CSS – 1
SOLID STATE
C1
Solids, as is well known, are characterised by incompressibility, rigidity and mechanical strength. The
indicates that the molecules, atoms or ions that make up a solid are closely packed, i.e, they are held
together by strong forces and cannot move at random. Thus, in solids there is well-ordered arrangement of
molecules, atoms or ions. The properties of solids not only depend upon the number and kind of
constituents but also on their arrangements.
C2A There are four different crystal types closely related to the nature of the forces which hold the structural
units together. These are summarized in Table.
C2B
Type of crystal
Structural units at the
lattice sites
Bonding force
Example
Ionic
Ions
Electrostatic
CsCl, NaCl etc.
Covalent
Atoms
Sharing of electron pairs
(covalent)
Diamond, Graphite,
Abestos
Metallic
Metal ions
Electrostatic attraction between
metals ions and electrons
surrounding the metal ions.
Fe, Cu, Ag etc.
Molecular
Molecules
Van der Waals, dipole dipole,
hydrogen bonds.
Ice, solid carbon
dioxide (dry ice).
Solids, basing on the interval structure are classified into
(1)
Amorphous solid
(2)
Crystalline solids.
Amorphous solids or Pseudo Solids
Crystalline solids
(i)
There is random arrangement of atoms
molecules or ions
There is a fixed arrangement of
atoms, molecules or ions.
(ii)
They are of short range order
They are of long range order
(iii)
They are isotropic i.e., same physical properties
in all the directions
They are anisotropic i.e.,
different physical properties in
different directions.
(iv)
Do not possess sharp melting point
Possess sharp melting point.
(v)
Also known as super cooled liquids
These are true solids.
(vi)
Are not very rigid, these can be distorted by bending
or by applying external forces
Are rigid and there shape is not
distorted by mild distorting
forces.
(vii)
Examples of Amorphous solid are Starch, Protein,
Glass, Plastics, Rubber etc.
Almost all metals have fine
granular shapes.
C2C Basic Crystal System
There are seven basic crystal shapes are possible from geometrical point of consideration, it is also known
as seven Bravias Crystal Systems, these are
Crystal System
Axial distance or
Edge lengths
Axial angles
Cubic
a=b=c
 =  =  = 900
Tetragonal
a=bc
 =  =  = 90
0
Orthorhombic
abc
 =  =  = 900
Monoclinic
Hexagonal
abc
a=bc
Examples
Rhombic sulphur, KNO3, BaSO4
0
Monoclinic sulphur, PbCrO2
0
0
Graphite, ZnO, CdS
 =  = 90 ;   120
 =  = 90 ;  = 120
0
 =  =   90
Triclinic
      900
Einstein Classes,
Sin (White tin), SnO2, TiO2,
CaSO4
0
Rhombohedral or a = b = c
Trigonal
abc
NaCl, Zinc blende, Cu
CaCO3(Calcite), HgS (Cinnabar)
K2Cr2O7, CuSO4.5H2O, H3BO3
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CSS – 2
Cubic :
Tetragonal :
Rhombohedral :
Hexagonal :
Triclinic :
Monochlinic :
Orthorombic :
C3A In case of cubic crystal they have three kinds of Bravais Lattices, these are :
C3B
1.
Simple or Primitive Cubic Lattice
3.
Face Centered Cubic
2.
Body Centered Cubic
Simple or Primitive Cubic Lattice : In which there are points at all the corners of the cubic unit cell.
In simple cubic, atoms of equal sized are in square colsed packing.
Since each atom and the corner is shared by eight similar cubes, contribution of each atom is
1
.
8
Relation between edge length ‘a’ and size of atom ‘r’ : a = 2r
Therefore effective number of atoms (Zeff) = 8 ×
Packing Fraction
(P.F) 
P .F 
1
= 1.
8
Volume occupied by effective atoms in a unit cell
Totalvolumeof unitcell
Z eff 4r 3
3 a3
......
[a is the edge length of unit cell, Zeff is
the effective atom in a unit cell]
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In case of simple cube (P.F) =

= 0.52....... [a = 2r in case of simple cube]
6
Void space in simple cube = 0.47. Coordination number in simple cube : 6
C3C Body Centered Cubic :
In this form each atom has eight nearest neighbours and six next nearest neighbours.
Effective atoms in bcc (Zeff) = 8 ×
1
+ 1 = 2.
8
Relation between edge length ‘a’ and size of atom ‘r’ : 3a = 4r.
Packing Fraction : P.F =
3
= 0.68 = 68% of the total volume is occupied and 32% of total volume is
8
vacant.
Practice Problems :
1.
A solid has a bcc structure. If the distance of closest approach between the two atoms is 1.73Å. The
edge length of the cell is
(a)
2.
200 pm
(b)
3/2 pm
(c)
142.2 pm
(d)
2 pm
CsBr has bcc structure with edge length 4.3. The shortest inter ionic distance in between Cs+ and
Br— is
(a)
3.72
(b)
1.86
(c)
7.44
(d)
4.3
[Answers : (1) a (2) a]
C3D Face Centred Cubic :
In which there are points at the corner as well as at the centre of each face.
In fcc there is a cubic closed packing (ccp). In a face centered cubic unit cell there are eight atoms at the
corners (each shared by eight unit cells) and six at the faces (each shared by two unit cells).
fcc is also known as ABCABC.... type of packing.
Effective atoms in fcc unit cell (Zeff) = 8 ×
1
1
+6×
= 4.
2
8
Relation between edge length ‘a’ and size of atom ‘r’ : 2a = 4r
Packing Fraction : P.F =

= 0.74
3 2

74% of the total volume is occupied and 26% of total volume is vacant.
Co-ordination Number or Nearest Neighbours in fcc is 12.
C3E
Density of solid : d 
mass of a unit cell
Z M
, d  eff 3
volume of a unit cell
N Aa
Practice Problems :
1.
Potassium has a bcc structures with nearest neighbour distance 4.52 Å. Its atomic weight is 39. Its
density will be
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(a)
2.
