Download 1 MATTER: Anything which occupies space , has volume and can

Class XII
Chemistry
states of matter [SOLID STATE]
MATTER:
Anything which occupies space , has volume and can be perceived by our
senses . E.g. solids , liquids, gases are forms of matter
Other states of matter are Plasma, BEC (Bose Einstein condensate).
PHYSICAL PROPERTIES OF SOLIDS LIQUIDS AND GASES:
Sr.
No
.
PROPERTIES
Related law
or formula
1
2
Mass
Shape
Units (kg)
Area :
Units (m2)
3
4
Volume
5
6
7
8
9
10
11
12
SOLIDS
definite
definite
GASES
Not possible
Definite
Definite
Acquires the Acquires the
shape of the shape of the
container
container
Definite
indefinite
Almost
Highly
negligible
compressible
Can flow
Can flow
Highly rigid
Slow
Less rigid
Fast
Units (m3 )
Compressibility
Z= PV
Factor
nRT
Fluidity
Viscocity
(μ)
Rigidity
Strength
Diffusion
r1
√d
No. of free Free sites
surfaces
Density
Mass
=
volume
units kg/m3
Packing
of Measure of
strength/den
particles
definite
Not possible
Inter particle
force
Strongest:
cohesive
forces>
thermal energy
sity
Spacing
between
molecules.
LIQUIDS
Not rigid
Very fast
Any no. of free Only one free None
surfaces
surface
High
Slightly lower
Very low
Most
packed
closely Less
closely Least closely
packed
packed.
Slightly
Negligible,
weaker than thermal
in solids.
energy
>cohesive
forces
More
than Most
solids
Translatory
Translator
Expansion on vaporisation Less
(KJ/mol)
heating
13 Motion
of Energy
Oscillatory
constituent
Joule (J)
particles
14 Kinetic
Energy (J) Least
Large
Very large
energy
of = 1mu2
particles
2
SOLID: Solid is a form of matter in which the constituting particles are arranged very
closely. The constituent particles can be atoms, molecules or ions.
PROPERTIES OF SOLIDS:
a. Definite mass, volume and shape.
b. Short Intermolecular distances , hence strong intermolecular forces (F  1/r2) .
c. Atoms/ ions occupies fixed positions and can only oscillate/vibrate about their mean
positions due to low thermal energy. Thermal energy  temperature
d. Closely packed, therefore low compressibility and rigid.
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Class XII
Chemistry
states of matter [SOLID STATE]
CLASSIFICATION OF SOLIDS ON BASIS OF ARRANGEMENT OF ATOMS/IONS:
a. Crystalline solids: definite arrangement (regular) of atoms/ ions e.g. metals (iron,
copper), non metals (diamond, graphite) all elements, compounds, alum, sugar.
Majority of solids are crystalline in nature.
b. Amorphous solids: no definite arrangement (irregular) of atoms / ions e.g. Glass,
plastics, rubber, fused silica, tar, pitch, high molecular weight polymers.
PROPERTIES OF CRYSTALLINE SOLIDS:
a. Definite geometrical shape or regular geometry.
b. A long range order, i.e. regular arrangement .
c. They have a sharp melting point.
d. They are anisotropic in nature
i.e. their physical properties show different
values when measured along different directions
in the same crystal.
e. Clean cleavage i.e. when cut with a sharp edged tool,
they split into two pieces and the newly generated
surfaces are plain and smooth.
f. Definite and characteristic heat of fusion
g. Also called true solids.
Anisotropy: In case of crystalline substances, properties like electrical properties,
refractive index, thermal expansion etc are having different values in different
directions.
POLYMORPHIC FORMS OR POLYMORPHS:
The different crystalline forms of a substance are known as polymorphic forms or
polymorphs. E.G. CaCO3 exists as Calcite and Aragonite.
NOTE: for elements it is called allotropy for example: graphite and diamond are
allotropes of Carbon.
ISOMORPHISM:
When two or more crystalline solids have similar chemical composition that exists in
same crystalline form or structure e.g. Na3PO4, Na3AsO4
ANNEALING: regular heating and cooling a glass of glass make the glass milky.
CHARACTERISTICS OF AMORPHOUS SOLIDS:
a. An irregular shape or no regular arrangement of particles.
b. A short range order i.e. regular arrangement of particle in small space.
c. Do not have definite heat of fusion. Gradually soften/ melt/ fuses over a range of
temperature. (not fixed)
d. Isotropic in nature i.e. their physical properties are the same in all directions.
e. Irregular cleavage, e.g. when cut with a sharp edged tool, they cut into two pieces
with irregular surfaces.
f. Also called as pseudo solids or super cooled liquids. This is because they have a
tendency to flow, though very slowly.
h. Also called pseudo solids (highly super cooled liquid of very high viscocity).
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Class XII
Chemistry
states of matter [SOLID STATE]
TYPES OF CRYSTALLINE SOLIDS:
Type of
Solid
Constituent
Particles
Bonding/
Attractive
Forces
Examples
1. Ionic
solids
+ve and –ve
Ions
Coulombic
or
electrostatic
force
of
attraction
NaCl,
MgO,
ZnS,
CaF2
Molecules
Dispersion
or London
forces
Dipole dipole
Interactions
2.
Molecular
solids
a.
Non
polar
Physical
Nature
(hardnes
s
/
brittlenes
s)
Hard but
brittle
Electrical
conductivity
Melting
point
Solubility
Insulators in
solid
state but
conductors
in molten
state and
in aqueous
solutions
High
Soluble in polar
and insoluble in
non polar
Ar, CCl4,
H2, I2,
CO2
HCl, solid
SO2, solid
NH3
Soft
Insulator
Very
low
Soluble as well as
insoluble in both
Soft
Insulator
Low
Soluble in polar
b. Polar
Molecules
c.
Hydrogen
bonded
3.
Covalent
or
network
solids
Molecules
Hydrogen
Bonding
H2O
(ice)
Hard
Insulator
Low
Soluble in polar
Atoms/
molecule
Covalent
Bonding
SiO2
(quartz),
SiC, C
(diamond
),
AlN
C(graphit
e)
Hard
Soft
Insulators
Conductor
(exception)
Very
High
Insoluble in polar
and
usually
soluble in non
polar solvents
4.
