CSOM – 1
STATES OF MATTER
Syllabus :
Classification of matter into solid, liquid and
gaseous state.
Gaseous State :
Measurable properties of gases; Gas laws - Boyle’s
law, Charle’s law, Charle’s law, Graham’s law of
diffusion, Avogadro’s law, Dalton’s law of partial
pressure; Concept of Absolute scale of temperature;
Ideal gas equation, Kinetic theory of gases (only
postulates); Concept of average, root mean square
and most probable velocities; Real gases, deviation
from Ideal behaviour, compressibility factor, van der
Waals equation, liquefaction of gases, critical
constants.
Liquid State :
Properties of liquids - vapour pressure, viscosity and
surface tension and effect of temperature on them
(qualitative treatment only).
Solid State :
Classification of solids : molecular, ionic, covalent
and metallic solids, amorphous and crystalline
solids (elementary idea); Bragg’s Law and its
applications; Unit cell and lattices, packing in
solids (fcc, bcc and hcp lattices), voids, calculations
involving unit cell parameters, imperfection in
solids; Electrical, magnetic and dielectric
properties.
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CSOM – 2
CONCEPTS
C1
The gaseous state of the matters is characterised by their following properties :
(1)
They have neither fixed volume not fixed shape. i.e., their volume and shape depends upon the
size and shape of the container.
(2)
They can expand indefinitely and uniformly to the shape available to them.
(3)
They may be compressed by the application of external pressure.
(4)
They can diffuse and mix with each other to form mixture of any composition
(5)
At high pressure and low temperature, they can be converted into their liquid state
(6)
They can exert pressure on the surrounding.
(7)
The properties of gases can be fully described in terms of pressure, temperature, volume and
amount of the gas.
All of these properties of the gaseous state are due to their very weak intermolecular force (nearly
negligible).
C2A GAS LAWS
The Laws which co-relates the variable while defining the behaviour of the gases are known as gas laws.
They are :
BOYLE’S LAW :
At constant temperature, the volume of definite mass of a gas is inversely proportional to its pressure.
Mathematically, it is expressed as
V  1/P or PV = constant
Graphical Representation of Boyle’s Law :
Isotherms : They refers the graphical variation of volume with pressure, when the temperature remains
constant.
CHARLE’S LAW :
At constant pressure for a fixed amount of a gas, the volume of a given gas is directly proportional to the
absolute temperature
V  T......
(temperature is in Kelvin),
V1 V2

