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Transcript of PMT Class XII Chemistry Solid State
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MEDICAL
THE SOLID STATE
C H E M I S T R Y S T U D Y M A T E R I A L
NARAYANA INSTITUTE OF CORRESPONDENCE COURSESF N S H O U S E , 6 3 K A L U S A R A I M A R K E T
S A R V A P R I Y A V I H A R , N E W D E L H I - 1 1 0 0 1 6PH.: (011) 32001131/32/50 FAX : (011) 41828320Websi te : w w w . n a r a y a n a i c c . c o mE-mai l : i n f o @ n a r a y a n a i c c . c o m
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2004 NARAYANA GROUP
This study material is a part of NARAYANA INSTITUTE OF CORRESPONDENCE COURSES for PMT, 2008-09. This is meant for
the personal use of those students who are enrolled with NARAYANA INSTITUTE OF CORRESPONDENCE COURSES,
FNS House, 63, Kalu Sarai Market, New Delhi-110016, Ph.: 32001131/32/50. All rights to the contents of the Package rest with
NARAYANA GROUP. No other Institute or individual is authorized to reproduce, translate or distribute this material in any form,
without prior information and written permission of the institute.
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PREFACE
Dear Student,
Heartiest congratulations on making up your mind and deciding to be a doctor to serve the society.
As you are planning to take various Pre-Medical Entrance Examinations, we are sure that this STUDY PACKAGE is
going to be of immense help to you.
At NARAYANA we have taken special care to design this package which will not only help but also guide you to compete
for various Pre-Medical Entrance Examinations including CBSE-PMT, DPMT, AIIMS, AFMC, JIPMER, BHU and other State
PMTs.
The salient features of this package include :
! Power packed division of units and chapters in a scientific way, with a correlation being there.
! Sufficient numbers of questions with solutions have been added in Physics and Chemistry to let the students have a
feel of Board and competitive examinations.
! Our revised edition of packages with exercises on new pattern includes Multiple-Choice Questions, Questions from
Competitive Examinations, True & False Questions, Fill in the Blanks, Assertion & Reason Type Questions andSubjective Questions.
These exercises are followed by answers in the last section of the chapter including Hints and Solutions to subjective
questions. This package will help you to know what to study, how to study, time management, your weaknesses and how to
improve your performance.
We, at NARAYANA, strongly believe that quality of our package is such that the students who are not fortunate enough to
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We feel that there is always a scope for improvement. We would welcome your suggestions and feedback.
Wish you success in your future endeavours.
THE NARAYANA TEAM
ACKNOWLEDGEMENT
While preparing the study package, it has become a wonderful feeling for the NARAYANA TEAM to get the wholehearted
support of our Staff Members including our Designers. They have made our job really easy through their untiring efforts and
constant help at every stage.
We are thankful to all of them.
THE NARAYANA TEAM
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CONTENTSCONTENTSCONTENTSCONTENTSCONTENTS
THE SOLID STATE
Theory
Exercises
True and False Statements
Fill in the Blanks
Assertion-Reason Type Questions
Multiple Choice Questions
MCQs Asked in Competitive Examinations
Subjective Questions
Answers
C
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CONTENTS
!!!!! Introduction
!!!!! Classification of Solids
!!!!! Types of Symmetry
!!!!! Types of Unit Cells
!!!!! Crystal Defects
!!!!! Radius Ratio
!!!!! Structure of Simple Ionic
Compounds
!!!!! Closest Packing
!!!!! Properties of Solids
!!!!! Metal Silicates
!!!!! Exercise
!!!!! Answers
THE SOLID STATE
M ost of the things around us are solidshaving different shapes and size. In a solid,
the molecules have fixed positions and their
motion is restricted to just vibrations. The
constituent particles in a solid are closely
packed and this leads to the properties like
incompressibility, rigidity, non-fluidity, slow
diffusion and mechanical strength. In this
unit, we will classify solids on the basis of
binding forces, understand structure of solids
describe the imperfections in solids and their
effects on properties and their application in
industries.
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A substance is said to be solid if its melting point is above the room temperature. Some of the importantcharacteristics of solids are as follows :
(i) Solids have definite shapes, size and volume
(ii) Solids are almost incompressible and generally have high density
(iii) The diffusion of solid is negligible or rather very slow as the particles have permanent positions fromwhich they do not move easily.
CLASSIFICATION OF SOLIDS
Solids can be classified broadly into two types.
1. Crystalline Solids 2. Amorphous Solids
Difference Between Crystalline and Amorphous Solids
Crystalline Solids Amorphous Solids
1.They have definite and regular geometry due to definite andorderly arrangement of atoms, ions or molecules in threedimensional space
They do not have any pattern of arrangementof atoms, ions or molecules and thus, do nothave any definite geometrical shape.
2. They have sharp melting points and change abruptly into liquidsAmorphous solids do not have sharp melting
points and do not change abruptly into liquids.
3. Crystalline solids are anisotropic. Some of their physicalproperties are different in different directions
Amorphous solids are isotropic. Their physicalproperties are same in all directions.
4. These are considered as true solids.These are considered pseudosolids orsupercooled liquids.
5.Crystalline solids are rigid and their shape is not distorted bymild distoring forces.
Amorphous solids are not very rigid. These canbe distorted by bending or compressing forces.
6.
Crystals are bound by plane faces. The angle between any twofaces is called interfacial angle. For a given crystalline solid, it isdefinite angle and remains always constant no matter how thefaces develop. When a crystalline solid is hammered, it breaksup into smaller crystals of the same geometrical shape.
Amorphous solids do not have well definedplanes.When an amorphous solid is broken, thesurfaces of the broken pieces are generally notflat and intersect at random angles.
7.Crystals have some sort of symmetry. (i) plane of symmetry,(ii) axis of symmetry or (iii) centre of symmetry
Amorphous solids do not have any symmetry.
INTRODUCTION
Gases and liquids belong to fluid state. The fluidity of liquids and
gases is due to relative free motion of their molecules. In solids,
molecules or atoms or ions oscillate around their fixed positions
due to strong intermolecular or interatomic or interionic forces. This brings
rigidity and long range order in solids. Solids are classified as crystalline
and amorphous. In crystalline solids atoms, ions or molecules are held in
an orderly array. In amorphous solid there is only short range order.
Industrial applications of solids in based on their electrical, magnetic and
dielectric properties.
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TYPES OF CRYSTALLINE SOLIDS
Crystalline Solid
Ionic crystal Covalent crystal Metallic crystal Molecular crystal
Characteristics of the Different types of crystalline Solids
Characteris tics Ionic Crys tal Covalent Crys tal Metallic Crytsal Molecular crys tal
1Constituentparticles
Cations (positiveions) and anions(negative ions)
Atoms Atoms Small molecules
2Bindingforces
Strong electrostaticforces
Covalent forcesElectrostatic attraction
between positive ions andmobile electrons
Weak van der Waal'sforces
3Physical
properties
Brittle, high m.p. andb.p., good conductorsof electricity in moltenstate and aqueoussolution, high heats offusion.
Generally hard,high m.p. and b.p.,
poor conductors ofheat, high heats offusion.
Hard and high m.p., verygood conductors of heatand electricity, exhibitproperties such as lustre.malleability ductility etc.,moderate heats of fusion.
Generally soft, low m.p.volatile, poorconductors of electricity(insulators) , poorconductors of heat, lowheats of fusion.
4 ExamplesNaCl, KF, CuSO
4,
NaNO3etc.
Diamond, siliconcarbide (SiC)quartz (SiO
2) etc.
Cu, Ag, Fe, Na, K etc.Solid CO
2, Iodine,
Naphthalene, Argonice etc.
TYPES OF SYMMETRY IN CRYSTALS
(i) Centre of Symmetry : It is such an imaginary point within the crystal that any line drawn throughit intersects the surface of the crystal at equal distances in both directions. A crystal always possesses
only one centre of symmetry.
(ii) Plane of symmetry :It is an imaginary plane which passes through the centre of a crystal and
divides it into two equal parts such that one part is exactly the mirror image of the other.
A cubical crystal like NaCl possesses, in all, nine planes of symmetry; three rectangular planes
of symmetry and six diagonal planes of symmetry. One plane of symmetry of each of the above
is shown in fig.(a) and (b).