454 kg/m3
(b)
804 kg/m3
(c)
852 kg/m3
(d)
908 kg/m3
Gold crystallizes in a face centred cubic lattice. If the length of the edge of the unit cell is 407 pm, the
density of gold is [Atomic mass of gold = 197 amu.]
(a)
22.20 g/cm3
(b)
14.44 g/cm3
(c)
17.20 g/cm3
(d)
19.4 g/cm3
[Answers : (1) d (2) d]
C4A Interstitial sites in ccp or voids in ccp :
In the closed packing of spheres there is some empty space left in between the spheres, this is known as
Iterstitial sites or voids. We come across two common interstitial sites tetrahedral and octrahedral sites
in the closed packed lattices.
1.
Tetrahedral void
2.
Octahedral void
Tetrahedral void : If one sphere is placed upon three other spheres which are touching one other
tetrahedral structure results.
Since the four sphere touch each other at one point only they leave us small space in between known as
tetrahedral void. The size of the void is much smaller than that of the spheres.
In ccp near every corner inside the unit cell there is one tetrahedral void.
Total and effective tetrahedral voids = 8.
Co-ordiantion number of tetrahedral void atom = 4.
There would be two tetrahedral voids associated with each sphere in fcc.
Octahedral void : This interstitial site or void is formed at the centre of six spheres, the centres of which
lie at the apices of a regular octahedral.
The figure shows two layers of closed-packed sphered. The full circles represent the
spheres in one plane while the dotted circles represent those in second plane. Two triangles have been
drawn. One of these joins the cetres of these spheres in one plane while the second (dotted lines) joins the
centres of three spheres in the second plane. The octahedral sites, marked by x, are formed where two
triangles of different layers are superimposed one above the other. In this position the apices of these
triangles point in opposite direction, as shown. Thus, each octahedral site is created by two equilateral
triangles with apices in opposite direction.
Effective octahedral void in fcc = 4. Coordination number of octahedral void = 6.
General relation for ccp and hcp packings
Effective atom = effective octahedral voids. Effective tetrahedral void = 2 × octahedral void.
Practice Problems :
1.
A mineral having the formula AB2 crystallises in the cubic close-packed lattice, with the A atoms
occupying the lattice points. The co-ordination number of the A atoms, that of B atoms and the
fraction of the tetrahedral sites occupied by B atoms are
(a)
8, 4, 100%
Einstein Classes,
(b)
2, 6, 75%
(c)
3, 1, 25%
(d)
6, 6, 50%
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CSS – 5
2.
3.
In the closest packing of atoms, there are
(a)
one tetrahedral void and two octahedral voids per atom
(b)
two tetrahedral voids and one octahedral voids per atom
(c)
two of each tetrahedral and octahedral voids per atom
(d)
one of each tetrahedral and octahedral voids per atom
If the anions (A) form hexagonal closest packing and cations (C) occupy only 2/3 octahedral voids in
it, then the general formula of the compound is
(a)
CA
(b)
CA2
(c)
C 2A 3
(d)
C 3A 2
[Answers : (1) a (2) b (3) c]
C4B
Hexagonal Closed Packing
In hcp it is known as ABAB..... type of packing.
Effective atoms in hcp (Zeff) =
1
1
 12  3   2 = 6. Coordination number in hcp = 12
6
2
Relation between ‘a’ and size of atom ‘r’ : a = 2r
Packing fraction in hcp = 6 
4r 3
3  24 2 r 3
...... (volume of hexagon = base area × height,
base area = 6 ×
C4C (a)
(b)
C5
2
3
× a2 , height ‘h’ = 4r
3
4
Types of structures of Simple Ionic Compounds (AB-Type) :
(i)
NaCl type or Rock salt type : It has fcc arrangement of ions. Na+ and Cl– ions have
6 : 6 coordination.Each unit cell has four NaCl units. For example : Halides of Ag,
halides of alkali metals, oxides and sulphides of alkaline earths.
(ii)
CsCl type structure : It has bcc arrangement. Cs+ and Cl– ions have 8 : 8 coordination.
Each unit cell has only one CsCl unit. For example : CsI, CsBr, CsCN, TII, TlCl, and
TlCN.
(iii)
Zinc blende (ZnS) type structure : It has fcc arrangement of S2– ions. Zn2+ occupies
the alternate position of tetrahedral void. Zn2+ and S2– ions have 4 : 4 coordination. For
example : CuCl, CuI, CuBr, AgI and BeS.
Structure of the Ionic Compounds (AB2 and A2B Type) :
(i)
Calcium fluorite (CaF2) type structure : It has ccp/fcc arrangement of Ca2+ ions.F–
occupies all the position of tetrahedral voids.Ca2+ and F– ions have 8 : 4 coordination
no.
(ii)
Antifluorite structure : Na2O has this type of structure. Oxide ions has fcc
arrangement whereas Na+ occupies all the tetrahedral position. Na+ and O2– ions have
4 : 8 coordination number.
Limiting Radius Ratios in Ionic Solids :
The radius ratio of cation to the radius ratio of anion is known as radius ratio of ionic solid.
i.e.,
r n
r m
 radius ratio
It was found that greater is the radius ratio, greater will be the C.N.
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r+n/r–m
C.N.
Structural Unit
Example
(a)
0.155 – 0.225
3
plane triangular
B2O3
(b)
0.224 – 0.414
4
Tetrahedral (bcc)
ZnS, CuX, AgI
(c)
0.414 – 0.732
6
Octahedral (fcc)
NaCl, NaI, KCl, RbI, FeO
(d)
0.732 – 1.00
8
Cubic
CsCl
Large number of ionic compounds obey this rule although there are many exceptions.
Practice Problems :
1.
KCl crystallizes in the same type of lattice as NaCl does. Given that
rNa 
rCl 
(a)
2.
rNa 
rK 
1.143
 0.7 the ratio of the side of the unit cell for KCl to that for NaCl is
(b)
11.43
(c)
1143.3
(d)
114.3
Iron has body centred cubic lattice structure. The edge length of the unit cell is found to be 286 pm.
