fortheAnnual Examinations in an independent manner, a large number of and have beenincorporated in the book with proper Answers/Hints. for self-practice have been categorised into various types,so as to enable the students to choose the appropriate formula with ease. Further, detailed Hints/Solutions have been provided to UnsolvedNumerical Problems. offers a special feature of the book. It contains real and

It is aimed to provide intensive understanding and deep insight of the subject to the students, so that they get the feel of thetype of questions asked in competitive examinations, such as I.I.T.

in each unit forms another special feature of the book and consists of three parts. of thecontents of a unit is for easy and handy reference of various physical laws, principles, terms and formulae in that unit and for its quick revision.

such as and have been provided with solutions byadopting a novel technique in the form of Armed with this technique, the students will be able to attack the otherwisebrainteasing and seemingly incomprehensible numerical problems with great ease. set in various competitiveexaminations, such as etc have been thoroughly covered in the book. For the sake of easypreparation, these questions have been categorised into and Multiple Choice Questions. The author is of the firmopinion that the learning is a continual and gradual process. With the Competitive Examination Files on different units at their disposal, thestudents would be able to master them steadily all through the academic year, while preparing for admission to professional courses.

The jargonWatch out

Self-StudyQuestions,

Solved Numerical ProblemsVery Short Answer Questions Short Answer Questions

Unsolved Numerical Problems

Techie-Stuff Conceptual Numerical Problems Conceptual ShortAnswer Questions.

The Competitive Examination File Revision at a Glance

Numerical Problems from Competitive Examinations, I.I.T., Roorkee I.S.M., DhanbadThought Process.

Multiple Choice QuestionsC.B.S.E., A.I.I.M.S., A.F.M.C., M.N.R., C.P.M.T., I.I.T.,

Text-Based Thought-Based

MODERN'SOF

ABOUT THE BOOK

ABOUT THE BOOK

ABOUT THE BOOK

PHYSICS

CHEMISTRY

MATHEMATICS

BIOLOGY

India'sIndia's

No.No.11

C M

Y K

C M

Y K

MODERN'SOF +

MODERN'SOF +

MODERN'SOF +

CONTENTS

Chapter Pages

C M Y K

Chemistry of -Block Elementsp

1. The Solid State

2. Solutions and Colligative Properties

3. Chemical Thermodynamics and Energetics

4. Electrochemistry

6. General Principles and Processes of Isolation of Elements

7.

8. Chemistry of -and -Block Elements

9. Coordination Compounds

10. Haloalkanes and Haloarenes

11. Alcohols, Phenols and Ethers

12. Aldehydes, Ketones and Carboxylic Acids

13. Compounds Containing Nitrogen

14. Biomolecules

15. Polymers

16. Chemistry in Everyday Life

d f

1/1—1/29

2/1—2/36

3/1—3/41

4/1—4/39

5/1—5/37

6/1—6/20

7/1—7/39

8/1—8/24

9/1—9/29

10/1—10/40

11/1—11/41

12/1—12/54

13/1—13/35

14/1—14/23

15/1—15/17

16/1—16/12

5. Chemical Kinetics

UNIT REVISION TEST

UNIT REVISION TEST

5/38—5/40

9/30—9/32

PRACTICE PAPERS (1–5) PP/1—PP/20

Solids are the substances which have definite volumes and definite shapes. They are rigid and almostincompressible. In solid state, constituent particles (atoms, ions or molecules) are held together by strongintermolecular forces. These particles have fixed positions and can only oscillate about their mean positions.Among all the states of matter, solid state is the most ordered state.

CLASSIFICATION OF SOLIDS

Solids can be classified into two classes :1. Crystalline solidsThe substances whose constituents are arranged in a definite orderly arrangement are called crystalline

solids. For example, elements like copper, silver, iron, sulphur, phosphorus, iodine, sodium chloride, zincsulphide, quartz, etc., are crystalline solids. They possess long range order.

2. Amorphous substancesThe substances whose constituents are not arranged in an orderly arrangement are called amorphous

substances. e.g. glass, rubber, quartz glass, plastics, etc. It may be noted that some amorphous solids havesome orderly arrangement but it is not extended to more than a few Angstrom units. Thus, amorphous solidsare said to have short range order.

The important points of distinction between crystalline and amorphous solids are given below :

Crystalline Solids Amorphous Solids

1. Crystalline solids have definite shape. 1. Amorphous solids have irregular shapes.2. They have sharp melting points. 2. They melt over a range of temperature.3. They exhibit anisotropy i.e. they have different 3. They exhibit isotropy i.e. they exhibit similar

physical properties in different directions. properties in all directions.4. They are true solids. 4. They are pseudo solids.5. They can be cut along definite planes to give 5. They can not be cut along definite planes. They

smooth surfaces. yield irregular surfaces when cut intotwo pieces.

