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CHEMISTRY
Part 1
Class 12
Syllabus : Higher Secondary - Second Year Chemistry Volume - I
INORGANIC CHEMISTRY
Unit 1 - Atomic Structure -II
Dual properties of electrons - de-Broglie relation - Heisenberg’s
uncertainty principle - Wave nature of an electron - Schrodinger wave equation
(only equation, no derivation) - Eigen values and Eigen function- significance
only - molecular orbital method
...
Unit 2 - Periodic classification-II
Review of periodic properties - Calculation of atomic radii - Calculation
of ionic radii - Method of determination of Ionisation potential - Factors affecting
ionisation potential - Method to determine the electron affinity - Factors affecting
EA - Various scales on electro negativity values
...
General trends - Phosphorous
- Allotropes and extraction - Compounds of phosphorous - Group - 16
...
- Group - 17 General
characteristics
...
Unit 4 d - BLOCK ELEMENTS
General characteristics of d-block elements - First transition series Occurrence and principles of extraction - chromium, copper and zinc - Alloys Second transition series - Occurrence and principles of extraction of silver Third transition series - Compounds - K2Cr2O7, CuSO45H2O, AgNO3, Hg2Cl2,
ZnCO3, Purple of cassius
...
(v)
Unit 6 - Coordination Compounds and Bio-coordination Compounds
An introduction - Terminology in coordination chemistry - IUPAC
nomenclature of mononuclear coordination compounds - Isomerism in
coordination compounds - Structural isomerism - Geometrical isomerism in
4 - coordinate, 6 – coordinate complexes - Theories on coordination compounds
- Werner’s theory (brief) - Valence Bond theory - Crystal field theory - Uses of
coordination compounds - Biocoordination compounds
...
Unit 7 - Nuclear chemistry
Nuclear energy nuclear fission and fusion - Radio carbon dating - Nuclear
reaction in sun - Uses of radioactive isotopes
...
Unit 9 - Thermodynamics - II
Review of I law - Need for the II law of thermodynamics - Spontaneous
and non spontaneous processes - Entropy - Gibb’s free energy - Free energy
change and chemical equilibrium - Third law of thermodynamics
...
Unit 11 - Chemical Kinetics -II
First order reaction and pseudo first order reaction - Experimental
determination of first order reaction - method of determining order of reaction temperature dependence of rate constant - Simple and complex reactions
...
(vi)
Unit 13 – Electrochemistry – I
Conductors, insulators and semi conductors - Theory of electrical
conductance - Theory of strong electrolytes - Faraday’s laws of electrolysis Specific resistance, specific conductance, equivalent and molar conductance Variation of conductance with dilution - Kohlraush’s law - Ionic product of water,
pH and pOH - Buffer solutions - Use of pH values
...
Unit 15 – Isomerism in Organic Chemistry
Geometrical isomerism - Conformations of cyclic compounds - Optical
isomerism - Optical activity - Chirality - Compounds containing chiral centres D-L and R-S notation - Isomerism in benzene
...
(glycol) - Properties - Uses - Methods of preparation of trihydric
alcohols - Properties - Uses - Aromatic alcohols - Methods of preparation of
benzyl alcohol - Properties - Uses - Phenols - Manufacture of phenols - Properties
- Chemical properties - Uses of Phenols
...
Unit – 18 Carbonyl Compounds
Nomenclature of carbonyl compounds - Comparison of aldehydes and
ketones - General methods of preparation of aldehydes - Properties - Uses
Aromatic aldehydes - Preparation of benzaldehyde - Properties - Uses - Ketones
- general methods of preparation of aliphatic ketones (acetone) - Properties Uses - Aromatic ketones - preparation of acetophenone- Properties - Uses preparation of benzophenone - Properties
...
Unit - 20 Organic Nitrogen Compounds
Aliphatic nitro compounds - Preparation of aliphatic nitroalkanes Properties - Uses - Aromatic nitro compounds - Preparation - Properties Uses - Distinction between aliphatic and aromatic nitro compounds - Amines Aliphatic amines - General methods of preparation - Properties - Distinction
between 1°, 2°, and 3° amines - Aromatic amines - Synthesis of benzylamine Properties - Aniline–preparation - Properties - Uses - Distinction between
aliphatic and aromatic amines - Aliphatic nitriles - Preparation - properties Uses - Diazonium salts - Preparation of benzene diazoniumchloride - Properties
...
Unit 22 - Chemistry in Action
Medicinal chemistry - Drug abuse - Dyes – classification and uses Cosmetics – creams, perfumes, talcum powder and deodorants - chemicals in
food - Preservatives artificial sweetening agents, antioxidants and edible colours
- Insect repellant – pheromones and sex attractants - Rocket fuels - Types of
polymers, preparation and uses
...
Detection of Nitrogen, Halogen and Sulphur in organic compounds
...
Detection of Functional groups present in organic compounds
...
Qualitative analysis
Determination of two cations and two anions in a given mixture
...
(Insoluble and interfering ions are to be excluded
...
IV
...
Titration of Oxalic acid Vs KMnO4
2
...
b) Dichrometry
1
...
2
...
Report should contain two acid radicals and two basic radicals, without
mentioning the name of the salt
...
(ix)
CONTENTS
UNIT NO
...
Inorganic Chemistry
1
Atomic Structure - II
1
2
Periodic Classification - II
38
3
p - Block Elements
59
4
d - Block Elements
99
5
f - Block Elements
133
6
Coordination Compounds and
Bio-Coordination Compounds
142
Nuclear Chemistry
167
7
Physical Chemistry
8
Solid State - II
188
9
Thermodynamics - II
205
10
Chemical Equilibrium - II
224
(x)
(xi)
P e rio d
7
6
5
4
3
2
3
4
57
5
7
8
9
10
11
12
Sr
38
37
56
Ba
8 5
...
9 1
A ctin id e s
Mn
(9 8 )
75
90
Th
2 3 2
...
12
89
59
Bh
(2 6 2 )
Sg
(2 6 3 )
60
Nd
(2 6 5 )
Hs
108
U
92
2 3 1
...
0 3
Pa
91
1 4 0
...
2 4
Pr
107
106
1 3 8
...
07
76
Ru
44
5 5
...
6 8
Ni
Pt
62
(2 3 7 )
Np
(24 4 )
Pu
94
15 0
...
25
Gd
64
U ub
65
(2 4 7 )
Bk
97
1 5 8
...
2
(25 1 )
Cf
98
16 2
...
5 9 20 4
...
7 2
49
7 2
...
9 6
Se
34
3 2
...
0 0
O
8
16
(2 5 2 )
Es
99
(25 7 )
Fm
10 0
1 6 4
...
26
Ho
67
U up
11 5
2 0 8
...
92
As
33
3 0
...
0 1
N
7
15
3 9
...
Ar
18
2 0
...
0 0 3
53
I
7 9
...
80
36
3 5
...
00
9
F
17
(25 9 )
10 2
No
Md
(2 5 8 )
71
Lu
(2 6 2 )
Lr
103
17 3
...
9 7
Yb
70
U uo
118
(22 2 )
101
1 6 8
...
4 1 114
...
7 1 1 2 1
...
60 1 26
...
29
80
81
83
85
82
84
86
Hg
Ti
Bi
Pb
Po
At
Rn
Cd
48
6 5
...
9 6
Eu
Uuu
63
111
19 6
...
87
Ag
47
6 3
...
2 2 1 9 5
...
9 1 1 0 6
...
9 3
Co
Pm
1 86
...
2 3
Re
W
1 8 3
...
9 4
74
Mo
43
5 2
...
9 4
Cr
Ce
(2 6 2 )
Db
105
1 8 0
...
91
Nb
41
5 0
...
49
Hf
72
9 1
...
8 7
39
Ti
4 4
...
91 1 3 7
...
62
Rb
4 0
...
1 0
Sc
21
20
Ca
K
19
2 8
...
0 1
C
6
14
2 6
...
3 1
2 2
...
8 1
9
...
9 4 1
25
6
5
1 3 /III
B
3
1
...
As Newlands did before him in 1863,
Mendeleyev classifies the elements, according to their atomic weights
and notices that they exhibit recurring patterns or periods of properties
...
ATOMIC STRUCTURE - II
Learning Objectives
( To study the dual property of electron and understand the
property through experiments
...
( To learn Heisenberg’s uncertainty principle
...
( To understand the concept of Hybridisation and Hybridisation of
s, p and d orbitals
...
Dalton(1808)
: Discovery of atom
2
...
Goldstein(1886)
: Discovered anode rays and proton
4
...
J
...
Thomson(1897)
: Discovered electron and determined
charge/mass(e/m) ratio for electron
5
...
MaxPlanck(1901)
: Proposed quantum theory of radiation
7
...
H
...
J
...
Niels Bohr(1913)
: Proposed a new model of atom
10
...
de-Broglie(1923)
: Established wave nature of particles
12
...
Werner Heisenberg(1927) : Uncertainty Principle
14
...
Anderson(1932)
: Discovery of positron
16
...
Hideki Yukawa(1935)
: Discovered mesons
18
...
Cork and Association(1956) : Discovered antineutron
2
Progress of Atomic Models
Ø
In 1803, John Dalton, proposed his atomic theory
...
Ø
J
...
Thomson proposed that an atom was a solid sphere of positively charged
material and negatively charged particles, electrons were embedded in it
like the seeds in a guava fruit
...
Ø
Rutherford suggested the planetary model, but this model was rejected
...
Based on the facts obtained from
spectra of hydrogen atom, he introduced the concept of energy levels of
atom
...
He defined sub energy levels for every major energy level
predicted by Bohr
...
Ø
From the study of quantum numbers, various rules are put forward for
filling of electrons in various orbitals by following
*
*
Pauli exclusion principle and
*
Ø
Aufbau principle
Hunds rule of maximum multiplicity
...
Further the nature of electron (s) is studied
...
1 DUAL PROPERTY OF AN ELECTRON
In case of light, some phenomena like interference, diffraction etc
...
However certain other
phenomena such as black body radiation and photo electric effect can be explained
only if it is believed to be a stream of photons i
...
, has particle character
...
Such studies on light were made by Einstein
in 1905
...
The wave associated with a particle is called a matter wave
...
1
...
PARTICLE
WAVE
1
...
a wave is spread out in space e
...
on throwing
position in space i
...
g
...
sand, a cricket ball etc
...
Thus a wave is
delocalized in space
...
When a particular space is occupied 2
...
cannot be occupied simultaneously
by any other particle
...
3
...
When a number of waves are present in a
present in a given region of space,
given region of space, due to interference, the
their total value is equal to their
resultant wave can be larger or smaller
sum i
...
than the individual waves i
...
interference may
be constructive or destructive
...
1
...
When this fine beam of accelerated electron is allowed to fall on a large single
crystal of nickel, the electrons are scattered from the crystal in different directions
...
1
...
D iffr action p attern
P h o to g ra p h ic
p la te
m
ea s
t b on
en ctr
c id le
In of e
Nickel crystal
Fig
...
1 Electron diffraction experiment by Davisson and Germer
Since X-rays have wave character, therefore, the electrons must also have
wave character associated with them
...
From the above discussion, it is clear that an electron behaves as a wave
...
P
...
He observed that if the beam of electrons after passing through
the thin foil of gold is received on the photographic plate placed perpendicular to
the direction of the beam, a diffraction pattern is observed as before (Fig
...
2)
...
5
Thin foil
of Gold
Fig
...
2 Diffraction of electron beam by thin foil of gold (G
...
Thomson
experiment)
b)
Verification of the particle character
The particle character of the electron is proved by the following different
experiments:i)
When an electron strikes a zinc sulphide screen, a spot of light known as
scintillation is produced
...
Therefore the striking electron which produces it, also must be localized
and is not spread out on the screen
...
Hence electron has particle character
...
J
...
e
...
iii)
The phenomenon of Black body radiation and Photoelectric effect also prove
the particle nature of radiation
...
2 de-Broglie Relation
The wavelength of the wave associated with any material particle was
calculated by analogy with photon as follows :In case of a photon, if it is assumed to have wave character, its energy is
given by
E = hν (according to the Planck’s quantum theory)
...
If the photon is supposed to have particle character, its energy is given by
E = mc2 (according to Einstein equation)
...
From equations (i) and (ii), we get
But
∴
or
hν
ν
h
...
The mass of the photon is replaced by the mass of the material particle
and the velocity “c” of the photon is replaced by the velocity v of the material
particle
...
The above equation is called de Broglie equation and ‘λ’ is called de
Broglie wavelength
...
Louis de-Broglie’s concept of dual nature of matter finds application in the
construction of electron microscope and in the study of surface structure of solids
by electron diffraction
...
,
7
Significance of de-Broglie waves
The wave nature of matter, however, has no significance for objects of
ordinary size because wavelength of the wave associated with them is too small
to be detected
...
i)
Suppose we consider an electron of mass 9
...
Its de-Broglie wavelength will be;
h
6
...
1 × 10-31 kg ×107 ms-1
= 0
...
27 × 10-11m
This value of λ can be measured by the method similar to that for the
determination of wave length of X-rays
...
Its de-Broglie wave length will be;
h
6
...
62 × 10-34m
mv
10-2 kg ×102 ms-1
This wavelength is too small to be measured, and hence de-Broglie relation
has no significance for such a large object
...
Problem 1
The kinetic energy of sub-atomic particle is 5
...
Calculate the
frequency of the particle wave
...
626 × 10-34 Js)
Solution
K
...
= ½ mv2 = 5
...
85 × 10-25 J
____________
6
...
77 × 109 s-1
Problem 2
Calculate the de-Broglie wavelength of an electron that has been accelerated
from rest through a potential difference of 1 kV
Solution
Energy acquired by the electron (as kinetic energy) after being accelerated
by a potential difference of 1 kV (i
...
609 × 10-19 J, (1 eV) = 1
...
diff
...
609 × 10-16 J
i
...
Kinetic energy
⎛1
2⎞
− 16
⎜ mv ⎟ = 1
...
1×10 − 31 v 2 = 1
...
536×1014
or
v = 1
...
626 ×10 −34
=
mv 9
...
88 × 10 7
= 3
...
1 × 10-31 kg)
moving with a velocity of 103m sec-1 (h=6
...
Solution
Here we are given
m = 9
...
626 × 10-34 kg m2 sec-1
h
6
...
1×10−31 ) ×103
= 7
...
55 × 10-25 joules of kinetic energy
...
1 × 10-31 kg and h = 6
...
Solution
Here we are given
Kinetic energy i
...
1
mv 2 = 4
...
1 × 10-31 kg
h = 6
...
1×10 − 31 )v 2 = 4
...
55 ×10 −25 × 2
= 10 6
− 31
9
...
626 ×10−34
=
mv (9
...
25 × 10-7 m
...
8 pm
...
11 × 10-31 kg, h = 6
...
Solution
According to de-Broglie equation,
ë=
∴
h
mv
v=
h
6
...
516 ×108 ms −1
−31
−12
më 9
...
8 ×10 m
or
v=
h
më
1
1
Kinetic energy = mv2 = × 9
...
516×108 ms−1 ) 2
2
2
= 10
...
047 × 10-14 J
Problem 6
Two particles A and B are in motion
...
Solution
According to de-Broglie relation,
ë =
h
p
or
p =
h
ë
h
For particle A, p A = ë
A
11
Here, pA and λA are the momentum and wavelength of particle A
...
1
pA
2
But,
pB =
∴
h 1 h
=
ëB 2 ëA
ëA
1
=
ëB
2
But
or
λB = 2λA
λA = 5 × 10-8 m
λB = 2λA = 2 × 5 × 10-8 m = 10 × 10-8 m = 10-7 m
...
Calculate the momentum of a particle which has a de-Broglie wavelength of
1A°
...
626 × 10-34 kg m2 s-1]
[Ans
...
63 × 10-24 kg ms-1]
2
...
626 × 10-34 Js]
[Ans
...
75 × 10-36 kg]
3
...
[Ans
...
92 × 10-39 m]
4
...
: 7
...
The wavelength of a moving body of mass 0
...
31 x 10-29m
...
626 x 10-34 Js)
...
Calculate the wavelength of a particle of mass m = 6
...
425 × 10-13 J (h = 6
...
[Ans
...
657 × 10-15 m]
7
...
[Ans
...
39 pm]
8
...
626 × 10-34 kg m2 s-1)
...
: 18
...
A moving electron has 4
...
Find out its de Broglie wavelength (Given h = 6
...
1 × 10-31 kg)
...
: 7 × 10-7 m]
1
...
Hence the
path or trajectories of such bodies can be predicted
...
Thus, it is not possible to talk of trajectory of an electron
...
”
Mathematically, uncertainty principle can be put as follows
...
Äp ≥
h
4ð
where, Δx = uncertainity in the position of the particle and
Δp = uncertainity in the momentum of the particle
...
13
Example 1
Calculate the uncertainty in the velocity of a wagon of mass 3000kg
whose position is known to an accuracy of ± 10 pm (Planck’s constant =
6
...
Solution : Ηere we are given
m
= 3000 kg
Δx = 10 pm
= 10 ×10-12 m = 10-11 m
∴ Βy uncertainty principle,
Äv =
h
4ð × m × Äx
6
...
76× 10-27ms-1
×
Example 2
Calculate the uncertainty in the position of an electron if the uncertainty in its
velocity is 5
...
626 × 10-34 kg m2 s-1, mass of the electron
= 9
...
Solution:
Here we are given
Δv = 5
...
1 × 10 −31 kg
h = 6
...
e
...
626 × 10 −34
22
× 9
...
7 × 105
7
= 1
...
e Uncertainty in position = ± 10-10 m
...
The approximate mass of an electron is 10-27 g
...
[Ans: 5
...
Calculate the product of uncertainity in position and velocity for an electron
of mass 9
...
[Ans: 5
...
Calculate the uncertainty in velocity (Δv ) of a cricket ball (mass = 0
...
e
...
[Ans: 3
...
Using uncertainity principle,calculate the uncertainty in velocity of an electron
if the uncertainty in position is 10-4 m
...
577 m sec-1]
5
...
Calculate the uncertainty in its velocity
...
25 x 10-28 m sec-1]
1
...
It has been clarified that matter is a collection of ultra
microscopic particles
...
However, this view point has became doubtful after the proposal of the Bohr
model of the atomic structure (Bohr’s quantum theory)
...
However, after the discovery of light quanta (photons), it was clarified that
15
the light has wave nature at one time and particle nature at another time
...
The idea of deBroglie wave nature waves or deBroglie matter waves is
based on the fact that light has both wave and particle nature
...
Einstein’s relations which connect the particle and wave aspects in light
quanta
h
(1)
ë
would be satisfied for de Broglie matter waves as well
...
(1), are often called Einstein-de Broglie’s relations
...
If we consider that the
electron in a hydrogen atom moves at constant speed along a circular orbit around
the nucleus (proton), the quantum condition in Bohr’s quantum theory is written
as Eq(2)
...
)
(2)
This equation means that the circumference of the circular orbit of the electron
must be a integral multiple of the wavelength of de Broglie wave
...
Therefore, we can easily understand the quantum condition that determines the
stationary states by considering the continuity of de Broglie waves (See the
following figure)
...
The condition
for stationary states
The circumference of the circular orbit of
the electron should be an integral multiple of
the wavelength of de Broglie wave,
otherwise the wave cannot be smoothly
continuous
...
By applying Schrodinger wave equation
to hydrogen atom, the energy of electron (En) was found as :
2ð 2 me4
En = − 2 2
(1)
n h
where n is the principal quantum number
...
Substituting the values of m, e and h in relation (1), we get
1312
kJ mol −1
2
n
Significance of negative electronic energy
En = −
(2)
The energy of an electron at infinity is arbitrarily assumed to be zero
...
When an electron moves and comes under the
influence of nucleus, it does some work and spends its energy in this process
...
, it
acquires a negative value
...
Calculate the wavelength of radiation emitted when the electron in hydrogen
atom makes a transition from n = 2 state to n = 1 state (Planck’s constant,
h = 6
...
0237 × 1023 mol-1)
...
E
...
E = -1312 kJ mol-1
Energy of hydrogen atom in the nth orbit (En) =
- 1312
kJ mol −1
2
n
1312
= -328 kJ mol-1
22
ÄE = E 2 − E1 = [−328 − (−1312)] kJ = 984 kJ mol−1
Energy of hydrogen atom in the second orbit (E2) = −
Energy released per atom =
ÄE
N
=
17
984×103 J/atom
6
...
626×10−34 Js × 3 ×108 ms−1 × 6
...
2 × 10-7 m
984 ×103 J
Example 2
The electron energy of hydrogen atom in the ground state works out to be
–2
...
Calculate what will happen to the position of the
electron in this atom if an energy of 1
...
Solution
Energy of H atom in the ground state = -2
...