The radius of an iron atom is
(a)
3.
 0.5 and
100 pm
(b)
124 pm
(c)
104 pm
(d)
140 pm
A solid A+ and B¯ had NaCl type close packed structure. If the anion has a radius of 250 pm, then the
ideal radius of the cation is
(a)
170.4 pm
(b)
134.4 pm
(c)
103.4 pm
(d)
184.4 pm
[Answers : (1) a (2) b (3) c]
C6
Defects in Solid
Three types of point defect : (i) Stoichiometric defect (ii) Non-stoichiometric defect (iii) Impurity defect.
(i)
Stoichiometric defects : The constituents in a solid are present in the same ratio as predicted by their
formula. These defects are also called intrinsic or thermodynamic defects.
Types of stoichiometric defects :
(a)
Vacancy defect : when some of the lattice sites are vacant, the crystal is said to be vacancy defect. As a
result the density of the solid decreases. This defect can also be develop upon heating.
(b)
Intestitial defect : When some constituent particles (atom or molecules) occupy an intenstial site, this
defect is created. This defect increases the density of the solid. Both defects are shown by non-ionic
compound. The ionic solids always exhibit electrical neutrialty and show schottky and frenkel defect.
(c)
Schottky defect : It is basically a vacancy defect in ionic solids. In order to maintain electrical neutrality
the number of missing cations and anions are equal. In this the density decreases. Scottky defect is shown
by those ionic compound in which the cation and anion are of almost similar sizes. e.g. NaCl, KCl, CsCl,
AgBr.
(d)
Frenkel defect : If smaller ions dislocate from lattice site to interstitial site then the vacancy defect arise at
lattice site and interstitial defect arises at new location. It does not change the density of the solid. This
defect shown by ionic solids in which there is large difference in the size of ions. e.g., AgCl, AgBr, AgI,
ZnS, ZnO etc. (AgBr shows both Frenkel and Schottky defect).
(ii)
Non-stoichiometric defects : are of two types (a) Metal excess defects (b) Metal deficiency defects.
(a)
Metal excess defects : arise in two ways
(i)
Due to anion vacancies : Alkali halides like NaCl, LiCl and KCl show this type of defect when
crystal of NaCl are heated in an atmosphere of sodium vapour, the sodium atoms are deposited
on the surface of the crystal. The Cl– ions diffuse to the surface of the crystal and combine with
Na+ ion to give NaCl. The loss of electron by sodium atom occupy anonic vacant site to maintain
the electrical neutrality called F-centre. F-centre impart the colour to the crystal. The colour
results by excitation of these electrons when they absorb energy from the visible light falling on
the crystals. NaCl impart yellow colour. Excess of lithium makes LiCl  pink colour. Excess of
potassium makes KCl  Violet.
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(ii)
Due to the presence of extra cations at interstitial sites : ZnO white in colour at room
temperature on heating it loses oxygen and turns yellow. The excess of Zn2+ ions move to
interstitial sites and to maintain the electrical neutrality electrons trap to neighbouring interstitial
sites.
ZnO

 Zn 2  
( white colour )
(b)
1
O 2   2e  .
2
Metal deficiency defect :
(i)
By cation vacancies : This defect is generally shown by transition metals capable of showing
variable oxidation states. In this defect some cations are missing from lattice sites and to
maintain the electrical neutrality their positive charges are balanced by the presence of extra
charges (i.e., higher oxidation state) on the neighbouring site. e.g., In FeO, Fe2+ ions missing
from lattice site & neighbour lattice site occupied by Fe3+.
(ii)
By the presence of extra anions at the interstitial sites
(iii)
Impurity Defects :
(a)
Impurity defect in ionic solids : e.g., NaCl doped with SrCl2 to produce one cation vacancy as to maintain
the electrical neutrality Sr2+ occupy one lattice site of Na+ vacancy and other Na+ lattice site is vacant. AgCl
doped with CdCl2 produce one cation vacancy.
(b)
Impurity defects in covalent solids :
(i)
n-type (electron rich impurities) : Group-14 element (like Si and Ge) which have four valence
electrons and doped with group-15 element like P or As which contain 5 valence electrons, they
occupy some of the lattice sites in silicon crystal. Four out of five electrons are used in the
formation of four covalent bonds with the four neighbouring silicon atoms. The fifth extra
electron is extra and delocalised. The fifth extra electron increase the conductivity due to
negatively charged electron. Silicon doped with electron rich impurity is called n-type
(n = negative) semiconductor
(ii)
p-type (electron deficient impurities) : Si or Ge doped with a group-13 element like B, Al or
Ga which contain only 3-valence electrons and hence the fourth valence electron is missing
called electron hole or electron vacancy which appear as it positively charged and are moving
towards negatively charged plate. This type of semiconductor is called p-type.
Practice Problems :
1.
2.
In a solid lattice the cation has left a lattice site and is located at an interstitial position. The lattice
defect is known as
(a)
Interstitial defect
(b)
Valency defect
(c)
Frenkel defect
(d)
Schottky defect
Ionic solids with Schottky defects contain in their structure
(a)
Equal number of cations and anion vacancies
(b)
Interstitial anions and anion vacancies
(c)
Cation vacancies only
(d)
Cation vacancies and interstitial cations
[Answers : (1) c (2) a]
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CSS – 8
SINGLE CORRECT CHOICE TYPE
1.
2.
3.
4.
5.
6.
7.
8.
Which of the following statements is correct in the
rock-salt structure of an ionic compounds ?
(a)
Both A and R are true and R is the
correct explanation of A
(a)
coordination number of cation is four
whereas that of anion is six
(b)
Both A and R are true and R is not the
correct explanation of A
(b)
coordination number of cation is six
whereas that of anion is four
(c)
A is true but R is false
(d)
Both A and R are false
(c)
coordination number of each cation and
anion is four
(d)
coordination number of each cation and
anion is six.