T1 T2
Graphical Representation of Charle’s Law :
GAY LUSSAC’S LAW :
An expression similar to Charles law exists between pressure and temperature of the gas at a fixed volume.
This law can be expressed as
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CSOM – 3
PT
or
P/T = constant
AVOGADRO’S HYPOTHESIS :
According to this equal volume of all the gases contain equal number of molecules, under similar conditions of temperature and pressure. It implies that
or
VN
(at constant temperature and pressure)
Vn
(where N is number of molecules and n is number of moles of
gas)
Dalton’s Law of Partial Pressure :
The total pressure exerted by a mixutre of non-reacting gases in a definite volume (closed container) is
always equal to the sum of the their individual pressures which each gas would exert if it occupies the same
space (volume) at a constant temperature.
If pA, pB, pC are individual pressures (partial pressures) of the non-reacting gases in the gaseous mixture and
PT is the pressure exerted by the mixture, then
PT = pA + pB + pC + .....
Also
pA = PTXA......
(XA is the mole fraction of A and PT is the total pressure of
the mixture)
Practice Problems :
1.
A flask contains 12 g of a gas of relative molecular mass 120 at a pressure of 100 atm was evacuated
by means of a pump until the pressure was 0.01 atm. Which of the following is the best estimate of
the number of molecules left in the flask (N0 = 6 × 1023 mol–1) :
(a)
2.
3.
(a)
remain unchanges
(c)
increase to four-fold
(c)
6 × 1017
(d)
6 × 1013
(b)
be doubled
(d)
be reduced to 1/4 th
Under constant pressure, a certain gas at 0 C was cooled until its volume was reduced to one-half.
The temperature at this stage is
0K
(b)
136.5 K
(c)
136.50C
(d)
–2730C
A gas of volume 100 ml is kept in a vessel at pressure 104 Pa maintained at temperature 240C. If now
the pressure is increased to 105 Pa, keeping the temperature constant, then the volume of the gas
becomes
10 ml
(b)
100 ml
(c)
1 ml
(d)
1000 ml
0
A vessel of 120 mL capacity contains a certain amount of gas at 35 C and 1.2 bar pressure. The gas
is transferred to another vessel of volume 180mL at 350C. The pressure would be
(a)
6.
6 × 1018
0
(a)
5.
(b)
If the absolute temperature of a gas is doubled and the pressure is reduced to one-half, the volume of
the gas will
(a)
4.
6 × 1019
0.3 bar
(b)
0.8 bar
(c)
1.8 bar
(d)
1.3 bar
0
N2 + 3H2  2NH3. 1 mol N2 and 4 mol H2 are taken in 15 L flask at 27 C. After complete conversion
of N2 into NH3, 5 L of H2O is added. Pressure set up in the flask is :
(a)
3  0.0821 300
atm
15
(b)
2  0.0821 300
atm
10
(c)
1 0.0821 300
atm
15
(d)
1 0.0821 300
atm
10
[Answers : (1) b (2) c (3) b (4) a (5) b (6) d]
C2B
IDEAL GAS EQUATION :
The combination of Boyle’s law and Charless law leads to the expression
PV = nRT.
where n is the amount of gas and R is universal gas constant whose value in different units are given below.
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CSOM – 4
R = 0.0821 L atm K–1 mol–1 or 8.314 J K–1 mol–1 or 2 cal mol–1 K–1
Also, ideal gas equation can be expressed in terms of density (d) and vapour density (D) as follows :
PM = dRT, 2PD = dRT
Practice Problems :
1.
A gas in an open container is heated from 270C to 1270C. The fraction of the original amount of gas
escaped from the container will be :
(a)
2.
(b)
1/2
(c)
1/4
(d)
1/8
(c)
2730C, 1 atm
(d)
2730C,2 atm
Then density of neon will be highest at
(a)
3.
3/4
S.T.P.
00C, 2 atm
(b)
At 00C; the density of a gaseous oxide at 2 bar is same as that of nitrogen at 5 bar. The molecular
mass of the oxide is
(a)
40
(b)
50
(c)
60
(d)
70
[Answers : (1) c (2) b (3) d]
C3A DIFFUSION AND EFFUSION :
The diffusion is defined “as a process of intermixing of two or more gases, irrespective of their density
without the help of any external agency.”
The diffusion of gases takes place due to rapid movement of gaseous molecules and due to the presence of
large intermolecular space available to them.
If a gas is taken in a pot and is allowed to leak through a small hole is diffuses into the atmosphere and the
process is known as Effusion.
Hence, effusion of a gas refers the passage of gas through a tiny opening into a large space.
C3B
GRAHAM’S LAW OF DIFFUSION AND EFFUSION :
The rate of diffusion of a gas is inversely proportional to the square root of its density or molar mass.
Mathematically, it is expressed as
r
1

or 
1
M ..... (P & T constant)
Comparison of the rate of diffusion of two gases when the pressure is not constant, then
r2 P1