938 × 10-18 J atom-1
Energy of electron in the excited state = (-2
...
938) × 10-18 J atom-1
= -0
...
18×10−18 J atom−1
− 0
...
18 × 10−18 J atom−1
= 9,
n =
−0
...
2
n=3
Example 3
Calculate the ionisation energy of hydrogen atom as well as energy needed
to promote its electron from first energy level to third energy level
...
e
...
The shape of this
region (electron cloud) gives the shape of the orbital
...
These two plots differ only slightly
...
Shape of s-orbitals
For s-orbitals, when l = 0, the value of m is 0 i
...
, there is only one possible
orientation
...
It should, therefore, be spherical
in shape
...
The size of an s-orbital depends upon value of the principal quantum number
n
...
Fig
...
3 Shapes of 1s and 2s-orbitals
19
An important feature of the 2s-orbital is that there is a spherical shell within
this orbital where the probability of finding the electron is zero (nearly)
...
In 2s orbital there is one spherical node
...
Shape of p-orbitals
For p-subshell l = 1, there are three values of m namely -1, 0, +1
...
These three p-orbitals
are equal in energy (degenerate state) but differ in their orientations
...
Depending
upon the orientation of the lobes, these are denoted as 2px , 2py and 2pz accordingly
as they are symmetrical about X,Y and Z - axis respectively
...
The boundary surface means the surface which
encloses 90 percent of the dots representing the electrons
...
For
example, for 2px orbital, YZ plane is the nodal plane x
...
1
...
1
...
The
probability of finding the electron is equal in both the lobes
...
Shape of d-orbitals
For d-subshell, l = 2, there are five values of m namely -2, -1, 0, 1, 2
...
These are represented by dxy, dyz,
dzx, dx2-y2 and dz2; for example, 3dxy, 3dyz, 3dzx, 3dx2-y2 and 3dz2
...
e
...
The dz2 orbital is symmetrical about Z-axis and has a dumb bell shape with a doughnut shaped electron cloud in the centre
...
The reason for the presence of four lobes in any nd orbital lies in the fact that
the d - orbitals have two nodes, and hence two changes in algebraic sign of ψ,
which lead to four lobes
...
1
...
5 MOLECULAR ORBITAL THEORY
Molecular orbital theory was put forward by Hund and Mullikan in 1932
...
This theory assume that in molecules,
atomic orbitals lose their identity and the electrons in molecules are present in
new orbitals called molecular orbitals
...
(ii) Molecular orbitals are formed by combination of atomic orbitals of equal
energies (in case of homonuclear molecules) or of comparable energies (in
case of heteronuclear molecules)
...
(iv) Two atomic orbitals can combine to form two molecular orbitals
...
The molecular orbital with lower energy is called bonding
molecular orbital and the other with higher energy is called anti bonding
molecular orbital
...
(vi) The bonding molecular orbitals are represented by σ (sigma), π (pi), δ (delta)
and the antibonding molecular orbitals are represented by σ∗, π∗, δ*
...
(Aufbau principle)
...
(Paul’s exclusion principle)
...
(Hund’s rule)
...
5
...
In the same manner the 2s and three 2p-orbitals of each atom
i
...
, eight atomic orbitals can give rise to eight new molecular orbitals viz
...
2s
2p
2p
2p
Atomic Structure and Chemical Bonding
Energy levels of these molecular orbitals have been determined
experimentally by spectroscopic studies
...
This energy diagram for the molecular orbitals is shown in Fig
...
7a
...
In case of
these elements, the order of energy levels of ó 2p z , ð2px and ð2py is reversed i
...
,
ó 2p z has lesser energy than ð 2p x or ð 2p y
...
*
ó 1s < ó 1s < ó 2s < ó * < ó 2p z < ð 2p x = ð 2p y < ð * x = ð * y < ó * z
2s
2p
2p
2p
22
This order of energies of various MOs is valid for molecules or ions like O2,
O (super oxide ion), O22- (peroxide ion), F2 and Ne2 (hypothetical)
...
1
...
2
Fig
...
7a Molecular orbital energy
level diagram for diatomic homonuclear
molecules of first and second period
(except O2, F2 etc
...
1
...
Molecular orbital
energy level diagram for
homonuclear diatomic
molecules of O2 and other
heavier elements
1
...
2 Electronic configuration of a molecule and its correlation with
molecular behaviour
The distribution of electrons among various molecular orbitals is called
electronic configuration of a molecule
...
1
...
From the electronic configuration it is
possible to find out the number of electrons in bonding molecular orbitals(Nb)
and number of electrons in antibonding molecular orbitals (Na)
...
(b) If Nb< Na, the molecule is unstable : This is again obvious because
in this case the influence of antibonding electrons will be more than the influence
23
of bonding electrons, resulting in a net force of repulsion
...
2
...
The stability of a
molecule or an ion can also be determined from another parameter called bond
order
...
e,
1
(N − N a )
2 b
The resulting molecule or ion will be stable if Nb > Na i
...
, if bond order is
positive
...
e, if bond
order is negative or zero
...
Relative stability of molecules or ions in terms of bond order : The
stability of a molecule or an ion is directly proportional to bond order
...
g
...
e
...
g
...
g
...
4
...
5
...
Greater the bond order, shorter the bond
length and vice versa
...
24
Table 1
...
Diamagnetic and paramagnetic nature of the molecule : If all the
electrons in the molecule are paired then the substance is diamagnetic in nature
...
1
...
3 Molecular orbital energy level diagrams of certain diatomic
homonuclear molecules and molecular ions
The filling of molecular orbitals is governed by the following principles
...
Now, let us consider some examples of homonuclear diatomic
molecules
...
Hydrogen molecule, H2
...
Each hydrogen atom in the ground state has one electron in 1s
orbital
...
According to Pauli’s exclusion
principle, these two electrons should have opposite spins
...
The molecular orbital energy level diagram of H2 molecule is given in
Fig
...
8
...
1
...
Here,
N b = 2 and Na = 0
25
∴
Bond order =
Nb − Na 2 − 0
=
= 1
...
ii)
Diamagnetic character : Since no unpaired electron is present in hydrogen
molecule, it is diamagnetic in nature
...
Diatomic helium molecule, He2 (Hypothetical)
...
As each helium atom
contains two electrons, there will be four electrons in He2 molecule
...
He2 : (σ1s)2 (σ*1s)2
...
1
...
Fig
...
9 Molecular orbital energy level diagram of He2 (hypothetical)
molecule
Here,
Nb = 2 and Na = 2
Nb − Na 2 − 2
=
= 0
...
∴
Bond order =
3
...
The electronic configuration of nitrogen (Z=7)
in the ground state is 1s 2 2s 2 2p1 2p1 2p1
...
These 14 electrons can be accommodated
in the various molecular orbitals in order of increasing energy
...
In calculating bond order, we can ignore
KK, as it includes two bonding and two antibonding electrons
...
1
...
Fig
...
10 Molecular orbital energy level diagram of N2
The bond order of N2 can be calculated as follows
...
Bond order = b
2
2
i) Nature of bond : A bond order of 3 means that a triple bond is present in
a molecule of nitrogen
...
4
...
The electronic configuration of oxygen (Z = 8)
in the ground state is 1s22s22p4
...
Therefore, the electronic configuration of O2 is
as follows
...
The molecular orbital energy level diagram of O2 molecule is given in Fig
...
11
...
1
...
6
Bond order =
Nb − Na 8 − 4
=
= 2
...
The new equivalent orbitals formed are known as the hybrid orbitals or
hybridized orbitals
...
28
Salient Features regarding Hybridisation
i)
ii)
iii)
iv)
v)
vi)
vii)
viii)
Orbitals involved in hybridization should have nearly the same energy
...
The number of hybrid orbitals formed is equal to the number of hybridizing
orbitals
...
A hybrid orbital which is taking part in bond formation must contain one
electron in it
...
The head on overlap of atomic orbitals give sigma (σ) bonds
...
1
...
1 Tips to Predict the Type of Hybridisation in a Molecule or Ion (Other
than Complex Ions)
Step 1 : Add the number of valence electrons of all the atoms present in the
given molecule/ion
...
Step 3 : (i) If the result obtained in step 2 is less than 8, divide it by 2 and find the
sum of the quotient and remainder
...
Divide the remainder R1 (if any) by 2 and find the
second quotient (Q2)
...
Let the final result obtained in (i) or (ii) be X
...
Total electrons = 17 + 1 = 18
18
2
= 2(Q1) + 2(R1) ;
= 1(Q2) + 0(R2) ; X = 2+1+0=3
8
2
∴ Hybridisation = sp2
ix) NO3- ion
Total valence electrons = 5 + 3 × 6 = 23; Charge = -1
∴ Total electrons = 23 + 1 = 24
x)
24
= 3(Q 1 ) + 0(R 1 ) ; X = 3
8
∴ Hybridisation = sp2
CO32Total valence electrons = 4 + 3 × 6 = 22; Charge = -2
∴ Total electrons = 22 + 2 = 24
24
= 3(Q 1 ) + 0(R 1 ) ; X = 3
8
∴ Hybridisation = sp2
xi) SO42Total valence electrons = 6 + 4 × 6 = 30; Charge = -2
∴ Total electrons = 30 + 2 = 32
32
= 4(Q 1 ) + 0(R 1 ) ; X = 4
8
∴ Hybridisation = sp3
xii) ICl4Total valence electrons = 7 + 7 × 4 = 35; Charge = -1
∴ Total electrons = 35 + 1 = 36
36
4
= 4(Q1 ) + 4(R 1 ) ; = 2(Q 2 ) + 0(R 2 ) ; X = 4 + 2 + 0 = 6
8
2
31
∴ Hybridisation = sp3d2
xiii) NH4+
Total valence electrons = 5 + 4 = 9; Charge = +1
∴ Total electrons in NH4+= 9 - 1 = 8
8
= 4(Q 1 ) + 0(R 1 ) ;
2
∴ Hybridisation is sp3
X = 4
Hybridisation in some Typical Molecules and Ions
Hybridisation
Examples
sp
Be F2, BeCl2, C2H2, CO2
sp2
SO2, BH3, BF3, NO2-, NO3-, CO32-
sp3
NH3, H2O, CH4, CCl4, SiCl4, H3O+,NH4+, ClO2-,
ClO3-, ClO4-,NF3
sp3d
PCl5, ClF3, SF4, XeF2
sp3d2
SF6, XeF4, XeOF4, BrF5
sp3d3
IF7, XeF6
1
...
Vander Waal (Dutch physicist, 1873) was the first to propose
the existence of attractive forces between the atoms of inert gases with fully filled
orbitals
...
The attractive interactions between the molecules are responsible for
bringing the molecules close together
...
The magnitude
of these forces is maximum in the solids and decreases on passing from solid to
liquids and from liquid to gaseous state
...
These forces are
purely electrostatic and thus physical in nature
...
Hydrogen bonding comes into existence as a result of
dipole-dipole interactions between the molecule in which hydrogen atom is
covalently bonded to a highly electronegative atom
...
ii) small size of the atom bonded to hydrogen so that it is able to attract the
bonding electron pair effectively
...
Only nitrogen, oxygen and fluorine form strong hydrogen bonds because
they have high value of electronegativity and small atomic size
...
It is a weak bond because it is merely an electrostatic
force and not a chemical bond
...
Since electronegativity of F > O >
N, the strength of H- bond is in the order H - F
...
H > HN
...
Hydrogen bonds are much weaker than covalent bonds
...
Types of Hydrogen bonds
There are two different types of hydrogen bonds as :
i) Intermolecular hydrogen bonding
...
Some examples
of the compounds exhibiting intermolecular hydrogen bonds are :
δ+ δ1
...
In the solid state, hydrogen fluoride consists
of long zig-zag chains of molecules associated by hydrogen bonds as shown
below :
F
F
F
H
H
H
H
Therefore, hydrogen fluoride is represented as (HF)n
...
Water H
...
The oxygen atom
due to its higher electronegativity acquires partial negative charge and the two
hydrogen atoms acquire partial positive charge
...
Each oxygen atom is tetrahedrally surrounded by four
hydrogen atoms as shown below :
Hydrogen bonding in water results in a hydrogen bridge (H-O-H) network
extending in three dimensions and the associated water molecule may be expressed
as (H2O)n
...
This type of bond is formed
between hydrogen atom and N, O or F atom of the same molecule
...
Intramolecular hydrogen bonding is possible when a six or
five membered rings can be formed
...
Importance of H-bonding
i)
Life would have been impossible without liquid water which is the result of
intermolecular H-bonding in it
...
iii) The cotton, silk or synthetic fibres also own their rigidity and tensile strength
to hydrogen bonding
...
v) Hydrogen bonding also exists in various tissues, organs, skin, blood and
bones
...
Choose the correct answer
1
...
313
...
84 to which value ‘n’ corresponds
n2
4
b)
3
c) 2
d) 1
En = −
Dual character of an electron was explained by
a) Bohr
3
...
b) Heisenberg
ë=
mv
h
b)
λ = hmv
c)
ë=
hv
m
d)
The value of Bohr radius for hydrogen atom is
a) 0
...
529 × 10-6 cm
b) 0
...
529 × 10-12 cm
35
ë=
h
mv
5
...
If the energy of an electron in the second Bohr orbit of H-atom is -E, what
is the energy of the electron in the Bohr’s first orbit?
a) 2E
b) -4E
c)
-2E
d)
4E
7
...
5
b) 1
c)
3
d)
2
The hybridisation in SF6 molecule is
a) sp3
b) sp3d2
sp3d
d)
sp3d3
a) −
8
...
4ð 2 me 4
n 2h 2
b) −
2ð 2 me 2
n 2h 2
c) −
c)
10
...
Answer in one or two sentences
11
...
State Heisenberg’s uncertainty principle
...
What is the significance of negative electronic energy?
14
...
15
...
Why He2 is not formed?
17
...
Define hybridisation
...
Answer not exceeding 60 words
19
...
20
...
What is its significance?
36
21
...
22
...
23
...
Summary
This chapter explains the dual nature of matter
...
Heisenberg uncertainty principle is explained
...
Molecular orbital theory and its application to certain homo diatomic and
hetero diatomic molecules are discussed
...
Different types of forces exist between
molecules are explained
...
S
...
2)
Selected topics in Inorganic Chemistry - V
...
D
...
D
...
- 2002
...
D
...
- 1977 and 5th Ed
...
37
2
...
( To learn the calculation of atomic and ionic radii using different
parameters
...
( To know how to explain the factors affecting ionisation potential with
specific examples
...
( To analyse the various scales of electronegativity values using simple
relations
...
1 REVIEW OF PERIODIC PROPERTIES
Repetition of properties of elements at regular intervals in the periodic table
is called periodicity in properties
...
Some of the properties are briefly
reviewed
...
1
...
As we move from left to right across a period, there is regular decrease in
atomic and ionic radii of the elements
...
On
moving down a group both atomic and ionic radii increase with increasing atomic
number
...
2
...
2 Ionisation Energy (Ionisation Potential)
The energy required to remove the most loosely bound electron from an
isolated atom in the gaseous state in known as Ionisation Energy
...
In a period, the value of ionisation potential increases from left to right with
breaks where the atoms have somewhat stable configurations
...
2
...
3 Electron affinity
Electron affinity or electron gain enthalpy is the amount of energy released
when an isolated gaseous atom accepts an electron to form a monovalent gaseous
anion
...
Electron gain enthalpies generally decrease on moving down the group
...
1
...
In a period, electronegativity increases on moving from left to right
...
In a group, electronegativity decreases on moving down the group
...
2
...
5 Anomalous periodic properties
A few irregularities that are seen in the increasing values of ionisation potential
along a period can be explained on the basis of the concept of half-filled and
completely filled orbitals
...
2 CALCULATION OF ATOMIC RADIUS (COVALENT RADIUS)
Atomic radius is the distance from the centre of the nucleus to the point
where the electron density is effectively zero
...
Homonuclear diatomic molecules
In case of homonuclear diatomic molecules of A2 type (e
...
F2, Cl2, Br2, I2
...
) the bond length, d(A-A) is given by
d(A – A) =
r(A) + r(A)
d(A – A) =
2 × r(A)
r(A)
=
d(A–A)
______
2
The above equation shows that in the case of homonuclear diatomic molecule
of A2 type, the covalent radius of an atom A, r(A) is equal to one half of the internuclear distance, d(A-A)
...
40
Example
1
...
98Å
...
d(Cl − Cl) 1
...
99Å
2
2
Diamond
The value of d(C–C) distance as found experimentally in a variety of saturated
hydrocarbons is 1
...
r(C) =
Thus
b
...
54
=
= 0
...
Example
i)
CCl4 molecule
The experimental value of d(C – Cl) is 1
...
76 – r(Cl)
Thus the covalent radius of carbon atom can be calculated by subtracting
the covalent radius of Cl atom from d(C–Cl) bond length
...
ii)
SiC
The experimental value of d(Si-C) is 1
...
Thus,
d(Si – C) =
r(Si) + r(C)
41
=
=
=
=
d(Si – C) – r(C)
1
...
93 – 0
...
77 Å]
1
...
Molecule
i)
ii)
iii)
iv)
v)
Bond
H2
F2
Cl2
Br2
H3C-CH3
Bond length (Å)
H–H
F–F
Cl–Cl
Br–Br
C–C
0
...
44
1
...
28
1
...
2
...
In each
ionic crystal the cations and anions are isoelectronic with inert gas configuration
...
i) The cations and anions of an ionic crystal are assumed to be in contact
with each other and hence the sum of their radii will be equal to the inter nuclear
distance between them
...
i
...
r(C + ) á
1
Z (C + )
(2)
r(A − ) á
1
Z (A − )
(3)
*
*
where,
Z*(C+) & Z*(A–) are the effective nuclear charges of cation (C+) and anion
(A-) respectively
...
Slater rules
The value of screening constant (S) and effective nuclear charge (Z*) can
be calculated by using Slater’s rules
...
i) Write down the complete electronic configuration of the element and
divide the electrons into the following orbital groups starting from the inside of
the atom
...
etc
...
For this
calculation add up the contributions to S for the other electrons according to the
following rules
...
35
(or 0
...
85
1
...
Solution
The electronic configuration of K atom is
K 19 = (1s2) (2s2 2p6) (3s2 3p6) 4s1
Effective nuclear charge (Z*) = Z – S
Z* = 19
= 19
– [(0
...
of electrons in (n –1)th shell) +
(1
...
85 × (8) + (1
...
20
Example 2
Calculate the effective nuclear charge of the last electron in an atom whose
configuration is 1s2 2s2 2p6 3s2 3p5
Z = 17
Z* = Z – S
= 17 – [(0
...
of other electrons in nth shell)
+ (0
...
of electrons in (n –1)th shell)
+ (1
...
35 × 6) + (0
...
9 = 6
...
The internuclear
distance between K+ an Cl- ions are found to be 3
...
Solution
r(K+) + r(Cl–) = d(K+–Cl–) = 3
...
The effective nuclear charge
for K+ and Cl– can be calculated as follows
...
35 × 7) + (0
...
25 = 7
...
35 × 7) + (0
...
25 = 5
...
75
=
=
=0
...
75
r(K+) = 0
...
74 r(Cl-) + r(Cl–) = 3
...
74 r(Cl–) = 3
...
14Å
= 1
...
74
45
(3)
From (2)
r(K+) = 0
...
74 × 1
...
33 Å
r(K+) = 1
...
81 Å
Ionisation potential
Ionisation energy of an element is defined as the amount of energy required
to remove the most loosely bound electron from isolated neutral gaseous atom in
its lowest energy state
...
Successive ionisation potentials
In addition to first ionisation potential (I1) defined above, second, third
...
ionisation potentials are also known
...
e
...
e
...
Each successive ionization potential or energy is greater than the previous
one, since the electron must be removed against the net positive charge on the
ion
...
The ionization energy depends upon
the following factors:
(a) Size of atom or ion
...
The larger the size of atom, lesser is the ionization energy
...
e
...
Hence ionization energy is
lower for larger atoms and higher for smaller atoms
...
E of Be (At
...
4) is greater than that of Li (At
...
3) because the
nuclear charge of Be (Z=4) is greater than Li(Z=3)
...
Hence, the first I
...
of Be is than that of Li
...
E
...
This can be explained
as follows:
Boron atom (Z = 5; 1s2 2s2 2px1 2py0 2pz0) is having one unpaired electron
in the 2p-subshell
...
As the fully filled 2s-subshell in Be-atom is more stable than B-atom due to
symmetry, more energy would be needed to remove an electron from Be-atom
...
E
...
The I
...
of carbon (At
...
6) more than that of boron (At
...
5)
Reason: Carbon (Z = 6; 1s2 2s2 2px1 2py1 2pz0) is having more nuclear
charge than boron (Z = 5; 1s2 2s2 2px1 2py0 2pz0)
...