(iii) Axis of symmetry : It is an imaginary straight line about which, if the crystal is rotated, it will
present the same appearance more than once during the complete revolution. The axes ofsymmetry are called diad, triad, tetrad and hexad, respectively, if the original appearance is
repeated twice (after an angle of 180), thrice (after an angle of 120), four times (after an
angle of 90) and six times (after an angle of 60) in one rotation. These axes of symmetry are
also called two-fold, three-fold, four fold and six-fold, respectively.
In general if the same appearance of a crystal is repeated on rotating through an angle of360
,n
around an imaginary axis, the axis is called an n-fold axis.
In all, these are 13 axes of symmetry possessed by a cubical crystal like NaCl as shown in fig.(c),
(d) and (e).
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(iv) Elements of symmetry:The total number of planes, axes and centre of symmetry possessed by acrystal are termed as elements of symmetry. A cubic crystal possesse a total of 23 elements ofsymmetry.
Planes of symmetry = (3 + 6) = 9 Fig (a) and (b)
Axes of symmetry = (3 + 4 + 6) = 13 Fig (c), (d) and (e)
Centre of symmetry = 1 Fig. (f)
The total number of symmetry elements = 23
CRYSTAL LATTICE
The regular three dimensional arrangement of the points in a crystal is called space lattice. The spacelattice is also called crystal lattice. The locations of the points in the space lattice are called lattice pointsor lattice sites.
UNIT CELL
The smallest but complete unit in the space lattice which when repeated over and over again in the threedimensions, generates the crystal of the given substance.
TYPES OF UNIT CELLS
Unit cells are basically of two types. These are primitive and non primitive.
Primitive unit cells :A unit cell is called primitive unit cell if it has particles (or points) only at thecorners. It is also called simple unit cell.
Non-primitive unit cells :In this type of unit cells, particles (or points) are present not only at thecorners but also at some other positions. These are of three types :
(i) Face centred :Particles (or points) are located at the corners and also in the centre of each face.
(ii) Body centred : Particles (or points) are located at the corners and also at the centre within thebody.
(iii) End centred :Particles (or points) are located at the corners and also at the centres of the endfaces.
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Simple or Primitive unit cell Body centred unit cell Face centred unit cell End centred unit cell
Fig.Primitive and non-primitive unit cells
SEVEN CRYSTAL SYSTEMS
It we take into consideration the symmetry of the axial distances (a, b, c) and also the axial anglesbetween the edges ( , , ) , the various crystals can be divided into seven systems. These are alsocalled crystal habits. These crystals alongwith suitable examples are listed in the table
The seven crystal systems
System Axial distance Axial angles Examples
Cubic a = b = c === 90 NaCl, KCl, Diamond, Cu, Ag
Tetragonal a = b c === 90 White tin, SnO2Orthorhombic a b c === 90 Rhombic sulphur, KNO3, K2SO4, BaSO4, PbCO3Monoclinic a b c == 90,90 Monoclinic sulphur, Na2SO4.10H2OHexagonal a = b c 90 , 120 = = = Graphite, ZnO, CdS
Rhombohedral a = b = c 90 = = Calcite (CaCO3), quartz, NaNO3Triclinic a
b
c
=90 CuSO
4.5H
2O, K
2Cr
2O
7
BRAVAIS LATTICES
The seven crystal systems listed above can be further classified on the basis of the unit cells present.Although each crystal system is expected to have four different unit cells, but actually all of them cannotexist in each case. Bravaisin 1848 has established fourteen different types of lattices called Bravaislattices according to the arrangement of the points in the different unit cells involved listed below:
Crystal System Types of Lattices
Cubic Simple, Face centred, Body centred
Tetragonal Simple, Body centred
Orthorhombic Simple, Face centred, Body centred, End centred
Monoclinic Simple, End centred
Rhombohedral Simple
Triclinic Simple
Hexagonal Simple
In various unit cells, there are three kinds of lattice points : points located at the corners, points in the facecentres and points that lie entirely within the unit cell. In a crystal, atoms located at the corner and face-centre of a unit cell are shared by other cells and only a portion of such an atom actually lies within agiven unit cell.
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(i) A point that lies at the corner of a unit cell is shared among eight unit cells and, therefore, only one-eighth of each such point lies within the given unit cell.
(ii) A point along an edge is shared by four unit cells and only one-fourth of it lies within any one cell.
(iii) A face-centred point is shared by two unit cells and only one half of it is present in a given unit cell.
(iv) A body-centred point lies entirely within the unit cell and contributes one complete point to the cell.
Types of lattice point Contribution to one unit cell
Corner 1/8
Edge 1/4
Face-centre 1/2
Body-centre 1
Total number of constitutent units per unit cell
81= occupied corners + 41 occupied edge centres + 21 occupied face centres + occupiedbody centre.
Determination of Number of Constitutent units per unit cell :
Let edge length of cube = a cm
Density of substances = d g cm3
Volume of unit cell = a3cm3
Mass of unit cells = volume density = (a3 d) g
Number of mole per unit cell =M
da3 ;
where M = molar massNumber of molecules per unit cell = Number of mole Avogadros Number
3a d NZ
M
=
EXAMPLES - 1
The density of KCl is 1.99 g cm3and the length of a side of unit cell is 6.29 as determined by X-raydiffraction. Calculate the value of Avogadros number.
SOLUTION :
KCl has face-centred cubic structure,
i.e., Z = 4
Avogadros number =Vd
MZ
Given, d = 1.99; M = 74.5; V = (6.29 108)3cm3
Avogadros number =23
8 3
4 74.56.01 10
1.99 (6.29 10 )
=
EXAMPLES - 2
An element occurs in bcc structure with a cell edge of 288 pm. The density of element is 7.2 g cm3. Howmany atoms does 208 g of the element contain?
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SOLUTION :
Volume of the unit cell = (288 1010)3= 23.9 1024cm3
Volume of 208 g of the element =3cm88.28
2.7
208=
Number of unit cell in 28.88 cm3= 24109.23
88.28
= 12.08 1023unit cells
So, total no. of atoms in 12.08 1023unit cells = 2 12.08 1023 [since each bcc unit cell contain 2atoms]
= 24.16 1023
EXAMPLES : 3
The density of potassium bromide crystal is 2.75 g cm3and the length of an edge of a unit cell is 654 pm.The unit cell of KBr is one of three types of cubic unit cells. How many formula units of KBr are there
in a unit cell? Does the unit cell have a NaCl or CsCl structure?SOLUTION :
We know that
=
A3 N
M
a
N
M
NaN A
3 =
10 3 232.75 (654 10 ) 6.023 103.89 4
119
= = !
Number of mass points per unit cell = 4
Thus ,, has NaCl type crystal is fcc structure.
EXAMPLES : 4
A cubic solid is made up of two atoms A and B. Atoms A are present at the corners and B at the centreof the body. What is the formula of the unit cell?
SOLUTION :
Contribution by the atoms A present at eight corners = 1/8 8 = 1
Contribution by the atom B present in the centre of the body = 1
Thus, the ratio of atoms of A : B = 1 : 1
Formula of unit cell = AB
EXAMPLES : 5If three elements P, Q and R crystallise in a cubic unit cell with P atoms at the corners. Q atoms at thecubic centre and R atoms at the centre of each face of the cube, then write the formula of the compound.
SOLUTION :
Contribution made by the atoms of P present at all the eight corners of the tube = 1/8 8 = 1
Contribution made by the atoms of Q present in the centre of the body = 1
Contribution made by the atoms of R present at the centre of all the six faces = 1/2 6 = 3
Formula of the compound = PQR3
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RELATION BETWEEN EDGE LENGTH (A) AND THE RADIUS OF SPHERE (R)
We have calculated the number of atoms (or spheres since an atom or particle may be regarded as thesphere) in different types of cube unit cells. Now, let us calculate the relation between the edge length (a)and the radius (r) of the spheres present in these unit cells in case of pure elements i.e. elements composedof the same type of atoms.
(a) Distance between the centres of the spheres present on the corners of the edges of the cubic iscalled edge length (a).