Metallic
solids
Positive
ions (+) in a
sea of
delocalised
electrons (-)
Metallic
Bonding
(electrostatic
attraction
between
cations
and
sea
of
electrons)
Fe, Cu,
Ag, Mg
Hard but
malleable
and
ductile
Conductors
in solid
state as
well as in
molten
state
Fairly
High
Insoluble in both
V.IMP. Q. DIFFERENCE BETWEEN ANY TWO CAN BE ASKED IN BOARD EXAMS.
V.IMP Q. DIFFERNENCE BETWEEN CRYSTALLINE AND AMORPHOUS SOLIDS? (2)
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Class XII
PROPERTY
Definition
Shape
Melting Point
Compressibility
Chemistry
CRYSTALLINE SOLID
Definite
arrangement
(regular) of atoms/ ions.
They
have
definite
shape
and
regular
geometrical form.
They
have
sharp
(definite) melting point.
They are rigid and
incompressible.
Cutting with a sharp They
give
clean
edged tool
cleavage, i.e. they break
into two pieces with plane
surfaces.
Heat of fusion
They have definite heat
of fusion.
Isotropic/
Anisotropic i.e. their
anisotropic
mechanical and electrical
properties depend on the
direction
along
which
they are measured.
Example
All
elements
and
compounds Cu, Ni, NaCl
etc
states of matter [SOLID STATE]
AMORPHOUS SOLIDS
No
definite
arrangement
(irregular) of atoms / ions.
They do not have definite
shape and regular geometrical
form.
They melt over wide range of
temperature.
They too are usually rigid
and cannot be compressed
to any appreciable extent.
However graphite is soft
because
of
its
unusual
structure.
They
give
irregular
cleavage, i.e. they break into
two pieces with irregular
surfaces.
They do not have a definite
heat of fusion.
Isotropic i.e. they have
similar physical properties in
all directions because the
constituents are arranged in
random manner.
Rubber , glass etc.
NOTE:
Any material can be made amorphous or glassy either by rapidly cooling its melt or
freezing its vapours. E.g. silica (SiO2) crystallises as quartz in which SiO4 tetrahedra are
linked in a regular manner but on melting and then rapid cooling, it gives glass in which
SiO4 tetrahedra are randomly joined to each other. thus quartz is crystalline SiO2
whereas silica glass is amorphous solids SiO2.
CRYSTAL LATTICE/ SPACE LATTICE:
Space lattice is a regular repeating arrangement of points in space.
4
Class XII
Chemistry
states of matter [SOLID STATE]
(a) Two dimensional lattices:
Regular arrangement of atoms in the plane of paper.
Sr.
No
1
2
3
4
5
Lattice
Unit cell
Square lattice
Rectangular lattice
Parallelogram lattice
Rhombic lattice
Hexagonal lattice
Square
Rectangle
Parallelogram
Rectangular with interior point
Rhombus with an angle of 600
(b) 3dimensional crystal lattice: A regular orderly arrangement of constituent
particles (atoms/ ions/molecules) in three dimensional space.
LATTICE POINTS OR LATTICE SITES:
The fixed positions on which the constituent particles (atoms/ ions / molecules) are
present are called lattice points or lattice sites.
SCC: lattice points: 8 , BCC: 9, FCC = 14
CRYSTAL LATTICE:
A group of lattice points which when repeated over and over again in 3 dimensions give
the complete crystal lattice.
UNIT CELL:
Smallest repeating unit in space lattice which
when repeated over and over again generates
the complete crystal lattice.
The crystal can consist of an infinite number of unit cells.
V.IMP Q. DEFINE UNIT CELL? CRYSTAL LATTICE? LATTICE POINTS, SPACE
LATTICE (1)?
PARAMETERS WHICH CHARACTERIZE A UNIT CELL:
a. Dimensions of the unit cell along the three edges,
a, b and c: These edges may or may not be mutually
perpendicular (900).
b. Inclination (angle) of the edges to each other:
This is denoted by the angle between the edges , ,
and respectively. is the angle between the edges b
and c, is the angle between the edges a and c, and 
is the angle between a and b.
TYPES OF CRYSTAL SYSTEMS:
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Class XII
S.
n
o
1
System
Chemistry
Maximu
m
symmetr
y
elements
9 planes
13 axes
1 centre
2
Cubic
or
regular
(most
symmetrical)
Tetragonal
3
Hexagonal:
7 planes
7 axes
4
Rhombohedr
al
or
trigonal:
Orthorhombi
c
Or Rhombic
Monoclinic:
7 planes
7 axes
5
6
7
5 planes
5 axes
3 plane
3 axis
1 plane
1 axis
Triclinic
No plane
(most
un No axis
symmetrical)
Axes and
angles
states of matter [SOLID STATE]
Examples
= = = NaCl, Cu, KCl, CaF2, ZnS, Cu2O, Diamond,
Alums, Pb, Ag,, Au, Hg.
90° ,
a=b=c
= = =
90° ;
a = b c
= = 90°,
=120°;
a = b c
=
=
90°;
a=b=c
= = =
90°;
a b c
== 90°,
90°;
a b c
SnO2,ZnO2,TiO2,CaSO4,NiSO4,ZrSiO4,PbWO4
,White Sn.
ZnO, PbI2,CdS, AgI, cinnabar(
Graphite, Ice, Beryl, Mg, Zn, Cd
HgS),
NaNO3, CaSO4, Calcite ,ICI, Quartz, As, Sb,
Bi
KNO3,
K2SO4,
PbCO3,
BaSO4,CaCO3,
Rhombic Sulphur, MgSO4.7H2O
Na2SO4.10H2O,
Na2B4O7.10H2O,
CaSO4.2H2O, Monoclinic Sulphur
90° CuSO4.5H2O, K2Cr2O7, H3BO3
; a b c
TYPES OF UNIT CELLS:
a. Primitive or simple unit cells:
6
Class XII
Chemistry
states of matter [SOLID STATE]
Have constituent particles only at its corners.
b. Centred unit cells are those unit cells: in which one or more constituent particles
are present at positions in addition to those present at the corners.