Carbon is having more nuclear charge
than boron
...
Thus, first I
...
of carbon would be more than that of
boron
...
The higher the nuclear charge of
protons in the nucleus, the higher is the ionization energy
...
For instance, magnesium has higher nuclear
charge (12 protons) as compared to sodium (11 protons)
...
Similarly the I
...
of fluorine is more than that of oxygen
...
(i) F (Z = 9; 1s2 2s2 2px2 2py2 2pz1) is having more nuclear charge than
oxygen (Z = 8; 1s2 2s2 2px2 2py1 2pz1)
...
As fluorine is having more nuclear charge
than oxygen, it means that the nucleus of fluorine will attract the outer 2p-electrons
more firmly than oxygen
...
E
...
(c) Effect of number of electrons in the inner shells
...
The attractive force exerted by the nucleus on the most loosely
bound electron is atleast partially counterbalanced by the repulsive forces exerted
by the electrons present in the inner shells
...
Thus, the electron in
the valence shell experiences less attraction from the nucleus
...
This is another reason why ionization energy decreases in
moving down a group
...
The shape of orbital also influences the
ionization potential
...
The electronic configuration of magnesium is [Ne]3s2 and that of aluminium is
[Ne] 3s2 3p1
...
But it is easier to remove the p electron than
the s-electron
...
(e) Effect of arrangement of electrons
...
As the noble gases have the
stablest electronic arrangements, they show maximum ionization energy
...
E of Ne is greater than that of F
...
Greater
the nuclear charge, greater would be the force of attraction between nucleus and
outermost electron
...
E
...
Electron Affinity or Electron gain enthalpy (E
...
)
The electron affinity of an element may be defined as amount of energy
which is released when an extra electron enters the valence orbital of an isolated
neutral atom to form a negative ion
...
Thus, ionisation potential measures the
tendency of an atom to change into a cation (M → M+ + le-) whereas the electron
affinity measures the tendency of an atom to change into anion (X + e- → X-)
...
As more than one electron can be
introduced in an atom, it is called second electron affinity for the addition of two
electrons and so on
...
A
...
A
...
For example, the overall E
...
for the formation of oxide or sulphide
ions has been found to be endothermic to the extent of 640 and 390 kJ mol-1
respectively
...
This is so due to the accommodation
of the electron in the higher p-orbital (Zn = -87 kJ mol-1, Cd = -56 kJ mol-1)
...
The reason for this is
that by picking up an electron halogens attain the stable noble gas electronic
configuration
...
49
Change of Electron Affinity along a Group
...
Consequently the atom will possess less tendency to
attract additional electrons towards itself
...
In case of halogens the decrease in electron
affinity from chlorine to iodine is due to steady increase in atomic radii from
chlorine to iodine
...
Thus, the electron
affinity of Cl should be less than F
...
The reason for this is probably
due to small size of fluorine atom
...
The
repulsive forces between electrons results in low electron affinity
...
As these atoms possess ns2np6
configuration in their valence shells, these are stablest atoms and there are no
chances for the addition of an extra electron
...
Electron affinities of beryllium and nitrogen are almost zero
...
As these are stable electronic configurations, they
do not have tendency to accept electrons and therefore, the electron affinities for
beryllium and nitrogen are zero
...
On moving across a period,
the size of atoms decreases and nuclear charge increases
...
Consequently, the atom will possess a greater tendency to attract the additional
electron, i
...
, its electronic affinity would increase as we move from left to right
...
Of all the metals, the E
...
of gold is comparatively high (222
...
This value may be attributed to the higher effective nuclear charge and poor
shielding of the nucleus by d electrons
...
As the size of
atom increases, the effective nuclear charge decreases or the nuclear attraction
for adding electron decreases
...
Therefore
...
In general, electron affinity decreases in going down the group and increases
in going from left to right across the period
...
(2) Shielding or Screening Effect
Electron affinity á
1
Shielding effect
Electronic energy state, lying between nucleus and outermost state hinder
the nuclear attraction for incoming electron
...
(3) Electronic Configuration - The electronic configurations of elements
influence their electron affinities to a considerable extent
...
This is because their atoms have
stable ns2 np6 configuration in their valence shell and there is no possibility for
addition of an extra electron
...
This is attributed to extra stability of the fully completed s-orbitals in them
...
Example 1
The electron affinities of Be, Mg and N are almost zero because both
Be (Z = 4; 1s2 2s2) and Mg (Z = 12; 1s2 2s2 2p6 3s2) are having s orbital fully
51
filled in their valence shell
...
Therefore, these elements would be having least tendency to accept electron
...
N (Z = 7 ; 1s2 2s2 2px1 2py1 2pz1) is having half filled 2p-subshell
...
Thus, nitrogen is having
least tendency to accept electron
...
Example 2
Electron affinity of fluorine is less than that of chlorine
...
71 Å) is very small
and has only two shells, i
...
, n = 1, 2 (9F = 1s2 2s2 2px2 2py2 2pz1)
...
There occurs repulsion among electrons of the valence shell and also with electron
to be added
...
ii) Because of small size of fluorine there occurs large crowding of electrons
around the nucleus
...
Because of this,
effective nuclear charge gets decreased
...
Hence electron affinity gets decreased
...
5 ELECTRONEGATIVITY SCALES
Electronegativity scale is an arbitary scale
...
Most commonly used scales are
1
...
Mulliken’s scale
3
...
Alfred and Rochow’s scale
Pauling’s scale (1932)
This scale is based on an empirical relation between the energy of a bond
and the electronegativities of bonded atoms
...
Let the bond energies of A-A, B-B and A-B bonds be represented as
EA-A, EB-B and EA-B respectively
...
e
...
208
Δ
E A − B − E A − A × E B− B
(X A − X B ) 2
= XA – XB
Here, XA and XB are the electronegativities of A and B respectively
...
208 arises from the conversion of Kcals to electron volt
...
1, Pauling
calculated electronegativities of other elements with the help of this equation
...
Problem
Calculate the electronegativity of chlorine from the following data
EH-H = 104 K cal mol-1 ; ECl-Cl = 36 K cal mol-1
EH-Cl = 134 K cal mol-1
According to Pauling’s equation
Ä
=
E HCl − E H − H × E Cl − Cl
Ä
=
=
134 − 104 × 36
134 – 61
...
82
53
⇒
=
X Cl − X H
X Cl − X H
=
=
=
=
=
0
...
82
1
...
77
1
...
1
3
...
1
XCl
2
...
208 Ä
[
XH = 2
...
According to this
method electronegativity could be regarded as the average of the ionization energy
and electron affinity of an atom
Electronegativity =
I
...
+ E
...
8 times higher than Pauling values
...
48 kJ mol-1
...
E
...
A
I
...
+ E
...
8 × 96
...
Disadvantage
Although Mulliken’s scale is less empirical than Pauling Scale, yet it suffers
from a serious disadvantage that electron affinities with the exception of a few
elements are not reliably known
...
4 eV/atom (Electron affinity) F =
3
...
0 ev/atom and (EA)Cl = 4
...
8
=
Electronegativity of fluorine
17
...
62
21
...
75
5
...
6
=
(IP) Cl + (EA) Cl
2 × 2
...
0 + 4
...
6
5
...
03
Nature of bond
The concept of electronegativity can be used to predict whether the bond
between similar or dissimilar atoms is non-polar covalent bond, polar covalent
bond (or) ionic bond
...
e
...
eg
...
ii)
When XA is slightly greater than XB, i
...
XA - XB is small, the A-B bond is
polar covalent bond and is represented as Aδ--Bδ+
...
g
...
iii)
When XA >> XB, i
...
, XA - XB is very large, A-B bond is more ionic or
polar bond and is represented as A--B+, Since XA >> XB
...
55
2
...
7, the amount of ionic character in Aδ-–Bδ+ bond is
50% and that of covalent character is also 50%
...
(ii) When (XA–XB) < 1
...
Thus Aδ-–Bδ+
bond is predominantly covalent and hence is represented as A–B
...
7, the amount of ionic character in Aδ-–Bδ+ bond is
more than 50% and that of covalent character is less than 50%
...
SELF EVALUATION
A
...
2
...
4
...
6
...
34Å
b) 1
...
54Å
d) 1
...
8
...
10
...
12
...
14
...
B
...
17
...
74
b) 1
...
98
d) 2
...
a) Li, Be, B
b) N, O, F
c) C, N, O, F
57
18
...
19
...
Which element of the following pairs of elements has higher ionisation energy?
Justify your answer
a) K or Ca b) Be or B c) I or Ba d) F or Cl
e) N or O
21
...
Justify your answer
...
Answer the following questions
a) Which element has the most positive value of electron affinity?
b) Which element has low electronegativity?
23
...
24
...
Answer not exceeding 60 words
25
...
Explain Pauling method to determine ionic radii
...
Explain the variation of IE along the group and period
...
Explain the various factors that affect electron affinity
...
How electronegativity values help to find out the nature of bonding between
atoms?
Summary
The periodicity in properties of elements are reviewed briefly
...
The factors
governing ionisation potential are explained with specific examples
...
The different scales of
electronegativity values are explained in detail
...
D
...
2) Selected Topics in Inorganic Chemistry, Wakid V
...
D
...
D
...
58
3
...
( Understands the preparation, properties and the uses of potash alum
...
( Understand the extraction of lead, properties and the uses of lead
...
( Recognize the anomalous behavior of Fluorine
...
( Know about the isolation of noble gases
...
( Appreciates the application of noble gases
...
There are 44 main group elements
...
p-block elements play dominant part in all natural processes
...
Carbon is the backbone of all organic
compounds
...
Nitrogen acts as a building
block of life
...
60
General characteristics of p-block elements
1
...
2
...
Most of them form covalent compounds
...
These elements possess relatively higher ionisation energy and the value
tends to increase along the period but decrease down the group
...
Most of the elements show negative (except some metals) as well as positive
oxidation states (except Fluorine)
...
One of the familiar characteristic of p-block elements is to show inert pair
effect i
...
the tendency of being less availability for ns electron in bonding
...
Group 13 elements – The Boron family
The group 13(IIIA) elements are Boron, aluminium, gallium, indium and
thallium
...
001% of the
earth’s crust by mass
...
Ø
Gallium is remarkable for its unusually low melting point (29
...
Its most important
use is in making gallium arsenide
...
Ø
Indium is also used in making semi conductor devices, such as transistors
and electrical resistance thermometers called thermistors
...
61
3
...
Table 3
...
1
...
Al2 (SO4)3
...
From Alunite: Alunite or alum stone is K2SO4
...
4Al(OH)3
...
When a little more potassium sulphate in
calculated amount is added, the alum is crystallised
...
Potash alum is a white crystalline solid
...
It is soluble in water but insoluble in alcohol
...
The aqueous solution is acidic due to the hydrolysis of Al2 (SO4)3
...
When heated, it melts at 365K and on further heating loses the whole of its
water of crystallisation and swells up
...
Uses
1
...
2
...
62
3
...
Ø
Carbon is an essential constituent of the molecules on which life is based
...
Ø
Both silicon and germanium are used in making modern solid – state electronic
devices
...
General Trends
Electronic configuration: The elements of this group possess ns2 np2
electronic configuration
...
2 Electronic Configuration of Group 14 elements
63
3
...
1 Silicones – structure and uses
The silicones are a group of organosilicon polymers
...
The complete hydrolysis of SiCl4 yields silica SiO2, which has a very stable
three-dimensional structure
...
S
...
R
Cl
Si
R
Si
R
OH
Si
Cl
R
R
HO
R
H 2O
Si = 0
OH
R
R
OH + HO
R
Si
R
OH
HO
R
Si
R
O
R
Si
OH
R
The starting materials for the manufacture of silicones are alkyl-substituted
chlorosilanes
...
R
R
Si
R
O
Si
R
R
R
The dialkyldichlorosilane R2SiCl2 on hydrolysis gives rise to straight chain
polymers and, since an active OH group is left at each end of the chain,
polymerisation continues and the chain increases in length
...
|
|
O
O
R
Si
O
Si
R
O
O
R
Si
O
Si
R
O
O
Uses
1)
Silicones act as excellent insulators for electric motors and other appliances
as they can withstand high temperatures
...
They
are water repellent because of the organic side group
...
3)
Silicone rubber retain their elasticity even at low temperatures and resist
chemical attack
...
4)
Silicone resins, a cross-linked polymer used as non-stick coating for pans
and are used in paints and varnish
...
Hence used
for high temperature oil bath, high vacuum pump etc
...
2
...
2
...
4
...
Galena
contains lead sulphide and small quantities of silver
...
Concentration: The ore is concentrated by froth floatation process
...
Smelting in a Reverberatory furnace: The concentrated ore is roasted
in a reverberatory furnace at a moderate temperature
...
During roasting, galena is partly oxidized
to lead monoxide and partly to lead sulphate
...
3
...
The temperature is raised and simultaneously
the air supply is reduced
...
PbS+2PbO
PbS+PbSO4
→ 3Pb+SO2
→ 2Pb+2SO2
Thus in this process roasting and smelting are carried out in the same furnace,
at two different temperatures
...
Lead is
recovered from the slag by heating with lime and powdered coke
...
It is refined by the following processes
...
Liquation
The impure metal is heated on a sloping hearth
...
The infusible impurities remain on the hearth
...
Desilverisation
Silver is removed by either Pattinson’s process or Park’s process
...
Electrolytic refining
Very pure lead is obtained by this process
...
67
Physical properties
1
...
3
...
Lead is a bluish grey metal with a bright luster
...
It is not a good conductor of heat and electricity
...
Chemical properties
1
...
When heated in air or oxygen, lead is oxidized to litharge (PbO) and red
lead (Pb3O4)
ii)
2Pb + O2 → 2PbO
3Pb + 2O2 → Pb3O4
2
...
This phenomenon is called Plumbo solvency
...
Action of acids
i)
ii)
Dilute H2SO4 and HCl have no action on lead
...
H2SO4 liberates SO2 but the reaction is retarded by the formation
of an insoluble layer of lead sulphate
...
2
...
4
...
6
...
Problem
An element A belongs to 14th group and occupies period number 6
...
HCl
...
A is used to prepare C which is used
as an antiknock in automobiles
...
Solution
1
...
2
...
HCl gives B
Pb + 4 HCl
→ H2PbCl4 + H2
∴ Compound B is chloroplumbic acid
...
Compound C is tetraethyl lead
...
3 GROUP – 15 ELEMENTS – THE NITROGEN FAMILY
The group 15 (VA) elements are nitrogen, phosphorus, arsenic, antimony
and bismuth
...
It makes up 78% of the earth’s atmosphere by volume
...
10% of the mass of the earth’s crust
...
Ø
Bismuth is a silvery solid
...
Ø
The natural abundance of As, Sb and Bi in the earth’s crust is relatively low
...
Table 3
...
3
...
Phosphorus chlorides are more important
...
I
...
P4 + 6Cl2 →
4PCl3
Dry white phosphorus is placed in the retort and gently heated on a water bath
...
The phosphorus
trichloride formed being volatile distils over and is collected in a water cooled
receiver
...
3
...
This is removed by distilling the PCl3 over white phosphorus
...
2
...
Colourless low boiling liquid
It fumes in moist air
It has pungent odour
...
It is violently hydrolysed by water giving phosphorus acid and hydrochloric
acid gas
...
PCl3 + 3CH3COOH
Acetic Acid
PCl3 + 3C2H5OH →
Ethyl alcohol
2
...
71
PCl3 + Cl2 →
PCl5
PCl3 + SO2Cl2 → PCl5 + SO2
3
...
It reacts with SO3 to form phosphorus oxychloride and SO2
SO3 + PCl3 → POCl3 + SO2
Structure: PCl3 molecule has a pyramidal shape, which arises from sp3
hybridisation of phosphorus atom
...
P
Cl
Cl
Cl
Cl
xx
x Px
x
Cl
Cl
x - E lectro n o f p
- E lectro n o f C l
II
...
PCl3 + Cl2 → PCl 5
Physical properties
1
...
Phosphorus pentachloride is a yellowish white crystalline solid
...
Chemical properties
1
...
PCl5
2
...
72
insufficient water
PCl5 + H2O
POCl3 + 2HCl
Excess of water
PCl5 + 4H2O
3
...
In all these cases, the hydroxyl group is replaced by chlorine
...
H3PO4 + 5HCl
→ C2 H5Cl + POCl3 + HCl
Ethyl Chloride
It reacts with metals on heating to give corresponding chlorides
...
Cl
Cl
Cl
x
x
Cl
Cl
x
P
x
x
P
Cl
Cl
Cl
x - E lectro n o f p
- E lectro n o f C l
Cl
Cl
b)
Oxides of phosphorus
I
...
4P + 3O2 → 2P2O3
Physical properties
1
...
It has a garlic odour
...
It reacts with cold water, gives phosphorus acid
...
It reacts with hot water vigorously to form inflammable phosphine
...
Phosphorus pentoxide P2O5 or P4O10
Phosphorus pentoxide can be prepared by burning phosphorus with sufficient
supply of air
...
Chemical properties
1
...
P4O10 + 2H2O → 4HPO3
When the solution is boiled, the metaphosphoric acid is changed to
orthophosphoric acid
...
Phosphorus pentoxide extracts water from many inorganic compound
including sulphuric acid, nitric acid and several organic compounds
...
P4O10
H2SO4
SO3
- H2O
P4O10
2HNO3
N2O5
- H2O
P4O10
RCONH2
Amide
-H2O
RCN
Nitrile
Use: It is used as a dehydrating agent
...
Phosphorus acid - H3PO3
It is prepared by the action of cold water on phosphorus (III) oxide or
phosphorus (III) chloride
...
Chemical Properties
1
...
H3PO3 + NaOH →
H3PO3 + 2NaOH
NaH2PO3 + H2O
Sodium dihydrogen Phosphite
→ Na2HPO3 + 2H2O
Disodium hydrogen Phosphite
2
...
Δ
4H3 PO3 → 3H3PO4 + PH3
3
...
It reduces silver
nitrate solution into silver
...
Ortho phosphoric Acid, H3PO4
Preparation
1
...
P2O5 + 3H2O
→
2H3PO4
2
...
Iodine acts as a catalyst
...
P+5HNO3 → H3PO4 +5NO2 +H2O
Physical properties
1
...
2
...
Chemical properties
1
...
It combines with alkalies like NaOH to form three series
of salts
...
On heating it gives pyrophosphoric acid at 523 K and at 589 K gives
metaphosphoric acid
523K
H3PO4
3
...
H3PO4 + 3AgNO3 → Ag3PO4+3HNO3
76
Uses
1
...
3
...
It is used in the preparation of HBr and HI as a substitute for sulphuric acid
...
It is used in the preparation of phosphate salts of sodium, potassium and
ammonium
...
Structure
Being a tribasic acid, the structure of phosphoric acid is represented as
O
H
O
x x
O x P x O
x
O
H
H
x - E lectro n o f P
- E lectro n o f O
- E lectro n o f H
P
HO
OH
OH
III
...
Pyrophosphoric acid, H4 P2 O7
Preparation: Pyrophosphoric acid is prepared by heating orthophosphoric
acid to 523 K – 533 K
...
Chemical Properties
1
...
When heated strongly, it yields metaphosphoric acid
H4P2O7
2ΗPO3 + H2O
77
Structure: The Structure of pyrophosphoric acid is represented as:
O
x x
H
O
x P
x
O
O
x x
x
O x
H
d)
P
x
O
O
O
O
x
H
HO
P
O
OH
P
OH
OH
H
Phosphine - PH3
Phosphine is the best known hydride of phosphorus
...
4P + 3NaOH + 3H2O
→ PH3 + 3NaH2PO2
Sodium hypophosphite
Phosphine so obtained is impure
...
PH4I is heated with KOH or NaOH, pure
phosphine is obtained
...
Chemical properties
1
...
4PH3
2
...
4PH3 + 8O2 →
P4O10 + 6H2O
78
3
...
PH3 + 3Cl2
→
PCl3 + 3HCl
PH3 + 4Cl2
→
PCl5 + 3HCl
4
...
When it is passed
through the salt solutions, corresponding metal is formed
...
Smoke screens
When PH3 burns it produces smoke which is dense enough to serve as
smoke screens
...
Holme’s signal : Containers which have a perforated bottom and a hole at
the top are filled with calcium phosphide and calcium carbide
...
Water enters the container through the bottom and reacts with calcium
carbide and calcium phosphide to give acetylene and phosphine
...
Thus a bright red flame is produced which is accompanied by huge smoke due to
the burning of phosphine
...
Ca3P2 + 6H2O → 2 PH3 ↑
CaC2
+ 3Ca(OH)2
+ 2H2O → C2H2 ↑ + Ca(OH)2
Problem
An element ‘A’ occupies group number 15 and period number 3 reacts
with chlorine to give B which further reacts with chlorine to give C at 273 K
...