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11

U

N

I

T

U

N

I

T

THE SOLID STATE

Classification of solids based on different forces; molecular, ionic, covalent and metallic solids, amorphousand crystalline solids (elementary idea), unit cell in two dimensional and three dimensional lattices,calculation of density of unit cell, packing in solids, voids, number of atoms per unit cell in a cubic unitcell, point defects, electrical and magnetic properties, Band theory of metals, conductors andsemiconductors and insulators and n and p type semiconductors.

R NE ISION OTESR NE ISION OTES

1/1

It may be noted that liquids and gases are isotropic.

Unit-1.pmd 8/16/2016, 12:27 PM1

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1/2 MODERN'S abc OF OBJECTIVE CHEMISTRY (MH-CET)

Types of Crystals

Characteristics

Molecular Ionic Covalent Metallic

Constituent Small Positive and Atoms Positive ions in aparticles molecules negative ions sea of electronsBinding Van der Waals Strong electro- Covalent Metallic bondsforces forces static forces bond forces (electric attraction

between +ve ionsand electrons).

Properties Soft, low melting Hard and brittle, Very hard, Soft to hard,point, volatile high m.p., poor high m.p., poor moderate togood insulators, conductors of heat conductors of high m.p., goodlow heats of fusion and electricity, very heat and electri- conductors, metallic

high heats of fusion city, high heats lustre, ductile andof fusion malleable, moderate

heats of fusionExamples Solids CO2, H2,I2, Salts like NaCl, Diamond, SiC, Common metals

ice, SO2, CCl4 KNO3, LiF, BaSO4 quartz, SiO2 (Cu, Na, Fe) andsome alloys.

Types Of Crystalline SolidsThe crystalline solids may be classified on the basis of type of constituent particles as shown below :

Crystalline Solids

Molecular Ionic Covalent or Network Metallic

Constituent Constituent Constituent particles Constituent particlesparticles- particles-ions -Atoms -positive ions in aMolecules sea of electrons

(Kernels)

The important characteristics of different types of solids are given in Table 1.Table 1 : Important characteristics of crystals

Molecular solids may furthur be classified into three types depending upon the types of forces presentbetween the molecules. These are :

(a) Non-Polar Molecular Solids : In these solids, constituent particles are either atoms of noble gases(like Ne, Ar, etc.) or non-polar molecules (Solid CO

2, I

2, etc.). Constituent particles are held together by

disperson forces or London forces.

(b) Polar Molecular Solids : The constituent particles are molecules having polar covalent-bonds (SolidHCl, Solid SO2, etc.). Constituent particles are held together by dipole-dipole forces.

(c) Hydrogen Bonded Molecular Solids : The constituent particles are molecules having hydrogenbonds (for example, ice). Constituent particles are held together by hydrogen bonds.

KNOWLEDGE PLUSKNOWLEDGE PLUS!!• Quartz and Quartz glass

Both quartz and quartz glass are forms of silica (SiO2). Quartz is crystalline while quartz glass isamorphous.

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THE SOLID STATE 1/3

UNIT CELL AND SPACE LATTICESpace Lattice or Crystal LatticeThe arrangement of points showing how constituent particles (atoms, ions

or molecules) of a crystal are arranged at different positions in a threedimensional space is called space lattice or crystal lattice..

Unit CellThe smallest repeating unit in space lattice which when repeated over

and over again gives the crystal of the given substance is called unit cell.The unit cell is characterized by the distances a, b and c along the three

edges of the unit cell and the angles α, β and γ between the pair of edges.

TYPES OF UNIT CELLS AND CRYSTAL SYSTEMSIn all, there are seven types of basic or primitive unit cells or crystal

systems. These are given in Table 2.

Table 2 : Seven types of basic crystal systems

System Axial distance Axial angles Examples

Cubic a = b = c α = β = γ = 90° NaCl, KCl, zinc blende, Cu, AgTetragonal a = b ≠ c α = β = γ = 90° White tin, SnO2, TiO2Orthorhombic a ≠ b ≠ c α = β = γ = 90° Rhombic sulphur, KNO3, PbCO3Monoclinic a ≠ b ≠ c α = γ = 90°, β ≠ 90° Monoclinic sulphur, CaSO4 . 2H2ORhombohedral a = b = c α = β = γ ≠ 90° Calcite, quartz, As, Sbor TrigonalTriclinic a ≠ b ≠ c α ≠ β ≠ γ = 90° K2Cr2O7 . H3BO3Hexagonal a = b ≠ c α = β = 90°, γ = 120° Graphite, ZnO, CdS

Unit cells can also be classified on the basis of arrangement of atoms inthe lattice. There are basically two types of unit cells constituting differentcrystal systems. These are :

1. Simple or Primitive : Constituent particles are present at allcorners of the unit cell.

2. Centred : These may further be classified into the following types :

(a) Face-centred unit cells : Constituent particles are present at allthe corners as well as at the centre of each face.