(b) Distance between the centres of the two nearest spheres is called nearest neighbour distance (d)and is equal to r + r = 2r. (Here r is the radius of the spheres or the atomic radius)
FOR SIMPLE CUBIC LATTICE
In a simple cubic unit cell, there is one atom (or one sphere) per unit cell. If r is the radius of the sphere,
volume occupied by one sphere present in the unit cell = 343r
Since the spheres at the corners touch each other;
Edge length (a) = r + r = 2r = d
No of atom on the corners = 188
1=
aTotal no. of atoms in a unit cell = 1
Radius of an atom = r
Distance between nearest neighbour (d) = a
Volume of cube = a3= (2r)3= 8r3
Packing fraction =cubeofVolume
sphereofVolume = 524.06r8
r3/4 3
3
==
% volume occupied by atom = 52.4
% volume unoccupied = 100 52.4 = 47.6
FOR FACE CENTRED CUBIC LATTICE
No. of atoms on the corners = 188
1=
a
B
C
A
No. of atoms on the faces = 362
1=
Total no. of atoms in a unit cell = 1 + 3 = 4
Volume occupied by atoms =3r
3
44
From ,ABC AC2 = AB2+ BC2
AC2 = a2+ a2= 2a2
AC = 2 a.
Distance between nearest neighbour (d) =AC a 2 a
2 2 2= =
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Radius of the atom (r) =d
2=
22
a
Volume occupied by atoms in a unit cell
3
22
a
3
4
4
=
Thus packing fraction 74.0a
22
a
3
44
3
3
=
=
% volume occupied by unit cell = 74
% volume unoccupied by unit cell = 100 74 = 28
FOR BODY CENTRED CUBIC LATTICE
No. of atoms on the corners = 188
1
=No. of atoms in the body = 1
Total no. of atoms in a unit cell = 1 + 1 = 2
a
B
C
A
DFrom ADC , AD2= AC2+ CD2= ( ) 222 a3aa2 =+
AD = 3a
Distance between nearest neighbour (d) =2
3a
2
AD=
Radius of an atom (r) = d/2 = 4
3a
Volume occupied by atoms =
34 a 3
23 2
Packing fraction =cubeof.Vol
atomsbyoccupied.Vol
3
3
4 a 32
3 20.68
a
= =
% volume occupied in a unit cell = 68
% volume unoccupied in a unit cell = 32
Lattice type Radius (r) Distance between Nearest neighbour Packing fraction
Simple cube 2/ar= d = 2r = a 0.524
Face centred r =22
a2d a= 0.74
Body centred
4
3ar=
2
3ad= 0.68
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EXAMPLES : 6
The radius of copper atom is 128 pm. If it crystallises in a face centred cubic lattice (fcc). what will bethe length of the edge of the unit cell?
SOLUTION :
For face centred cubic crystal
Edge length (a) = pm362)pm128()4142.1(2r22 ==EXAMPLES : 7
Xenon crystallises in a face centred cubic lattice and the edge of the unit cell is 620 pm. Calculate thenearest neighbouring distance and the radius of the xenon atom?
SOLUTION :
Nearest neighbour distance (d) = pm4.4384142.1
)pm620(
2
a==
Radius of xenon atom (r) = pm2.2194142.12
)pm620(
22
a
==
EXAMPLES : 8
The face diagonal of a cubic closed packed unit cell is 4. What is its face length?
SOLUTION :
Face diagonal = 2 2a a a 2+ = ; 83.24142.1
)4(
2
4
2
diagonalFacea ====
EXAMPLES : 9
An element A crystallises in face-centred cubic lattice with edge length of 100 pm. What is the radius ofits atom A and the nearest neighbour distance in the lattice? What is the length of its face-diagonal?
SOLUTION :
For a face centred cubic lattice (fcc)
Radius (r) = pm4.354142.12
)pm100(
22
a=
=
Nearest neighbour distance (d) = 2r = 2 35.4 pm = 70.8 pmLength of face diagonal = 4r = 4 35.4 pm = 141.6 pm
CRYSTAL DEFECTS OR IMPERFECTIONS IN SOLID
Crystal defects
Electronicimperfection
Atomic imperfectionor
point defect
Stoichiometric defect Non-stoichiometric defect Impurity defect
Schottky defect Frenkel defect Metal excess Metal deficiency
Anion vacancy defect Extra cation Impurity defect
in covalent solid
Impurity defect
in Ionic solid
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SCHOTTKY DEFECT
It consists of a pair of holes in the crystal lattice i.e one positive and one negative ion are absent fromtheir proper position. This sort of defect occurs mainly in highly ionic compounds where the +ve and -ve
ions are of similar size and hence co-ordination no. is high e.g.NaCl, CsCl, KCl, KBr etc.
No of Schottky defect formed for cm3(ns) is given by
ns=swN.exp.
2kT
where N = No. of sites per cm3that could be left vacant
A B A B A
A+
B B A B
A+
B AA A
A+
B AA B B
+ + +
+
+ + +
+ + +
WS= work necessary to form a schottky defect.
K = gas constantT = absolute temp.
At room temp. NaCl has one defect in 1015lattice site.At 500C, the value rising to one in 106sites.
FRANKEL DEFECT
It consists of a vacant lattice site (a hole) and the ionwhich ideally should have occupied the site nowoccupies an interstitial position.
In this type of defect metal ions are generaly smaller
than the anion i.e. large difference in size and co-ordination number is low. e.g. AgCl, AgBr, AgI.
The no. of Frenkel defect formed per cm3(Nf) isgiven by
ff
wN NN ' exp
2kT
=
A+
B
A+
A
+
B
B
A
+
A+
B
A+
A+
A+
B
A+
A+
B
B
B A
+A
+
A+
B
A+
B
A+
where N is the number of sites per cm3that could be left vacantNis the no. of alternative interstitial per cm3.wfis the work necessary to form a Frenkel defect, K is the gas constant and T is absolute
temp.
METAL EXCESS DEFECT
F-centres :This arises due to absence of anion from its Latticesite, leaving a hole which is occupied by an electron therebymaintaining the electrical balance. When compound such as
NaCl, KCl LiH or TiO are heated with excess of theirconstituent metal vapours or treated with high energy radiation,they become deficient in negative ion and their formula may be
1AX , where = small fraction.
Anion sites occupied by electrons in this way are called Fcentres.
More the no. of F-centres, more is the colour of compound.
A+
B A
A+
B B A
A+
AA
A+
B AA B
B
B
A
+
+ +
++
+ +
e
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Interstitial ions and electrons :Metal excess defect also occur when an extra +ve ion occupies oninterstitial position in the lattice, the electrical neutrality is maintained by the inclusion of an interstitialelectron. The composition may be represented by general formula XA1 + . The example include ZnO,CdO, Fe2O3and Cr2O3.
METAL DEFICIENCY DEFECT
This type of compound is represented by the formula XA1 . This deficiency can occur in two ways :
(i) Positive ion absent : If a positive ion is absent from its lattice site, the charges can be balanced by
an adjacent metal ion having an extra +ve charge. e.g. FeO, NiO, -TiO, FeS and CuI.
A B A
A+B B A
A+B AA
A+B AA B
B
B
+ +
+
2++
+ +
A B A
A+
B B A
A+
B AA
A+
B AA B
B
B
A
+ +
+ +
+ 2+
+ +
B
Positive ion absent Extra interstitial negative ion
(ii) Extra interstitial negative ion : It is possible to have an extra negative ion in an interstitialposition and to balance the charges by means of an extra charge on an adjacent metal ion.
RADIUS RATIO
The ratio of the radius of cation to that of anion in an ionic crystalline solid, is called radius ratio.
Radius ratio =)r(anionofRadius
)r(cationofRadius
+
The relation between the radius ratio (r+ / r) and co-ordination number is quite often known as Radiusratio rule.This helps in predicting the structures of ionic crystals. The same has been depicted in thetable.
Radius ratio Possible Co-ordination Structural Examples
(r+ / r) number of cation Arrangement
155.0 2 Linear
0.155 0.225 3 Trigonal planar B2O30.225 0.414 4 Tetrahedral ZnS, CuCl, CuBr, HgS
0.414 0.732 6 Octahedral NaCl, KBr, MgO, CaO, CaS, NaBr
0.732 1 8 Cubic CsCl, CsBr, CsI, NH4Br
Larger the cation more will be its co-ordination number due to increase of its radius ratio. As the radius ratio increases more and more beyond 0.732, anion more farther and farther apart and
cubic void are generated.
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EXAMPLES : 10
In the crystal of an ionic compound, the anions B2form a closed packed lattice and cations A+occupy allthe tetrahedral voids. What is the simplest formula of the compound ?
SOLUTION :
We know that in a close packed lattice involving anions B2having tetrahedral voids, there are twotetrahedral sites for each anion and both are occupied by the cations A+.
Thus, for one anion (B2), there are two cations (A+)
Formula of the compound = A2B.
EXAMPLES : 11
A solid AB has the NaCl structure. If radius of the cation A+is 120 pm. Calculate the maximum valueof the radius of the anion B.