Which of the following expressions is correct in the
case of sodium chloride unit cell (edge length, a) ?
9.
10.
The unit cell with crystallographic dimentions
a = b  c,  =  =  = 900 is
(a)
Cubic
(b)
Tetragonal
(c)
Monoclinic
(d)
Hexagonal
Bragg’s equation is
(a)
rc + ra = a
(b)
rc + ra = a/2
(a)
n = 2 sin 
(b)
n = 2d sin 
(c)
rc + ra = 2a
(d)
rc + ra = 2 a
(c)
2n = d sin 
(d)
 = (2d/n)sin
The edge length of unit cell of sodium chloride is
564 pm. If the size of Cl¯ ion is 181 pm, the size of
Na+ ion would be
11.
The number of atoms/molecules contained in one
face centred cubic unit cell of a monoatomic
substance is
(a)
101 pm
(b)
167 pm
(a)
4
(b)
6
(c)
202 pm
(d)
383 pm
(c)
8
(d)
12
Which one of the following schemes of orderin, close
packed sheets of equal sized spheres do not
generate close packed lattice.
(a)
ABCABC
(b)
ABACABAC
(c)
ABBAABBA
(d)
ABCBCABCBC
The co-ordination no. of cation and anion in
Fluorite CaF2 and Rutile TiO2 are respectively
12.
13.
Which of the following statements are ture
(a)
Piezoelectricity is due to net dipole
moment
(b)
Ferro electricity is due to alignment of
dipoles in same direction
(c)
Pyroelectricity is due to heating polar
crystals
(d)
All
(a)
8 : 4 and 6 : 3
(b)
6 : 3 and 4 : 4
The three states of matter are solid, liquid and gas.
Which of the following statements are correct about
them
(c)
6 : 6 and 8 : 8
(d)
4 : 2 and 2 : 4
(a)
Gases and liquids have viscosity as a
common property
(b)
The molecules in all the three states
possess random translational motion
(c)
Gases cannot be converted into solids
without passing through the liquid phase
(d)
Solids and liquids have vapour pressure
as a common property
A solid is formed and it has three types of atoms X,
Y and Z, X forms a fcc lattice with Y atoms
occupying all the tetrahedral voids and Z atoms
occupying half the octahedral voids. The formula
of the solid is
(a)
X2Y4Z
(b)
XY2Z4
(c)
X4Y2Z
(d)
X4YZ2
A solid has a structure in which W atoms are
located at the corners of a cubic lattice, O atom at
the centre of edges and Na atom at centre of the
cube. The formula for the compound is
(a)
NaWO2
(b)
NaWO3
(c)
Na2WO3
(d)
NaWO4
Assertion (A) : In crystal lattice the size of the
cation is larger in a tetrahedral hole than in an
octahedral hole.
14.
For an ionic crystal of the general formula
A+B— and co-ordinates number 6, the radius ratio
will be
(a)
Greater than 0.73
(b)
Between 0.73 and 0.41
(c)
Between 0.41 and 0.22
(d)
Less than 0.22
Reason (R) : The cations occupy more space than
anions in crystal packing.
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15.
16.
17.
18.
19.
20.
21.
22.
23.
The pure crystalline substance on being heated
gradually first forms a turbid liquid at constant
temperature and still at higher temperature
turbidity completely disappears. The behaviour is
a characteristic of substance forming
(a)
Allotropic crystal
(b)
Liquid crystals
(c)
Isomeric crystals
(d)
Isomorphous crystals
24.
(A)
(B)
(C)
The maximum proportion of available volume that
can be filled by hard spheres in diamond is
(a)
0.52
(b)
0.34
(c)
0.32
(d)
0.68
(D)
Which of the following will show anisotropy
(a)
Glass
(b)
BaCl2
(c)
Wood
(d)
Paper
An alloy of copper, silver and gold is found to have
copper constituting the ccp lattice. If silver atoms
occupy the edge centres and gold is present at body
centre, the alloy has a formula
(a)
Cu4Ag2Au
(b)
Cu4Ag 4Au
(c)
Cu4Ag3Au
(d)
CuAgAu
Which is an example of ferroelectric compound
(a)
Quartz
(b)
(b)
Barium titanate (d)
PbCrO4
None
How many ‘nearest’ and ‘next nearest’ neighbours
respectively potassium have in bcc lattice
(a)
8, 8
(b)
8, 6
(c)
6, 8
(d)
8, 2
Ionic compounds for Schottky’s defects
(a)
NaCl, KCl, CsCl
(b)
AgCl, AgBr, AgI
(c)
NaCl, KCl, ZnS
(d)
KCl, CsCl, ZnS
Na and Mg crystallize in BCC and FCC type
crystals respectively, then the number of atoms of
Na and Mg present in the unit cell of their
respective crystal is
(a)
2 and 4
(b)
14 and 9
(c)
4 and 2
(d)
9 and 14
Graphite is a soft solid lubricant extremely
difficult to melt. The reason for this anomalous
behaviour is that graphite
(a)
has carbon atoms arranged in large
plates of rings of strongly bound carbon
atoms with weak interplate bonds
(b)
is a non-crystalline substance
(c)
is an allotropic form of diamond
(d)
has molecules of variation molecular
masses like polymers
Einstein Classes,
25.
Match the crystal system/unit cells liested in
Column I with their characterstic features in
Column II.
Column I
Column II
sc and fcc
(P)
a = b = c and  =  = 
cubic and
(Q)
are two crystal systems
rhombohedral
cubic and
(R)
have only two
tetragonal
crystallographic angles
of 900
hexagonal and
(S)
belong to same crystal
monoclinic
system
(a)
A-P, S; B-P, Q; C-Q; D-Q, R
(b)
A-P; B-P, Q; C-Q; D-Q, R
(c)
A-P, S; B-P; C-Q; D-Q, R
(d)
A-P, S; B-P, Q; C-Q; D-Q
Assertion-Reason Type
Each question contains STATEMENT-1 (Assertion)
and STATEMENT-2 (Reason). Each question has
4 choices (A), (B), (C) and (D) out of which ONLY
ONE is correct.