r1 P2
M1
......(T constant) M1 and M2 are molecular mass of two gases.
M2
(Rate of diffusion of gas)
rd =
V( volume of gas diffused)
t(time of diffusion)
rd 
A(area of cross  section of tube)  d
.....
t(time of diffusion)
(d is the distance
travelled by the gas till
the point of diffusion)
Practice Problems :
1.
2.
A bottle of dry ammonia and a bottle of dry hydrogen chloride connected through a long tube are
opened sumultaneously at both ends, the white ammonium chloride ring first formed will be
(a)
at the centre of the tube
(b)
near the hydrogen chloride bottle
(c)
near the ammonia bottle
(d)
throughout the length of the tube
Equal volumes of two gases A and B diffuse through a porous pot in 20 and 10 seconds respectively.
If the molar mass of A be 80, the molar mass of B is
(a)
20
Einstein Classes,
(b)
30
(c)
40
(d)
50
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CSOM – 5
3.
240 ml of a hydrocarbon diffuses through a porous membrane in 20 min while 120 ml of SO2 under
identical conditions diffuses in 20 min. The molecular weight of hydrocarbon is
(a)
4.
6.
(b)
46
(c)
30
(d)
16
20 dm of SO2 diffuse through a porous partition in 60s. The volume of O2 will diffuse under similar
conditions in 30s is
(a)
5.
44
3
14.14
(b)
25
(c)
0.707
(d)
0.25
X ml of H2 gas effuses through a hole in a container in 5 seconds. The time taken for the effusion of
the same volume of the gas specified below under identical condition is
(a)
10 seconds : He
(b)
20 seconds : O2
(c)
25 seconds : CO
(d)
55 seconds : CO2
One mole of nitrogen gas at 0.8 atm takes 38s to diffuse through a pinhole, whereas one mole of an
unknown compound of xenon with fluorine at 1.6 atm takes 57 s to diffuse through the same hole.
The molecular formula of the compound is [At. Mass Xe = 131.30, F = 19.0]
(a)
XeF4
(b)
XeF2
(c)
XeF6
(d)
none
[Answers : (1) b (2) a (3) d (4) a (5) b (6) c]
C4A KINETIC THEORY OF GASES :
Following are the different assumptions of kinetic theory which true only for the ideal gases :
C4B
(i)
The volume occupied by a molecule is negligible in comparison to the total volume.
(ii)
The molecules are in state of rapid, straight-line motion, colliding with each other. The collision
between them is perfectly elastic.
(iii)
The absolute temperature of gas is a measure of the average K.E. of all the molecules present in
it.
KINETIC GAS EQUATION :
PV 
1
mnC 2 ......
3
[P  Pressure of gas
V  Volume of gas
m  mass of a molecule
C  root mean square velocity]
For one mole of a gas PV = RT and n = NA (Avogadro’s Number)
PV 
Kinetic energy per mole =
1
1
mN A C 2 , RT  MC 2
3
3
3
3
RT , Average kinetic energy = kT
2
2
Practice Problems :
1.
At what temperature will hydrogen molecules have the same KE as nitrogen molecules at 280 K :
(a)
280 K
(b)
40 K
(c)
400 K
(d)
50 K
[Answers : (1) a]
C4C DIFFERENT TYPES OF VELOCITIES OF GASEOUS MOLECULES :
ROOT MEAN SQUARE SPEED :
u rms
 u 2  u 22  ....  u 2N
 1

N

Einstein Classes,
½

 and is given by the expression u
rms 


3RT
M
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CSOM – 6
AVERAGE SPEED :
u av 
u 1  u 2  ....  u N
, u av 
N
8RT
M
MOST PROBABLE SPEED :
u mp 
2RT
, Comparing the three speeds listed above, we find that u rms : u av : u mp :: 3 : 8 /  : 2
M
urms > uav > ump
Practice Problems :
1.
The average velocity of gas molecules is 400 m/sec. The rms velocity at the same temperature is
(a)
2.
(b)
300 m/sec
(c)
4.34 m/sec
(d)
3.00 m/sec
0
The average velocity of an ideal gas molecule at 27 C is 0.3 m/sec. The average velocity at 9270C will
be
(a)
3.
434 m/sec
0.6 m/sec
(b)
0.3 m/sec
(c)
0.9 m/sec
(d)
3.9 m/sec
The average speed at T1K and the most probable speed at T2K of CO2 gas is 9.0 × 104 cm/s. The
values of T1 and T2 are respectively
(a)
T1 = 1000 K, T2 = 2000 K
(b)
T1 = 1682 K, T2 = 2143.4 K
(c)
T1 = 1082 K, T2 = 2140.4 K
(d)
T1 = 3282 K, T2 = 4140.4 K
[Answers : (1) a (2) a (3) b]
C5A VAN DER WAAL’S EQUATION OF STATE :
Van der Waal pointed out that the deviations shown by real gases are due to the following two facts.
1.
The volume occupied by the molecules is not negligible in comparison to the total volume of the gas.
2.
There exist forces of attraction between the molecules.
Van der Waal systematically corrected the ideal gas equation in the light of the above two facts. The corrected equation is


 P  a ( Vm  b)  RT
2 

vm 

Units of a and b :
Unit of a : We know that
P
an 2
(P = pressure connection), a 
V2
PV 2
n2
so, that unit of a is atmL2mol–2.
Unit of b : It is the volume correction per mole of gas, so its unit is litres mol–1.
C5B
Significance of a and b :
(i)
The value of a is a measure of the intermolecular forces of attraction. An easily liquefiable has
has greater intermolecular foces of attraction. So, greater is the value of a for a gas, greater will
be the ease of its liquification. The easily liquefiable gases (SO2 > NH3 > H2S > CO2) have
higher values of a than the permanent gases like N2, H2, O2 and He. The constant a is dependent
on temperature.
(ii)
The value of b has constant value over a wide range of temperature and pressure, this indicates
that the gas molecules are incompressible.
VANDER WAAL’S EQUATION AT LOW PRESSURE :
Van der Waals equation accounts for the behaviour of real gases. At low pressures, the gas equation can be
written as