C is a better
chlorinating agent because it chlorinates metals also
...
B has a pyramidal shape
...
Identify the element A and the compounds B and C
...
1
...
Therefore A is phosphorus
...
Therefore compound B is phosphorus trichloride and it has a
pyramidal shape
...
PCl3 further reacts with Cl2 to give PCl5
...
PCl3 + Cl2
3
...
So, both
reacts with C2H5OH gives C2H5Cl
...
PCl5 is a better chlorinating agent
...
PCl5 + 2Cu
5
...
PCl3 + SO 3
→
POCl3 + SO2
3
...
The first four elements are collectively called CHALCOGENS or ore
forming elements, because many metal ores occur as oxides and sulphides
...
It constitutes 46
...
Ø
Sulphur is less abundant and occurs in free and combined states
...
Ø
Polonium, a radioactive element that occurs in trace amounts in uranium
ores
...
General Trends
Electronic configuration
All of these elements have ns2np4 configuration, just two electrons short of
an octet configuration
...
4 Electronic configuration of Group 16 elements
Element
Atomic
Number
Electronic Configuration
Group
Number
Period
Periodic
Number
Numbe
Oxygen
8
[He] 2s22p4
16
2
Sulphur
16
[Ne] 3s23p4
16
3
Selenium
34
[Ar] 3d10 4s2 4p4
16
4
Tellurium
52
[Kr] 4d10 5s2 5p4
16
5
Polonium
84
[Xe] 4f14 5d10 6s2 6p4
16
6
3
...
These are collectively known as HALOGENS
...
Because most of them exist in Sea water
...
Table 3
...
Oxidising power: An important feature of the halogen is their oxidising
property which is due to high electron affinity of halogen atoms
...
Fluorine is the strongest oxidising agent
...
F + 2 X– → 2F – + X (X– = Cl –, Br –, I–)
2
2
Halogen of low atomic number oxidises the halide ion of higher atomic number
...
Solubility: Halogens, being non-polar molecules, do not dissolve to a
considerable extent in a polar solvent like water
...
2F2 + 2H2O
→ 4HF + O2
3F2 + 3H2O
→ 6HF + O3
Chlorine, bromine and Iodine are more soluble in organic solvents such as
CCl4, CHCl3 and produce yellow, brown and violet colour
...
Hydrides of the Halogens (Hydrogen halides):
i)
All halogens react with hydrogen to form volatile covalent hydrides of formula
HX
...
iii)
The activity of halogens towards hydrogen decreases from fluorine to iodine
...
It combines
with chlorine in the presence of sunlight and with bromine on heating
...
iv) Hydracids are the reducing agents
...
HF is a liquid because of inter
molecular hydrogen bonding
...
H –F
...
H–F
vi) The acidic character of HX are in the following order
...
82
Anamalous Nature of Fluorine
1
...
This is due to the
minimum value of F–F bond dissociation energy
...
Fluorine decomposes cold dilute alkalies liberating OF2 and with conc
...
Under similar conditions, the other halogens will give rise to
the hypohalites and halates respectively
...
It has the greatest affinity for hydrogen, forming HF which is associated
due to the hydrogen bonding
...
...
H– F
...
4
...
NaF and NaHF2
...
The salts of HF differ from the corresponding salts of other hydracids
...
6
...
7
...
8
...
Thus we have Ι3 –, Br3–, Cl3– ions but no F3– ion
...
CaF2 + H2SO4→ CaSO4 + 2HF
...
Aqueous HF thus obtained is stored in wax bottles
...
Na2 SiO3 + 6HF → Na2SiF6 + 3H2O
SiO2 + 4HF
→ SiF4 + 2H2O
83
The action of hydrofluoric acid on silica and silicates is used for etching
glass
...
The design to be
etched is now drawn on the waxed surface and is then exposed to the action of
hydrofluoric acid
...
The wax is finally
washed off with turpentine
...
5
...
It occurs in the combined form
...
In this fluorine is prepared by the electrolysis of fused sodium or potassium
hydrogen fluoride (perfectly dry) Electrolysis is carried out between graphite
electrodes in a V-shaped electrically heated copper tube
...
The copper tube is thickly lagged to prevent loss of heat
...
3
...
This removes the hydrogen fluoride vapours coming with fluorine
...
2
...
Fluorine is a gas and has pale greenish yellow colour
...
It is heavier than air
...
1
...
H2 + F2 → 2HF
2
...
C + 2F2 → CF4
Tetra fluoromethane
Si + 2F2 → SiF4
Silicon tetrafluoride
2P + 5F2 → 2PF5
Phosphorus pentafluoride
3
...
2Ag + F2 → 2AgF
2Al + 3F2 → 2AlF3
4
...
Br2 + 3F2 → 2Br F3
Ι2 + 5F2 → 2 ΙF5
85
Uses
1
...
These non-toxic, non-combustible and volatile liquids are used as
refrigerants in refrigerators, deep freezers and air conditioners
...
2
...
3
...
4
...
5
...
6
...
INTERHALOGEN COMPOUNDS OR INTERHALOGENS
Each halogen combines with another halogen to form several compounds
known as interhalogen compounds
...
In naming also, the less electronegative element is mentioned first
...
AX
AX3
AX5
AX7
CIF
BrF
BrCl
ICl
IBr
ClF3
BrF3
ICl3
BrF5
IF5
IF7
They can all be prepared by direct combination or by the action of a halogen
on a lower interhalogen, the product formed depends on the conditions
...
The interhalogens are generally more reactive than the halogens (except F)
because the A-X bond is weaker than the X–X bond in the halogens
...
Hydrolysis gives halide and oxyhalide ions,
the oxyhalide ion being formed from the larger halogen present
...
1
...
As excepted, the compounds of the type AX are linear
...
Electronic structure of Chlorine atom, in the ground state and hybridised
state is represented as in Fig
...
4
...
4 Linear structure of the interhalogen compounds
of the type AX
87
Although the spatial arrangement of the four electron pairs (bp = 1 and
lps = 3) round the central chlorine atom is tetrahedral, due to the presence of
three lone pairs of electrons in three hybrid orbitals, the shape of AX molecule
gets distorted and become linear
...
Type AX3 Compounds of the type AX3 have trigonal bipyramidal structure,
Fig
...
5 for the ClF3 molecule
...
3
...
The three dotted arrows indicate
electrons contributed by the three fluorine atoms (without lone pair it is T-shaped)
...
3
...
3
...
Type AX5 (ΙF5, BrF5, etc
...
3
...
88
Fig
...
7 Structure of IF5
4
...
This compound has a pentagonalbipyramidal structure
Ι
since this is formed by sp3d3 hybridisation
...
3
...
A reacts with water forms a mixture of B, C and acid D
...
A also reacts with hydrogen violently even in dark to
give an acid D
...
write the reactions
...
ii) Fluorine reacts with water and forms a mixture of B and C
2F2 + 2H2O → 4HF + O2
89
3F2 + 3H2O →
6HF + O3
Therefore, B is Oxygen and C is Ozone
...
F2 + H2 → 2HF
D is Hydrofluoric acid
...
6 GROUP 18 NOBLE GASES OR INERT GASES
Group 18 of the periodic table consists of helium, neon, argon, krypton,
xenon and radon
...
All of them (except Rn) are present in air in traces
...
On account of their very minute quantities in atmosphere, they were named
as rare gases
...
A number of xenon compounds and two krypton fluorides were prepared
and thus they were named as noble gases
...
The differentiating electron
enters into p-sub shell and thus are included in p-block elements
...
6 Electronic Configuration of Group 18 elements
Element
Atomic
Number
Electronic Configuration
Group
Number
Period
Periodic
Number
Numbe
Helium
2
1s2
18
1
Neon
10
1s22s22p6
18
2
Argon
18
1s22s22p63s2 3p6
18
3
Krypton
36
1s22s22p63s23p63d104s2 4p6
18
4
Xenon
54
1s22s22p63s23p63d104s2
4p64d105s2 5p6
18
5
Radon
86
1s22s22p63s23p63d104s2 4p6
4d104f145s2 5p65d106s2 6p6
18
6
90
ISOLATION OF NOBLE GASES
The noble gases are isolated from air by removing oxygen and nitrogen
from air free from carbon-di-oxide, water vapour, dust particles, etc
...
In the chemical
method, the unwanted gases are removed by means of compound formation
while in the physical method, these are removed by the fractional evaporation of
liquid air
...
The second step
is to separate the various constituents from one another taking advantage of the
fact that they can be adsorbed on activated charcoal at different temperatures
...
Two platinum electrodes
are introduced and a discharge from a transformer of about 6000 - 8000 volts is
passed by the action of which nitrogen and oxygen rapidly combine to form
oxides of nitrogen
...
N2+ O2
→ 2 NO
2 NO + O2 → 2NO2
2NO2 + 2NaOH → NaNO3 + NaNO2 + H2O
Oxygen if any is removed by introducing alkaline pyrogallol in the globe
...
91
To Tr a n sfo rm e r
Soda out
G a se s in
S o d a in
Fig
...
9 Chemical method for isolation of noble gases
Step 2 Separation of noble gases (DEWAR’S METHOD)
The mixture of noble gases obtained by the above method is separated into
individual constituents by the use of coconut charcoal which adsorbs different
gases at different temperatures
...
3
...
It is allowed to
remain in contact with the charcoal for about half an hour
...
These are pumped out and collected
...
The charcoal at 173K containing argon, krypton and xenon is placed in
contact with another charcoal at the temperature of the liquid air when argon
diffuse into the other charcoal
...
173K) still containing krypton
and xenon is raised to 183K when krypton is set free while xenon remain adsorbed
in the charcoal
...
XENON FLUORIDE COMPOUNDS
Xenon forms three binary Fluorides XeF2, XeF4, and XeF6 by the direct
union of elements under appropriate experimental conditions
...
They are powerful fluorinating agents
...
For example
...
XeF2 and XeF4 have the linear and square planar structure
respectively
...
93
USES OF NOBLE GAS
(A) HELIUM
1
...
2
...
3
...
It is much less soluble in blood than N2
...
4
...
5
...
pt 4
...
6
...
(B) NEON
1
...
2
...
3
...
4
...
(C) ARGON
1
...
2
...
(D) KRYPTON AND XENON
1
...
2
...
(E) RADON
1
...
94
SELF EVALUATION
A
...
b) Si
c) Ge
d) Sn
b) NO2
c) P2 O3
d) SO3
The compound with garlic odour is
a) P2 O3
5
...
B
...
c) Ge
Which of the following is most abundant in earth’s crust?
a) C
3
...
b) trigonal bipyramidal
d) tetrahedral
The compound used as smoke screen
a) PCl3
7
...
c) PH3
Which shows only -1 oxidation state?
a) fluorine
8
...
Halogens belong to the group number
a) 14
b) 15
c) 17
95
d) 18
11
...
12
...
Which is not known?
a) XeF6
b) XeF4
c) XeO3
d) ArF6
14
...
Which of the following has highest first ionisation energy?
a) He
b) Ne
c) Ar
d) Kr
...
Answer in one or two sentences
16
...
17
...
Why?
18
...
Prove that P2O5 a powerful dehydrating agent
...
Why H2O is a liquid while H2S is a gas?
21
...
Fluorine atom is more electronegative than iodine atom yet, HF has lower
acid strength than HI
...
What are interhalogen compounds? How are they formed?
24
...
25
...
Why do noble gases form compounds with fluorine and oxygen only?
96
27
...
Answer not exceeding 60 words
28
...
Give an account of manufacture of lead
...
30
...
How fluorine is isolated from their fluorides? Describe its important properties
and uses
...
Give a detailed account of the interhalogen compounds with special reference
to the compounds involving iodine
...
33
...
Give an account of various types of compounds which are formed by xenon?
D
...
An element A belongs to 14th group is a metal, which can be cut with a
knife
...
A in pure state does
not react with water but air dissolved water forms hydroxide
...
2
...
A reacts with caustic soda a to give B which
is having rotten fish odour
...
Identity A, B and C
...
3
...
The compound B on hydrolysis gives a dibasic
acid C
...
Identify the element A, compound B, C and D
...
4
...
Element A reacts with another element B, Which
occupies group number 17 and period number 4, to give a compound C
...
Identify the elements A and B and the compound C
...
SUMMARY
Ø
Groups 13-18 of the periodic table are known as p-block elements
...
Potash alum is manufactured from
alunite
...
Silicones are organo silicon
polymers
...
Ø
Group 15 is known as nitrogen family
...
Ø
Group 16 is known as oxygen family
...
Fluorine has different behaviour than
other halogens
...
Fluorine is the most electronegative of all elements
...
Ø
Group 18 is known as rare gases or inert gases, with closed valence shell
electronic configuration exhibit low chemical reactivity
...
References
1
...
L
...
2
...
3
...
D
...
4
...
98
4
...
(To learn the general methods of extraction of metals
...
( To study about the methods of preparation of compounds, their
properties and uses
...
this is my
belief, it progresses, it grows stronger, this is worth living for, this is worth
waiting for
...
99
27
Co
40
Zr
46
Pd
110
Uun
The d-block elements are located in the middle of the periodic table and
consists of metals only
...
In these elements, the last electron enters the d orbital of the penultimate
shell i
...
the last electron goes to (n-1) d orbital
...
These elements have partly filled d-subshells in their
elementary form or in their simple ions
...
Classification of d-block Elements
Based on whether the last electron goes to 3d,4d,5d or 6d orbital, d-block
elements are classified into four series
...
100
Electronic configuration of d-block Elements
In the transition elements, d-orbitals of penultimate shell are successively
filled
...
It starts from
scandium (Z=21) and goes up to zinc (Z=30)
...
The third transition series involves filling of 5d-orbitals
...
It is followed by fourteen elements called
lanthanides which involve the filling of 4f-orbitals
...
The general electronic configuration of transition elements is
(n-1)d1-10 ns1-2
...
1 General characteristics of d-block elements
4
...
1 Atomic and Ionic Radii
The atomic and ionic radii of transition elements are smaller than those of pblock elements and larger than those of s-block elements
...
1
...
1 Atomic radii of elements of first transition series
Ele me nt
Atomic
(pm) radii
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
144
132
122
118
117
117
116
115
117
125
The atomic radii of first transition series decreases from Sc to Cr and
remains almost constant till Cu and then increases towards the end
...
These two effects oppose each other resulting in increase in nuclear
charge
...
It has been observed that Zirconium and Hafnium have almost equal
atomic radii
...
4
...
2 Metallic character
All the transition elements are metals, since the number of electrons in the
outermost shell is very small, being equal to 2
...
The presence of partially d-orbitals favours covalent bonding
...
4
...
3 Formation of coloured ions
Most of the transition metal compounds are coloured in their solid or solution
form
...
Hence very small amount of energy is required for excitation
of electrons from one energy level to the other
...
The colour observed corresponds to the complementary colour
of the light absorbed
...
2
...
Sc3+ ions are also colourless because of the absence of d-electrons
...
1
...
The catalytic activity of transition metals is due to the following reasons
...
ii) They are also capable of forming interstitial compounds which can adsorb
and activate the reacting species
...
ii) Vanadium pentoxide (V2O5) is used for catalytic oxidation of SO2 to SO3
iii) TiCl4 is employed as a catalyst in the manufacture of polythene
...
1
...
This property is due to the following reasons
...
ii)
The energies of (n – 1)d and ns orbitals are fairly close to each other
...
The elements which exhibit the maximum number of oxidation states occur
either in or near the middle of the series
...
2
...
The elements at the end of the series exhibit fewer
oxidation states, because they have too many d electrons and hence fewer
vacant d-orbitals can be involved in bonding
...
The transition elements in lower oxidation states (+2 and +3) generally form
ionic bonds and in higher oxidation state form covalent bonds
...
The highest oxidation state shown by any transition metal is +8
...
5
...
Ni(CO)4 and Fe(CO)5 are common examples
...
1
...
There are two main type of substances
...
Paramagnetic substances are the substances which are attracted by
magnetic field
...
Diamagnetic substance are the substances which are repelled by the magnetic
field
...
Greater the number of unpaired electrons in the
substance greater is the paramagnetic character, The magnetic character of a
substance is expressed in terms of magnetic moments
...
Hence
µ = 1(1 + 2)BM =
3 = 1
...
M
104
Larger the value of magnetic moment, the greater is the paramagnetic character
...
These substances are called ferromagnetic substances
4
...
7 Complex formation
The cations of d-block elements have strong tendency to form complexes
with certain molecules (e
...
CO, NO, NH3
...
g
...
etc) called ligands
...
i)
Small size and high positive charge density
...
Examples of some complex compounds are,
[Cu(NH3)4]2+, [Ag(NH3)2]+, [Fe(CN)6]4-,
...
4
...
8 Formation of alloys
Transition metals form alloys with each other
...
Eg
...
etc
...
2 FIRST TRANSITION SERIES
4
...
1 Occurrence and principles of extraction of copper
Atomic mass : 63
...
Occurrence
Copper was known to the earliest races of mankind
...
Copper
is found in the native state as well as in the combined state
...
In India, copper is mainly found in
Singhbhum (Bihar), Khetri and Darbia (Rajasthan) and in Tamilnadu
...
Cuprite or Ruby copper, Cu2O
...
The chief ore of copper is copper pyrite
...
Extraction from copper pyrites
Extraction of copper from copper pyrites involves the following steps
...
Crushing and concentration
The ore is crushed and then concentrated by froth-floatation process
...
Roasting
The concentrated ore is heated strongly in the reverberatory furnace, in
excess of air
...
The volatile impurities are removed
...
→
S + O2
P4 + 5O2 →
4As + 3O2 →
SO 2
2P 2O5
2As2O3
iv) The copper pyrite is partly converted into sulphides of copper and iron
...
Cu2S + 2FeS + SO2
2FeO + 2SO2
Smelting
The roasted ore is mixed with powdered coke and sand and is heated in a
blast furnace
...
Hot air at
800°C is introduced from the tuyers near the base of the furnace
...
2FeS + 3O2
→
2FeO + 2SO2
106
FeO
2Cu2S
Cu2O
FeO
+
+
+
+
SiO2
3O2
FeS
SiO2
→
→
→
→
FeSiO3 (fusible slag)
2Cu2O + 2SO2
Cu2S + FeO
FeSiO3 (fusible slag)
As a result of smelting, two separate molten layers are formed at the bottom of
the furnace
...
It chiefly consists of cuprous sulphide and some
unchanged ferrous sulphide
...
Bessemerisation
The molten matte is transfered to a Bessemer converter as shown in the Fig
...
2
...
Any sulphur, arsenic and antimony still present escape as their respective
oxides
...
The impure
metal thus obtained is called blister copper and is about 98% pure
...
→
→
2Cu2O
6Cu
+ 2SO 2
+ SO 2
Refining
Blister copper contains about 2% of impurities and it is purified by electrolytic
refining
...
For electrolytic
refining of copper,
i)
ii)
iii)
A block of impure copper metal acts as anode
A thin plate of pure copper metal acts as cathode
Copper sulphate solution acidified with sulphuric acid is taken as electrolyte
...
Properties
Physical properties
Copper is a reddish brown metal, with high lustre, high density and high
melting point 1356°C
...
CuCO3
(Green) Copper Carbonate
Action of Heat
Copper when heated to redness (below 1370K) in the presence of oxygen
or air, first it gets converted to black cupric oxide and further heating to above
1370K, it gets converted into red cuprous oxide
...
HCl and H2SO4
Dilute acids such as HCl and H2SO4 have no action on these metals in the
absence of air or an oxidising agent
...
HNO3
Copper reacts with dil
...
3Cu + 8HNO3(dil) → 3Cu(NO3)2 + 2NO↑ + 4H2O
108
c)
With con
...
H2SO4
Copper reacts with con
...
H2SO4 with the liberation of
NO2 and SO2 respectively
...
Cu + Cl2
v)
→
CuCl2
Action of alkalis
Copper is not attacked by alkalies
...
2
...
4
...
It is used for making utensils, containers, calorimeters, coins,
...
It is used in electroplating
...
Alloys of Copper
Alloy
% composition
Uses
i)
Brass
Cu = 60-80, Zn = 20-40 For making utensils, condenser
tubes, wires,
...
ii)
Bronze
Cu = 75-90, Sn = 10-25 For making cooking utensils,
statues, coins
...
iii)
Gun metal Cu = 87, Sn = 10, Zn = 3 For making gun barrels, gears,
castings etc
...
2
...
99
Valency : 0,1,2,3,4,5,6
Atomic number : 24
Symbol : Cr
Position in the periodic table : Period Number -4, Group Number -6
...
N
...
It was named chromium because it forms
coloured compounds [Greek word - chroma - colour]
Occurrence
Metallic chromium does not occur in the native state
...