(A)
Statement-1 is True, Statement-2 is True;
Statement-2 is a correct explanation
for Statement-1
(B)
Statement-1 is True, Statement-2 is True;
Statement-2 is NOT a correct
explanation for Statement-1
(C)
Statement-1 is True, Statement-2 is False
(D)
Statement-1 is False, Statement-2 is True
Statement-1 : In any ionic solid [MX] with
Schottky defects, the number of positive and
negative ions are same.
Statement-2 : Equal number of cation and anion
vacancies are present.
(a)
A
(b)
B
(c)
C
(d)
D
ANSWERS (SINGLE CORRECT
CHOICE TYPE)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
d
b
a
c
a
a
b
d
b
b
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
a
d
a
b
b
b
b
c
c
b
21.
22.
23.
24.
25.
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
a
a
a
a
d
CSS – 10
EXCERCISE BASED ON NEW PATTERN
1.
2.
3.
4.
5.
6.
7.
COMPREHENSION TYPE
Comprehension-1
Atoms in a solid can be closely packed resulting into
either cubical-closest packing or hexagonal closest
packing. There exists two types of voids,
tetrahedral and octahedral.
The number of tetrahedral and octahedral voids
per atom respectively are
(a)
2 and 1
(b)
1 and 2
(c)
1 and 1
(d)
2 and 2
The sizes of voids follow the expression
(a)
tetrahedral > octahedral
(b)
tetrahedral < octahedral
(c)
tetrahedral = octahedral
(d)
tetrahedral > or < or = octahedral
In between the two closest packed layers,
octahedral voids are situated
(a)
near the top of the bottom layer
(b)
near the bottom of the top layer
(c)
at the centre of two layers
(d)
randomly between the two layers
Comprehension-2
Atoms in solid can be closely packed resulting into
either cubical-closest packing or hexagonal closest
packing
Which of the following arrangements of occupancy
of sites is correct in cubical-closest packing ?
(a)
ABABAB...
(b)
ABCABCABC...
(c)
ABBAAB...
(d)
ABCACBABC...
The unit cell of cubical-closest packing is
(a)
primitive cubic
(b)
body-centred cubic
(c)
face-centred cubic
(d)
not known
The maximum size of atoms that can be fitted into
octahedral voids in terms of radius r of atoms
closely packed is
(a)
0.414 r
(b)
0.225 r
(c)
0.372 r
(d)
0.680 r
Comprehension-3
In a cubical packing of atoms, the arrangement of
atoms may result into primitive (P), body-centred
(I) and face-centred (F) cubical unit cell. Identify
the correct choice in the following.
The percentage of void volume follow the
expression
(a)
P : I : F :: 47.64 : 31.98 : 25.95
(b)
P : I : F :: 47.64 : 25.95 : 31.98
(c)
P : I : F :: 31.98 : 47.64 : 25.95
(d)
P : I : F :: 31.98 : 25.95 : 47.64
Einstein Classes,
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
The sizes of atoms in terms of the respective
edge-length of the unit cell follow the expression
(a)
P : I : F :: a/2 : (2/4)a : (3/4)a
(b)
P : I : F :: a/2 : (3/4)a : (2/4)a
(c)
P : I : F :: (2/4)a : a/2 : (3/4)a
(d)
P : I : F :: (2/4)a : (3/4)a : a/2
The coordination numbers of atoms follow the
expression
(a)
P : I : F :: 12 : 6 : 8
(b)
P : I : F :: 12 : 8 : 6
(c)
P : I : F :: 6 : 8 : 12
(d)
P : I : F :: 6 : 12 : 8
Comprehension-4
Potassium crystallizes in a body-centered lattice
with a unit cell, a = 5.20 Å. [Given : At. Mass K=39]
The distance between nearest neighbour is
(a)
9.01 Å
(b)
5.2 Å
(c)
4.50 Å
(d)
3.677 Å
The distance between next-nearest neighbours is
(a)
9.01 Å
(b)
5.2 Å
(c)
4.50 Å
(d)
3.677 Å
The nearest neighbours does K atom have
(a)
6
(b)
8
(c)
12
(d)
4
The next nearest neighbours does each K atom have
(a)
6
(b)
8
(c)
12
(d)
4
The calculated density of crystalline K is
(a)
0.925 g/cm3
(b)
9.25 g/cm3
3
(c)
92.5 g/cm
(d)
none
Comprehension-5
Metallic gold crystallises in the face centered
lattice, the length of the cubic unit cell is a = 4.07 Å.
[Give At. Mass Au = 197]
The closest distance between gold atoms is
(a)
2.87 Å
(b)
4.07 Å
(c)
7.049 Å
(d)
3.52 Å
The “nearest neighbours” does each gold atom have
at the distance calculated in above question is
(a)
6
(b)
8
(c)
12
(d)
4
The density of gold is
(a)
19.4 g/cc
(b)
1.94 g/cc
(c)
0.194 g/cc
(d)
194 g/cc
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CSS – 11
(A)
(B)
(C)
(D)
(A)
(B)
(C)
(D)
(A)
(B)
(C)
(D)
1.
MATRIX-MATCH TYPE
Matching-1
Column - A
Column - B
A unit cell of NaCl has
(P)
181.5 pm
four formula unit. The
edge length of the unit
cell is 564 pm. The
density of NaCl is
For a cubic crystal, the
(Q)
2.165 g cm–3
face diagonal is 3.50 Å.