SOLUTION :Since NaCl has octahedral structure.
The limiting ratioA
B
r0.414
r
+
=
or pm290414.0
120
414.0
rr AB ===
+
STRUCTURE OF SIMPLE IONIC COMPOUNDS
Anions (large in size) form a close packed arrangement but cations (smaller in size) occupy the interstitialsites resulting from the close packing of the anions e.g. NaCl, ZnS, CsCl etc.
In CaF2, Ca2+ions from the close packed arrangement while the Fions fit in the interstitial sites.
In a close packed arrangement, the number of octahedral sites is half the number of tetrahedral sites. For the radius ratio more than a minimum value or limiting value, close packed arrangement may open
slightly to accommodate the anions that are of bigger size.
Greater the co-ordination number (CN) more is the stability of ionic crystal due to greater forces ofattraction.
The different ionic compounds can have the following type of structures.
(i) Ionic compounds of AB type
(ii) Ionic compounds of AB2type
(iii) Ionic compounds of A2B type
STRUCTURES OF IONIC COMPOUNDS OF AB TYPE
Ionic compounds of the type AB suggest that in their crystal lattice A+and Bions are present in the ratio of1 : 1. These compounds have following three types of structures.
Rock salt (NaCl) type structures.
Cesium chloride (CsCl) type structure.
Zinc blende (ZnS) type structure.
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ROCK SALT (NaCl) TYPE STRUCTURE
Cl ion surrounded
octahedrally by six Na ions
+
Na surrounded
octahedrally by six Cl ions
+
ion
Fig.Sodium Chloride Structure
Characteristics:
NaCl is composed of Na+ion and Clions. The ionic radius of Na+ and Cl ion are 95 pm and 181 pm respectively. Thus radius ratio is
Na
Cl
r 950.525
r 181
+
= =
The structure is octahedral in which Clion constitute a cubic closed packed and Na+ions occupythe octahedral voids.
The coordination number is 6 : 6 ie. each Na+ion is surrounded by six Clions and vice versa.
A unit cell of NaCl consists of four NaCl units. Other examples having similar structure are
(i) Halide of Li, Na, K, Cs.
(ii) Halide of Ammonium.
(iii) Oxides and sulphides of Mg, Ca, Sr & Ba.
(iv) Halides of Ag (Except AgI)
(v) CaCO3and CaC2
CESIUM CHLORIDE (CsCl) TYPE STRUCTURES
Characteristics :
(i) The crystal is composed of Cs+and Clions and their ionic radii are 160 pm and 181 pm respectively.The radius ratio for the crystal is :
r Cs 160 pm0.889
r Cl 181pm
+
= =
The radius ratio suggests a body centred cubic structure. However, it is more than the ideal ratio which is0.732.
(ii) The Clions are present at the eight corners of the unit cell while one Cs+ion is present in the centreof the body as shown in the fig.(a)
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(iii) Each Cs+ion is surrounded by 8 Clions and each Clion by 8 Cs+ions. Thus, the coordinationnumber of each ion is 8 or these are in 8 : 8 co-ordination.
(iv) The contribution by all the 8 Clions at the corners is 1 while by the Cs+ion present in the centre ofbody is also 1. Therefore, the unit cell of CsCl consist of one Cs+ion and one Clion and the formula
of cesium chloride is CsCl.
= Cs+
= Cl
(a) (b)
Fig.Cesium chloride structure
ZINC BLENDE (ZnS) TYPE STRUCTURE
Fig.Structure of Zinc Blende (ZnS)
Characteristics :
(i) The sulphide ions (S2) with radius = 184 pm adopt cubic close packed arrangement (ccp) i.e. S2
ions are present at all the corners as well as at the centre of each face.
(ii) The radii of Zn2+and S2ions are 74 pm and 184 pm respectively. Therefore, the radius ratio is :
2
2
Zn
S
r 74pm0.402
r 184 pm
+
= =
(iii) Each Zn2+ion present occupies a tetrahedral site. As there are two tetrahedral sites per atom in aclose packed arrangement one half of these sites are occupied by Zn2+ions whereas the remaininghalf are vacant. Thus Zn2+ions are present at the alternative sites as shown in fig.
(iv) Each Zn2+ion is surrounding by four S2ions which are directed towards the four corners of atetrahedron. Similarly each S2ion is also surrounding by four Zn2+ions. Thus, coordination numberof both these ions is 4 and these are also present in the ratio of 4 : 4.
(v) Other examples of this type are : Chloride, bromides and iodides of copper, silver iodide, beryllium sulphideetc.
STRUCTURES OF IONIC COMPOUNDS OF AB2TYPE
These are the crystalline ionic solids in which cations and anions are in the ratio 1 : 2. Most of these
compounds have calcium fluoride (CaF2) type structure which is also knows as fluorite structure.
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= Ca2+
= F
Fig.Structure of Calcium Fluoride (Fluorite Structure)
Characteristics
(i) The number of Fions is twice the number of Ca2+ions i.e. Ca2+and Fions are in the ratio 1 : 2.The radii of Ca2+and Fions are 99 pm and 136 pm respectively. Therefore, radius ratio is :
Ca
F
r (99pm)0.73
r (136pm)= =
2+
(ii) The radius ratio suggests a body centred cubic structure in which the Ca2+ions are present at thecorners and at the centre of each face of the cube.
(iii) The radius ratio suggests that each Zn2+ion present occupies a tetrahedral site. As there are twotetrahedral sites per atom in a close packed arrangement, one half of these sites are occupied byZn2+ions whereas the remaining half are vacant. Thus, Zn2+ions are present at the alternative sites.
(iv) Each Zn2+ ion is surrounded by four S2 ions which are directed towards the four corners of atetrahedron. Similarly each S2ion is also surrounded by four Zn2+ions. Thus, coordination numberof both these ions is 4 and these are also present in the ratio of 4 : 4.
The other example of this type of Lattice are: BaF2, BaCl2, SrF2, SrCl2, CdCl2, PbF2etc.
ANTIFLUORITE STRUCTURE IN IONIC COMPOUNDS OF A2B TYPE
= Na
O ion(in fcc arrangement)
2
Na ion
(at tetrahedral hole)
+
= O
+
2
Fig.Structure of Na2O (Antifluorite Structure)
(a) The compounds of A2B type structure is known as antifluorite structure.
(b) In this structure, the positions of the cations and anions as compared to fluorite structure get reversedi.e. the smaller cations occupy the position of fluoride ions while the anions with bigger size occupythe positions of calcium ions.
(c) Sodium oxide (Na2O) is an example of antifluorite structure.
(d) The oxide ions (O2) have ccp arrangement and Na+ions occupy the tetrahedral voids.
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(e) Thus there are two tetrahedral sites per atom in close packed arrangement and these are occupiedby Na+ions.
(f) Each Na+ion is surrounded by four O2ions and each O2ions is surrounded by eight Na+ions. TheC.N. is 4 : 8.
(g) There are several oxides and sulphides of the metals with antifluorite structure e.g. Li2O, K2O,Rb2O and Rb2S.
EFFECT OF PRESSURE AND TEMPERATURE ON CRYSTALSTRUCTURE
At ordinary temperature and pressure, chlorides, bromide and iodides of alkali metals lithium, sodium,potassium and rubidium and halides of silver have NaCl type structure with 6 : 6 co-ordination.
NaCl structure transform to CsCl structure at high pressure in which the ions are in the ratio 8 : 8. When high temperature of 760 K is applied on cesium chloride (CsCl), with co-ordination of 8 : 8, it
changes sodium to chloride with co-ordination of 6 : 6.
BRAGGS LAW
When X-rays are incident on a crystal face, they penetrate into the crystal and strike the atoms indifferent planes. From each of these planes, X-rays are deflected. Bragg presented a relationship betweenthe wavelength of the X-rays and the distance between the planes, such as
= sind2nwhere n is an integer such as, 1, 2, 3, ..., is the wavelength, d is the distance between repeating planesof particles and the angle of deflection or glancing angle. Two parallel layers of the crystals are shownwith the help of horizontal lines separated from each other by distance d. Two waves comprising X-raysthat are in plane strike against the crystal and after reflection emerge from the crystal.