(b) Body-centred unit cells : Constituent particles are present at allthe corners as well as at the centre of each cube.

(c) End-centred unit cells : Constituent particles are present at allthe corners and at the centres of two opposite faces.

If all types of space lattices are counted, fourteen lattices are obtained,which are known as Bravais lattices.RELATIONSHIP BETWEEN NEAREST NEIGHBOUR DISTANCE (d) ;RADIUS OF ATOM (r) AND THE EDGE OF UNIT CELL (a).

The relationship between nearest neighbour distance, d ; radius of atom, r and edge of unit cell, a fordifferent unit cells is summed up as follows :

KNOWLEDGE PLUSKNOWLEDGE PLUS!!• Diamond, graphite and Buckminster fullerene are allotropes of carbon and are examples of covalent

or network solids.• Diamond contains only sigma bonds and each carbon is sp3 hybridised. However, graphite contains

both sigma and pi-bonds and each carbon is sp2 hybridised.• The formula of Buckminster fullerene is C60. Like graphite, all carbon atoms in Buckminster fullerene

are sp2 hybridised.

Fig. 1. Three dimensional lattice

Space Lattice and Unit Cell

KNOWLEDGE PLUSKNOWLEDGE PLUS!!Contribution of atoms in a unitcell

Atoms at corners = 18

Atoms at edges = 14

Atoms in the face = 12

Atoms in the body = 1No. of atoms in different unitcells

Simple cubic = 1Body centred cubic = 2Face centred = 4(fcc or ccp)hcp = 6

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1/4 MODERN'S abc OF OBJECTIVE CHEMISTRY (MH-CET)

MemoriseUnit cell d r

Simple cubic a2a

Face centered cubic2

a

2 2

a

Body centered cubic32a 3

4a

� Example 1. Which of the following statement is nottrue about crystalline solids ?

(A) Polar molecular solids have higher enthalpiesof vaporisation than those of non-polar molecularsolids.

(B) Graphite, though covalent solid, is a goodconductor of electricity.

(C) Ionic solids are conductors in molten state.(D) Non-polar molecular solids have London forces

between the constituents and have highermelting points than polar molecular solids.

Strategy/Solution : It is a conceptual question andwe are to select the wrong statement. Choices (A), (B) and(C) are true but choice (D) is not true. The non polarmolecular solids have London forces or dispersion forcesbetween their molecules. These are weak forces andtherefore, the solids have low melting points. On the otherhand, polar solids have strong dipole-dipole forces betweentheir molecules and have melting points higher than thosein non-polar solids.

Correct Answer : (D)

� Example 2. A solid compound contains P, Q and Ratoms in cubic lattice with P atoms occupying thecorners, Q atoms in the body centred position andR atoms at the centres of faces of the unit cell. Theempirical formula of the compound is

(A) PQ2 R3 (B) PQR

(C) PQR3 (D) P2Q2R3

Strategy : We know that in a cubic lattice, there are 8atoms at the corners (each contributing 1/8), 1 at the bodycentre (contributing 1) and 6 at the face centres (eachcontributing 1/2).

Solution: Atoms of P per unit cell = × =1

8 18

Atoms of Q per unit cell = 1

Atoms of R per unit cell = × =1

6 32∴ Empirical formula PQR3

Correct Answer : (C)

Packing Efficiency. It is thepercentage of total space filled bythe particles.

Structure Packingefficiency

1. Hexagonal closedpacked (hcp) 74%

2. Cubic close packed(ccp) or face centredcubic (fcc) 74%

3. Body centredcubic (bcc) 68%

4. Simple cubic 52.4%

� Example 3. Sodium crystallises in bcc structure withedge length of 4.27 Å. What is the radius of a sodiumatom ?

(A) 2.42 Å (B) 1.85 Å(C) 5.73 Å (D) 4.72 Å

Strategy : For bcc structure, the radius and edgelength are related by the formula

r = 34

a

Solution : a = 4.27 Å ; r = ?For bcc structure,

r = = ×3 3

4.274 4

a = 1.85 Å

Correct Answer : (B)� Example 4. A metal crystallizes in a ccp structure.