The face length will be
Sodium metal crystallises (R)
2.475 Å
in bcc lattice with cell edge
429 pm. The radius of
sodium atom
The unit cube length for (S)
186 pm
LiCl (NaCl structure) is
514 pm. Assuming
anion-anion contact, the
anionic radius is
(T)
90.5 pm
Matching-2
Column - A
Column - B
Tetragonal
(P)
abc
      900
Orthorhombic
(Q)
a = b  c,
 =  =  = 900
Rhombohedral
(R)
abc
 =  =  = 900
Triclinic
(S)
a=b=c
 =  =   900
(T)
abc
 =  = 900
  900
Matching-3
Column - A
Atomic Crystal
Column - B
(P)
Vander Waal’s
forces
Molecular Crystal
(Q)
Metallic bond
Covalent Crystal
(R)
London
dispersion
force
Metallic Crystal
(S)
Covalent
bonds
(T)
Ionic bonds
MULTIPLE CORRECT CHOICE TYPE
In Fe3O4
(a)
Oxide ions have C.C.P. arrangement
(b)
Fe2+ ions occupy Octahedral voids
(c)
Fe3+ ions occupy tetrahedral voids
(d)
50% voids are occupied by cations
Einstein Classes,
2.
3.
4.
5.
Which of the following paramgnetic substances lose
their magnetism in absence of magnetic field
(a)
TiO
(b)
MnO
(c)
MnO2
(d)
CuO
Which of the following statements are correct ?
(a)
The coordination number of each type of
ion in CsCl is 8
(b)
A metal that crystallizes in bcc structure
has a coordinates number of 12
(c)
A unit cell of an ionic crystal shares some
of its ions with other unit cells
(d)
The length of unit cell in NaCl is 552 pm
(r Na+ = 95 pm; rCl– = 181 pm)
In the crystal lattice of sodium chloride,
(a)
there are six neighbouring anions around
a cation.
(b)
there are twelve neighbouring cations
around a cation.
(c)
there are eight next nearest neighbouring
anions around a cation.
(d)
there are six next nearest neighbouring
cation around a cation.
For a compound AB crystallises in a body-centred
cubic lattice with a as the edge length of unit cell,
(a)
the minimum distance between A and A
is a
(b)
the mean distance between A and B is
3a / 2
(c)
the density of crystal is given by

2 M AB
a3 NA
(d)
6.
the minimum distance between B and B
is a/2.
For a compound AB crystallises in a face-centred
cubic lattice with edge length a of a unit cell.
(a)
the minimum distance between A and A
is a/2.
(b)
the minimum distance between B and B
is a/2.
(c)
the minimum distance between A and B
is a/2
(d)
the density of the crystal is given by

1 M AB
a3 NA
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CSS – 12
1.
2.
3.
4.
Assertion-Reason Type
Each question contains STATEMENT-1 (Assertion)
and STATEMENT-2 (Reason). Each question has
4 choices (A), (B), (C) and (D) out of which ONLY
ONE is correct.
(A)
Statement-1 is True, Statement-2 is True;
Statement-2 is a correct explanation
for Statement-1
(B)
Statement-1 is True, Statement-2 is True;
Statement-2 is NOT a correct
explanation for Statement-1
(C)
Statement-1 is True, Statement-2 is False
(D)
Statement-1 is False, Statement-2 is True
Statement-1 : In crystal lattice the size of the
cation is smaller in a tetrahedral hole than in an
octahedral hole.
Statement-2 : The cations occupy more space than
anions in crystal packing.
Statement-1 : Copper is conducting as such while
copper sulphate is conducting only in molten state
or in aqueous solution.
Statement-2 : Copper conducts due to free
electrons. It is the ions which conduct electricity
in ionic compounds.
Statement-1 : Frenkel defects not found in pure
alkali halides.
Statement-2 : There is large difference between
the size of cation and anion, the cation cannot
occupy interstitial position.
Statement-1 : When ferrimagnetic Fe3O4 is heated
to 850 K. It becomes paramagnetic.
Statement-2 : The randomisation of spins takes
place.
5.
6.
7.
8.
9.
10.
Statement-1 : Glass and quartz both conain SiO44–
units.
Statement-2 : Glass is amorphous whereas quartz
is crystalline.
Statement-1 : In rock structure all the octrahedral
voids in the close packing of anions are occupied
by cations.
Statement-2 : In rock salt structure, the distance
of closest approach between two anions equal to
half the face diagonal of the unit all.
Statement-1 : bcc arrangement is less closely
packed them ccp arrangement.
Statement-2 : In ccp arrangement, the two atoms
at the corners of the unit cell are touching each
other whereas in bcc arrangement, they are not
touching each other.
Statement-1 : In any ionic solid (MX) with
schottky defects, number of positive and negative
ions are same.
Statement-2 : Equal number of cation and anion
vacancies are present.
Statement-1 : For ionic solids exhibiting Frenkel
defects, the density remains unaltered.
Statement-2 : Doping of group 14 element with
suitable element of group 13 produces p-type of
semi-conductors.
Statement-1 : Ferromagnetic substances are
strongly attacted by magnetic field.
Statement-2 : Ferromagnetism arises due to
spontaneous alignment of magnetic moments of
ions or atoms in the same direction.
(Answers) EXCERCISE BASED ON NEW PATTERN
COMPREHENSION TYPE
1.
a
2.
b
3.
c
4.
b
5.
c
6.
a
7.
a
8.
b
9.
c
10.
c
11.
b
12.
b
13.
a
14.
a
15.
a
16.
c
17.
a
2.
[A-Q; B-R; C-S; D-P]
3.
[A-R; B-P; C-S; D-Q]
MATRIX-MATCH TYPE
1.
[A-Q, B-R, C-S, D-P]
MULTIPLE CORRECT CHOICE TYPE
1.
a, b
6.
a, b, c
2.
a, d
3.
a, c, d
4.
a, b, c, d
5.
a, b
B
6.
ASSERTION-REASON TYPE
1.
C
2.
A
3.
A
4.
A
7.
C
8.
A
9.
B
10.
A
Einstein Classes,
5.
B
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CSS – 13
INITIAL STEP EXERCISE
(SUBJECTIVE)
1.
A metallic element exists as a cubic lattice. Each
edge of the unit cell is 2.88 Å. The density of the
metal is 7.20 g/cm3. How many unit cells will be
present in 100 gm of the metal ?
2.