The centred unit cells are of three types:
i. Face centred unit cell:
A face centred unit cell consists of one constituent particle present at the centre of each
face in addition to those present at the corners.
ii. Body centred unit cell:
A body centred unit cell consists of a one constituent particle is present at its body
centre in addition to those present at the corners.
iii. End centred unit cell:
An end centred unit cells consists of one constituent particle present at the centre of any
two opposite faces in addition to those present at the corners.
CONTRIBUTION OF PARTICLES AT DIFFERENT LATTICE POSITIONS:
a. Corner: If an atom is present at any one corner, it is shared by eight unit cells. So,
only one eighth of an atom actually belongs to the unit cell.
CONTRIBUTION =1/8
b. Face centre: If an atom is present at the centre of the face, it is shared by two unit
cells. So, only half of the atom actually belongs to the unit cell.
CONTRIBUTION =1/2
c. Body centre: If an atom is present at the body centre, it is not shared by any other
unit cell. So, that one atom completely belongs to the same unit cell.
CONTRIBUTION =1
d. Edge centre: If an atom is present at the edge centre, it is shared by four unit cells.
So, only one fourth of an atom belongs to the unit cell.
CONTRIBUTION =1/4
TOTAL NUMBER OF ATOMS IN DIFFERENT UNIT CELLS FOR ONE UNIT CELL:
a. Primitive unit cell (SCC): 1 atom (8 x 1/8)
=1
b. Face centred unit cell (FCC): 4 atoms (8 x 1/8+ 6 x ½)
c. Body centred unit cell (BCC): 2 atoms (8 x 1/8+ 1)
=4
=2
7
Class XII
Chemistry
states of matter [SOLID STATE]
Coordination number is the number of nearest neighbours of any particle.
CN for simple cubic (SCC) =6, BCC =8, FCC (hcp or ccp) = 12.
Q1. A compound formed by elements A and B has a cubic structure in which A
atoms are at the corners of the cube and B atoms are at the face centres.
Derive the formula of the compound?
Ans. As A atoms are present at the 8 corners of the cube, therefore, number no. of A
atoms in a unit cell = 1/8 x 8 =1
As B atoms are present at the face centres of the 6 face of the cube, therefore , no. of B
in the unit cell: 1/2x6 = 3
Therefore ratio of atoms A:B = 1:3
Hence the formula is AB3.
Q2. a cubic solid is made up of two elements X and Y. Atoms Y are present at
the corners of the cube and atoms x at the body centre. What is the formula of
a compound? What are the coordination number of X and Y?
Ans. As atoms Y are present at the 8 corners of the cube, therefore no of atoms of Y in
the unit cell = 1/8 x 8 =1
As atoms X are present at the body centre, therefore no of atoms of X, in the unit cell
=1, Therefore ratio of atoms X:Y ::1:1
Hence, the formula of the compound is XY.
Coordination no of each of X and Y is 8
DO ASSIGNMENT NO:1 (GIVEN SEPARATELY)
CLOSE PACKING IN CRYSTAL STRUCTURES:
a. Close packing in one dimension: Each sphere is in contact with two of its
neighbours. Coordination number is two.
b. Close packing in two dimensions: It is generated by stacking the rows of close
packed spheres in two ways:
i. Square close packing: AAA ----- TYPE
When the spheres of the second row are placed exactly above those of the first row. This
way the spheres are aligned horizontally as well as vertically. The arrangement is AAA
type. Coordination number is 4.
8
Class XII
Chemistry
states of matter [SOLID STATE]
ii. Hexagonal close packing: ABAB----type
When the spheres of the second row are placed above the first one in a staggered
manner in such a way that its spheres fit in the depression of the first row. The
arrangement is ABAB type. Coordination number is 6.
Space occupied 74% , empty space =26%
c. Close packing in three dimensions:
They can be obtained by stacking the two dimensional layers one above the other. It can
be obtained in two ways:
i. Three dimensional close packing from two dimensional square close packed
layers: AAAA --- pattern
Here, The spheres of the upper layer are placed
exactly over the first layer such the spheres
of the layers are perfectly aligned horizontally
and vertically. It has a AAAA.. type pattern.
The lattice is simple cubic lattice.
ii. Three dimensional close packing from two dimensional hexagonal close
packed layers:
There are two steps involved as:
i. Placing the second layer over the first layer: if a two dimensional layer is considered as
A, the second layer which is placed above the first layer in such a way that the spheres
of the second layer (considered as B) are placed in the depressions of the first layer.
This gives rise to two types of voids: tetrahedral voids and octahedral voids.
ii. Placing the third layer over the second layer: There are two possibilities:
TYPES OF VOIDS:
a. Tetrahedral voids: ABAB-- pattern
This type of void is formed at the centre when four spheres are joined in the form
of a tetrahedron. Called as triangular void.
Radius of tetrahedral voids r = 0.225 R Where R is radius of sphere in close packing
For cations in the voids and anions in the packing r+ = 0.225r-
9
Class XII
Chemistry
states of matter [SOLID STATE]
b. Octahedral void: ABCABC-- pattern
This type of void is surrounded by six spheres.
Double triangular void surrounded by six spheres is an octahedral voids.
Radius of octahedral voids r = 0.414 R where R is radius of sphere in close packing
For cations in the voids and anions in the packing r+ = 0.414ra. Covering the tetrahedral voids (hexagonal close packing):
Here, tetrahedral voids of the second layer may be covered by the spheres of the third
layer. It gives rise to ABABAB… type pattern.
The three dimensional structure is called hexagonal close packed structure. Coordination
number is 12. Example: Mg, Zn
10
Class XII
Chemistry
states of matter [SOLID STATE]
b. Covering the octahedral voids (cubic close packing):
Here, octahedral voids of the second layer may be covered by the spheres of the third
layer. It gives rise to ABCABCABC… type pattern. The three dimensional structure is
called cubic close packed structure or face centred cubic structure.