C
DB
A
d
E
Fig.X-ray diffraction by regularly spaced constituents in a crystal
Further, the wave emerging along path AE is ahead of the wave that follows the path CD. The differencein the distance travelled by the two rays after reflection also called path difference is :
Path difference = BC + CD = AC sin + AC sin = 2d sin This must be an integral multiple of wave length i.e. n . Therefore.
n 2d sin = (Bragg Equation)
EXAMPLES : 12
X-rays of wavelength 1.54 strike a crystal and are observed to be deflected at an angle of 22.5A.Assuming that n = 1. Calculate the spacing between the planes of atoms that are responsible for thisreflection.
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SOLUTION :
According to Braggs law,
= sind2n
Given n = 1, ,54.1= = 22.5 A01.2
383.02
54.1
5.22sin2
54.1d =
=
=
EXAMPLES : 13
If NaCl is doped with 103mol percent of SrCl2what is the concentration of cation vacancy?
SOLUTION :
Na+ Cl Na+ Cl
Cl Cl Na+
Sr2+ Cl Na+ Cl
Cl
Na
+
Cl
Na
+
Number of cationic vacancies per mol 18233
10023.6100
10023.610=
=
vacancies per mol.
CLOSEST PACKING
In crystals the atoms, ions or molecules are arranged in aregular way in a three-dimensional space. The arrangementhas minimum energy and hence maximum stability. Formaximum stability, a constituent in the aggregate must besurrounded by the maximum number of neighbours. Formaximum number of contacts, each constituent of the
crystal must be packed as closely as possible. In a two-dimensional plane the closest arrangement being that inwhich each sphere is in contact with six other spheres.This layer
is denoted by A.
To obtain the closest packing in space, the spheres areplaced over the layer obtain the closest packing in space,the spheres are placed over the layer A in a regular manner.There are six vacant sites or triangular pockets aroundany sphere in layer A. In fig around a sphere X these sitesare labelled as 1, 2, 3, 4, 5 and 6. We can place only three
spheres touching each other in alternate vacant sites, e.g.either in 1, 3 and 5 or in 2, 4 and 6.
Suppose the second layer is constructed by placing thespheres in the vacant sites labelled as 1, 3 and 5 then thesites marked 2, 4 and 6 are left unoccupied. This layer isdenoted by B.
A
B
A
C
B
A
B
A
B
A
Y
6
5
4
3
2
1
Close packed structures(a) Hexagonal closest packing(b) Cubical closest packing
(a)(b)
x
The second layer again has two types of vacant sites.Around any atom in the second layer, one set of vacantsites lies just above the vacant sites 2, 4 and 6 of the firstlayer and the other set lies above
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the centres of the spheres of the first layer. Thus after the second layer is complete, there are twodifferent ways of placing the spheres in the third layer. If the spheres in the third layer are placed in thevacant sites which are above the centres of the spheres of the first layer, then the third layer repeats thearrangement of the first layer and we have only two types of layers, viz., ABAB .... etc. This is called a
hexagonal closest packing, hcp. fig (a) On the other hand, if the spheres are placed in the vacant sitesabove 2, 4 and 6, a new arrangement (C) of the spheres in produced. Fig (b) The entire arrangement isnow of the type ABCABC ........... etc., and is referred to as a cubical closest packing, ccp or facecentred cubical closest packing, fcc.
The number of nearest neighbours (known as the coordination number) in each arrangement is twelve;six in the same layer three in the layer above and three in the layer below it.
In the fcc structure,there are two types of vacant sites or holes : tetrahedral and octahedral. A tetrahedralhole is surrounded by four spheres while an octahedral hole is the empty space surrounded by six spheres.These vacant sites can accommodate other smaller atoms or molecules giving rise to a variety of differentstructures.
(a) (b)
Fig.(a) Tetrahedral holes and (b) Octahedral holes in an fcc structure
In an fcc structure, there are eight tetrahedral holes per unit cell as shown in fig.(a) The number ofoctahedral holes in the unit cell of the fcc structure is four as can be deduced from the followingconsiderations. In fig(b) each (X) mark represent a vacant site and there are twelve such vacant sites atthe edges of unit cubic lattice. The vacancy at an edge is common to four unit cells and hence the number
of such vacant sites per unit cell is12
34
= . In addition to these vacancies there is one octahedral hole at
the centre of the unit cell. Thus the total number of octahedral holes per unit cell is four.
Fig.Body centered cubical close-packed structure
Another less packed arrangement of spheres is the body centered cubic arrangement as shown in figureabove in which each sphere has eight nearest neighbours; four in the same plane, two above and two
below it in the adjacent layers. Since the coordination number in fcc or hcp structure is greater than thatin the bcc structures, the latter are therefore less denser.
PROPERTIES OF SOLIDS
There are three main types of properties.
1. ELECTRICAL PROPERTIES
Based on their conductivity, solids can be divided into three categories :
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(i) Conductors
(ii) Insulators
(iii) Semi-conductors
(i) Conductors : The solids through which electricity can flow to a large extent are called conductors.They are either metallic conductors or electrolytic conductors. In the metallic conductors, the flowof electricity is due to flow of electrons (electronic conductors) without any chemical change occurringin the metal. While in electrolytic conductors like NaCl, KCl etc, the flow of electricity takes placeto a good extent only when they are taken in the molten state or in aqueous solution. The flow ofelectricity is due to flow of ions.
(ii) Insulators: The solids which almost do not allow the electricity to pass through them are calledinsulators e.g. S, P, plastics, wood, rubber etc.
(iii) Semi-conductors :The solids whose conductivity lies between those of metallic conductors andinsulators are called semi-conductors. The electrical conductivity of semi-conductors and insulators
is due to the presence of impurities and defects. Their conductivity increases with increase oftemperature as the defects (such as holes) increase with increase of temperature.
(a) Pure substances that show conductivity similar to that of silicon and germanium are calledintrinsic semiconductors.
(b) Electrical conductivity of semiconductors increases with increase of temperature becausewith increase of temperature, large number of electrons can jump from valence band toconduction band.
(c) Electrical conductivity of silicon and germanium is very low at room temperature. It can beincreased by doping with elements of Group 15 or Group 13.
(d) Similar to combination of Group 14 elements with those of Group 15 or Group 13, semiconductors
have been prepared by combination of elements of Group 13 and 15 (e.g. InSb, AlP and GaAs)or Group 12 and 16 (e.g. ZnS, CdS, CdSe and HgTe)
(e) n- and p-type semiconductors are combined suitably to form a number of electronic components.
2. MAGNETIC PROPERTIES
Solids are divided into following categories on the basis of magnetic properties.
(i) Diamagnetic substances : Substances which are weakly repelled by the external magnetic fieldare called diamagnetic substances e.g. TiO2, NaCl, benzene etc. The property thus exhibited iscalled diamagnetism. This property is shown only by those substances which contain fully filledorbitals i.e. no unpaired electron is present.
(ii) Paramagnetic substances : Substances which are attracted by the external magnetic field arecalled paramagnetic substances. The property thus exhibited is called paramagnetism. This propertyis shown by those substances whose atoms, ions or molecules contain unpaired electrons e.g. O2,Cu2+, Fe3+etc. These substances, however, lose their magnetism in the absence of the magneticfield.
(iii) Ferromagnetic substances : Substances which show permanent magnetism even in the absenceof the magnetic field are called ferromagnetic substances, e.g. Fe, Ni, Co, CrO2show ferromagnetism.Such substances remain permanently magnetised, once they have been magnetised. This type ofmagnetism arises due to spontaneous alignment of magnetic moments due to unpaired electrons inthe same direction, as shown in fig.(a)
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(c) Ferrimagnetism(a) Ferromagnetism (b) Anti-ferromagnetism(iv) Anti-Ferromagnetic substances : Substances which are expected to possess paramagnetism or
ferromagnetism on the basis of unpaired electrons but actually they possess zero net magneticmoment are called anti-ferromagnetic substances, e.g. MnO. Antiferromagnetism is due to the
presence of equal number of magnetic moments in the opposite directions as shown in fig.(b)
(v) Ferrimagnetic Substances : Substances which are expected to possess large magnetism on thebasis of the unpaired electrons but acutally have small net magnetic moment are called ferrimagneticsubstances e.g. Fe3O4, ferrites of the formula M
2+Fe2O4where M = Mg, Cu, Zn etc. Ferrimagnetismarises due to the unequal number of magnetic moments in opposite direction resulting in some netmagnetic moment, as shown in fig(c). Each ferromagnetic substance has a characteristic temperature
above which no ferromagnetism is observed. This is known as curie temperature.