Iron occurs as a body centred as well as face centred
cubic unit cell. If the effective radius of an atom of
iron is 124 pm, compute the density of iron in both
these structures.
3.
NH4Cl crystallizes as a body centred cubic lattice
with a unit distance of 387 pm. Calculate (a) The
distance between the oppositely charged ions in the
lattice and (b) The radius of the NH4+ ion if the
radius of the Cl¯ ions is 181 pm.
4.
An element crystallize in a structure having f.c.c.
unit cell of an edge 200 pm. Calculate the density if
200 g of this element contains 24 × 1023 atoms.
5.
The density of KBr is 2.75 g cm–3. The length of the
edge of the unit cell is 654 pm. Show that KBr has
face centred cubic structure.
12.
Density of NaCl = 2.165 g cm–3
Distance between Na+ and Cl¯ in NaCl = 281 pm
13.
A unit cell of sodium chloride has four formula
units. The edge length of the unit cell is 0.564 nm.
What is the density of sodium chloride ?
14.
The edge length of a cubic unit cell of an element
(atomic mass = 95.54) is 313 pm and its density is
10.3 g cm–3. Calculate the atomic radius.
15.
Cesium may be considered to form interpenetrating simple primitive cubic crystals. The edge length
of unit cell is 412 pm. Determine : (a) the density of
CsCl, and (b) the ionic radius of Cs+ if the ionic
radius of Cl¯ is 181 pm.
Given M(Cs) = 133 g mol–1.
16.
An element ‘A’ (At. wt. = 100) having BCC
structure has unit cell edge 400 pm. Calculate the
density of A and the number of unit cells and the
number of atoms in 10 g. of A.
17.
Calculate the radius of the atom which are
occupying the body centred cubic unit cell. Length
of the cube = 4 × 10–10 m.
18.
Uranium forms a simple cubic lattice and its atomic
radius is 1.38 × 10–10 m. If the atomic volume of
uranium is 12.5 × 10–3 dm3 mol–1 calculate the ratio
of theoretical to actual volume.
19.
Vanadium metal has a body centred cubic
structure and the length of the unit cell is 3.04 Å. (i)
Calculate the volume of the unit cell in cubic
centimeters (ii) If vanadium has a density of
5.96 g/cm3, calculate the mass of one unit cell. (iii)
How many vanadium atoms are in each unit cell of
the element ?
20.
Calculate the atomic volume of an element whose
atoms have a radius 1.414 Å when the atoms are in
cubic closest packed structure. What is the length
of an edge of the unit cell ?
21.
The edge of the unit cell of KCl is 6.29 × 10–10 m,
and the density of KCl is 1.99 × 10–3 kg m–3. If the
KCl unit cell is face centred cubic and contains 4K+
ions and 4Cl-ions per unit cell, then calculate the
number of K+ ions in one kg of KCl.
22.
Iron occurs as BCC as well as FCC unit cell. If the
effective radius of an atom of iron is 124 pm,
compute the density of iron in both these structures.
23.
The two ions A+ and B¯ have radii 88 and 200 pm
respectively. In the close packed crystal of
compound AB, predict the co-ordination number
of A+.
At mass : K = 39, Br = 80)
6.
A crystal of lead (II) sulphide has NaCl structure.
In this crystal the shortest distance between the Pb2+
ion and S2– ion is 297 pm. What is the length of edge
of the unit cell in lead sulphide ? Also calculate the
unit cell volume.
7.
The effective radius of an iron atom is 1.42 Å. It
has rock salt like structure. Calculate its density
(Fe = 56 amu).
8.
If length of the body diagonal for CsCl which
crystallises into a cubic structure with Cl¯ ions at
the corners and Cs+ ions at the centre of the unit
cells is 7 Å and the radius of the Cs+ ion is 1.69 Å,
what is the radii of Cl¯ ion ?
9.
In a cubic closed packed structure of mixed oxides
the lattice is made up of oxide ions, one eighth of
tetrahedral voids are occupied by divalent ions (A2+)
while one half of the octahedral voids occupied
trivalent ions (B3+). What is the formula of the oxide ?
10.
Cesium chloride forms a body centred cubic
lattice. Cesium and chloride ions are in contact
along the body diagonal of a cell. The length of the
side of the unit cell is 412 pm and Cl¯ ion has a
radius of 181 pm. Calculate the radius of Cs+ ion.
11.
Xenon crystallises in the face-centred cubic lattice
and the edge of the unit cell is 620 pm. What is the
nearest neighbour distance and what is the radius
of xenon atom ?
Einstein Classes,
Calculate the value of Avogadro’s number from the
following data :
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CSS – 14
24.
CsCl has bcc arrangement and its unit edge length
is 400 pm. Calculate the interionic distance in CsCl.
25.
Sodium has a bcc structure with nearest neighbour
distance 365.9 pm. Calculate its density (Atomic
mass of sodium = 23)
26.
27.
Tungsten has body centered cubic lattice and each
lattice point is occupied by one atom. Calculate the
radius of metallic tungsten if density of tungsten is
19.03 g cm–3 and at. wt. is 183.9.
28.
KF has NaCl structure. What is the distance
between K+ and F¯ in KF if the density of KF is
2.48 g cm–3 ?
29.
An element occurs in BCC structure with a cell
length of 288 pm. The density of the element is
7.2 g cm–3. How many atoms of the element does
208 g. of the element contain ?
30.
Polonium crystallises in a simple cubic unit cell. It
has an atomic mass = 209 and density = 91.5 kg
m–3. What is the edge length of its unit cell ?
A compound formed by element X and Y crystallises
in the cubic structure, where X is at the corners of
a cube and Y is at the face center. What is the
formula of the compound. If side length is 5 Å, estimate the density of the solid assuming at. wt. of X
and Y as 60 and 90 respectively.
FINAL STEP EXERCISE
(SUBJECTIVE)
1.
2.
3.
4.
5.
6.
Iron (II) oxide, FeO, crystal has a cubic structure
and each edge of the unit cell is 5 Å. Taking density
of the oxide as 4 g/cc, calculate the number of
Fe2+ and O2– ions present in each unit cell.