 P  a  Vm  RT
2 

vm 

Einstein Classes,
or
Z
pVm
a
 1
RT
Vm RT
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CSOM – 7
where Z is known as compressibility factor.
C5C VANDER WAAL’S EQUATION AT HIGH PRESSURE :
At high pressure the real gas equation can be written as :
P(Vm – b) = RT, Z 
*
PVm
Pb
 1
RT
RT
At very low pressure and at very high temperature real gases behave ideally i.e., Z = 1.
Practice Problems :
1.
2.
3.
The compressibility factor of gas is less than unity at STP, therefore :
(a)
Vm (molar volume) > 22.4 L
(b)
Vm < 22.4 L
(c)
Vm = 22.4 L
(d)
Vm = 44.8 L
For the non-zero value of force of attraction between gas molecules, gas equation will be
(a)
PV  nRT 
(c)
PV = nRT
n 2a
V
(b)
PV = nRT + nbP
(d)
P
nRT
Vb
The compression factor (compressibility factor) for one mole of a van der Waals gas at 00C and 100
atmosphere pressure is found to be 0.5. Assuming that the volume of a gas molecular is negligible,
the van der Waals constant(a) is
(a)
30.55
(b)
1.253
(c)
25.55
(d)
20.22
[Answers : (1) b (2) a (3) b]
C6
*
Important Points for Gases (Real Gases)
All gases do not follow the ideal gas behaviour (PV = nRT) over a range of pressure and
temperature. They show the ideal behaviour at high temperature and very low pressure. But at high
pressure and low temperature they are not obeying the behaviour of ideal gases.
All gases condense at high pressure and sufficiently low temperature.
The deviation of real gas from ideal gas is measured by compressibility factor (Z)
Z
1.
2.
3.
PVm
PV
V
or
(Vm = molar volume =
)
nRT
RT
n
Graphs are at 273 K
Conclusions from Graph :
Z = 1 for ideal gas (No force of attraction or repulsion)
All gases have 1 at very low pressure.
Z > 1 means repulsive forces dominant and the gases are more difficult to compress. Z > 1 is at high
pressure.
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CSOM – 8
4.
5.
1.
At intermediate P, some of the gases have Z < 1 indicating that attractive forces are dominant and
favour compression.
For H2, Z > 1 at all ‘P’ at 273 K. Although below 273K, for H2 Z is less than 1
Liquification of gases and Critical Point :
Critical Temperature : The maximum temperature at which a gas can be liquified i.e. the
temperature above which a liquid cannot exist.
2.
Critical Pressure and Critical Volume : The minimum pressure necessary to liquify and gas at its
critical temperature is called critical pressure (PC) and the corresponding volume by one mole of gas
is called critical volume (VC)
3.
4.
5.
At critical point densities of a substance in liquid and gaseous state is same.
Generally gases below their TC are called vapours.
Boyle’s Temperature : Temperature at which real gases obeys the gas laws over a wide range of
pressure is called Boyle’s temperature.
Gases which are easily liquified have a high Tb whereas which are difficult to liquify have a low Tb.
Practice Problems :
1.
Under critical states of a gas for one mol of a gas, compressibility factor is :
(a)
3
8
(b)
8
3
(c)
1
(d)
1
4
[Answers : (1) a]
Solid State
C7
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.
C8A 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.
C8B
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
Einstein Classes,
(2)
Crystalline solids.
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CSOM – 9
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.
C8C 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
a=bc
 =  =  = 90
0
 =  =  = 90
0
Tetragonal
Orthorhombic
Monoclinic
Hexagonal
Rhombohedral
Triclinic
abc
Examples
0
abc
Copper, Zinc blende, KCl
Sin (White tin), SnO2, TiO2
Rhombic sulphur, CaCO3
 =  = 90 ;   90
0
a=bc
0
 =  = 90 ;  = 120
a=b=c
abc
Monoclinic sulphur, PbCrO2
0
Graphite, ZnO
 =  =   90
0
CaCO3(Calcite), HgS (Cinnabar)
      90
0
K2Cr2O7, CuSO4.5H2O
C9A In case of cubic crystal they have three kinds of Bravais Lattices, these are :
C9B
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]
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CSOM – 10
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
C9C 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]
C9D 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.
C9E
Density of solid : d 
Einstein Classes,
mass of a unit cell
Z M
, d  eff 3
volume of a unit cell
N Aa
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CSOM – 11
Practice Problems :
1.
Potassium has a bcc structures with nearest neighbour distance 4.52 Å. Its atomic weight is 39. Its
density will be
(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) a (2) d]
C10A 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,m 25%
(d)
6, 6, 50%
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CSOM – 12
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]
C10B 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 ×
C11
2
3
× a2 , height ‘h’ = 4r
3
4
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.
r n
i.e.,
r m
 radius ratio
It was found that greater is the radius ratio, greater will be the C.N.
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.
 0.5 and
rNa 
1.143
rK 
 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)
100 pm
Einstein Classes,
(b)
124 pm
(c)
104 pm
(d)
140 pm
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CSOM – 13
3.
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]
C12
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.
(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
( white colour )
(b)