Its metallic radius is 141.5 pm. The number of unitcells in 64 cm3 of metal are(A) 1 × 1024 (B) 1 × 1022

(C) 1 × 1023 (D) 1 × 1032

Strategy : To calculate number of unit cells in 64cm3, we need to calculate volume of one unit cell. Forthis, we need edge length (A), which can be calculatedfrom the radius of metal atom.

Solution : For ccp structure

a = 2 2 r = 2 × 1.414 × 141.5

= 400 pm

or = 400 × 10–10 cm

Volume of = (400 × 10–10 cm)3

unit cell

= 6.4 × 10–23 cm3

No. of unit cells

in 64 cm3 =23

64

6.4 10−× = 1 × 1022 unit cells

Correct Answer : (B)

CLOSED PACKED STRUCTURES

The closed packed structures are those in which maximum available space is occupied leaving minimumvacant space. These are summarised ahead :

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THE SOLID STATE 1/5

Close packing

In two dimensions In three dimensions

⎯→ Square close packing or AAA..... arrangement. ⎯→ Square close packing or AAA..... arrangement.• Packing efficiency is 52.4% • Packing efficiency is 52.4%.• Co-ordination number is 4 • Co-ordination number is 6.

⎯→ Hexagonal close packing (hcp) or ABAB..... ⎯→ Hexagonal close packing (hcp) or ABAB...........arrangement arrangement• Packing efficiency is 60–40%. ⎯→ Cubic close packing (ccp) or ABCABC.....• Co-ordination number is 6 arrangement

• In both hcp & ccp, packing efficiency is 74%.

• Co-ordination number is 12

It may be noted that co-ordination number is the number of nearest neighbours in contact with agiven sphere.

INTERSTITIAL SITESTwo important interstitial sites are :(i) Tetrahedral site. When a sphere in the second layer is placed above

three spheres which are touching one another, a tetrahedral site is formed.(ii) Octahedral site. This type of site is formed at the centre of six spheres

and is produced by two sets of equilateral triangles which point in oppositedirections.

RADIUS RATIO RULE FOR IONIC COMPOUNDSThe ratio of the radius of the cation to the radius of the anion is called radius ratio. This is very important

in determining the structure of ionic solids. The limiting radius ratios in different crystals and their coordinationnumbers are given in Table 3.

Table 3 : Relation between radius ratio and Co-ordination number.

Radius ratio Possible coordination Structural Examples(r+/r–) number arrangement

0.225 – 0.414 4 Tetrahedral ZnS0.414 – 0.732 6 Octahedral NaCl0.732 – 1.0 8 Cubic CsCl1 12 Close packing Metals

DENSITY OF A CRYSTAL

Density of a crystal can be calculated by knowing the edge length of the unit cell.

Density =Na3

Z Mg cm3,

where a = length of the edge of unit cell in cm

Z = number of atoms per unit cell

M = atomic mass of element or formula mass of the compound

N = Avogadro number

• If ‘a’ is in pm, then

Edge length = a pm

= a × 10–12 m

= a × 10–10 cm

and a3 = a3 × 10–30 cm3

Memorise• There are two tetrahedral

sites for each sphere.• There is one octahedral site

for each sphere.

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1/6 MODERN'S abc OF OBJECTIVE CHEMISTRY (MH-CET)

• Z is the number of atoms per unit cell or formula units per unit cell.e.g. for fcc, Z = 4

for bcc, Z = 2 for simple cubic, Z = 1

� Example 5. The total number of tetrahedral andoctahedral voids in the face centred unit cell is

(A) 12 (B) 8

(C) 6 (D) 3

Strategy : For fcc unit cell, there are four atomsper unit cell. For each atom in a close packedarrangement, there are two tetrahedral voids and oneoctahedral void.

Solution: For fcc unit cell, number of atoms perunit cell = 4

No. of tetrahedral voids = 4 × 2 = 8No. of octahedral voids = 4 × 1 = 4Total voids = 8 + 4 = 12Correct Answer : (A)

� Example 6. In a spinel structure of mixed oxides,oxide ions are cubic close packed whereas 1/8th oftetrahedral voids are occupied by cations A2+ and1/2 of octahedral voids are occupied by cations B3+.The general formula of the compound having spinelstructure is

(A) A2BO4 (B) AB2O4

(C) A2B4O (D) A4B2O

Strategy : In a close packed arrangement, thereis one octahedral and two tetrahedral voidscorresponding to each atom constituting the lattice.In a ccp arrangement, there will be four oxide ions perunit cell.