Ores
The important ore of chromium is
Chromite or chrome ore, FeO Cr2O3
The chief ore of chromium is chromite ore
...
1
...
2
...
During this process, chromite ore is converted into soluble
sodium chromate
...
Cr2O3) + 8Na2CO3 + 7O2 (from air)
↓ 900-1000°C
Chromite ore
8Na2CrO4 +2Fe2O3 + 8CO2
Soluble
Insoluble
Conversion of Na2CrO4 into Na2Cr2O7
The solution containing Na2CrO4 is treated with a calculated quantity of
H2SO4, Na2CrO4 is converted into Na2Cr2O7
...
→
Na2Cr2O4 + 3CO↑
Na2Cr2O4 + H2O →
Cr2O3↓ + 2 NaOH
Na2Cr2O7 + 3C
Reduction of Cr2O3 to chromium metal
Aluminothermic process
Fig
...
1 Aluminothermic process
1
...
Fireclay crucible
4
...
Magnesium ribbon,
5
...
A mixture of barium peroxide and Mg powder
is placed over this
...
The mixture is ignited by a piece of Mg ribbon
...
111
The molten chromium is collected in the crucible and aluminium oxide is removed
as slag
...
6 kJ
Properties of Cr
Physical Properties
1
...
2
...
It melts at 2113K
...
Action of air : It is unaffected by air at ordinary temperatures
...
4Cr + 3O2 → 2Cr2O3
2
...
However it
decomposes steam at red heat to give chromic oxide and hydrogen
...
Action of Acids : It dissolves in dilute hydrochloric acid and sulphuric acid
to liberate hydrogen and forms chromous salts
...
With hot concentrated sulphuric acid it gives chromic sulphate and liberates
sulphur dioxide
...
Dilute nitric acid does not attack the pure metal while concentrated acid
renders it inactive or passive i
...
, it does not show its usual reactions
...
Action with Halogens : Chromium combines directly with fluorine and dry
chlorine to give chromium (III) halides
...
1
...
During electrolysis chromium deposits on the article
(cathode)
...
2
...
g
...
3
...
4
...
5
...
Alloys of chromium
Alloy
% composition
i)
Ferrochrome
Cr = 65% Fe = 35%
It is used in manufacture of
chrome steel, burglar proof safe
ii)
Stainless steel
Cr = 11-13%
C = 0
...
4%
Fe = 73% Ni = 8%
It is used for cutlery and house
hold wares
...
25%
C = 0
...
5%
It is used in cutlery, surgical
instruments,
...
iv) Stellite
113
Uses
4
...
3 Occurrence and principles of extraction of zinc
Zinc
Atomic mass : 65
...
The ancient used an alloy of Zn and Cu not very different from brass without
knowing its actual composition
...
It is commonly
called jast
...
However,
in the combined state, zinc is widely distributed
...
In India large deposits of zinc blende occur
in Zawar mines near Udaipur in Rajasthan
...
1
...
2
...
Δ
2 ZnS + 3O2 → 2ZnO
3
...
114
ZnO + C ⎯ 1673⎯ →
⎯ K
Zn + CO
Purification
Zinc is purified by electrolytic refining
...
The electrolyte is ZnSO4 solution
containing a little of dil
...
On passing electric current, pure zinc get deposited
at the cathode
...
It is malleable and ductile
...
773 K
2Zn + O2 ⎯ ⎯ ⎯⎯ →
2ZnO
ii)
Action of water
Pure zinc does not react with water but impure zinc (Zn-Cu couple)
decomposes steam quite readily evolving H2 gas
...
However, impure zinc reacts with
dilute acids with the liberation of H2
...
H2SO4
Zinc reacts with hot con
...
Zn + 2H2SO4
→ ZnSO4 + SO2 ↑ + 2H2O
115
v)
Action of HNO3
Zinc reacts with HNO3 at various concentrations and it gives different
products
...
Zn + 2NaOH + 2H2O → Na2ZnO2 + H2O
Sodium zincate
Uses of zinc
1
...
3
...
It is widely used for galvanizing iron sheets
...
Zinc plates and rods are used in batteries and dry cells
...
4
...
3
...
Occurrence
Silver occurs both in the native as well as in the combined state
...
Sb2 S3
The chief ore of silver is Argentite
...
Silver is obtained to a small extent from the gold ores in the kolar
fields in Karnataka and in the Anantapur mines
...
The various steps involved in this process are as follows
...
Concentration
The crushed ore is concentrated by froth-floatation process
...
Treatment of the ore with NaCN
The concentrated ore is treated with 0
...
6% solution of sodium cyanide
for several hours
...
Ag2S + 4NaCN
3
...
2Na [Ag(CN)2] + Zn → Na2[Zn(CN)4] + 2Ag↓
4
...
The impure silver
is made the anode while a thin sheet of pure silver act as the cathode
...
On passing electricity
pure silver gets deposited at the cathode
...
Extraction of silver from silver coins
Silver coins are Ag-Cu alloys
...
The solution is boiled to expel excess of nitric acid,
then the solution is treated with con
...
AgCl is
separated and converted to silver by fusing with excess Na2CO3
...
Properties
Physical properties
i)
ii)
It is a white lustrous metal
...
iii) Molten silver absorbs about twenty times its volume of oxygen which it
again expels on cooling
...
This is called
“spitting of silver”
...
Chemical properties
i)
With dilute HNO3
Dilute HNO3 reacts with silver liberating nitric oxide
...
HNO3 or con
...
HNO3 or con
...
Ag + 2 HNO3 →AgNO3 + NO2↑ + H2O
2Ag + 2H2SO4 → Ag2SO4 + SO2↑ + 2H2O
iii) Action with chlorine
Chlorine attacks silver, forming silver chloride
...
Uses of silver
i)
ii)
iii)
Silver salts are used in silvering of mirrors
AgBr is used in photography
Silver salts are used in silver plating
...
v) Silver leaf is used in medicine, while silver amalgam is employed in dental
filling
...
4 THIRD TRANSITION SERIES
4
...
1 Occurrence and principles of extraction of gold
Gold
Atomic mass : 196
...
Occurrence
Gold is mainly found in the native state either mixed with quartz in ancient
rocks (vein gold) or scattered in sand and gravel (placer gold) or in the alluvial
sand (alluvial gold)
...
India occupies 8th position among the gold producing countries of
the world and account for about 2% of the total world production
...
97% of our gold output
...
The extraction of gold from the sulphide or telluride ore involves the following
steps :
i)
Crushing, grinding and concentration
The gold ore is crushed and powdered, and then concentrated by frothfloatation process
...
119
iii) Treatment with KCN
The finely powdered roasted ore is taken in large vats made up of cement
or iron and having false wooden bottoms with holes
...
5%) solution of KCN in presence of excess of air for 24 hours
...
4Au + 8KCN + 2H2O + O2 → 4K[Au(CN)2] + 4KOH
Pot
...
Gold being less electropositive
than zinc, it gets precipitated
...
It is further purified by electrorefining
...
The impure gold is
made the anode while a thin sheet of pure gold acts as the cathode
...
On
passing current pure gold gets deposited on the cathode
...
ii)
It is the most malleable and ductile of all metals
...
Chemical properties
i)
Gold is unaffected by dry or moist air or oxygen
...
120
iii) With Aquaregia
Gold dissolves in aquaregia (3 parts of con
...
HNO3) to
form auric chloride
...
2Au + 3Cl2 → 2AuCl3
Uses
i)
ii)
iii)
It is used in coinage, jewellery and ornamental vessels
...
“Purple of cassius” which is a combination of gold with colloidal stannic
hydroxide is used for making ruby red glass and high class pottery
...
Pure gold is 24 carats
...
4
...
5
...
4FeO
...
H2SO4,
when sodium chromate is converted to sodium dichromate
...
Na2Cr2O7 + 2KCl
→ K2Cr2O7 + 2NaCl
Properties
Physical properties
1
...
It forms orange red crystals which melt at 396°C
...
Chemical properties
1
...
4K2Cr2O7 → 4K2CrO4 + 2Cr2O3 + 3 O2
2
...
K2Cr2O7 + 2KOH → 2K2CrO4 + H2O
Orange-red
Yellow
on acidifying the yellow colour of K2CrO4 again changes to orange red due to
reformation of K2Cr2O7
...
Cr2O72- + H2O
Orange red
3
...
In presence of dil
...
K2Cr2O7 + 4 dil
...
i)
It liberates I2 from KI
K2Cr2O7 + 7H2SO4 + 6KI → 4 K2SO4 + Cr2 (SO4)3 + 3 I2 + 7H2O
ii)
It oxidises ferrous to ferric salt
K2Cr2O7 + 7H2SO4 + 6 FeSO4 →
K2SO4 + Cr2 (SO4)3 + 3 Fe2 (SO4)3+ 2H2O
iii) It oxidises H2S to sulphur
K2Cr2O7 + 4H2SO4 + 3H2S → K2SO4 + Cr2 (SO4)3 + 7H2O + 3S
4
...
H2SO4
reddish brown vapours of chromyl chloride are obtained
...
Uses
i)
ii)
iii)
iv)
It is used in volumetric analysis
In chrome tanning in leather industry
In calico printing and dyeing
In photography and in hardening gelatin film
...
5
...
5H2O); Blue vitriol
Preparation
In laboratory it is prepared by dissolving cupric oxide (or) cupric hydroxide
(or) cupric carbonate in dilute H2SO4
CuO + H2SO4
→ CuSO4 + H2O
Cu(OH)2 + H2SO4 → CuSO4 + 2H2O
CuCO3 + H2SO4
→ CuSO4 + H2O + CO2↑
123
Properties
Physical properties
1
...
The anhydrous salt is colourless but the hydrated salt is blue in colour
...
Chemical Properties
1
...
5H2O loses its water of crystallization and decomposes
at 720°C to give cupric oxide and sulphur trioxide
...
5 H 2 O
(B lu e)
SO 3 + CuO
2
...
H 2 O
-H 2 O 2 3 0 °C
CuSO 4
(W h ite)
Action of ammonia
Copper sulphate gives deep blue colour with NH4OH forming complex
compound
...
Action of KI
When KI is added to a solution of CuSO4, a white precipitate of cuprous
iodide is produced
...
Action with KCN
A yellow precipitate of cupric cyanide is first formed with KCN and it
decomposes to give cyanogen gas
...
Action of alkalies
With alkalies, a pale blue precipitate of copper hydroxide is formed
...
Action with H2S
With H2S it gives a black precipitate of copper sulphide
...
It is used as a germicide and insecticide in agriculture
...
A mixture of copper sulphate and lime, commonly known as Bordeaux
mixture, is used as fungicide
...
It is used in electroplating, calicoprinting and in electrical batteries
...
5
...
3Ag + 4HNO3 → 3AgNO3 + 2H2O + NO↑
Properties
Physical
1
...
It is soluble in water
...
Chemical Properties
1
...
723 K
980 K
2A gN O 2 + O 2
S ilv er n itrite
Ag
+ NO2
With NaCl
AgNO3 gives white precipitate of AgCl with soluble ionic chlorides
...
Silver nitrate is strongly caustic and oxidizing in nature
...
With KBr & KI
AgNO3 gives pale yellow precipitate of AgBr with bromides and yellow
AgI with iodides
...
Action with organic compounds
Ammoniacal silver nitrate (Tollen’s reagent) is reduced to silver mirror by
compounds like formic acid, formaldehyde or glucose
2AgNO3 + 2NH4OH → Ag2O + 2NH4NO3 + H2O
Ag2O + HCOOH → 2Ag↓ + H2O + CO2
Uses
1
...
3
...
5
...
It is used in silvering mirrors
...
It is used for silver plating
...
126
4
...
4 Zinc carbonate (ZnCO3)
Zinc carbonate occurs in nature as calamine
...
ZnSO4 + 2NaHCO3 → ZnCO3 + Na2SO4 + H2O + CO2↑
Properties
Physical properties
It is a white powder, insoluble in water
...
It is used in the preparation of cosmetics
...
4
...
5 Purple of cassius
Purple of cassius is only a form of colloidal gold
...
Preparation
It is prepared by mixing very dilute solution of gold chloride with stannous
chloride solution
...
SnCl4 + 4H2O → Sn(OH)4 + 4HCl
Uses
It is used in making ruby-red glass and high class pottery
...
Choose the correct answer
1
...
Formation of coloured ions is possible when compounds contains
a) paired electrons
b) unpaired electrons
c) lone pairs of electrons
d) none of the above
3
...
The colour of Ti(H2O)6 ion is due
a) d-d transistion
b) Presence of water molecules
c) Inter atomic transfer of electrons
d) None of the above
5
...
Paramagnetism is the property of
a) paired electrons
b) completely filled electronic subshells
c) unpaired electrons
d) completely vacant electronic subshells
7
...
The correct electronic configuration of copper atom is
a) 3d10 4s1
b) 3d10 4s2
c) 3d9 4s2
d) 3d5 4s2 4p4
9
...
Silver salt used in photography is
a) AgCl
b) AgNO3
c) AgF
d) AgBr
11
...
Excess of sodium hydroxide reacts with zinc to form
b) Na2 ZnO2
c) ZnO
d) Zn(OH)2
a) Zn H2
13
...
Which of the ions will give colourless aqueous solution?
a) Ni2+
b) Fe2+
c) Cu2+
d) Cu+
15
...
In the extraction of Cu, the reaction which does not take place in the
Bessemer converter is
a) 2CuFeS2 + O 2 → Cu2S + FeS + SO2
b) 2Cu2S + 3O2 → 2Cu2O + 2SO2
c) 2Cu2O + Cu2S → 6Cu + SO2
d) 2FeS + 3O2 → 2FeO + 2SO2
17
...
Choose the wrong statement regarding K2Cr2O7
a) It is a powerful oxidizing agent
b) It is used in tanning industry
c) It is soluble in water
d) It reduces ferric sulphate to ferrous sulphate
19
...
The correct statement in respect of d-block elements is
a) They are all metals
...
c) They form coloured ions and complex salts
...
21
...
Which of the following has the maximum number of unpaired electrons?
a) Mn2+
b) Ti3+
c) V3+
d) Fe2+
23
...
b) Argentite and cuprite are oxides
...
d) Malachite and azurite are ores of copper
...
The chemical composition of slag formed during the smelting process in
the extraction of copper is
a) Cu2O + FeS
b) FeSiO3
c) CuFeS2
d) Cu2S + FeO
25
...
Which transition element show highest oxidation state
a) Sc
b) T
i
c) Os
d) Zn
B
...
28
...
30
...
What are “d”-block elements?
How d-block elements are classified?
Explain why d-block elements exhibit variable oxidation states?
Why transistion elements form complexes?
Why does Mn(II) show maximum paramagnetic character among the bivalent
ions of the first transistion series?
32
...
[Ti (H2O)6]3+ is coloured while [Sc (H2O)6]3+ is colourless
...
34
...
9 BM
...
Explain why the melting and boiling points of Zn,Cd,Hg are low?
36
...
Write two alloys of copper and their uses
...
Write short notes on alumino thermic process?
39
...
40
...
41
...
What happens when KI solution is added to an aqueous solution of copper
sulphate?
43
...
Answer not exceeding 60 words
44
...
45
...
Explain how it is extracted from its alluvial gavel
...
List the ores of silver
...
Briefly explain the extraction of zinc from zinc blende
...
Explain how dichromate is extracted from its chromite are
...
D
...
The chief ore of Zinc, on roasting gave a compound A, which on reduction
by carbon, gives B
...
50
...
This compound is also called as Blue
vitriol
...
131
51
...
Its chief ore (A) on roasting with molten alkali gives compound (B)
...
Compound C on treatment
with KCl gave compound D
...
Explain
with proper chemical reactions
...
General methods of extraction of metals, purification and properties
are thoroughly discussed
...
References
1
...
D
...
2
...
L
...
132
27
Co
40
Zr
46
Pd
110
Uun
133
5
...
( To study the comparison of properties of Lanthanides and Actinides
( To know the general method of extraction of Lanthanides
...
Sweedish chemist Carl mosander discovered the metallic element Lanthanum
in 1839
...
Glean seaborg (University of California) discovered Californium
...
The elements in which the extra electron enters ( n- 2 )f orbitals are called
f- block elements
...
The f-block
elements are also known as rare earth elements
...
i) The Lanthanide series (4f-block elements)
ii) The Actinide series (5f- block elements )
5
...
1
...
e
...
Lanthanum and Lutetium have no partly filled 4f- subshell but
have electrons in 5d-subshell
...
However, all these elements closely resemble lanthanum and hence are
considered together
...
Electronic configuration
The electronic configuration of Lanthanides are listed in the table 5
...
The
fourteen electrons are filled in Ce to Lu with configuration [54 Xe ]4f1-14 5d1 6s2
2
...
Some elements exhibit +2 and
+4 states as uncommon oxidation states
...
Radii of tripositive lanthanide ions
The size of M3+ ions decreases as we move through the lanthanides from
lanthanum to lutetium
...
Cause of lanthanide contraction
The lanthanide contraction is due to the imperfect shielding of one 4f electron
by another in the same sub shell
...
However, due to imperfect shielding, the effective nuclear charge increases causing
a contraction in electron cloud of 4f-subshell
...
i)
Basicity of ions
Due to lanthanide contraction, the size of Ln3+ ions decreases regularly with
increase in atomic number
...
Since the order of size of Ln3+ ions
are
La3+ > Ce3+
...
iii)
Regular decrease in their tendency to act as reducing agent, with increase in
atomic number
...
v) Due to lanthanide contraction, these elements occur together in natural
minerals and are difficult to separate
...
1
...
The fifteen elements from actinium to lawrencium constitute the actinide
series of the periodic table
...
2
...
Out
of these, +4 oxidation state is most common state
...
Radii of M3+ and M4+ ions
The ionic radii of actinide elements decrease gradually as we move along
the actinide series
...
Cause of actinide contraction
Cause of actinide contraction is the imperfect shielding by 5f-electrons
...
Hence as the
atomic number increases, the inward pull experienced by 5f-electrons increase
...
136
5
...
3 Extraction of Lanthanides from Monazite sand
The method used for extraction of lanthanides from monazite sand consists
of the steps which have been shown in flowsheet
...
The
anhydrous fluorides and chlorides are heated under argon atmosphere in presence
of calcium at 1270 K to get the individual metal
...
5
...
Both show close resemblance
because these involve filling of f-subshells
...
However, these
also differ from each other as shown in the following table
...
i)
Binding energies of 5f electrons are
lower
...
g
...
Uranium
exhibits +6 oxidation state in UF 6
and UO 2Cl2
iii)
4f electrons have greater
shielding effect
...
iv)
Most of their ions are
colourless
...
v)
They are also paramagnetic but
their magnetic properties are very
difficult to interpret
...
vi)
They have much greater tendency
to form complexes
...
vii)
All of them are radioactive
...
viii) Their compounds are less
basic
...
ix)
They form oxocations such as
UO 22+ , UO + , N pO 2+ , PuO 2+
...
3 Uses of Lanthanides and actinides
Use of lanthanides
1
...
2
...
138
3
...
4
...
Lanthanido - thermic processes can yield sufficiently pure
Nb, Zr, Fe, Co, Ni, Mn, Y, W, U, B and Si
...
Alloys of Lanthanides are known as mish - metals
...
Mish-metals are used
for the production of brands of steel like heat resistant, stainless and
instrumental steels
...
Uses of Actinides
1
...
2
...
SELF EVALUATION
A
...
The electronic configuration of Lanthanides is
a)
c)
2
...
b)
c)
The electronic configuration of Actinides is
a)
c)
3
...
Lanthanides are extracted from
a) Limonite b) Monazite
6
...
c) Magnetite d) Cassiterite
s block elements
d block elements
b)
d)
p block elements
f block elements
The Lanthanides contraction is due to
a) Perfect shielding of 4f electron b) Imperfect shielding of 4f electron
c) Perfect shielding of 3d electron d) Imperfect shielding of 3d electron
8
...
b) tracer bullets c) gas lamp materials d) none of the above
_______is used in gas lamp material
a) MnO2
b) CeO2
c) N 2O 5
d) Fe2O3
10
...
Metallothermic processes involving Lanthanides are called as
a) Aluminothermic process
c) Reduction process
b) Lanthanido-thermic process
d) Oxidation process
12
...
Maximum oxidation state exhibited by Lanthanides is
a) +1
b) +2
c) +3
d) Alkalimetals
d) +4
14
...
Answer in one or two sentences
15
...
140
16
...
What are lanthanides? Give the various oxidation states of lanthanides
...
What are mish metals? Give their uses
...
Write the uses of Lanthanides and Actinides
...
Answer not exceeding 60 words
20
...