 Zn 2  
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 normality Sr2+ occupy one lattice site of Na+ vacancy and other Na+ lattice site is vacant. AgCl
doped with CdCl2 produce one cation vacancy.
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CSOM – 14
(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 gp-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 increase silicon atoms. 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 gp-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]
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CSOM – 15
INITIAL STEP EXERCISE
1.
Which of the following statements is correct in the
rock-salt structure of an ionic compounds ?
(a)
(b)
(c)
(d)
2.
3.
4.
5.
6.
7.
8.
coordination number of cation is four
whereas that of anion is six
Assertion (A) : In crystal lattice the size of the
cation is larger in a tetrahedral hole than in an
octahedral hole.
Reason (R) : The cations occupy more space than
anions in crystal packing.
coordination number of cation is six
whereas that of anion is four
(a)
coordination number of each cation and
anion is four
Both A and R are true and R is the
correct explanation of A
(b)
coordination number of each cation and
anion is six.
Both A and R are true and R is not the
correct explanation of A
(c)
A is true but R is false
(d)
Both A and R are false
Which of the following expressions is correct in the
case of sodium chloride unit cell (edge length, a) ?
(a)
r c + ra = a
(b)
rc + ra = a/2
The unit cell with crystallographic dimentions
a = b  c,  =  =  = 900 is
(c)
rc + ra = 2a
(d)
rc + ra = 2 a
(a)
Cubic
(b)
Tetragonal
(c)
Monoclinic
(d)
Hexagonal
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
(a)
101 pm
(b)
167 pm
(c)
202 pm
(d)
383 pm
9.
10.
Which one of the following schemes of orderin, close
packed sheets of equal sized spheres do not
generate close packed lattice.
11.
A quantity of gas is collected in a graduated tube
over the mercury. The volume of the gas at 200C is
50.0 mL and the level of the mercury in the tube is
100 mm above the outside mercury level. The
barometer reads 750 mm. Volume at STP is
(a)
39.8 mL
(b)
40 mL
(c)
42 mL
(d)
60 mL
(a)
ABCABC
(b)
ABACABAC
(c)
ABBAABBA
When CO2 under high pressure is released from a
fire extinguisher, particles of solid CO2 are formed,
despite the low sublimation temperature (–770C) of
CO2 at 1.0 atm. It is due to :
(d)
ABCBCABCBC
(a)
the gas does work pushing back the
atmosphere using KE of molecules and
thus lowering the temperature
(b)
volume of the gas is decreased rapidly
hence temperature is lowered
(c)
both correct
(d)
none of correct
The co-ordination no. of cation and anion in
Fluorite CaF2 and Rutile TiO2 are respectively
(a)
8 : 4 and 6 : 3
(b)
6 : 3 and 4 : 4
(c)
6 : 6 and 8 : 8
(d)
4 : 2 and 2 : 4
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
A 0.20 mol sample of a hydrocarbon CxHy after
complete combustion with excess O 2 gas yields,
0.80 mol of CO 2 and 1.0 mol of H 2 O. Hence
hydrocarbon is :
(c)
X4Y2Z
(d)
X4YZ2
(a)
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
Einstein Classes,
12.
13.
C4H10
(b)
C4H 8
(c)
C4H 5
(d)
C8H16
At low pressure, the vander Waals equation is
written
as
a 