Solution : No. of oxide ions per unit cell = 4No. of tetrahedral voids per unit cell = 4 × 2 = 8

No. of A2+ ions = × =1

8 18

No. of octahedral voids per oxide ion in lattice= 4 × 1 = 4

No. of B3+ ions = × =1

4 22

Formula of the compound = AB2O4

Correct Answer : (B)

� Example 7. Two ions A+ and B– have radii 88 pmand 200 pm respectively. What is the co-ordinationnumber of A+ in a close-packed structure ?

(A) 4 (B) 6

(C) 8 (D) 12

Strategy : The co-ordination number of an ion in asolid depends on the value of radius ratio. Therefore,we first calculate radius ratio.

Solution : +

−

rr = =

880.44.

200

For octahedral void, the radius ratio lies between0.414 to 0.732. Therefore, the co-ordination numberis 6.

Correct Answer : (B)

� Example 8. An element crystallizes in a structurehaving fcc unit cell of edge length 200 pm. If 200gof this element contain 24 × 1023 atoms, what is thedensity of element in g cm–3 ?

(A) 41.6 g cm–3 (B) 21.4 g cm–3

(C) 60.4 g cm–3 (D) 82.8 g cm–3

Strategy : Density of a crystal, d, is given by 3ZM

Na

where MN

is the mass of 1 atom, Here, mass of 1 atom

can be calculated by dividing 200 g by given numberof atoms (24 × 1023 atoms).

Solution : d = 3Z M

N a

×

×

= 23 10 34 200

(24 10 )(200 10 )−

×

× ×

= 41.6 g cm–3

Correct Answer : (A)

STRUCTURES OF SIMPLE SUBSTANCESIn simple ionic solids ccp or hcp types of arrangement are generally present. The large ions (anions) adopt

these arrangements and the smaller ions (cations) occupy interstitial sites. The summary of some commoncrystals is given ahead :

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THE SOLID STATE 1/7

Table 4. Some Common types of structures of ionic compounds

Compound Description Coordination number Other Examples

NaCl ccp arrangement of Cl–, Na+ Na+ = 6 Li, Na, Kin all the octahedral sites Cl– = 6 halides, AgCl,

AgBr, MgO, CaOZnS ccp arrangement of S2–, Zn2+ Zn2+ = 4 BeS, CuCl,(Zinc blende) in alternative tetrahedral sites S2– = 4 CuBr, CuICsCl Simple cubic arrangement of Cs+ = 8 CsBr, CsI

Cl–, Cs+ in cubic sites Cl– = 8CaF2 ccp arrangement of Ca2+, F– Ca2+ = 8 SrF2, BaF2(Fluorite structure) occupy all tetrahedral sites F – = 4Na2O ccp arrangement of O2–, Na+ Na+ = 4 Li2O, K2S(Antifluorite structure) occupy all tetrahedral sites O2– = 8

POINT DEFECTS IN CRYSTALSIdeal crystals with perfect arrangement of constituents are found only at 0

K. Above this temperature, all crystalline solids have some defects in thearrangement of its unit cells. An ideal crystal of A+B– type may be representedas shown in Fig. 2.

Defects in the crystals may give rise to

(A) Stoichiometric and

(B) Non-stoichiometric structures.

(A) Stoichiometric structures

The compound A+B– is stoichiometric if it contains equal number of atomsA+ and B– as suggested by the chemical formula of the compound. There aretwo types of defects in stoichiometric structures :

(i) Schottky Defect. In this defect, equal number of cations and anions are missing from their normallattice sites. It is predominant in compounds with high coordination number and where the ions are of similarsize. For example, NaCl, KCl, CsCl, etc.

Schottky defect is shown by solids having• low co-ordination number• large size difference between cation and anion.

(i) Schottky defect (ii) Frenkel defect

Fig. 3. Schottky and Frenkel defects in crystals

(ii) Frenkel Defect. This defect consists of vacancies at cation sites in which the cation moves to anotherposition in between two layers called interstitial sites. This defect is most predominant in compounds whichhave low coordination number and ions of different sizes. For example, AgI, AgCl, ZnS, etc.

Frenkel defect is shown by solids having• high co-ordination number• small size difference between cation and anion.

Fig. 2. Ideal crystal A+ B–

Unit-1.pmd 8/16/2016, 12:27 PM7

MOD ABC Of Objective Chemistry MH-CET

Publisher : MBD GroupPublishers

ISBN : 9789351845386Author : Dr. S. P. Jauhar,Dr. Sheenu Jauhar

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