3. DIELECTRIC PROPERTY
(i) Piezoelectricity or Pressure electricity : Electricity is produced due to displacement of ions,when mechanical stress is applied on such crystal. The electricity thus produced is calledpiezoelectricity and the crystals are called piezoelectric crystals.
+ + + + + + +
+ + +
+ + +
Electric field
Dipoles aligned inan ordered manner
Voltmeter
Mechanical stress
Mechanical stressFig.Production of electricity on applying mechanical stress on piezoelectric crystals
A few examples of piezoelectric crystals include titanates of barium and lead, lead zirconate (PbZrO3),ammonium dihydrogen phosphate (NH4H2PO4) and quartz. These crystals are used as pick-ups inrecord players where they produce electrical signals by application of pressure. They are also usedin microphones, ultrasonic generators and sonar detectors.
(ii) Pyroelectricity : Some piezoelectric crystals when heated produce a small electric current. Theelectricity thus produced is called pyroelectricity (pyre means heat).
(iii) Ferroelectricity : In some of the piezoelectric crystals, the dipoles are permanently polarized evenin the absence of the electric field. However on applying electric field, the direction of polarizationchanges. This phenomenon is called ferroelectricity due to analogy with ferromagnetism. Someexamples of the ferroelectric solids are barium titanate (BaTiO3), sodium potassium tartarate (Rochellesalt) and potassium dihydrogen phosphate (KH
2
PO4
). It may be pointed out here that all ferroelectricsolids are piezoelectric but the reverse is not true.
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(iv) Anti-ferroelectricity : In some crystals, the dipoles align themselves in such a way that alternately,they point up and down so that the crystal does not possess any net dipole moment. Such crystalsare said to be anti-ferroelectric. A typical example of such crystals is lead zirconate (PbZrO3).
METAL SILICATESMetal silicates have wide range of structures ranging from simple monomers to complex polymers,however, the basic unit of all the silicate ions is tetrahedron nature. These tetrahedral occur singly, or bysharing oxygen atoms in small groups, in small cyclic groups, in infinite chains or infinite sheets givingvarious structures.
1. Simple orthosilicates : In these silicates SiO44tetrahedra do not share oxygen atoms with one
another and exist as discrete SiO44orthosilicate anions e.g. Mg2SiO4, Fe2SiO4and Zn2SiO4. In all
these compounds SiO44ion is linked to M2+ions through co-ordinate bond. [See Fig. (a)]
2. Pyrosilicates or islands : In these two SiO44tetrahedral share a common oxygen, island structures
having formula (Si2O7)6
e.g. Pyrosilicates. [See Fig. (b)]
3. Ring or cyclic silicate anions : Ring anions are obtained when two oxygen atoms of eachtetrahedron are shared with others. (Si3O9)
6and (Si6O18)12. General formula of any such anion
must be (SinO3n)2n. [See Fig. (c)&(d)]
(a) SiO44
(b) (Si O )2 75
(c) (Si O )3 96
(d) (Si O )6 1812
(e) (SiO ) chain3 n2
(f) (Si O ) layer 2 5 n2
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4. Infinite chain silicates : These are of two types the pyroxenes which contain single strand chainsof composition (SiO3
2)nand amphiboles which contain double strand cross-linked chains of composition(Si4O11
6)n. General formula of the anion in a pyroxene is the same as in a silicate with cyclic anione.g. Diopside (MgCaSi2O6). Various asbestos minerals are examples of chain silicates (amphiboles).
5. Infinite sheet Silicates : When three oxygen atoms of each SiO4tetrahedron are shared, the twodimensional sheet structures with general formula of (Si2O5
2)n are obtained e.g. clay.[See Fig. (e)]
6. Framework silicate : Three dimensional network is obtained when all the four oxygen atoms ofeach of SiO4tetrahedron are shared. e.g. Quartz and zeolites. The general formula of frameworksilicate is SiO2. [See Fig. (f)]
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EXERCISE
Section A : (i) True and False Statements
(ii) Fill in the Blanks
(iii) Assertion-Reason Type Questions
Section B : (i) Multiple Choice Questions
(ii) MCQs asked in Competitive Examinations
Section C : Subjective Questions
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EXERCISES
SECTION - A
TRUE AND FALSE STATEMENTS
1. The number of atoms in a unit cell having f.c.c. structure is 2
2. The electricity produced by applying mechanical stress on a crystal is called piezoelectricity.
3. Quartz is amorphous whereas silica is crystalline.
4. Frankel defect not found in pure alkali metal halides.
5. Fe0.95O is called stoichiometric oxide of iron.
6. Crystalline solids are anisotropic.
7. Silicon carbide is a covalent crystal.
8. In (hcp) systems, each sphere is surrounded by eight other spheres.
9. The number of atoms per unit cell of body-centred cube is 2.
10. Schottky defect in crystal results in appreciable increase in its density.
FILL IN THE BLANKS
1. On applying __________________ pressure, NaCl structure changes to CsCl type strucgture.
2. As the radius of cation increases, C.N. __________________
3. No. of tetrahdral voids is __________________ the no. of octahidal void.
4. Iron, Nickel, copper, silver and gold crystallises in __________________ structure.
5. It is observed that NaCl crystal at room temperature there are about 10
22
ions and 10
6
schottky pairs percm3. It means that there is one schottky pair per __________________ ions.
6. The crystal structure of CsCl is __________________ .
7. A face-centred point is shared by __________________ unit cells.
8. In face-centred cubic structure the empty space is __________________ %.
9. In hcp mode of packing, a sphere has coordination number __________________ .
10. The presence of a trace of arsenic in germanium makes it __________________ .
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ASSERTION-REASON TYPE QUESTIONS
I n each of the foll owing questions two statements are given as Asserti on A and Reason R. Examine the
statements carefull y and answer the questions according to the instructions given below :
"
Mark(1)if both Aand R are correct and Ris the correct reason of A." Mark (2)if both Aand Rare correct and Ris not the correct reason of A." Mark (3)if Ais correct and Ris wrong." Mark (4)if both Aand Rare wrong.
1. Assert ion A : In crystal lattice the size of the cation is larger in octahedral hole than in tetrahedral hole.Reason R: Cations occupy more space than anions in crystal packing.(1) (2)(3) (4)
2. Assert ion A : In a ccp lattice made up of Yions, all the tetrahedral voids are occupied by X+ions;hence the formula of the solid is X2Y.Reason R: In a ccp lattice number of tetrahedral voids formed is twice the number of ions making thelattice.
(1) (2)(3) (4)
3. Assert ion A : NaCl crystal unit cell has fcc arrangement.Reason R: There are 4 units of NaCl present per unit cell.(1) (2)(3) (4)
4. Assert ion A : A crystal having fcc structure is more closely packed than a crystal having bcc structure.Reason R: Packing fraction for fcc structure is double that of bcc structure.(1) (2)(3) (4)
5. Assert ion A : The two ions A+and Bhave radii 88 and 200 pm respectively. Coordination number ofA+will be 6.Reason R: When r+/r= 0.414 0.732, the coordination number is 6.(1) (2)(3) (4)
6. Asser tion A : Covalent crystals have the highest melting point.Reason R: Covalent bonds are stronger than ionic bonds.(1) (2)(3) (4)
7. Assert ion A : CsCl as body-centred cubic arrangement.Reason R: CsCl has one Cs+ion and 8 Clions in its unit cell.(1) (2)(3) (4)
8. Assert ion A : Triclinic system is the most unsymmetrical system.Reason R: No axial angle is equal to 90 in triclinic system.(1) (2)(3) (4)
9. Asser tion A : Antiferromagnetic substances on heating to high temperature become paramagnetic.Reason R: On heating, randomisation of spins occur.(1) (2)(3) (4)
10. Assert ion A : In any ionic solid, [MX], with Schottky defects, the number of positive and negative ionsare same.Reason R: Equal number of cation and anion vacancies are present.(1) (2)
(3) (4)
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SECTION - B
MULTIPLE CHOICE QUESTIONS
1. A mineral having the formula AB2crystallizes in the c.c.p. lattice, with A atoms occupying the latticepoints. The CN of A is 8 and that of B is 4. What percentage of the tetrahedral sites is occupied by Batoms?