Coordination number is 12. Example: Cu, Ag
Cubic close packing ccp:
ABCABCABC type space occupied 74% empty 26%
Body centred cubic packing BCC space occupied 68% empty 32%
Scc packing space occupied 52.4%
In hcp (ABAB pattern) or ccp (ABCABC pattern) arrangement,
octahedral and tetrahedral voids are present. The number of octahedral voids present in
a lattice is equal to the number of close packed particles. The number of tetrahedral
voids is twice the number of octahedral voids.
Example: If the number of close packed particles = n
Number of particles present in octahedral voids = n
Number of particles present in tetrahedral voids = 2n
NO OF VOIDS FILLED AND FORMULA OF A COMPOUND:
No .of octahedral voids = n, where n is no. of particles present in close packing
Effective no of octahedral voids in ccp structure
One octahedral void is present at the body centre, 12 octahedral voids are
present on the centre of 12 edges of the cube = 1+ 1/12 x ¼ =1+3 =4
Note: Each edge centre is shared by 4 unit cell therefore contribution to one is ¼ for edge
centre voids.
11
Class XII
Chemistry
No .of tetrahedral voids = 2 x n
states of matter [SOLID STATE]
where n is no. of octahedral voids
In ccp, total no. of voids per unit cell = 4 (octahedral) + 8 (tetrahedral) =12
In hcp, total no. of voids per unit cell = 6 (octahedral) + 12 (tetrahedral) =12
Assuming n particles of B are present in the packing and 1/3 rd of octahedral voids are
occupied by particles A, then ratio of A:B = n/3 : n :: 1/3: 1
Hence formula is A3B.
SIZE OF VOIDS:
Octahedral void: For an atom to occupy an octahedral void, its radius must be 0.414
times the radius of the sphere. r = 0.414 R
Tetrahedral void: For an atom to occupy a tetrahedral void, its radius must be 0.225
times the radius of the sphere. (no. of tetrahedral voids= double the no. of spheres)
r = 0.225R
TRIGONAL VOIDS:
Radius of trigonal void is r = 0.155R where r is radius of spherical trigonal site, R is
radius of closely packed sphere.
CUBIC VOIDS: r= 0.732 R
Derivation of radius of voids as per class lecture.
Imp. DIFFERENCE BETWEEN TETRAHEDRAL VOIDS AND OCTAHEDRAL VOIDS:
Sr. No
Tetrahedral voids
Octahedral voids
1
2
r = 0.225R
r = 0.414 R
Much smaller than the size of the
sphere in the packing
3
Coordination no.4
Smaller than the size of the sphere
in the packing but larger than the
tetrahedral voids.
Coordination no.6
12
Class XII
4
Chemistry
In hcp, ccp packing, each sphere
is in contact with 3 spheres in the
layer below and three above it,
thus forming one tetrahedral void.
Hence there are 2 tetrahedral
voids per sphere. i.e. no of
tetrahedral voids is double the no.
of spheres in the packing.
states of matter [SOLID STATE]
As
octahedral
voids
is
a
combination of two voids of the two
layer, no of octahedral voids is
equal to half the no of tetrahedral
voids and hence equal to no of
spheres in the packing.
No .of tetrahedral voids = 2 x no. of No .of octahedral voids = no. of
octahedral voids
particles present in close packing
Coordination number is the number of nearest neighbours of any particle.
CN for simple cubic =6, Bcc =8, fcc (hcp, ccp) = 12.
PACKING EFFICIENCY: It is the percentage of total space occupied by constituent
particles (atoms, molecules or ions)
Packing efficiency= Volume occupied by spheres x 100%
Total volume of unit cell
a. Packing efficiency for face centred unit cell = 74%
b. Packing efficiency for body centred cubic unit cell = 68%
c. Packing efficiency for simple cubic unit cell = 52.4%
13
Class XII
Chemistry
states of matter [SOLID STATE]
Derivations: As per class lecture.
Q3. Formula of a Compound and Number of Voids Filled A compound is formed
by two elements X and Y. Atoms of the element Y (as anions) make ccp and
those of the element X (as cations) occupy all the octahedral voids. What is the
formula of the compound?
Ans The ccp lattice is formed by the element Y. The number of octahedral voids
generated would be equal to the number of atoms of Y present in it. Since all the
octahedral voids are occupied by the atoms of X, their number would also be equal to
that of the element Y. Thus, the atoms of elements X and Y are present in equal
numbers or 1:1 ratio. Therefore, the formula of the compound is XY.
Q4. Atoms of element B form hcp lattice and those of the element A occupy
2/3rd of tetrahedral voids. What is the formula of the compound formed by the
elements A and B?
Ans The number of tetrahedral voids formed is equal to twice the number of atoms of
element B and only 2/3rd of these are occupied by the atoms of element A.
Hence the ratio of the number of atoms of A and B is 2 × (2/3):1
or 4:3 and the formula of the compound is A4B3.
Q5. In a crystalline solid, anions B are arranged in a cubic close packing.
Cations A are equally distributed between octahedral voids and
tetrahedral voids. If all the octahedral voids are occupied, what is the
formula of the solid?
Ans. Suppose the no of anions B =n
Then no. of octahedral voids = n
No of tetrahedral voids =2n
As octahedral voids and tetrahedral voids are equally occupied by cations A and
all the octahedral voids are occupied (given), therefore n cations A are present in
octahedral voids and n cations A are present in tetrahedral voids. In other words
, corresponding to n anions B, there are n+n= 2n cations A.
Thus cations A and anions B are in ratio of 2n : n :: 2:1
Thus formula A2B
Q6. In the mineral, spinel, having the formula MgAl2O4, oxide ions are
arranged in the cubic close packing, Mg+2 ions occupy the tetrahedral
voids while Al+3 ions occupy the octahedral voids.
(i) what %age of tetrahedral voids is occupied by Mg+2 ions?
(ii) what %age of octahedral voids is occupied by Al+3 ions?