21
...
i) +3 oxidation state of lanthanides is the most stable
...
iii) Lanthanides are grouped together
...
Comparing La(OH)3 and Lu(OH)3, which is more basic and explain why?
23
...
24
...
Summary
The elements in which the extra electron enters (n-2) f orbitals are called
f-block elements
...
The two series of f-block elements are lanthanide series and Actinide series
...
Cause and consequences of lanthanide
contraction are discussed
...
A
flow chart is given for extraction of lanthanides from monazite sand
...
References
1)
Advanced Inorganic Chemistry F
...
Cotton, G
...
A
...
Bochmann, John Wiley & Sons, 2003
...
I Gurdeep Raj, Goel Publishing House,
2002
...
COORDINATION COMPOUNDS AND
BIO-COORDINATION COMPOUNDS
Learning Objectives
( To understand the nature of simple salts, double salts and complex
salts
...
( To learn the latest nomenclature of coordination compounds as per
IUPAC rules
...
( To learn briefly about, Werner’s theory and valence bond theory
...
142
6
...
There are different
types of salts
...
KOH
→
+ HCl
KCl + H2O
Normally, a simple salt ionises in water and produces ions in solution
...
b)
Molecular (or) addition compounds
i)
Double salts
These are molecular compounds which are formed by the evaporation of
solution containing two (or) more salts in stoichiometric proportions
...
Double salts retain their properties only in solid state
...
Example
K2SO 4
...
24H2O
- Potash alum
FeSO4
...
6H2O
- Mohr’s salt
K2SO 4
...
24H2O → 2K+ + 2Al3+ + 4SO42- + 24H2O
The double salts give the test of all their constituent ions in solution
...
The molecular compounds, do not dissociate into its constituent
ions in solution are called coordination compounds
...
4KCN (or) K4[Fe(CN)6]
143
Ferrous cyanide
Fe(CN)2
...
The metal of
the complex ion is not free in solution unlike metal in double salt in solution
...
K4 [Fe(CN)6]
ii)
[Fe (CN)6 ]4complex anion
A cationic complex contains complex cation and simple anion
[Co(NH3 )6] Cl3
iii)
4K+
+
simple cation
[Co(NH3)6]3+ + 3Clcomplex cation
simple anion
In the case of a complex compound, [Cr(NH3 )6] [Co(CN)6], it gives both
complex cation and complex anion
[Cr(NH3)6]3+ + [Co(CN)6]3complex cation complex anion
[Cr(NH3)6] [Co(CN)6]
6
...
(b) Lewis Base
All electron donors are lewis base
...
The acceptor is usually a metal / metal ion to which one (or) more of
neutral molecules (or) anions are attached
...
Hence, central metal cation in a complex serves as a
lewis acid
...
It is oxidation number that denotes the charge, if the
central metal atom would have if all the ligand in the complex were removed
along with their electron pairs that were shared with the central atom
...
(e) Ligand (Latin word meaning to bind)
A ligand is an ion (or) a molecule capable of functioning as an electron
donor
...
These coordination
groups or ligands can donate a pair of electrons to the central metal ion (or)
atom
...
Types of ligands
When a ligand is bound to a metal ion through a single donor atom, as with
Cl , H2O or NH3, the ligand is said to be unidentate
...
Such
ligands are called polydentate ligands
...
For example, ethylenediamine
is a bidentate ligand because it has two amino groups each of which can donate
a pair of electrons
...
H - C H - C H -
...
NH2 - NH3+ hydrazinium
This ligand, though positive can bind through the uncharged nitrogen
...
But water
is written as ‘aqua : Ammonia is written as ammine
...
Negative Ligands
Negative ligands end in suffix ‘O’
...
Chelates
If a ligand is capable of forming more than one bond with the central metal
atom (or) ion then the ring structures are produced which are known as metal
chelates
...
2 N H 2C H 2C H 2N H 2 + C u
H 2C
N
...
H
2+
NH2
2
Cu
2
...
This represents
a single constituent unit
...
146
[M(L)n](n-) (or) (n+)
[Fe(CN)6]4- ,
[Cu(NH3)4]2+
These ions do not ionise to give the test for constituent ions
...
Numerically
coordination number represents the total number of the chemical bonds formed
between the central metal ion and the donor atoms of the ligands
...
Charge on the complex ion
Charge on the complex ion is equal to the sum of the charges on the metal
ion and their ligands
...
[Cu(NH3)4]2+ can be written as [Cu2+(NH3)4]2+ since NH3 ligand is neutral
...
This can be determined as shown below
Charge on the metal ion (Cu2+) = +2
Charge on the ligand (NH3) = 4 × 0 = 0
∴ Net charge on the complex ion = +2 + 0 = +2
2
...
Charge on the metal ion (Fe2+) = +2
Charge on the ligand (CN-)
= 6 × (-1) = –6
∴ Net charge on the complex = +2 – 6 = –4
147
6
...
In naming the entire complex, the name of the cation is given first and the
anion second (just as for sodium chloride), no matter whether the cation or
the anion is the complex species
...
In the complex ion, the name of the ligand or ligands precedes that of the
central metal atom (This procedure is reversed for writing formulae)
...
Ligand names generally end with ‘O’ if the ligand is negative (‘chloro’ for
Cl-, ‘cyano’ for CN-, ‘hydrido’ for H-) and unmodified if the ligand is neutral
(‘methylamine’ for MeNH2)
...
4
...
) indicates the number
of each ligand (mono is usually omitted for a single ligand of a given type)
...
For example, [Ni(PPh3)2Cl2] is named dichlorobis(triphenylphosphine)
nickel(II)
...
A Roman numeral or a zero in parentheses is used to indicate the oxidation
state of the central metal atom
...
If the complex ion is negative, the name of the metal ends in ‘ate’ for example,
ferrate, cuprate, nickelate, cobaltate etc
...
If more than one ligand is present in the species, then the ligands are named
in alphabetical order regardless of the number of each
...
Some additional notes
i)
Some metals in anions have special names
B Borate
Au Aurate
Ag
Pb Plumbate Sn Stannate
Cu
148
Argentate
Cuprate
Fe
Ni
Ferrate
Nickelate
ii)
Use of brackets or enclosing marks
...
Examples
[Co(en)3]Cl3
tris(ethylenediamine)cobalt(III) chloride
[Co(NH3)3(NO2)3]
triamminetrinitrocobalt (III)
K2[CoCl4]
potassiumtetrachlorocobaltate(II)
note that it is not necessary to enclose the halogens in brackets
...
3H2O
Q1) What is the central metal ion?
A1) Central metal is Chromium
Q2) What is its oxidation state?
A2) O
...
is III
Q3) What is its electronic configuration? A3) electronic configuration is d3
Q4) What is its coordination number?
A4) C
...
is 6(3 bidentate ligands
present)
Q5) What is the shape of the ion?
A5) structure is octahedral
Q6) Can the structure have isomers?
A6) Yes, optical isomers are possible
Q7) What is the IUPAC name of the
complex?
A7) Potassiumtris(oxalato)
chromate(III) trihydrate
IUPAC Nomenclature of mono nuclear coordination compounds
[CoIII (NH3)5Cl]2+
- pentaamminechlorocobalt(III) ion
[CoIII (NH3)6]Cl3
- Hexaamminecobalt(III) chloride
[CrIII (en)3]Cl3
- Tris (ethylenediamine)chromium(III) chloride
K4[FeII(CN)6]
- Potassium hexacyanoferrate(II)
[NiII(CN)4]2-
- Tetracyanonickelate(II) ion
[CuII(NH3)4]2+
- Tetraamminecopper(II) ion
[PtII Cl2(NH3)2]
- Diamminedichloroplatinum(II)
149
6
...
This phenomenon is known as isomerism
...
Each of
these is further classified as shown below
...
4
...
In such a case the distribution of ligands between the two coordination
spheres can vary, giving rise to isomers called the coordination isomers
...
This isomerism is illustrated by
the following pairs of complexes where the complex cation and anion contain
different metal centres
...
[CoIII(NH3)6] [Cr(CN)6] and [CrIII(NH3)6] [CoIII(CN)6]
Hexammine hexacyano
Hexamine
hexacyano
cobalt(III) chromate(III)
chromium (III) cobaltate (III)
2
...
This property is known as
ionisation isomerism
...
[Co(NH3)4Cl2]NO2
and
[Co(NH3)4 NO2Cl]Cl
Tetraamminedichlorocobalt(III) nitrite
[Co(NH3)5NO3]SO4
Tetraamminechloronitrocobalt(III) chloride
and
[Co(NH3)5 SO4]NO3
pentaamminenitratocobalt(III) sulphate
c)
pentaamminesulphatocobalt(III) nitrate
Hydrate isomerism or Solvate isomerism
The best known examples of this type of isomerism occurs for chromium
chloride “CrCl3
...
1
...
2H2O - Bright green
Tetraaquadichlorochromium(III) chloride dihydrate
2
...
H2O - grey-green
Pentaaquachlorochromium(III) chloride monohydrate
3
...
d)
Linkage isomerism
Linkage isomerism occurs with ambidentate ligands
...
The best known cases involve the
monodentate ligands SCN-/NCS- and NO2-/ONOFor example
[Co(NH3)5ONO]Cl2 the nitrito isomer - red colour
pentaamminenitritocobalt(III) chloride - O attached
[Co(NH3)5 NO2]Cl2 the nitro isomer - yellow colour
pentaamminenitrocobalt(III) chloride - N attached
151
e)
Ligand isomerism
Ligand isomerism arises from the presence of ligands which can adopt
different isomeric forms
...
H2N - CH2 - CH2 - CH2 - NH2 or
H2N - CH2 - CH - CH3
|
NH2
6
...
2 Stereoisomerism (space isomerism)
Consider two compounds containing the same ligands attached to the same
central metal ion, but the arrangement of ligands in space about the central metal
ion are different, then these two compounds are said to be stereoisomers and
this phenomenon is known as stereoisomerism
...
a) Geometrical isomerism or b) Optical isomerism
...
In a cis-isomer two identical (or) similar groups
are adjacent to each other whereas in a trans-isomer they are diametrically
opposite to each other
...
Example
of this type of complexes are [Pt (NH3)2 Cl2] and [Pd(NH3)2 (NO2)2]
...
They are (1,2) (1,3) (1,4) (1,5) (2,6) (3,6) (4,6) (5,6) (3,4) (4,5) (2,5) and
(2,3)
...
There are three trans positions; they are (1,6) (2,4) and (3,5)
...
An octahedral complex of the type [Ma4b2] where a and b are monodentate
ligands, exists as two geometrical Isomers:
n+
b
a
b
a
M
a
n+
b
a
M
a
a
a
cis-iso m e r
a
b
tran s-isom er
153
A specific example for such Isomerism is [Co(NH3)4 Cl2]+ which exists as
two geometrical isomers
...
A specific example for this is [Co(H2N-CH2-CH2NH2)2 Cl2]+
H 2C
H2
N
+
Cl
Cl
Cl
CH2
NH2
NH2
Co
H 2C
N
H2
CH2
Co
NH2
H 2N
H 2C
CH2
H 2N
NH2
CH2
CH2
Cl
cis-iso m e r
tran s-iso m e r
The octahedral complex of the type, [Ma3b3]n±, where a and b are
monodentate ligands also exist as geometrical isomers, For example, [Rh(py)3
Cl3] exist as cis-(1,2,3 trichlorocomplex) and trans-(1,2,6-trichloro complex)
isomers as represented below
Cl
Cl
py
py
Cl
Rh
Rh
Cl
tran s-iso m er
py
cis-iso m ers
b)
py
py
Cl
py
Cl
Optical Isomerism
This is a phenomenon in which certain organic or inorganic compounds
have the property of rotating plane polarised light
...
The optical isomers of a compound have
identical physical and chemical properties
...
In a
coordination compound of type [PtCl2(en)2]12+, two geometrical isomers are
possible
...
Among these two isomers, cis isomer shows
optical activity because the whole molecule is asymmetric
...
5 THEORIES OF COORDINATION COMPOUNDS
6
...
1 Werner’s Theory
Alfred Werner (1866-1919) French born Swiss chemist founded the modern
theory on coordination compounds
...
Werner
was the first inorganic chemist to be awarded the nobel prize in chemistry
...
Brief concepts of Werner’s theory of coordination compounds
Alfred Werner studied the structure of coordination complexes and put
forward his ideas in the year 1893 which were known as ‘Werner’s coordination
theory
...
The primary valency of the metal ion is always satisfied by negative ions
...
The secondary valencies may be satisfied by either negative
ions or neutral molecules
...
5)
The ligands which satisfy secondary valencies must project in definite
directions in space
...
6)
The ligands have unshared pair of electrons
...
Such compounds
are called coordination compounds
...
In this representation, the primary valency (dotted lines) are satisfied by the three
chloride ions
...
NH3 Cl
H 3N
Cl
NH3
Co
H 3N
Cl
NH3
NH3
Defects of Werner’s theory
Werner’s theory describes the structures of many coordination compounds
successfully
...
6
...
2 Valence bond theory (VB Theory)
Valence bond theory, primarily the work of Linus Pauling regarded bonding
as characterized by the overlap of atomic or hybrid orbitals of individual atoms
...
156
2)
These vacant orbitals form covalent bonds with the ligand orbitals
...
This complete overlap leads to the formation of a metal
ligand, σ (sigma) bond
...
This maximum overlapping is possible only when the metal
vacant orbitals undergo a process called ‘hybridisation’
...
The following table gives the coordination number, orbital hybridisation and
spatial geometry of the more important geometrics
...
It is attracted by an external field
...
ìs = n(n+ 2) BM
μs = spin-only magnetic moment
n = number of unpaired electrons
BM = Bohr magneton, the unit which expresses the magnetic moment
...
Number of unpaired electrons
Spin-only moment (BM)
1
1(1 + 2) = 3 = 1
...
83
157
3
3(3 + 2) = 15 = 3
...
90
5
5(5 + 2) = 35 = 5
...
83 BM
Since the hybridisation is sp3, the geometry of the molecule is tetrahedral
...
4s
4p
3d
2+
N i io n
The ligand CN- is a powerful ligand
...
Hence this complex ion does not contain unpaired electrons
...
158
4s
3d
[N i(C N ) 4 ]
2-
N
N
4p
N
N
-
NN
NN
-
-
CN CN CN CN
-
2
d sp h y brid isatio n
The geometry of the molecule is square planar
...
90 BM
The molecule is paramagnetic
...
2)
Fe+2 ion
4s
3d
4p
In [Fe(CN)6]4- complex the CN- ligand is a powerful ligand, it forces the unpaired
electrons in 3d level to pair up inside
...
The molecule is diamagnetic
...
Defects of Valence bond theory
Although VB theory was the principal way in which chemist visualized
coordination compounds until the 1950s, it has fallen into disfavour due to its
inability to account for various magnetic, electronic and spectroscopic properties
of these compounds
...
6 Uses of coordination compounds
1
...
It is a complex of hydroxyanthraquinone
...
2
...
a) Colour Tests : Since many complexes are highly coloured they can be used as
colourimetric reagents e
...
formation of red 2,2’-bipyridyl and 1,10-phenanthroline
complexes as a test for Fe2+
b) Gravimetric Analysis : Here chelating ligands are often used to form
insoluble complexes e
...
Ni(DMG)2 and Al(oxine)3
...
g
...
By careful adjustment of
160
the pH and using suitable indicators, mixtures of metals can be analysed, e
...
Bi3+
in the presence of Pb2+
...
3
...
g
...
4
...
Therapeutic chelating agents are used as antidotes for heavy metal poisoning
...
(e
...
Pb2+)
...
5
...
6
...
The chelating agent sequesters hard-water cations, rendering them incapable of
interfering with the surfactant
...
7 BIO COORDINATION COMPOUNDS
Coordination compounds play an important role in many biological processes
in plants and animals
...
Name
1
...
3
...
161
6
...
1 Haemoglobin
Haemoglobin in the red blood cells carries oxygen from the lungs to the
tissues
...
When the
oxygen has been released for cell respiration, haemoglobin loses its bright red
colour and becomes purple
...
Nature of haemoglobin and myoglobin
Both are having the same structure excepting the fact that myoglobin is a
monomer and haemoglobin is a tetramer
...
These are biocoordination complexes formed between porphyrin and iron
in its +2 oxidation state (Fe2+)
...
Each haemoglobin molecule consists of
four subunits, each unit is being a folded chain
...
The sixth octahedral site is available to bind oxygen molecule
...
7
...
The magnesium is at the
centre of the modified porphyrin ring septeon (corrin)
...
The modified porphyrin acts as the ligand
...
In plants, chlorophyll ‘a’ is the pigment directly responsible for the
transformation of light energy to chemical energy
...
The conversion of atmospheric
carbondioxide and atmospheric moisture into carbohydrate and molecular oxygen
in the presence of sunlight, by the plant is called as photosynthesis
...
xCO 2 + yH 2O ⎯chlorophyll → Cx(H 2 O)y + O 2
⎯ ⎯⎯
sunlight
162
Photosynthesis requires, in addition to chlorophyll, the help of four other metal
complexes, a manganese complexes, two iron complexes and a copper complex
...
But chlorophyll helps in the conversion of atmosphere CO2 into
molecular oxygen which act as a fuel for human cell
...
Choose the correct answer
1
...
3
...
Which a double salt
a) K2SO4
...
24H2O b) NaCl c) K4[Fe(CN)6] d) KCl
An example of a complex compound having coordination number 4
...
b) Chloro
d) triangular
b) -1
c) +2
d) -2
[Cu(NH3)4]Cl2
K3[Fe(CN)6]
b)
d)
K4[Fe(CN)6]
[NiCl4]2-
b)
d)
Square planar
Octahedral
The geometry of [Ni(CN)4]2- is
a)
c)
9
...
b) square planar
The oxidation number of Nickel in the complex ion, [NiCl4]2- is
a) +1
7
...
c) Bromo
Tetrahedral
Triangular
An example of an ambidentate ligand is
a) CN-
c) NO2-
b) Cl163
d) I-
10
...
In [FeII(CN)6]4-, the central metal ion is
a) Fe
b) Fe+2
c) Fe+3
d) CN-
12
...
The name of [PtIV(NH3)2Cl2]2+ is
a) Diamminedichloroplatinum(IV) ion
b) Diamminedichloroplanitate(IV)
c) Diamminedichloroplatinum
d) Dichlorodiammineplatinum(IV) ion
14
...
A metal ion from the first transition series forms an octahedral complex with
magnetic moment of 4
...
The metal ion is
a) Fe2+
b) Co2+
c) Mn2+
d) Ni2+
c) BM
d) ergs
16
...
The type of isomerism found in the complexes [Co(NO2)(NH3)5]SO4 and
[Co(SO4)(NH3)5] NO2
a) Hydrate isomerism
c) Linkage isomerism
b) Coordination isomerism
d) Ionisation
18
...
Answer in one or two sentences
19
...
164
d) colour
20
...
21
...
What are ligands and coordination number?
23
...
24
...
a) [Fe(NH3)4Cl2] NO3
b) Na[B(NO3)4]
25
...
Write the formula structure of the following
a) tris(ethylenediamine)cobalt(III) ion
b) pentaamminesulphatocobalt(III) chloride
27
...
What are chelates? Give one example
...
Answer not exceeding 60 words
29
...
30
...
For the complexes K4[Fe(CN)6], [Cu(NH3)4] SO4 mention
a) Name b) Central metal ion c) Ligands d) Coordination number
32
...
In what way [FeF6]4- differs from [Fe(CN)6]4-
...
[Ni(CN)4]2- diamagnetic, whereas [NiCl4]2- is paramagnetic
...
165
35
...
Explain
...
Explain the limitations of VB theory
...
Taking [FeF6]4- as an example, discuss geometry, nature of d-orbital splitting
and magnetic property using VB theory
...
Mention the function of haemoglobin in natural process
...
How chlorophyll is important in environmental chemistry? Mention its function
...
The
terminology used in coordination chemistry and types of isomerism found in
coordination compounds are explained with specific examples
...
The scope and limitations of these theories are also compared
...
The function
and role of haemoglobin and chlorophyll are explained
...
Coordination Chemistry, S
...
A
...
2
...
D
...
166
7
...
( Know about the nuclear reactions and chemical reactions
...
( Learn about principle of nuclear reactors
...
( Learn about the application of radioactive isotopes including tracer technique
adopted in understanding reaction mechanisms
...
In the same year Pierre Curie (France) and Mary Curie (France) were
awarded for their research on radiation phenomenon
...
In 1938, Enrico Fermi (Italy) was awarded
Nobel Prize for the discovery of nuclear reactions induced by slow neutrons
...
Radioactivity is a nuclear phenomenon and it is not affected by external
factors such as temperature, pressure etc
...