 P  2  (V) = RT, hence
V 

compressibility factor is :
(a)
1
a
RTV
(b)
1
RTV
a
(c)
1
a
RTV
(d)
1
RTV
a
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CSOM – 16
14.
15.
16.
The molecular weight of O2 and SO2 are 32 and 64
respectively. If one litre of O2 at 150C and 750 mm
presure contains N molecules, the number of
molecules, the number of molecules in two litres of
SO2 under the same conditions of temperature and
pressure will be
(a)
N/2
(b)
N
(c)
2N
(d)
4N
When an ideal gas undergoes unrestrained
expansion, no cooling occurs because the molecules
(a)
are above the inversion temperature
(b)
exert no attractive force on each other
(c)
do work equal to loss in kinetic energy
(d)
collide without loss of energy
(c)
18.
19.
20.
21.
23.
In Van der Wall’s equation of state for a non-ideal
gas, the term that accounts for intermolecular force
is
(a)
17.
22.
(V – b)
(b)
2
[p + (a/V )]
(d)
RT
(RT)
–1
(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
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
The value of Van der Waal’s constant ‘a’ for the
gases O2, N2, NH3 and CH4 are 1.360, 1.390, 4.170
and 2.253 litre2atm.mole–2. The gas which can most
easily be liquefied is
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)
O2
(b)
N2
(a)
Allotropic crystal
(c)
NH3
(d)
CH4
(b)
Liquid crystals
(c)
Isomeric crystals
(d)
Isomorphous crystals
16 g of oxygen and 3 g of hydrogen are mixed and
kept in 760 mm pressure at 00C. The total volume
occupied by the mixture will be nearly ?
24.
The three states of matter are solid, liquid and gas.
Which of the following statements are correct about
them
(a)
22.4 l
(b)
33.6 l
25.
Which arrangement of electrons leads to ferromagnetism
(c)
448 litres
(d)
44800 ml
(a)

(b)

(c)