Ans According to the formula MgAl2O4:
If there are 4 oxide ions, there will be 1 Mg+2 ions and 2Al+3 ions. But if thw 4O2ions are in ccp arrangement , there will be 4 octahedral voids and 8 tetrahedral
voids.
Thus 1Mg+2 ions is present in one of the 8 tetrahedral voids.
Therefore %age tetrahedral voids occupied by Mg+2 = 1/8 x 100= 12.5%
Similarly , 2Al+3 ions are present in two octahedral voids out of 4 available.
14
Class XII
Chemistry
states of matter [SOLID STATE]
Therefore %age of octahedral voids occupied by Al+3 ions = 2/4 x100= 50%
SUMMARY OF STRUCTURE AND PACKING OF SOLIDS:
Property
Hexagonal
Cubic
close Body
centred
close packing
packing
cubic (bcc)
Arrangement of Close packed
Closed packed
Not close packed
packing
Type of packing
ABABAB..
ABCABCA...
AB AB AB A
Available
space 74%
74%
68%
occupied
Coordination
12
12
8
number
Malleable
and Less malleable , Malleable and
ductility
hard and brittle
ductile
Examples
Be, Mg, Ca, Cr, Cu, Ag, Au, Pt
Alkali metals, Fe
Mo, V, Zn
RADIUS RATIO:
Ratio of radius of cation to that of anion is called radius ratio.
Radius ratio = Radius of cation (r+)
Radius of anion (r-)
Larger is the radius ratio, larger is the size of cation and hence greater is its coordination
no.
Radius
Coordination ratio
number
3
0.155-0.225
4
0.225- 0.414
6
0.414- 0.732
8
0.732-1.00
Geometrical shape i.e. Examples
position
of
anions
around cations
Planar triangle
B2O3
Tetrahedral
ZnS (sphalerite), CuCl,
CuI, BaS, HgS
Octahedral
NaCl (rock salt )
NaBr, KBr, MgO, MnO,
CaO, CaS, NH4Br
Body centred cube
CsCl (caesium chloride),
CsBr, TlBr, TlCl
Density of a unit cell is same as the density of the substance.
RADIUS RATIO in an octahedral void:
For an atom to occupy an octahedral void, its radius must be 0.414 times the radius of
the sphere. r = 0.414 R
Radius ratio for tetrahedral void:
For an atom to occupy a tetrahedral void, its radius must be 0.225 times the radius of
the sphere. (no. of tetrahedral voids= double the no. of spheres) r = 0.225R
TRIGONAL VOIDS: radius ratio of trigonal void is r = 0.155R r is radius of spherical
trigonal site, R is radius of closely packed sphere.
CUBIC VOIDS: r= 0.732 R
DECREASING ORDER OF SIZE OF VARIOUS VOIDS IS:
Cubic> octahedral> tetrahedral > trigonal
15
Class XII
Chemistry
states of matter [SOLID STATE]
Q7. A solid A+B- has NaCl type close packed structure. If the anion has a radius
of 241.5pm, what should be the ideal radius of the cation? can a cation C+
having a radius of 50pm e fitted into the tetrahedral hole of the crystal A+B-?
Solution : As A+B- has NaCl structure, A+ ions will be present in the octahedral voids.
Ideal radius of the cation will be equal to the radius of octahedral void because in that
case, it will touch the anions and the arrangement will be close packed. Hence
Radius of octahedral void = rA+ = 0.414rB- = 0.414x 241.5pm = 100pm
Radius of tetrahedral void = 0.225x rB- = 0.225 x241.5pm = 54.3pm
As the radius of the cation C+ (50pm) is smaller than the size of the tetrahedral hole, it
can be placed into the tetrahedral void (but not exactly fit into it).
Q8. the two ions A+ and B- ions have radii 88 and 200pm respectively.
In the close packed structure of compound AB, predict the coordination
no of A+ ?
Ans.
r+ = r(A+) = 88pm
= 0.44
rr(A-)
200pm
it lies in the range of 0.414-0.732
hence coordination no of A+ is 6
RELATIONSHIP BETWEEN RADIUS OF CONSTITUENT PARTICLE (R) AND EDGE
LENGTH (A) AND NEAREST NEIGHBOUR DISTANCE(d):
a. Simple cubic unit cell: a= 2r
b. Face centred unit cell: a 2 √2 r
c. Body centred unit cell: a = 4r/√3
(or r= a/2)
d =2r
(or r= a/2√2)
(or r= √3a/4)
d = 2r
d = 2r
Volume of a unit cell = (edge length)3 = a3
a. Simple cubic unit cell: Volume= (2r)3
b. Face centred unit cell: Volume(2√2 r)3
c. Body centred unit cell: Volume=( 4r√3 )3
Number of atoms in a unit cell (z):
a. Simple cubic unit cell: z = 1
(8 x 1/8)
b. Face centred unit cell: z = 4
(8 x 1/8+ 6 x ½)
c. Body centred unit cell: z = 2
(8 x 1/8+ 1)
16
Class XII
Chemistry
states of matter [SOLID STATE]
SUMMARY OF UNIT CELLS AND THEIR PROPERTIES:
Property
No of atoms per
unit cell (Z)
Packing
efficiency
Relation
between r and
(a) edge length
Nearest
neighbour
distance d=2r
Volume (a3)
Coordination
number
examples
SCC
BCC
FCC
1
2
4
52.4%
68%
74%
r= a/2
r= √3a/4
r= a
2√2
d=2r= a
d=2r= 2 x √3a
4
d=2r = 2 r= 2xa
2√2
(2r)3
(2√2 r)3
( 4r√3 )3
6
8
12 (hcp, ccp)
Mn
BCC
metals)
(alkali NaCl
ionic
compound,ZnS
DENSITY OF UNIT CELL:
Density of unit cell ()g/cm3 = Z M
a3 N a
Z no. of formula units in the unit cell, (1 for sc, 4 for fcc, and 2 for bcc)
M is atomic mass of the element and
a is length of the unit cell in pm
for cubic crystal of ionic compound , Z= No. of formula units per unit cell,
M = formula mass Remember, for ionic compounds, A+B- having fcc structure
(e.g. Na+Cl-) Edge (a) = 2 x distance between A+ and B- ions.