(1) 25% (2) 50%
(3) 75% (4) 100%
2. CsBr has b.c.c. structure with edge length 4.3. The shortest inter ionic distance in between Cs+and Bris
(1) 3.72 (2) 1.86
(3) 7.44 (4) 4.3
3. The formula for determination of density of cubic unit cell is
(1) 303
cmgMz
Na
(2)3
30 cmg
aMNz
(3)3
0
3
cmgNz
Ma
(4)3
30
cmgaN
Mz
4. The closest-packing sequence ABAB ..... represents
(1) Primitive cubic packing (2) Body-centred cubic packing
(3) Face-centred cubic packing (4) hexagonal packing
5. In Braggs equation for diffraction of X-rays n represents
(1) The number of mole (2) Quantum number (3) The order of reflection (4) Avogadros number
6. Tetragonal crystal system has the following unit cell dimensions
(1) a = b = c and === 90(2) ==== 90andcba(3) === 90andcba
(4) ==== 120,90andcba
7. In a solid lattice, the cation has left a lattice site and is located at interstitial position, the lattice defect is
(1) Interstitial defect (2) Vacancy defect
(3) Frenkel defect (4) Schottky defect
8. Each unit cell of NaCl consists of 13 chloride ions and _______.
(1) 13 Na+ions (2) 14 Na+ions
(3) 6 Na+ions (4) 8 Na+ions
9. Germanium or silicon becomes semi-conductor due to
(1) Schottky defect
(2) Chemical impurity
(3) Frenkel defect
(4) None of these
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10. Which one of the following is correct?
(1) Schottky defect lowers the density
(2) Frenkel defect increases the dielectric constant of the crystals
(3) Stoichiometric defects make the crystals good electrical conductors(4) All of these
11. The electrons trapped in anion vacancies in metal excess defects are called
(1) Mobile electrons (2) Trapped electrons
(3) Valence electrons (4) F-centres
12. The correct statement regarding F-centre is
(1) Electrons are held in the voids of crystals
(2) F-centre provides colour to the crystals
(3) Conductivity of the crystal increases due to F-centre
(4) All of these13. Crystals where dipoles may align themselves in an ordered manner, so that there is a net dipole moment,
exhibit
(1) Pyro-electricity (2) Piezo-electricity
(3) Ferro-electricity (4) Anitferro electricity
14. Some of the polar crystals, on heating produce a small electric current called
(1) Pyro-electricity (2) Piezo-electricity
(3) Ferro-electricity (4) Anitferro-electricity
15. Germanium is an example of
(1) An intrinsic semiconductor (2) An n-type semiconductor (3) A p-type semiconductor (4) Insulator
16. Which type of semiconductor is obtained on mixing the arsenic into the silicon?
(1) n-type (2) p-type
(3) Internal (4) both (1) and (2)
17. If indium is added in small quantity of Ge metal, we get
(1) An n-type semiconductor (2) A p-type semiconductor
(3) Rectifier (4) Insulator
18. When n- and p-type semiconductors are allowed to come into contact
(1) Some electrons will flow from n to p (2) Some electrons will flow from p to n
(3) The impurity element will flow from n to p (4) The impurity element will flow from p to n
19. The electrical conductivity of semiconductors
(1) Increases with temperature (2) Decreases with temperature
(3) Remains constants on heating (4) None of the above
20. Super conductors are substances which
(1) Conduct electricity at low temperature (2) Conduct electricity at high temperatures
(3) Offer high resistance to the flow of current (4) Offer no resistance to the flow of current
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21. The phenomenon of superconductivity was first discovered by
(1) Einstein (2) Soddy
(3) Hund and Mulliken (4) Kammerlingh Onnes
22. Which substances possess zero resistance at 0 K?(1) Conductors (2) Semiconductors
(3) Superconductors (4) Insulators
23. Amorphous solids are
(1) Solid substances in real sense (2) Liquids in real sense
(3) Supercooled liquids (4) Substances with definite M.P.
24. Which of the following is not an example of molecular crystal?
(1) Hydrogen (2) Iodine
(3) Ice (4) Sodium chloride
25. The pure crystalline substance on being heated gradually first forms a turbid liquid at constant temperatureand still at higher temperature turbidity completely disappears. The behaviour is a characteristic ofsubstance forming
(1) Allotropic crystals (2) Liquid crystals
(3) Isomeric crystals (4) Isomorphous crystals
26. The co-ordination number of a body-centred atom in cubic structure is
(1) 4 (2) 6
(3) 8 (4) 12
27. The radius of an ion in a body centred cube of edge ais
(1)2a (2) 2
4a
(3)3
4
a(4) a
28. The fraction of the total volume occupied by atoms in a simple cube is
(1)2
(2)
8
3
(3) 6
2
(4) 6
29. A substance AxBycrystallizes in an f.c.c. lattice in which atoms of A occupy each corner of the cubeand atoms of B occupy the centres of each face of the cube. Identify the correct composition of thesubstance AxBy.
(1) AB3 (2) A4B3(3) A3B (4) Composition cannot be specified
30. In a closed packed lattice, the number of octahedral sites as compared to tetrahedral ones will be
(1) Equal (2) Half
(3) Double (4) None of these
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31. CsCl on heating to 760 K changes into
(1) Liquid (2) NaCl structure
(3) ZnS structure (4) None of these
32. Compounds having NaCl type structure when subjected to high pressure changes into(1) Unstable compound (2) CsCl type structure
(3) ZnS type structure (4) None of the three
33. In a face centred cubic cell, an atom at the face centre is shared by
(1) 4 unit cells (2) 2 unit cells
(3) 1 unit cell (4) 6 unit cells
34. Ionic solids are characterised by
(1) Good conductivity in solid state (2) High vapour pressure
(3) Low melting point (4) Solubility in polar solvents
35. Silicon is a(1) Conductor (2) Semiconductor
(3) Non conductor (4) Metal complex
36. On adding a little phosphorus to silicon, we get a/an
(1) n-Type semiconductor (2) p-Type semiconductor
(3) Metallic conductor (4) Insulator
37. Piezoelectric crystals are used in
(1) Radio (2) TV
(3) Record player (4) Refrizerator
38. Glass is
(1) Supercooled liquid (2) Crystalline solid
(3) Liquid crystal (4) None of these
39. In a crystal pair of ions are missing from normal sites. This is an example of
(1) F-centres (2) Interstitial defect
(3) Frenkel defect (4) Schottky defect
40. Frenkel defect generally appears in
(1) AgBr (2) ZnS
(3) AgI (4) All of these
41. Missing of one cation and one anion from the crystal lattice is called :
(1) Ionic defect (2) Crystal defect
(3) Schottky defect (4) Frenkel defect
42. In a closed packed array of N spheres, the number of tetrahedral holes are
(1) N/2 (2) N
(3) 4N (4) 2N
43. For an ionic crystal of general formula AX and co-ordination number 6, the value of radius ratio willbe
(1) Greater than 0.73 (2) In between 0.732 and 0.414
(3) In between 0.41 and 0.22 (4) Less than 0.22
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44. Close packing is maximum in the crystal lattice of:
(1) Face-centred cubic (2) Body-centred cubic
(3) Simple-centred cubic (4) None of these
45. In the crystal of CsCl, the nearest neighbours of each Cs ion are(1) Six chloride ions (2) Eight Cs ions
(3) Six Cs ions (4) Eight chloride ions
46. Schottky defect in crystals is observed when
(1) An ion leaves its normal site and occupies an interstitial site
(2) Equal number of cations and anions are missing from the lattice
(3) Unequal number of cations and on ions are missing from the lattice
(4) Density of the crystal is increased
47. Pick up the correct statement
(1) The ionic crystal of AgBr does not have Schottky defect
(2) The unit cell having crystal parameters, a=b c, = = 90, =120 is hexagonal
(3) In ionic compounds having Frenkel defect, the ratio r+/ris high
(4) The coordination number of Na+ion in NaCl is 4
48. When NaCl is dopped with MgCl2the nature of defect produced is
(1) Interstitial defect (2) Frenkel defect
(3) Schottky defect (4) None of these
49. The second order Bragg diffraction of X-rays with 1.00 = from a set of parallel planes in a metaloccurs at an angle of 60. The distance between the scattering planes in the crystal is