(d)
None
Bragg’s equation is
(a)
n = 2 sin 
(b)
n = 2d sin 
(c)
2n = d sin 
(d)
 = (2d/n)sin
26.
The number of atoms/molecules contained in one
face centred cubic unit cell of a monoatomic
substance is
(a)
4
(b)
6
(c)
8
(d)
12
27.
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)
28.
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
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)
Cu4Ag 2Au
(b)
Cu4Ag 4Au
(c)
Cu4Ag 3Au
(d)
CuAgAu
All
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CSOM – 17
29.
30.
31.
Which is an example of ferroelectric compound
(a)
Quartz
(b)
(b)
Barium titanate (d)
32.
Ionic compounds for Schottky’s defects
PbCrO4
(a)
NaCl, KCl, CsCl
None
(b)
AgCl, AgBr, AgI
(c)
NaCl, KCl, ZnS
(d)
KCl, CsCl, ZnS
The phenomenon in which crystals on subjecting
to a pressure or mechanical stress produce
electricity is called
(a)
Pyroelectricity
(b)
Piezoelectric effect
How much time would it take to distribute one
Avogadro number of wheat grains, if 1010 grains
are distributed each second ?
(c)
Ferro electricity
(a)
6.02 × 1023
(b)
6.02 × 1013
(d)
Ferri electricity
(c)
6.02 × 103
(d)
none
How many ‘nearest’ and ‘next nearest’ neighbours
respectively potassium have in bcc lattice
33.
34.
The total kinetic energy of the molecules in Joules
in 8.0 g of methane at 270C is
(a)
8, 8
(b)
8, 6
(a)
1222.22 J
(b)
1870.56 J
(c)
6, 8
(d)
8, 2
(c)
2321.44 J
(d)
4425.33 J
FINAL STEP EXERCISE
1.
2.
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%
(b)
2, 6, 75%
(c)
3, 1,m 25%
(d)
6, 6, 50%
4.
5.
The curve shows Maxwell’s distribution of
velocities with increase of temperature
6.
3.
(a)
most probable velocity increases
(b)
Fraction of molecules possessing most
probable velocity decreases
(c)
A true, B wrong
(d)
A and B are true.
An iron cylinder contains helium at a pressure of
250 kPa at 300 K. The cylinder can withstand a
pressure of 1 × 106 Pa. The room in which cylinder
is placed catches fire. Predict whether the cylinder
will blow up before it melts or not.
(M.P. of the cylinder = 1800 K).
7.
Through the two ends of a glass tube of length
200 cm hydrogen chloride gas and ammonia are
allowed to enter. The distance at which ammonium
chloride will first appear is
(a)
80.8 cm
(b)
50 cm
(c)
95 cm
(d)
119 cm
An evacuated glass vessel weighs 50.0 g when empty,
148.0 g filled with a liquid of density 0.98 g ml–1
and 50.5 when filled with an ideal gas at 760 mm
Hg at 300 K. The molecular weight of the gas is
(a)
200.12 g mol–1
(b)
123.15 g mol–1
(c)
184.22 g mol–1
(d)
333.21 g mol–1
If the van der Waals constant a = 3.592 dm6 atm
mol–2. The pressure exerted by one mole of CO2 gas
at 273 K is
(a)
P = 0.9935 atm (b)
P = 2.99 atm
(c)
P = 4.56 atm
P = 5.333 atm
(d)
A mixture of ethane (C 2H 6) and ethene (C 2H 4)
occupies 40 L at 1.00 atm and at 400 K. The
mixture reacts completely with 130 g of O 2 to
produce CO 2 and H 2 O. Assuming ideal gas
behaviour, the mol fractions of C2H6 and C2H4 in
the mixture is
(a)
X C 2 H 6  0.67, X C 2 H 4  .33
(b)
X C 2 H 6  0.33, X C 2 H 4  .67
(a)
cylinder will blow up
(b)
it will not effect
(c)
X C 2 H 6  0 .2 , X C 2 H 4  . 8
(c)
cylinder will not blow
(d)
none
(d)
cannot be predicted
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CSOM – 18
8.
9.
10.
When 0.5 lt. of carbon dioxide is passed over red
hot coke the volume is found to be 700 mL.
Assuming that all the measurements are made at
NTP, the composition of the product.
(a)
CO2 = 300 mL, CO = 400 mL
(b)
CO2 = 400 mL, CO = 300 mL
(c)
CO2 = 500 mL, CO = 200 mL
(d)
CO2 = 600 mL, CO = 100 mL
Pressure of 1 g of an ideal gas A at 270C is found to
be 2 bar when 2 g of another ideal gas B is
introduced in the same flask at same temperature
the pressure becomes 3 bar. Then the ratio of
molecular mass of A to B is
(a)
7/4
(b)
4/7
(c)
9/7
(d)
7/9
ANSWERS (INITIAL STEP
EXERCISE)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
d
b
a
c
a
a
b
d
b
a
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
a
a
c
c
b
c
c
d
b
a
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
d
a
b
b
a
b
b
c
c
b
31.
32.
33.
34.
b
a
b
b
The radius of an Xe atom is 1.3 × 10 –8 cm. A
100 cm3 container is filled with Xe at a pressure of
1 atm and a temperature of 273 K. The fraction of
the volume that is occupied by Xe atoms is
(a)
(c)
2.47%
–2
2.47 × 10 %
(b)
0.247 %
(d)
none
ANSWERS (FINAL STEP
EXERCISE)
1.
2.
3.
a
d
a
6.
7.
8.
a
a
a
4.
5.
a
b
9.
10.
a
c
AIEEE ANALYSIS [2002]
1.
2.
3.
Value of gas constant R is
4.
(a)
83 erg mol–1K–1
(b)
0.082 litre atm.
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
(c)
8.3 J mol–1K–1
(a)
2 and 4
(b)
14 and 9
(d)
0.987 cal mol–1K–1
(c)
4 and 2
(d)
9 and 14
Kinetic theory of gases proves
(a)
only Charle’s law
(b)
only Avogadro’s law
(c)
only Boyle’s law
(d)
all of these
For an ideal gas, numer of moles per litre in terms
of its pressure P, gas constant R and temperature T
is
(a)
PT/R
(b)
RT/P
(c)
P/RT
(d)
PRT
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CSOM – 19
AIEEE ANALYSIS [2003]
5.
6.
7.
A pressure cooker reduces cooking time for food
because
8.
How many unit cells are present in a cube-shaped
ideal crystal of NaCl of mass 1.00 g
(a)
cooking involves chemical changes helped
by a rise in temperature
[Atomic masses : Na = 23, Cl = 35.5]
(a)
1.71 × 1021 unit cells
(b)
heat is more evently distributed in the
cooking place
(b)
2.57 × 1021 unit cells
(c)
boiling point of water involved in
cooking is increased
(c)
5.14 × 1021 unit cells
(d)
1.28 × 1021 unit cells
(d)
the higher pressure inside the cooker
crushes the food material
9.
Glass is a
What volume of hydrogen gas, at 273 K and
1 atm. pressure will be consumed in obtaining
21.6 g of elemental boron (atomic mass = 10.8) from
the reduction of boron trichoride by hydrogen ?
(a)
polymeric mixture
(a)
22.4 L
(b)
89.6 L
(b)
micro-crystalline solid
(c)
67.2 L
(d)
44.8 L
(c)
super-cooled liquid
(d)
gel
10.
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
According to the kinetic theory of gases, in an ideal
gas, between two sucessive collisions a gas molecule
travels
(a)
with an accelerated velocity
(b)
In a circular path
(c)
in a wave path
(d)
in a straight line path
AIEEE ANALYSIS[2004/2005/2006]
11.
As the temperature is raised from 200C to 400C, the
average kinetic energy of neon atoms changes by a
factor of which of the following ?
(a)
1/2
(b)
(c)
313/293
(d)
13.
What type of crystal defect in indicated in the
diagram below ?
Na2 Cl– Na+ Cl– Na+ Cl–
Cl–
313 / 293
2
Cl– Na+
Na+ Cl–
Na+
Cl– Na+ Cl–
[2004]
12.
Cl– Na+ Cl– Na+
Na+
In van der Waals equation of state of the gas law,
the constant b is a measure of
(a)
Frenkel defect
(a)
intermolecular repulsions
(b)
Schottky defect
(b)
intermolecular attraction
(c)
interstitial defect
(c)
volume occupied by the molecules
(d)
Frenkel and Schottky defects
(d)
intermolecular collisions per unit volume
[2004]
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
[2004]
CSOM – 20
14.
An ionic compound has unit cell consisting of A ions
at the corners of a cube and B ions on the centres of
the faces of the cube. The empirical formula for
this compound would be
(a)
AB
(b)
A2B
(c)
AB3
(d)
A3B
18.
[2005]
15.
16.
Total volume of atoms present in a face-centred
cubic cell of a metal is (r is atomic radius)
(a)
16 3
r
3
(b)
20 3
r
3
(c)
24 3
r
3
(d)
12 3
r
3
Lattice energy of an ionic compound depends upon
[2006]
(a)
Charge on the ion only
(b)
Size of the ion only
(c)
Packing of ions only
(d)
Charge on the ion and size of the ion
19.
An ideal gas is allowed to expand both reversibly
and irreversibly in an isolated system. If Ti is the
initial temperature and Tf is the final temperature,
which of the following statements is correct ?
[2005]
(a)
Which one of the following statements is NOT true
about the effect of an increase in temperature on
the distribution of molecular speeds in a gas ?
Tf = Ti for both reversible and
irreversible processes
(b)
(Tf)irrev > (Tf)rev
(c)
Tf > Ti for reversible process but Tf = Ti
for irreversible process
(d)
(Tf)rev = (Tf)irrev
(a)
The most probable speed increases
(b)
The fraction of the molecules with the
most probable speed increases
(c)
The distribution becomes broader
(d)
The area under the distribution curve
remains the same as under the lower
temperature
[2006]
[2005]
17.
Based on lattice energy and other considerations
which one of the following alkali metal chlorides is
expected to have the highest melting point ?
(a)
LiCl
(b)
NaCl
(c)
KCl
(d)
RbCl
[2005]
AIEEE ANALYSIS [2007]
20.
Equal masses of methane and oxygen are mixed in
an empty container at 250C. The fraction of the total
pressure exerted by oxygen is
(a)
1
3
(b)
1
2
(c)
2
3
(d)
1 273