X RAY DIFFRACTION AND CRYSTAL STRUCURE:
Brag equation : nλ = 2dsinθ where n =1,2,3,4,-----d/λ = n/ 2sinθ
Where:
λ is wavelength, d is distance between the successive atomic planes
and the angle of reflection (θ).
SUMMARY OF STRUCTURE OF METALS
Crystal structure
Arrangement of ions
AB type
Rock salt
Cl- = fcc arrangement
Na+ ions = edge centres
and body centre
Cl- = 4
Bcc arrangement
Cs+ = 8
Cl = corners
S2- = 4
Cs+ = body centre
Caesium chloride
Coordination
no
Na+ = 4
Formula
units
4
1
17
Class XII
Chemistry
states of matter [SOLID STATE]
Zinc blende (ZnS) Ccp arrangement
type
S2- = fcc ,
Zn+2=in tetrahedral voids
AB2 type
Ccp arrangement
(i) Fluorite (CaF2) Ca+2
=fcc,
F=
all
type
tetrahedral voids
(ii)
Antifluorite B2- = ccp , A+ = half
(A2B)
tetrahedral voids
Zn+2 = 4
S2- = 4
4
Ca+2 =8
F- = 4
4
Na+ = 4
O2- = 8
4
EFFECT OF TEMPERATURE ON CRYSTAL STRUCTURE:
Increase of pressure increase coordination no
While increase of temperature decreases coordination no.
NaCl Structure
 High pressure CsCl
(6:6 coordination)
760K
(8: 8 coordination)
Q9. An element has a body-centred cubic (bcc) structure with a cell edge of 288
pm. The density of the element is 7.2 g/cm3. How many atoms are present in
208 g of the element?
Solution: Volume of the unit cell = (288 pm)3
= (288×10-12 m)3 = (288×10-10 cm)3
= 2.39×10-23 cm3
Volume of 208 g of the element = mass = 208 g = 28.88 cm3
density
7.2 g/cm3
Number of unit cells in this volume = 28.88 cm3
= 12.08×1023 unit cells
2.39 x10-23 cm3/ unit cell
Since each bcc cubic unit cell contains 2 atoms,
therefore, the total number of atoms in 208 g = 2 (atoms/unit cell) × 12.08 × 1023 unit
cells
= 24.16×1023 atoms
Q10. X-ray diffraction studies show that copper crystallises in an fcc unit cell
with cell edge of 3.608×10-8 cm. In a separate experiment, copper is
determined to have a density of 8.92 g/cm3, calculate the atomic mass of
copper.
Solution : In case of fcc lattice, number of atoms per unit cell, z = 4 atoms
Therefore, M = dNA a3
Z
= 8.92g/cm3x 6.022x1023 atoms/mol x (3.608x10-8 cm)3
4atoms
= 63.1 g/mol
Atomic mass of copper = 63.1u
Q11. Silver forms ccp lattice and X-ray studies of its crystals show that the
edge length of its unit cell is 408.6 pm. Calculate the density of silver (Atomic
mass = 107.9 u).
Solution : Since the lattice is ccp, the number of silver atoms per unit cell = z = 4
Molar mass of silver = 107.9 g mol–1 = 107.9×10-3 kg mol–1
Edge length of unit cell = a = 408.6 pm = 408.6×10–12 m
Density, d = z.M
a3 .NA
= 4 x107.9x10-3 kgmol-1
(408.6 x10-12m)3 x 6.022x 1023 mol-1 = 10.5×10-3 kg m–3 = 10.5 g cm-3
18
Class XII
Chemistry
states of matter [SOLID STATE]
IMPERFECTONS OR DEFECTS IN SOLIDS:
Crystal defects: Irregularities or departure from perfectly ordered arrangement of
constituent particles.
Defects are of two types:
a. Point defects - Point defects are the irregularities or deviations from ideal
arrangement around a point or an atom in a crystalline substance.
b. Line defects - Line defects are the irregularities or deviations from ideal
arrangement in entire rows of lattice points.
Point defects are of three types:
a. Stoichiometric or intrinsic or thermodynamic defects: These are the point
defects that do not disturb the stoichiometry of the solid.
b. Non–stoichiometric defects: These are the point defects that disturb the
stoichiometry of the solid.
c. Impurity defects: These are the defects in ionic solids due to the presence of
impurities present in them.
Stoichiometric defects for non- ionic solids are of two types:
VACANCY DEFECT
INTERSTITIAL DEFECT
A crystal is said to have vacancy defect A crystal is said to have interstitial defect
when some of the lattice sites are when some constituent particles (atoms
vacant.
or molecules) occupy an interstitial
site.
This results in decrease in density of the This results in increase in density of the
substance.
substance.
Stoichiometric defects for ionic solids are of two types:
SCHOTTKY DEFECT
FRENKEL OR DISLOCATION DEFECT
In this defect equal number of cations and In this defect, the smaller ion (usually
anions are missing
cation) is dislocated from
its normal site to an interstitial site.
It is basically a vacancy defect in ionic It creates a vacancy defect at its original
solids.
site and an interstitial defect at its new
location.
It decreases the density of a solid
It does not change the density of the solid
Schottky defect is shown by
Frenkel defect is shown by
(a) ionic substances in which the cation (a) ionic substance in which there is a large
and anion are of almost similar sizes.
difference in the size of ions (cations and
(b) Occurs where coordination no. is high.
anions).
(b) Occurs where coordination No is low.
For example: NaCl, KCl, CsCl and AgBr
For example: ZnS, AgCl, AgBr and AgI
Substance having both schottky defects and frenkel defects : AgBr
19
Class XII
Chemistry
states of matter [SOLID STATE]
Non-stoichiometric defects are of two types:
a. Metal excess – This type of defect is due to excess of metal cations. These may be
due to:
vimp i. Anionic vacancies: A compound may have an
extra metal ion if the negative ion is absent from its
lattice site. This empty lattice site is called a hole.