*
To explain the spontaneous decay of radioactive elements, Rutherford and
Soddy put forward the theory of radioactive disintegration
...
Based on the above theory, the following equation is derived which confirms
that all radioactive reactions follow I order
t=
λ = decay constant
2
...
The half life period (t½) of a
radioactive substance is independent of initial concentration
...
t½ is used to
indicate the relative stability of radioactive substance
...
0
...
44 t½
ë 0
...
*
An α - particle is equal to the bundle of two protons and two neutrons and
hence it is equal to the Helium nucleus (2He4)
...
*
γ-radiation is a waver of very short wavelength with very high energy
...
There are about 4 decay series
...
Hence
for atom, the atomic mass is lower than the sum of masses of protons,
neutrons and electrons present
...
This is the measure of the binding energy of proton and neutron in
the nucleus
...
7
...
Only the electrons in the extranuclear part
of atoms take part in the chemical process
...
Such reactions in
which the nuclei of the atoms interact with other nuclei or lighter particles or
photons resulting in the formation of new nuclei and one or more lighter particles
are called nuclear reactions
...
These reaction involve some loss,
gain or overlap of outer orbital
electrons of the reactant atoms
...
Nuclear reactions involve emission of
alpha, beta and gamma particles from
the nucleus
...
A chemical reaction is balanced in
terms of mass only
2
...
3
...
reaction is very much less when
compared with nuclear reaction
...
4
...
4
...
5
...
5
...
The following facts are taken into account while expressing a nuclear reaction:
i)
Reactions are written like a chemical equation
...
ii)
Mass number is written as super script on the symbol of the element
...
iii)
In a chemical reaction the total number of atoms of various elements are
balanced on the two sides
...
iv) Symbols used for projectiles:
The bombarding particles are called projectiles
...
n1
H1 or p
1
He4 or α
2
H2 or 1D2
1
e0 or e
-1
e0
+1
0
-
neutron
proton
α particle
deuteron
electron or β-particle
positron
170
The nucleus to be attacked is called as target nucleus or parent
...
The particle ejected during a
nuclear reaction is called as ejected particle
...
Hence the above
reaction is represented as 7N14 (α,p) 8O17
...
Qvalue
where
mr
mp
=
(mp-mr) 931 MeV
- Sum of the masses of reactants
- Sum of the masses of products
In the case of energy absorbed then mp>mr, then Q value will be positive
...
In the case of energy released, mr>mp, and hence Q value will be
negative
...
2 TYPE OF NUCLEAR REACTION
1
...
29
2
...
This type of nuclear fission reaction was first observed by German
Chemists Otto Hahn, F
...
The process is usually accompanied by emission of neutrons
...
Mechanism of fission
In the fission process, the heavy nucleus absorbs a neutron and forms an
unstable compound nucleus
...
Example
A typical example of the fission process in the fission of uranium by neutrons
is explained by the following equation
...
In this way a chain reaction is set up resulting
into the liberation of an enormous amount of energy
...
92
140
92
U
235
+ 0n
1
92
U
93
1
B a + 36K r + 3 0n
144
+ 38 S r 90 + 2 0 n 1
5 4X e
144
90
1
+ 3 7R b + 2 0n
55C s
56
236
172
This fission process is self multiplying process and hence a tremendous amount
of energy is released in a very short interval of time
...
Atom bomb is based on nuclear fission process
...
118 amu
The isotopic mass of 42Mo95 = 94
...
95 amu
= 1
...
118 + 1
...
936 + 138
...
009
236
...
906 amu
∴ The mass converted into energy is
= (236
...
906) amu
= 0
...
213 × 931
...
The fission in both the cases is similar and
uncontrolled
...
Nuclear Power Generator
A nuclear reactor or nuclear power generator is a kind of furnace for carrying
out the controlled fission of a radioactive material like U235 for producing power
...
Heavy
water at high pressure takes heat away from the core
...
The steam is taken away to drive turbines that make
electricity
...
173
3)
Nuclear Fusion
When lighter nuclei moving at a high speed are fused together to form a
heavy nucleus, the process is called nuclear fusion
...
Thus, just like a fission reaction, the source of energy
in a fusion reaction is also the disappearance of mass, which gets converted into
energy
...
Therefore, this reaction is called thermonuclear reaction
...
018 amu and the corresponding energy released
is 1
...
Hydrogen Bomb
The highly destructive hydrogen bomb is also based on the fusion reactions
of hydrogen to form helium producing large amount of energy
...
Fission reaction provides the high
temperature necessary to start the fusion
...
i)
ii)
→
→
→
Fission
Li6 + 0n1
3
H2 + 1H3
1
heat + neutrons
H3 + 2He4 + 4
...
6 MeV
2
7
...
This method is based onthe fact that 6C14, radioactive
isotope of carbon is formed in the upper atmosphere by reaction with neutrons
(from cosmic rays)
...
Animals too consume C14 by
174
eating plants
...
Carbon14 begins to decay
...
Therefore by knowing either the amount of C14 or the
number of β-particles emitted per minute per gram of carbon at the initial and
final stages, the age of carbon material can be determined by the following equation
...
303 × t½
0
...
2)
It is very useful in understanding the evolution of life, and rise and fall of
civilizations
...
4 NUCLEAR REACTIONS TAKING PLACE IN SUN (STARS)
It has been estimated that the sun is giving out energy equally in all possible
directions at the rate of 3
...
The energy of the sun is supposed to
arise from the fusion of hydrogen nuclei into helium nuclei which in going on
inside it all the time
...
5 USES OF RADIOACTIVE ISOTOPES
a) Study of reaction mechanism
i) Mechanism of photosynthesis in plants
*
A small quantity of Radioactive C O 2 containing radioactive oxygen O18 is
mixed with ordinary carbondioxide and the process is carried out
...
Therefore O2 produced comes from water and not from carbondioxide
...
*
*
6C O 2 + 6H2O → C6H12 O 6 + 6O2
ii)
Study of hydrolysis of ester
By labelling oxygen, the mechanism of ester hydrolysis can be studied by
using water labelled with O18
...
OR
Radioactive isotopes which are useful in medicine
Isotope
Use
1
H3
Tritium
Measure water content of the body
6
C 11
Carbon - 11
Brain scan
6
C 14
Carbon - 14
Radio immunology
I131
Iodine - 131
Diagnosis of damaged heart muscles and hyper
thyroidism
Hg197
Mercury - 197
Kidney scan
15
P 32
Phosphorous-32
Detection of eye tumours
26
Fe59
Iron - 59
Diagnosis of anemia
27
Co60
Cobalt - 60
Treatment of cancer
53
80
176
Na24
Sodium - 24
Location of blood clots and circularity
disorders
Au198
Gold - 198
Curing of cancers
11
79
Radio isotopes which are useful in industry and in agriculture
38
Sr90
Strontium - 90 Thickness of coatings or levels of liquids in tanks
Practice Problems
1)
On neutron bombardment fragmentation of U-235 occurs according to the
equation
92
U235 + 0n1 →
42
Mo95 + 57La139 + x -1e0 + y 0n1
Calculate the values of x and y
...
92
U235 + 0n1 →
42
Mo98 + 54Xe136 + x -1e0 + y 0n1
Calculate the values of x and y
...
After 24 hours, only 0
...
what is half-life period?
Solution
N0 = 1g
∴ ë =
N = 0
...
303
N
log 0
t
N
2
...
125
= 0
...
693 0
...
99 hours
ë
0
...
Half-life period of a radioactive element is 100 seconds
...
How much time will it take
for 90% decay?
Solution
t½
∴ ë=
= 100 sec
0
...
693
= 0
...
3 sec
ë 0
...
303
0
...
2
...
303
0
...
3 sec
log
N
N
0
100
10
The half-life of cobalt - 60 is 5
...
Calculate the % activity remaining
after 4 years
...
26 years
ë
t
0
...
26
= 4 years
=
N
Here to find the % of activity (i
...
303
N
ë
log 0
=
t
N
178
log
ë×t
2
...
693
4
×
5
...
303
0
...
2288)
=
N0
N
=
N
N0
1
...
693
=
0
...
= 0
...
6 and 15
...
Calculate the age of the
artifact
...
303
log ⎢
⎥
ë
⎣ Amount of old wood ⎦
t½ =
0
...
693
t½
∴Age of the artifact =
⎡ Amount of fresh wood ⎤
2
...
693
⎣ Amount of old wood ⎦
2
...
2
log
0
...
6
= 5700 years
=
5
...
Find the value of
disintegration constant interms of second
...
693
0
...
693
1500 × 365 × 24 × 60 × 60 sec
0
...
1465 × 10-10 sec-1
=
6
...
90
Let ‘a’ and ‘b’ be the number of α β particles emitted during the change
Th232 → 82pb208 + a 2He4 + b -1e0
Comparing the mass numbers,
232 = 208 + 4a + b × 0
4a = 232 - 208
= 24
a = 6
90
Comparing the atomic numbers
90 = 82 + 2 × a + (-1)b
= 82 + 2a - b
2a-b= 90 - 82 = 8
2(6) - b = 8
b = 12-8 = 4
Number of α - particle emitted = 6
Number of β - particles emitted = 4
7
...
01823 amu, 4
...
00715 amu respectively
...
3
Solution
Mass of reactants = mass of Li + mass of H
= 7
...
00715
= 8
...
00 387
= 8
...
02538-8
...
01764 amu
∴ Energy evolved during reaction
= 0
...
423 MeV
8
...
also report the mass number and atomic number
of the product atom
...
9
...
Solution
0
...
693 τ
t½ =
ô =
( ∴ Average life, ô =
t½
0
...
693
= 202
...
The activity of a radioactive isotope falls to 12
...
Calculate
the half life and decay constant
...
5 t = 90 days
decay constant ë =
2
...
303
100
log
90
12
...
02558 log 8
= 2
...
693
0
...
99 days
ë
2
...
Calculate Q value of the following nuclear reaction 13Al27 + 2He4 → 14Si30 +
H1+ Q
...
9815 amu, 14Si30 is 29
...
0026 amu and 1H1 is 1
...
Δm =
=
Q =
=
(29
...
0078) - (26
...
0026)
-0
...
0025 × 931 MeV
2
...
Choose the correct answer
1
...
The most penetrating radiations are
a) α rays
b) β rays
c) γ rays
3
...
Which one of the following particles is used to bombard 13Al27 to give
p30 and a neutron
15
a) α particle
b) deuteron
c) proton
d) neutron
5
...
7
...
9
...
a) 3α and 3β
b) 5α and 3β
c) 3α and 5β
d) 5α and 5β
U235 nucleus absorbs a neutron and disintegrates into 54Xe139, 38Sr94 and
x
...
Which of the following is used as neutron absorber in the nuclear reactor?
a) Water b) Deuterium c) Some compound of uranium d) Cadmium
B
...
12
...
14
...
16
...
18
...
Define radio activity
...
Write two difference between chemical reaction and nuclear reaction
...
Give example for each type
...
What is Radio carbon dating?
State two uses of radio carbon dating
...
Complete the following nuclear reactions
i) 42Mo96 (
...
(α, 2n) 85At211
iii)
96
Cm246 + 6C12 →
...
v) 11Na23 (n, β)
...
vii) 27Co59 (d, p)
...
ix) 11Na23 +
...
Answer not exceeding 60 words
21
...
What is nuclear fission? What are controlled and uncontrolled fission
reactions? How can the energy released in such reactions be used for practical
purposes?
23
...
Differentiate chemical reactions from nuclear reactions
...
Explain the use of radioactive isotopes with specified examples
...
Problems
26
...
Calculate the decay constant for Ag108 if its half life is 2
...
28
...
4 × 1010 years
...
29
...
U238 undergoes a series of changes by emitting α and β particles and
finally 82pb206 is formed
...
92
31
...
In terms of its
radioactivity six α and four β particles are emitted
...
32
...
33
...
Ba137 + 36Kr93 +
Ni58 +
iv) 21H3 → 2He4 +
?
?
...
?
...
Predict the bombarding projectile in the following nuclear reactions
i)
13
Al27 + ? →
11
ii)
34
Se83 + ? →
34
iii)
7
Na24 + 2He4
Se84 + γ rays
N14 + ? → 8O17 + 1H1
35
...
31× 10-4 year-1 calculate the half life period
...
*
Difference about nuclear reaction and chemical reactions
...
*
Principle of nuclear reactors, nuclear reactions in sun and applications of
radioactive isotopes are thoroughly discussed
...
Inorganic Chemistry by Puri and Sharma
...
Inorganic Chemistry by P
...
Soni
...
”
186
MATHEMATICAL RELATION
Logarithms and exponentials
ln x + ln y +
...
ln x – ln y = ln (x/y)
a ln x = ln xa
Derivatives
d (f = g) = df + dg
d (fg) = fdg + gdf
d(f/g) = 1/g df – f/g2 dg
dxn/dx = nxn–1
deax /dx = aeax
d ln x /dx = 1/x
Integrals
∫ xndx = xn+1/ n+1
∫ (1/x) dx = ln x + constant
187
PHYSICAL CHEMISTRY
8
...
( To recognise the properties and types of ionic crystals
...
( To learn the nature of Amorphous solids
...
Solids are usually classified as either crystalline or amorphous
...
They possess a sharp melting point
...
, although possessing many characteristics
of crystalline solids such as definite shape, rigidity and hardness, but are devoid
of a regular internal structure and melt gradually over a range of temperature
...
Crystallography : A study of internal structure of crystals
...
They also demonstrated
the wave nature of X-rays
...
UNIT CELL
Unit cell is the smallest fundamental repeating portion of a crystal lattice
from which the crystal is built by repetition in three dimension
...
Hence the contribution of
1
each atom to the unit cell is
...
fcc
A face atom is shared equally between two unit cells and therefore a face
⎛ N
⎞
f
atom contributes only ⎜ 2 ⎟ to the unit cell
...
190
Nc Nf 8 6
+
= + = 1+ 3 = 4
8
2
8 2
Nf = Number of
atoms at the
faces
Nf
BCC
In a bcc lattice, the body centred atom belongs exclusively to the unit cell
...
The contribution from each edge atom is therefore
1/4
...
=
Nc Ne
+
8
4
=
8 12
+
= 1+3=4
8 4
Ne = Number of atoms at the edge centre
Ne
Ne
191
8
...
The wave nature
of X-rays is not confirmed by diffraction experiment, because a grating of about
40 million ruling per cm is required for diffraction experiment
...
At the same time, crystallographers believed
that atoms in crystals are regularly arranged with an interatomic distance of about
10-8 cm
...
Based on
this, Laue suggested that crystal can be used as a three dimensional diffraction
grating for X-rays
...
The photograph obtained is known as
Laue diffraction pattern
...
Though Laue diffraction pattern gives more information about
the symmetry of crystals, the interpretation of the pattern seems to be difficult
...
1
...
L
...
H
...
The scattering of X-rays
by crystals could be considered as reflection from successive planes of atoms in
the crystals
...
The fundamental
equation which gives a simple relation between the wavelength of the X-rays,
the interplanar distance in the crystal and the angle of reflection, is known as
Bragg’s equation
...
1
...
On the other hand, if we use a
192
crystal whose interatomic distance ‘d’ is known, then the wavelength of
X-rays can be calculated
...
3)
When the experiment is done, there will be a maximum reflection at a
particular angle θ
...
It corresponds to first order reflection
(n=1)
...
It corresponds to second order reflection (n=2)
...
The values of angles obtained are in accordance with the Bragg’s equation
...
8
...
3 Bragg’s spectrometer method
This method is one of the important method for studying crystals using
X-rays
...
The rotating
table is provided with scale and vernier, from which the angle of incidence, θ can
be measured
...
8
...
The rays reflected from the crystal enter into the ionisation
chamber and ionise the gas present inside
...
The current of ionisation is a
direct measure of intensity of reflected beam from the crystal
...
These values are plotted in the form of graph
...
9°, 11
...
15°
...
103, 0
...
312 which are in the ratio 1:2:3
...
The ratio confirms the correctness of Bragg’s equation
...
2 TYPES OF CRYSTALS
Crystals are classified into the following four types depending upon the
nature of the units which occupy the lattice points
...
Molecular Crystals
3
...
Covalent Crystals
4
...
2
...
The forces binding the molecules together are of two types
(i) Dipole-dipole interaction and (ii) Vanderwaal’s forces
...
g
...
The Vanderwaal’s
forces are more general and occur in all kinds of molecular solids
...
2
...
Diamond is a good example for this type
...
2
...
Thus, each electron belongs to a number of positive ions and
each positive ion belong to a number of electrons
...
This force of attraction is strong and is thus responsible for a compact
solid structure of metals
...
2
...
Each ion of a given sign is held by coulombic forces of attraction to all ions
of opposite sign
...
The ionic crystals have the following
characteristics
...
2
...
4
...
6
...
The heats of vapourisation of ionic crystals are high
...
The melting and boiling points of ionic crystals are very high
...
Ionic crystals are insulators in the solid state
...
Ionic solids are good conductors when dissolved in water
...
2
...
In general, ionic crystals are classified
into AB and AB2 type
...
Lattice type
Coordination number
AB
: CsCl
: 8
AB
NaCl
6
AB
FeS
6
AB
ZnO
4
AB AB
ZnS BN
4
3
Let us discuss the structure of CsCl for AB type
...
The chloride ions are at the corners of a cube where as Cs+ ion is at the
centre of the cube or vice versa
...
+
Cs
Cl
Number of chloride ions per unit =
Nc 8
= =1
8 8
Nb 1
= =1
1 1
Thus number of CsCl units per unit cell is one
...
For example Rutile (TiO2) has the following structure
...
3 IMPERFECTIONS IN SOLIDS
Almost all the crystals encountered in practice suffer from imperfections or
defects of various kinds
...
The term
imperfection or defect is generally used to describe any deviation of the ideally
perfect crystal from the periodic arrangement of its constituents
...
3
...
Such defects arise due to imperfect
packing during the original crystallisation or they may arise from thermal vibrations
of atoms at elevated temperatures
...
Comparatively less common point defects
are the metal excess defect and the metal deficiency defect
...
The points
which are unoccupied are called lattice vacancies
...
The existence of two vacancies, one due to a missing Na+ ion and the other due
to a missing Cl- ion in a crystal of NaCl is shown in Fig
...
2
...
8
...
Frenkel defects
This defect arise when an ion occupies an interstitial position between the
lattice points
...
AgBr is an example for this type of defect
...
8
...
+
Ag
Br
+
Ag
Br
-
-
+
Br Ag
+
Ag Br
Ag
Br
+
Ag Br
+
Br
+
Ag
Br
+
Ag
+
Ag
Br
+
Ag
Br
-
Br
+
Ag
Br
+
Ag
+
Ag
Br
+
Ag
Br
-
Br
+
Ag
Br
+
Ag
+
Ag
Br
+
Ag
Br
Fig
...
3 Frenkel Defects in a Crystal
The crystal remains neutral since the number of positive ions is the same as
the number of negative ions
...
This yellow colour is due to the formation of a non-stoichiometric compound of
NaCl in which there is a slight excess of sodium ions
...
Metal deficiency defects
In certain cases, one of the positive ions is missing from its lattice site and
the extra negative charge is balanced by some nearby metal ion acquiring additional
charges instead of original charge
...
FeO and FeS
show this type of defects
...
4 PROPERTIES OF CRYSTALLINE SOLIDS
Crystalline solids exhibit an important property called conductivity
...
Conducting materials are generally classified into three types namely
conductors, semi conductors and super conductors
...
However, they exhibit
appreciable conductivity upon addition of impurities as Arsenic and Boron
...
Semi conductors which exhibit
conductivity due to the flow of excess negative electrons are called n-type
semiconductors
...
Semiconductors find application in modern devices as recitifiers,
transitors and solar cells
...
4
...
This superconductivity state is a state in
which a material has virtually zero electrical resistance
...
N o rm al
m etal
S u p er co n d u cto r
Tc
T
The super conducting transition temperature ‘Tc’ of a material is defined as
a critical temperature at which the resistivity of the material is suddenly changed
to zero
...
At the extremely low temperatures, vibration of the nuclei of certain atoms
slow down so much and they synchronise with the passing waves of electrons in
a flow of electric current
...
8
...
2 Application of superconductors
1)
It is a basis of new generation of energy saving power systems
...
These generators consume very low energy
and so we can save more energy
...
3)
Superconducting solenoids are used in Nuclear Magnetic Resonance Imaging
equipment which is a whole body scan equipment
...
5 AMORPHOUS SOLIDS
Amorphous solids possess properties of incompressibility and rigidity to a
certain extent but they do not have definite geometrical forms
...
5
...
The viscosity of the
liquid increases steadily and finally a glassy substance is formed
...
On the other hand
glasses are optically isotropic and on heating without any sharp transition passes
into a mobile liquid
...