(a)
If titanium is described as occupying
holes in the BaO lattice, what type of the
hole does it occupy ?
(b)
What fraction of the holes of this type
does it occupy ?
(c)
Can you suggest a reason why it has
certain holes of this type but not the other
holes of the same type.
+
A solid A and B¯ had NaCl type close packed
structure. If the anion has a radius of 250 pm, what
should be the ideal radius of the cation ? Can a
cation C+ having a radius of 180 pm be slipped into
the tetrahedral site of the crystal of A+B¯ ? Give
reason for your answer.
7.
in this lattice ( rK  = 1.33 Å,
A metal crystallises into two cubic phases, face
centered cubic (fCC) and body centred cubic (bCC),
whose unit cell length are 3.5 and 3.0 Å,
respectively. Calculate the ratio of densities of fCC
and bCC.
Iron crystallizes in several modifications. At about
9100C, the bCC -form undergoes a transition to
the face centred cubic -form. Assuming that the
distance between nearest neighbours is the same in
the two forms at the transition temperature,
calculate the ratio of the density of iron of fCC form
to that of iron of bCC form at the transition
temperature.
Lithium metal has a body-centred cubic unit cell.
(a) How many atoms are there in a unit cell ? (b)
Krypton crystallizes in a structure that has four Kr
atoms in each unit cell and the unit cell is a cube.
The edge length of the unit cell is 0.558 nm.
Calculate the density of crystalline Kr, in kg/m3.
BaTiO3 crystallises in the pervoskite structure. This
structure may be described as a cubic lattice, with
barium ions occupying the corners of the unit cell,
oxide ions occupying the face centres and titanium
ions occupying the centres of the unit cells.
Einstein Classes,
The NaCl lattice has f.c.c. unit cell KBr crystallizes
rBr ¯ = 1.95 Å) . (a) How
+
many K ions and how many Br¯ ions are in each
unit cell ? (b) Assuming the additivity of ion radii,
what is a ? (c) Calculate the density of the perfect
KBr crystal. (d) What minimum value of r+/r– is
needed to prevent anion-anion contact in this
structure ? [At. mass Br = 80]
8.
The ZnS structure is cubic. The unit cell may be
described as a face centered sulphide ion sublattice
with zinc ions in the centres of alternating minicubes
made by partitioning the main cube into eight equal
parts.
(a)
How many nearest neighbours does each
Zn2+ have ?
(b)
How many nearest neighbours does each
S2— have ?
(c)
What angle is made by the lines
connecting any Zn2+ to any two of its
nearest neighbour ? (d) What minimum
r
r ratio is needed to avoid anion-anion
contact ? Closest cation-anion pairs are
assumed to touch.
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CSS – 15
9.
10.
(a)
X-ray diffraction give the unti cell edge
length of NaCl to be 0.5627 nm.
Calculate the density in gm/cm.
(b)
If the measured density is found to be
2.164 gm/cc. Find the percentage of Na+
and Cl¯ ions missing. Assume that
difference between theoretical and
observed densities to theoritical density
is lattice vacancies.
(a)
The figures given below show the
location of atoms in two crystallographic
planes in a FCC lattice. Draw the unit cell
for the corresponding structure and
identify these planes in your diagram.
(b)
In fcc arrangement of A & B atoms,
where A atoms are at corners of the unit
cell, B atoms at the face-centers, one of
the atoms are missing from the corner in
each unit cell then find the percentage of
void space in the unit cell.
11.
At room temperature sodium has a body centred
cubic structure with a cell edge length of 429.06 pm.
At –1950C the density is 4% larger but the edge
now measures 535 pm. Which type of cubic unit
cell does sodium have at –1950C ?
12.
An oxide of the formula TiO has the NaCl
structure with a cube edge of 4.18A. The density of
the oxide is 4.92 g/cc. Calculate the per cent
occupancy of Ti and O sites in the solid.
[At. wt. Ti = 47.9]
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CSS – 16
ANSWERS SUBJECTIVE (INITIAL STEP EXERCISE)
1.
3.
4.
6.
7.
8.
10.
13.
5.82 × 1023
2.
(a)
335.15 Pm
(b)
154.15 Pm
41.67 g cm–3
a = 5.94 × 10–8 cm, v = 2.096 × 10–22 cm3
5.74 g cm–3
1.81 Å
9.
175.8 Pm
11.
–3
2.166 gm cm
14.
15.
d = 4g/cc3, rcs  175.8 pm
16.
17.
18.
19.
5.188 g/cc, 3 × 1022 unit cells, 6 × 1022 atoms
1.7 × 10–10m
52.38 %
(i)
2.81 × 10–23 cm3
(ii)
(iii)
2
3
9.6 cm /mole
21.
8.59 g/cm3
6
24.
1.51g/cm3
26.
3
XY3, 4.38 g/cm
28.
24.16 × 1023
15.59 × 10–8 cm
20.
22.
23.
25.
27.
29.
30.
7.887 g/cm3, 8.59 g/cm3
AB2O4
438.5 Pm, 219.25 Pm
135.5 Pm
1.67 × 10–22 g/unit cell
8.07 × 1027
346.4 pm
1.37 Å
2.68 × 10–8
ANSWERS SUBJECTIVE (FINAL STEP EXERCISE)
1.
3.
4.
6.
7.
8.
9.
12.
Approx. 4
1.88 × 10–6
1.09
(a)
Octahedral holed
(b)
1/4
(c)
In case of other octahedral holes,
the proximity of Ba2+ and T14+
would be electrostatically
unfavourable.
(a)
four of each (b)
6.56 Å
(a)
4
(b)
4
(a)
2.18 g/cm3
(b)
0.734
84.6 %
Einstein Classes,
2.
103.4 Pm, NO
5.
(a)
(c)
(c)
11.
2.8 g/cm3
109028’
fcc
Two
(b)
3190 kg/m3
(d)
(d)
0.414
0.225
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