To maintain electrical neutrality this site is occupied
by an electron. The hole occupied by an electron
is called f-centre or Farbenzenter centre . F- centre
is responsible for the colour of the compound.
e.g. on heating NaCl in presence of Na ions, some anions (Cl-) leave lattice sites which
are occupied by electrons called F centres.
ii. Presence of extra cations: A compound is said to have extra cations if a cation is
present in the interstitial site. An electron is present in the interstitial site to maintain
the electrical neutrality. E.g. ZnO, O2 gas is lost, Zn+2 ions and e- occupy interstitial
sites.
b. Metal deficiency:
This defect arises because of absence of metal ions
from its lattice sites. The electrical neutrality is
maintained by an adjacent ion having a higher
positive charge. E.g 2A+ in lattice sites may be
replaced by A2+ in one lattice site and one lattice
site remains vacant. Fe2O ideal composition
composition (actual Fe0.95O)
CLASSIFICATION OF SOLIDS BASED ON THEIR ELECTRICAL CONDUCTIVITIES:
A. CONDUCTORS-The solids with conductivities () ranging between 104to 107 ohm–1m–
1
are called conductors.
B. INSULATORS - These are the solids with very low conductivities () ranging between
10–20 to 10–10 ohm–1m–1.
C. SEMI-CONDUCTORS - These are the solids with conductivities () in the
intermediate range from 10–6 to 104 ohm–1m–1.
Band theory – A metal is characterized by a band structure. The highest filled band is
called valence band and the lowest unoccupied band is called conduction band. The gap
between the two bands is called forbidden band.
20
Class XII
Chemistry
states of matter [SOLID STATE]
a. In case of conductors, the valence band and conduction band overlap
b. In case of insulators, the forbidden gap is very large and the electrons are
unable to excite to the conduction band.
c. In case of semiconductors, forbidden gap is small. Therefore, some electrons may
jump to conduction band and show some conductivity. Electrical conductivity of
semiconductors increases with rise in temperature, since more electrons can
jump to the conduction band.
TYPES OF SEMICONDUCTORS:
a. Intrinsic:
These are those semiconductors in which the forbidden gap is small. Only some
electrons may jump to conduction band and show some conductivity. They have very
low electrical conductivity. Example: Silicon, germanium
b. Extrinsic:
When an appropriate impurity is added to an intrinsic semiconductor. Their electrical
conductivity is high.
DIFFERENCE BETWEEN INTRINSIC SEMICONDUCTORS AND EXTRINSIC SEMI
CONDUCTIORS:
INTRINSIC SEMI CONDUCTORS
EXTRINSIC SEMI CONDUCTORS
It is a pure semi conductor and no impurity It is prepared by doping small quantity of
is added to it.
impurity atoms to the pure semi
conducting materials.
2. examples are crystalline forms of pure Si e.g.si and ge crystals doped with impurity
and germanium.
as As, Sb, P etc.
3. no of free electrons in the conduction
The no. of holes and valence electrons are
band and no. of holes in the valence band
never equal.
is exactly equal and very small indeed.
4. low electrical conductivity.
High electrical conductivity.
5. electrical conductivity is a function of
Electrical conductivity depends upon the
temperature alone.
temperature as well as the quantity of
impurity atoms doped in the structure.
DOPING: The process of adding an appropriate amount of suitable impurity to increase
the conductivity of semiconductors.
TYPES OF EXTRINSIC SEMI CONDUCTORS:
21
Class XII
Chemistry
N-TYPE SEMICONDUCTORS
1. It is obtained by doping the impurity
atoms of group 15th (Si) . to pure semi
conductor e.g. As, Sb, P.
2. The increase in conductivity is
due to the negatively charged
electrons.
3. the electrons are majority carrier
and holes are minority carriers. i.e. ne
>>nh
states of matter [SOLID STATE]
P –TYPE SEMICONDUCTORS
It is obtained by doping the impurity
atoms of group 13th . to pure semi
conductor of group 15 (Si).
The increase in conductivity is
due to the positively charged
holes.
Holes are majority carrier and
electrons are minority carriers.
i.e. nh >>ne
DIODE:
It is a combination of n-type and p-type semiconductors and is used as a rectifier.
e.g. LED light emitting diode, Photo diode used in solar cell
TRANSISTORS:
They are made by sandwiching a layer of one type of semiconductor between two layers
of the other type of semiconductor. npn and pnp type of transistors are used to
detect or amplify radio or audio signals.
12- 16 COMPOUNDS – These compounds are formed by the combination of group 12
and group 16 compounds. They possess an average valency of 4. Example: ZnS, CdS,
CdSe and HgTe
13- 15 COMPOUNDS – These compounds are formed by the combination of group 13
and group 15 compounds. They possess an average valency of 4. Example: InSb, AlP
and GaAs
Every substance has some magnetic properties associated with it. The origin of these
properties lies in the electrons. Each electron in an atom behaves like a tiny magnet. Its
magnetic moment originates from two types of motions (i) its orbital motion around the
nucleus and (ii) its spin around its own axis.
VARIATION OF RESISTANCE OF SOME ELECTRICAL MATERIAL AND
TEMPERATURE:
Electric resistance (R) – is the ratio of potential difference (V) across the ends of the
conductor to the current (I) flowing through it, i.e.
R =V Ohm (Ω)
I
Units of R is ohm (Ω)
22
Class XII
Chemistry
states of matter [SOLID STATE]
RESISTANCE :
Property of a substance by virtue of which it opposes the flow of current through it.
VARIATION OF RESISTANCE:
R= ρL where R is called resistance of the material, ρ is the resistivity of the
A
material, A is the area of cross-sectional of wire, L is the length of wire.
R also depends on Material of the conductor.
Rt = R0 (1 + t) Where Rt = Resistance at temperature toC
Ro = Resistance at temperature 0oC
(i) METALS: for metals Temperature coefficient of resistance ()>0. Therefore
resistance increases wrt Temperature.