3 298
ANSWERS AIEEE ANALYSIS
1.
c
2.
d
3.
c
4.
a
5.
c
6.
c
7.
a
8.
b
9.
c
10.
d
11.
c
12.
c
13.
b
14.
c
15.
d
16.
b
17.
b
18.
a
19.
b
20.
a
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CSOM – 21
TEST YOURSELF
1.
2.
3.
4.
5.
Ideal condition for the gas is
(a)
high temperature and low pressure
(b)
low temperature and high pressure
(c)
low temperature and low pressure
(d)
high temperature and high pressure
O2 diffuse how many times as compare to N2 ?
(a)
0.94
(b)
0.65
(c)
0.30
(d)
0.75
Which of the following is correct ?
(a)
Vrsm > Vav > Vmp
(b)
Vav > Vrms > Vmp
(c)
Vmp > Vav > Vrms
(d)
Vrms > Vmp > Vav
Compressibility factor at low pressure is
(a)
Z  1
b
Vm RT
(b)
Z  1
b
Vm RT
(c)
Z  1
a
Vm RT
(d)
Z  1
a
Vm RT
The temperature at which the pressure is 100 atm,
volume is 500 cm3 and number of moles is 2.
(a)
31.500
(b)
45.730
(c)
298 K
(d)
320.50 K
ANSWERS
Einstein Classes,
1.
a
2.
a
3.
a
4.
c
5.
a
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111