(1) 0.575 (2) 1.00
(3) 2.00 (4) 1.15
50. If we mix a pentavalent impurity in a crystal lattice of germanium, what type of semiconductor formationwill occur?
(1) p-Type (2) n-Type
(3) Both (1) and (2) (4) None of the two
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MCQS ASKED IN COMPETITIVE EXAMINATIONS
1. The fraction of total volume occcupied by the atoms present in a simple cube is: [CBSE PMT 2007]
(1) 4
(2) 6
(3)3 2
(4)
4 2
2. The appearance of colour in solid alkali metal halides is generally due to [CBSE PMT 2006]
(1) Interstitial positions (2) F-centres
(3) Schottky defect (4) Frenkel defect
3. CsBr crystallises in a body centred cubic lattice. The unit cell length is 436.6 pm. Given that the atomicmass of Cs=133 and that of Br = 80 amu and Avogadro number being 6.02 1023mol1, the density ofCsBr is
(1) 4.25 g/cm3 (2) 42.5 g/cm3
(3) 0.425 g/cm3 (4) 8.25 g/cm3 [CBSE PMT 2006]
4. In face-centred cubic unit cell, edge length is [DPMT 2005]
(1)4
r3
(2)4
r2
(3) 2r (4)3
r2
5. In a face-centred cubic lattice,unit cell is shared equally by how many unit cells?[CBSE P MT 2005]
(1) 4 (2) 2
(3) 6 (4) 86. If Z is the number of atoms in the unit cell that represents the closest packing sequence.... ABC
ABC....., the number of tetrahedral voids in the unit cell is equal to [A.I.I.M.S. 2005]
(1) Z (2) 2Z
(3) Z/2 (4) Z/4
7. What type of crystal defect is indicated in the diagram below? [A.I.I.M.S. 2003]
Na Cl Na Cl Na Cl+ + +
Cl Cl Na NaNa Cl Cl Na ClCl Na Cl Na Na
+ +
+ +
+ + +
(1) Frenkel defect (2) Schottky defect
(3) Interstitial defect (4) Frenkel and Schottky defects
8. The crystal system of a compound with unit cell dimensions a = 0.387, b = 0.387 and c = 0.504 nm and= = 90 and = 120 is [A.I.I.M.S 2004](1) cubic (2) hexagonal
(3) orthorhombic (4) rhombohedral
9. The liquefied metal expanding on solidification is [A.I.I.M.S 2004]
(1) Ga (2) Al
(3) Zn (4) Cu
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10. The pyknometric density of sodium chloride crystal is 2.165 103kg m3while its X-ray density is2.178 103kg m3. The fraction of the unoccupied sites in sodium chloride crystal is
[C.B.S.E. P.M.T 2002]
(1) 5.96 (2) 5.96 102
(3) 5.96 101 (4) 5.96 103
11. Which of the following metal oxides is anti-ferromagnetic in nature? [M.P.P.E.T. 2002]
(1) MnO2 (2) TiO2(3) VO2 (4) CrO2
12. To get n-type doped semiconductor, impurity to be added to silicon should have the following number ofvalence electrons [K.C.E.T. 2001]
(1) 1 (2) 2
(3) 3 (4) 5
13. Addition of arsenic to germanium makes the latter a [Tamilnadu, CET 2002]
(1) Metallic conductor (2) Intrinsic semiconductor
(3) Mixed conductor (4) Extrinsic semiconductor 14. Semiconductors are derived from compounds of [Kerala MEE 2002]
(1) p-block elements (2) Lanthanides
(3) Actinides (4) Transition elements
15. A semiconductor of Ge can be made p-type by adding [Manipal PMT 2002]
(1) Trivalent impurity (2) Tetravalent impurity
(3) Pentavalent impurity (4) Divalent impurity
16. Due to Frenkel defect the density of ionic solids [M.P.P.E.T. 2002]
(1) Decreases (2) Increases
(3) Does not change (4) Changes
17. Which of the following crystals does not exhibit Frenkel defect? [M.P.P.E.T. 2000](1) AgBr (2) AgCl
(3) KBr (4) ZnS
18. The packing fraction for a body centred cubic is [M.P.P.M.T. 2000]
(1) 0.42 (2) 0.53
(3) 0.68 (4) 0.82
19. The number of octahedral sites per sphere infcc structure is [M.P.P.M.T. 2000]
(1) 1 (2) 2
(3) 4 (4) 8
20. The interionic distance for cesium chloride crystal will be [M.P.P.E.T. 2002]
(1) a (2) a/2(3) 2/a3 (4) 3/a2
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SECTION - C
SUBJECTIVE QUESTIONS
1. What is the contribution of the atom when it is placed at the centre of the edge in a unit cell of a cube ?
2. What is the atomic radius of the atom in body-centred cubic structure ?
3. What is the coordination number in hcp and ccp arrangements ?
4. Explain, why ionic crystals do not conduct electricity in solid state ?
5. Name the two important kinds of holes normally encountered in a closed packed structure. How many
such holes are present per sphere in a close packed structure ?
6. Explain the term dislocation in relation to crystal.
7. Which type of crystals exhibit piezo-electricity.
8. What is the coordination no of each sphere in ccp of spheres in three dimensions.9. Mention one property which is caused due to presence of F-centre in solid.
10. What is the effect of pressure on NaCl type crystals.
11, A given solid can belong to one of the four crystal types, i.e., ionic, molecular, covalent and metallic.
Indicate the crystal type of the following :
(a) Diamond (b) Sodium chloride (c) Ice (d) Copper
(e) Boron nitride (f) Zinc oxide (g) Paraffin wax.
12. X-rays of wavelength 1.54 strike a crystal and are observed to be deflected at an angle of 22.5.
Assuming that n = 1, calculate the spacing between the planes of atoms that are responsible for this
reflection.
13. At room temperature, sodium crystallises in body-centred cubic lattice with a = 4.24. Calculate the
theoretical density of sodium (at. mass of Na = 23.0)
14. A compound formed by elements A and B crystallises in cubic structure where A atoms are at the
corners of a cube and B atoms are at the face centre. What is the formula of the compound ?
15. A solid AB has the NaCl structure. If radius of the cation A+is 120 pm, calculate the maximum value of
the radius of the anion B.
16. In a cubic lattice, the closed packed structure of mixed oxides of the lattice is made up of oxide ions; one
eighth of the tetrahedral voids are occupied by divalent ions (A2+) while one half of the octahedral voids
are occupied by trivalent ions (B3+). What is the formula of the oxides?
17. If NaCl is doped with 103mol percent of SrCl2, what is the concentration of cation vacancy?
18. A metal crystallises into two cubic phases, face-centred cubic (f.c.c.) and body-centred cubic (b.c.c.)
whose unit lengths are 3.5 and 3.0 respectively. Calculate the ratio of densities of f.c.c. and b.c.c.
19. Calculate the Miller indices of crystal planes which cut through the crystal axes at (i) (2a, 3b, c) (ii) (a,
b, c) (iii) (6a, 3b, 3c), (iv) (2a, 2b, )
20. The unit cell length of NaCl is observed to be 0.5627 nm by X-ray diffraction studies; the measured
density of NaCl is 2.164 g cm3. Correlate the difference of observed and calculated density and calculate
% of missing Na+and Clions.
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ANSWERS
SECTION - A
(True and False Statements)
1. False 2. True
3. False 4. True
5. False 6. True
7. True 8. False
9. True 10. False
(Fill in the Blanks)
1. High 2. increases
3. double 4. ccp
5. 1016 6. B.c.c.
7. Two 8. 26
9. 12 10. Semi-conductor
(Assertion-Reason Type Questions)
1. (3) 2. (1)3. (1) 4. (3)
5. (1) 6. (3)
7. (3) 8. (2)
9. (1) 10. (1)
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SECTION - B
(Multiple Choice Questions)
1. (4) 2. (1) 3. (4) 4. (4) 5. (3)6. (2) 7. (3) 8. (2) 9. (2) 10. (4)
11. (4) 12. (4) 13. (2) 14. (1) 15. (1)
16. (1) 17. (2) 18. (1) 19. (1) 20. (4)
21. (4) 22. (3) 23. (3) 24. (4) 25. (2)
26. (3) 27. (3) 28. (4) 29. (1) 30. (2)
31. (2) 32. (2) 33. (2) 34. (4) 35. (2)
36. (1) 37. (3) 38. (1) 39. (4) 40. (4)
41. (3) 42. (4) 43. (2) 44. (1) 45. (4)46. (2) 47. (2) 48. (3) 49. (4) 50. (2)
(MCQs asked in Competitive Examinations)
1. (2) 2. (2) 3. (1) 4. (2) 5. (3)
6. (2) 7. (2) 8. (2) 9. (1) 10. (4)
11. (1) 12. (4) 13. (4) 14. (1) 15. (1)
16. (3) 17. (3) 18. (3) 19. (1) 20. (3)
SECTION - C
(Subjective Questions)
Answers are given in the separate booklet.