Physical explanation: collision frequency of free electron with immobile +ve ions
increases.
(ii) SOLID NON METALS: For these  =0, so resistance is indepenedent with
temperature.
Physical explanation: complete absence of free electrons.
(iii) SEMI CONDUCTORS: For semi conductors  <0, i.e. resistance decreases with the
temperature rise.
Physical explanation: covalent bonds breaks, liberating more free electron and
conduction increases.
(IV) ELECTROLYTE: for electrolyte  <0, i.e. resistance decreases with temperature
rise.
Physical explanation: the degree of ionisation increases and solution and solution
becomes less viscous and more conducting.
(V) IONISED GAS: for ionised gas  <0, i.e. resistance decreases with temperature
rise.
Physical explanation: degree of ionisation increases.
(VI) ALLOYS: for alloys α has a small +ve value so with rise of temperature resistance
of alloys is almost constant. Further alloy resistance are slightly higher than the pure
metal resistance.
Alloys are used to made standard resistances, wires of resistance box, potentiometer
wire, meter bridge wire etc.
Commonly used alloys are : constantan, maganin, Nichrome etc.
(VII) SUPER CONDUCTORS: At low temperature , the resistance of certain substances
becomes exactly zero. (e.g. Hg below 4.2K or Pb below 7.2K)
These substance are called super conductors and phenomenon is super conductivity. The
temperature at which resistance becomes zero is called critical temperature and depends
upon the nature of the substance.
USES: (i) Helps in producing super computers.
(ii) Superconductor cables for electricity transmission without any loss.
(iii) To create strong magnetic field with small electrical power.
Therefore electrical conductivity of metals decreases with temperature (due to
vibtations of +ve ions which hinders the flow of electrons).
23
Class XII
Chemistry
states of matter [SOLID STATE]
Semi conductors increases with temperature (electron jump from valence band
to conduction band)
Some transition metal oxides and their conductivities:
(i) TiO, CrO2, ReO3 are metallic
(ii) MnO, FeO and CuO are insulators
(iii) VO, VO2, VO3 and TiO3 changes from metallic to insulators at a certain temperature.
MAGNETIC PROPERTIES OF SOLIDS:
CLASSIFICATION OF SOLIDS ON THE BASIS OF MAGNETIC PROPERTIES:
A. DIAMAGNETIC: Diamagnetic substances are weakly repelled by a magnetic field.
Diamagnetism is shown by those substances in which all the electrons are paired and
there are no unpaired electrons. E.g N2, TiO2, H2O, Zn, Cd, NaCl,Cu+, benzene etc.
B. PARAMAGNETIC: These are those substances which are weakly attracted by the
magnetic field. It is due to presence of one or more unpaired electrons and hence
magnetic moment. E.g. O2,NO, metal ions (Cu+2, Fe+3, Cr+3 ), metals (Cr, Mn, Ni, Co,
Fe) , metal ocide (CuO, VO2)etc.
C. FERROMAGNETIC: These are those substances which are attracted very strongly by
a magnetic field even in the absence of magnetic field. E.G. Fe, Ni, Co, Gd,CrO2
(magnetic tapes, recorders)
D. ANTIFERROMAGNETIC: Which are expected to possess paramagnetism or
ferromagnetism but actually have zero net magnetic moment. They have equal number
of parallel and anti parallel magnetic dipoles (domains) resulting in a zero net dipole
moment. E.g. MnO, Mn2O3, MnO2
E. FERRIMAGNETIC: which are expected to possess large magnetism but actually have
small net magnetic moment, e.g. magnetite (Fe3O4), ferrites (M2+Fe2O4). Because of
unequal number of parallel and anti parallel magnetic dipoles resulting in a net dipole
moment.
CURIE TEMPERATURE:
It is the temperature above which a ferromagnetic substance shows no ferromagnetism.
Some magnetic properties of common substance:
(a) TiO, VO, CuO are paramagnetic whereas MnO, CoO, NiO are antiferromagentic.
(b) TiO2 is diamagnetic, VO2 is paramagnetic, CrO2 is ferromagnetic whereas MnO2 is
antiferromagnetic.
24
Class XII
Chemistry
states of matter [SOLID STATE]
PIEZOELECTRICITY: When mechanical stress is applied on such crystals, electricity is
produced due to displacement of ions which is known as piezoelectricity.
Piezoelectric crystals: titanates of barium and lead, lead zirconate (PbZrO3), ammonium
dihydrogen phosphate (NH4H2PO4)
PYROELECTROCITY: Some piezoelectric crystals when heated produce a small electric
current. Electricity thus produced is called pyroelectricity (pyro means heat)
FERROELECTROCITY: In some piezoelectric crystals, the dipoles are permanently
polarised even in the absence of electric field. However on applying electric field,
direction of polarisation changes. This phenomenon is called ferro-electricity.
e.g. barium titanate (BaTiO3) , sodium potassium tartarate (Rochelle salt) and potassium
di hydrogen phosphate (KH2PO4). It may be pointed out here that all ferroelectric solids
are piezoelectric but the reverse is not true.
ANTIFERROELECTRICITY: In some crystals, the dipoles align themselves in such a
way that alternatively, they point up and down so that the crystals does not possess any
net dipole moment. Such crystals are said to be anti ferroelectric. E.g. lead zirconate.
Q12. analysis shows that a metal oxide has the empirical formula of M0.96O1.00 .
calculate the percentage of M2+ and M3+ ions in this crystal?
Ans. Formula of metal oxide = M0.96O1.00 means
M = 0.96
= 96
O 1.00
100
Thus if there are 100 O atoms M atoms would be 96
Charge on 100 O2- ions =100 x (-2) = -200
Suppose M atoms as M2+ = x
and M3+ = 96-x
total charge on M2+ and M3+ =(+2) x + (+3)x (96-x) = 288-x =200
or x= 88
therefore %age of M as M2+ = 88 x 100
= 91.7%
96
%age of M as M3+ = 100 - 91.7% = 8.3%
Curie temperature:
It is the temperature above which a ferromagnetic substance shows no
ferromagnetism.
25