Therefore, glasses are
regarded as amorphous solids or super cooled liquids as well
...
Solved Problems
1)
Determine the number of formula units of NaCl in one unit cell
...
Solution
1
−
In the fcc arrangement, there are Eight corners 8 × = Cl ion
8
1
6 × = 3Cl − ions
At the six faces
2
1
+
Along the 12 edges 12 × = 3Na ions
4
199
1× 1 = 1Na + ion
At the cube centre ___________________
Total 4 Cl– and 4Na+ ions
Hence, the unit cell contains four NaCl units
...
29A° gives
a first order reflection at 27°8′
...
29 A°
n = 1,
d=?
θ = 27°8′
on substitution, 1 × 2
...
29A° = 2d (0
...
29A°
= 2
...
456
In a fcc arrangement, the corner atoms are A type and those at face centres
are B type
...
A atoms are present at
the corners while B type are at face centres
...
The diffraction of a crystal with X-ray of wavelength 2
...
9′
...
[Ans : 2
...
Diffraction angle 2θ equal to 14
...
400 nm when second order diffraction was observed
...
[Ans : 0
...
Find the interplanar distance in a crystal in which a series of planes produce
a first order reflection from a copper X-ray tube (λ = 1
...
2°
...
9573 A°]
4
...
5A° are incident on a crystal having an interatomic
distance of 1
...
Find out the angles at which the first and second order
reflection take place
...
Calculate the angle at which (a) first order reflection and (b) second order
reflection will occur in an X-ray spectrometer when X-ray of wavelength
1
...
04A°
...
Choose the correct answer
1
...
(a) 12
2
...
(c)2λ = nd sinθ
(d) nλ = 2d sinθ
A regular three dimensional arrangement of identical points in space is called
(a) Unit cell
4
...
The crystal structure of CsCl is
(a) Simple cubic
(c) Tetragonal
6
...
An example for Frenkel defect is
(a) NaCl
(b) AgBr
10
...
12
...
14
...
(d) FeS
(b)
(d)
n-type semiconductors
Insulators
In the Bragg’s equation for diffraction of X-rays,’n’ represents
(a) The number of moles
(c) A quantum number
9
...
(b) face-centred cubic
(d) Body centred cubic
(b)
(d)
Avogadro number
Order of reflection
The number of close neighbours in a body centred cubic lattice of identical
spheres is
(a) 6
(b) 4
(c) 12
(d) 8
The crystals which are good conductors of electricity and heat are
(a) Ionic crystals
(b) Molecular crystals
(c) Metallic crystals
(d) Covalent crystals
In a simple cubic cell, each point on a corner is shared by
(a) One unit cell
(b) Two unit cell (c) 8 unit cell (d) 4 unit cell
The ability of certain ultra cold substances to conduct electricity without
resistance is called
(a) Semiconductor (b) Conductor (c) Superconductor (d) Insulator
The total number of atoms per unit cell is bcc is
(a) 1
(b) 2
(c) 3
(d) 4
Rutile is
(a) TiO2
(b) Cu2O
(c) MoS2
(d) Ru
Semiconductors are used as
(a) rectifiers (b) transistors
(c) solar cells (d) all the above
202
16
...
Answer in one or two sentences
17
...
18
...
19
...
Sketch the (a) simple cubic (b) face-centred cubic and (c) body centred
cubic lattices
...
How crystals are classified?
22
...
23
...
Give two example for AB and AB2 type ionic crystals
...
What is imperfection in solids?
26
...
Write a note on the assignment of atoms per unit cell in fcc
...
Write a short note on metallic crystals
...
How are glasses formed?
C
...
31
...
33
...
35
...
37
...
Write the properties of ionic crystals
...
What is super conductivity? Give its uses
...
How Bragg’s equation is used for determining crystal structure
...
Explain the nature of glass
...
Types of ionic crystals AB
and AB2 type are explained with examples
...
The properties of ionic crystals and nature and properties of glasses
are discussed
...
References
1)
C-Kittel, Introduction to solid state physics,Third edition, John Wiley,1966
...
S
...
C
...
3)
Problems in Physical chemistry, K
...
Sharma, Second Edition, 1997
...
204
9
...
( To learn the definition of second law of thermodynamics and the
mathematical statements
...
Also to learn the mathematical representation of entropy
‘S’
...
( To study the significance of ΔG
205
9
...
The mathematical expression which is used to represent the first law of
thermodynamics is ΔE = q-w, where ΔE refers to the internal energy change of
the system when ‘w’ amount of work is done by the system when it absorbs ‘q’
amount of heat and carries out ‘w’ amount of work
...
1
...
2
...
9
...
This is called as the Kelvin – Planck statement of II law of thermodynamics
...
This is called as the clausius statement of II law of thermodynamics
...
This statement is called as the entropy statement of II law of thermodynamics
...
A system always spontaneously
changes from ordered to a disordered state
...
iv) “Efficiency of a machine can never be cent percent”
...
Output can be in the form of any measurable energy or
temperature change while input can be in the form of heat energy or fuel
amount which can be converted to heat energy
...
Consider a heat engine which has
an initial temperature T1 and final temperature as T2, then if T1>T2 then when
some amount of heat is being converted into work, T2 is the lowered temperature
...
Hence, % efficiency can never be achieved as cent percent
...
9
...
Entropy function ‘S’ represents the ratio of the heat involved (q) to the
q
temperature (T) of the process
...
This relation is valid only for
T
reversible processes
...
Entropy (S) and the change in entropy of
207
the process (ΔS) are each state functions even though q and δq path functions
...
S universe = S system + S surroundings
= Constant
Entropy change (ΔS) can be derived for various thermodynamic processes
as below:
Isothermal process (T = constant)
1
q −q
∫ ä q rev = 2 T 1
T
2
ΔS
=
1
Isothermal and isobaric process (T and P = Constant)
2
ΔS
1
= T ∫ δ q p, rev
1
Isothermal, and isochoric process (T and V- Constant)
2
1
ΔS = T ∫ δ q v, rev
1
The term ‘natural process means that the process is spontaneous and does
not need to be induced
...
If this change is positive i
...
if the entropy of the universe
increases, the process will take place spontaneously and irreversibly
...
When the external pressure is less than the internal pressure of a gaseous
system, the gas expands spontaneously
...
Thus, spontaneous
processes are associated with increase in disorder
...
Entropy is a measure of microscopic disorder in the system and also represents
spontaneity
...
For an irreversible (spontaneous) process
...
According to this rule, the heat of vaporisation (ΔHvap)in calories per mole divided
by the boiling point of the liquid in Kelvin is a constant equal to 21 cal deg-1
mole-1 and is known as the entropy of vapourisaiton
...
deg
−1
...
This equation is useful for estimating the molar heat of vaporization of a liquid of
known boiling point
...
ii)
Polar substances like water, alcohol which form hydrogen bonded liquids
and exhibit very high boiling points as well as high ΔHvap
...
Those liquids that obey Troutons rule are said to behave ideally
...
7oC, 61
...
6oC respectively
...
Since the liquids behave ideally, Trouton’s rule is obeyed
...
deg
−1
...
mole
...
184J) (273 + 76
...
71 kJ mol
...
mole-1
...
184 J) (273 + 61
...
376 kJ mol-1
ΔHvap of H S
= (21 cal mol-1
...
184 J) × (273–59
...
74 kJ
...
ii)
Entropy change ‘ΔS’ of a system under a process is defined as the constant
equal to the ratio of the heat change accompanying a process at constant
temperature to the temperature of the system under process
...
ΔSrev =
Äq rev
T(K)
Heat, q is not a state function , But for a reversible process Δq = (q2-q1)
divided by temperature (T) of the process is a state function
...
Entropy increases in all
spontaneous processes
...
iv) When a system undergoes a physical (or) a chemical process, there occurs
a change in the entropy of the system and also in its surroundings
...
For an isothermal process
(T=constant), the entropy change of the universe during a reversible process is
Zero
...
v) The energy of the universe remains constant although the entropy of the
universe tends to a maximum
...
ΔS is positive (ΔS > 0)
...
For a non spontaneous process,
ΔS is negative or (ΔS < 0)
...
The entropy is expressed as calories per degree which is
referred to as the entropy units (eu)
...
per
mole
...
K-1 denoted as eu
...
1 eu = 4
...
viii)
Entropy change is related to enthalpy change as follows:
For a reversible and isothermal process,
Äq rev
...
ΔS is also calculated from ΔH as ΔS =
where T
T
is the temperature of the process involving ΔH, amount of enthalpy change, at
constant pressure
...
The enthalpy of fusion per mole of ice
is 6008 J mol-1
...
mol −1
=
=
Tm(K)
(0 + 273) K
∴ΔSfusion = 22
...
211
Example 3
Calculate the change of entropy for the process, water (liq) to water (vapor,
373K) involving ΔHvap = 40850 J mol-1 at 373K
...
517 J
...
−1 mol−1
ΔS vap = Tb (K) =
373 K
ΔS vap = 109
...
mol-1
1 mole Sn (α, 13oC)
T
1 mole Sn (β, 13oC)
...
transition
ΔStrans
=
ÄH trans
2090 J
...
307 J
...
-1
Example 5
When does entropy increase in a process?
...
b)
In physical process, when a solid changes to liquid, when a liquid changes
to vapour and when a solid changes to vapour, entropy increase in all these
processes
...
Tsystem
= 273+55 = 328 K
...
212
ΔS univ
= ΔStotal = ΔSsystem + ΔSsurroundings
q system
− 75J
=
= − 0
...
260 JK −1
293K
∴ ΔSuniv = ΔStotal = (–0
...
26)
= + 0
...
1 atm and at a certain temperature
absorbs 3710J heat and expands to 2 litres
...
For 1 mole of an ideal gas,
PV = RT,
P = 4
...
1 atm × 2lit × 1 mole
0
...
q 3710 J
=
= = 37
...
T 100 K
ΔS of expansion = 37
...
Standard Entropy
The absolute entropy of a pure substance at 25°C (298 K) and 1 atm
pressure is called the standard entropy, S°
...
When the standard entropies, S° of various substances are known, the
standard entropy change of a chemical process or reaction is written as
ΔS° = Σ S°
products
– Σ S°
reactants
213
This ΔS° is the standard entropy change of the reaction
...
ΔS°f can be calculated for chemical compounds using the S° values
of elements from which the compound is formed
...
Given the
standard entropies of CO2(g), C(s), O2(g) as 213
...
740 and 205
...
ΔS°f
compound
= ΣS°compound – Σ S°elements
= S°CO2(g) – [S°C + S°O2(g)] JK-1
(s)
= 213
...
74 – 205
...
86 JK-1
2
When standard entropy of formation of compounds are known, the standard
entropy change for a stoichiometrically balanced chemical reaction can be written
as,
ΔS°reaction = Σ ΔS°f products – Σ ΔS°f reactants
Example 9
Urea hydrolysis in presence of water to produce ammonia and
carbondioxide
...
55, 16
...
06 and 46
...
mol-1
...
What is the standard entropy change for this reaction
...
ΔS°reaction = Σ S°( product) – Σ S°( reactants)
The reaction is,
CO(NH2)2(aq) + H2O(l) → CO2(g) + 2NH3(g)
ΔS°reaction = (S°CO (g) + 2S°NH3(g)) - (S°urea(aq) + S°H2O(l))
2
= (51
...
01 - 41
...
72) cal
...
81 cal
...
214
ΔS°reaction is a positive value
...
Hence urea hydrolysis
in water is a spontaneous reaction
...
4 Gibbs free energy ‘G’
According to II law of thermodynamics, inorder to predict the spontaneity
of a process entropy of universe is considered
...
It is difficult to determine ΔSsurroundings in most of the physical and
chemical processes
...
For this purpose, “a free energy function” has been introduced by II law of
thermodynamics
...
In an isothermal process, if ΔH and ΔS are the changes in enthalpy and entropy
of the system, then free energy change ΔG is given by,
ΔG = ΔH - TΔS
If 1 and 2 refer to the initial and final states of the system during the isothermal
process, then
ΔG = (G2-G1) = (H2-H1) - T(S2-S1)
from I law of thermodynamics
ΔH = ΔE + PΔV
Therefore
ΔG = ΔE + PΔV - TΔS
...
This is because in an exothermic process, the enthalpy of the final
state (H2) is lower than the enthalpy of the initial state (H1) so that (H2-H1) is
negative and the process take place spontaneously to attain the lower enthalpy
state
...
Combining negative ΔH and positive ΔS, in the
215
expression for free energy change ΔG, at constant temperature, the overall
magnitude of ΔG becomes negative for a spontaneous process
...
ΔG = ΔH – TΔS
Hence, criterion for the prediction of feasibility of a reaction (or) the prediction
of thermodynamic spontaneity of a process based on the free energy change
(ΔG) of the process is given as : when at constant temperature and pressure of
the system, if,
ΔG < 0,
ΔG is –ve, the process is spontaneous and feasible
i
...
ΔG = 0,
the process is in equilibrium
i
...
ΔG < 0,
ΔG is +ve, the process is non spontaneous and non feasible
...
All reversible processes are considered as equilibrium processes
...
T = temperature
...
ii)
G is an extensive property while ΔG = (G2-G1) which is the free energy
change between the initial (1) and final (2) states of the system becomes the
intensive property when mass remains constant between initial and final states
(or) when the system is a closed system
...
iv) G and ΔG values correspond to the system only
...
When, ΔG<0 (negative), the
process is spontaneous and feasible; ΔG = 0
...
v)
ΔG = ΔH – TΔS
...
∴ ΔG = q – w + PΔV – TΔS
But ΔS =
q
and TΔS = q = heat involved in the process
...
The decrease in free energy –ΔG, accompanying a process taking place at
constant temperature and pressure is equal to the maximum obtainable work
from the system other than work of expansion
...
∴ Net work = –ΔG = w – PΔV
...
Standard free energy (G°)
Like standard enthalpy of formation of substances, standard enthalpy change
of a reaction, standard free energy of formation of substances and standard free
energy change of reactions are considered
...
i
...
Hence,
standard free energy change of a reaction which is stoichiometrically balanced, is
217
equal to the difference between the total sum of the standard free energies of
products and the total sum of the standard free energies of reactants, at standard
conditions
...
ΔH°reaction
and ΔS°reaction can be calculated from H°f and S° values of respective product
and reactant molecules at the constant temperature and pressure
...
deg-1 mol-1 respectively
...
Predict the nature of the reaction
...
mole-1
at 400 K; ΔG = –10,000 – 20 × 400
= –18,000 cals
...
Therefore
the reaction is spontaneous (feasible) at both temperatures
...
mol-1 at 25°C
...
Given S°C2H2 = 200
...
8 JK-1 mol-1
...
8 – 3 × 200
...
8 JK-1
ΔG° = ΔH° – TΔS°; T = 298°K
ΔH°
= –631 KJ
...
8)
= –502
...
Since ΔG° is a –ve value, the reaction is spontaneous
...
094 kJ mol-1K-1
...
4 and
–137
...
mol-1 respectively
...
4 + 137
...
2 kJ mol-1
ΔG°reaction = – 257
...
2 – 300 (–0
...
4 kJ
...
Therefore the reaction is spontaneous
...
Therefore the reaction is exothermic
...
Choose the correct answer
1
...
b) ΔH
c) ΔS
d) ΔG
All the naturally occuring processes proceed spontaneously in a direction
which leads to
a) decrease of entropy
c) increase in free energy
b) increase in enthalpy
d) decrease of free energy
219
3
...
b) q = 0
c) ΔE = q
d) PΔV =0
When a liquid boils, there is
a) an increase in entropy
b) a decrease in entropy
c) an increase in heat of vapourisation d) an increase in free energy
5
...
Spontaneous
Reversible
b)
d)
Non-spontaneous
Equilibrium
Which of the following does not result in an increase in the entropy?
a) crystallisation of sucrose from solution
c) conversion of ice to water
7
...
ΔH > 0, ΔS > 0
ΔH > 0, ΔS < 0
b)
d)
ΔH < 0, ΔS > 0
ΔH < 0, ΔS < 0
Change in Gibb’s free energy is given by
a)
c)
9
...
Answer in one or two sentences
10
...
Predict the feasibility of a reaction when
i) both ΔH and ΔS increase
ii) both ΔH and ΔS decrease
iii) ΔH decreases but ΔS increases
12
...
Give Kelvin statement of second law of thermodynamics
...
How ΔG is related to ΔH and ΔS? What is the meaning of ΔG = 0?
15
...
C
...
State the various statements of second law of thermodynamics
...
What are spontaneous reactions? What are the conditions for the spontaneity
of a process?
Exercises
18
...
[Ans : 22
...
What is the entropy change of an engine that operates at 100°C when 453
...
cal of heat is supplied to it?
[Ans : ΔS= 1216
...
Calculate the entropy increase in the evaporation of 1 mole of a liquid when
it boils at 100°C having heat of vaporisation at 100°C as 540 cals\gm
...
06 cal K-1 mol-1]
21
...
9 kJmol-1
and ΔS°reaction = 66
...
calculate ΔG° reaction at 300K
...
07 kJ
...
Calculate the standard free energy change ( ΔG°) of the following reaction
and say whether it is feasible at 373 K or not ½H2(g) + ½ I2(g) → HI(g); ΔH°r
is +25
...
Standard entropies of HI(g), H2(g) and I2(g) are 206
...
6 and 116
...
[Ans : Spontaneous]
23
...
The standard enthalpy
of formation of H2O(l) is 285
...
5, 205
...
3 J
...
[Ans : -237
...
In the reaction ½N2(g) + H2(g) → NH3(g)
...
6, 130
...
5 JK-1 mol-1 respectively
...
67 kJ
...
[Ans : – 46
...
Predict whether the reaction CO(g) + H2O(g) → CO2(g) + H2(g) is spontaneous
or not
...
27, –228
...
38 kJ mole-1 respectively
...
51 kJ]
26
...
Standard
free energies of formation of NH3(g), NO(g) and H2O(l) are 16
...
61,
–237
...
mole-1 respectively
...
02 kJ mole-1]
27
...
83 kJ
...
Will the reaction be spontaneous at 25°C?
[Ans : ΔG°r = –ve; spontaneous]
28
...
2°C
...
pt
...
022 kJ
...
The standard entropy change ΔS°r for
CH4(g) + 2O2(g)
→
CO2(g) + 2H2O(l)
is -242
...
Calculate the standard reaction enthalpy for the
above reaction if standard Gibbs energy of formation of CH4(g), CO2(g) and
H2O(l) are -50
...
36 and - 237
...
[Ans : ΔH°r = –890
...
mol-1]
30
...
98
J
...
Calculate ΔG° of the reaction
...
6 kJ mole-1]
31
...
23 JK-1 at 25°C
...
78, – 394
...
13 kJ
...
[Ans : -2058
...
*
II law of thermodynamics can be stated in many ways
...
For a process with ΔS equals to positive
(or) zero (or) negative value becomes spontaneous (or) equilibrium (or)
non spontaneous respectively
...
ΔS°reaction = ΣS° products – ΣS° reactants
*
Entropy S, is a measure of ‘randomness’ or disorder of the process
...
*
Gibbs free energy function G = H – TS is a state function and for a system
at constant T, the free energy change ΔG = ΔH – TΔS
...
References
1)
Thermodynamics for Chemists by S
...
2)
Physical Chemistry by P
...
Alkins Oxford University Press
...
CHEMICAL EQUILIBRIUM - II
Learning Objectives
( To study the applications of law of mass action for equilibrium reactions
with gaseous phase reactants and products
...
( To learn the statement of Le Chatlier’s principle
...
224
10
...
One of
the reactions produces the products and is known as the forward reaction while
the other produces the reactants from products and is known as the reverse
reaction
...
At equilibrium, reactant and
product molecules are both present in the reaction mixture in definite amounts
...
Consider a general equilibrium reaction at constant temperature represented
by
kf
aA + bB
kr
cC + dD
According to law of mass action, the rate of forward reaction is
Rf = kf [A]a [B]b
and the rate of reverse reaction is
Rr = kr [C]c [D]d
where kf and kr are the rate constants of the forward and reverse reactions
respectively
...
Kc is the
equilibrium constant expressed in terms of molar concentrations of reactants and
225
products
...
Thus
Kc in equilibrium law equation becomes Kp which is the equilibrium constant
expressed in terms of partial pressure
...
1
...
aA + bB + cC +
...
M N
pa p b pe
...
In terms of molar
concentrations of reactants and products
Kc =
[L]l [M]M [N]n
...
For any gaseous component ‘i’ in a mixture, its partial pressure ‘pi’ is related
to its molar concentration ‘Ci’ as
Ci =
where
pi
RT
since pi =
ni
RT
V
ni
= Ci = number of moles of i per litre
...
V
Substituting, concentration terms by partial pressures,
Kc =
(p L /RT)l (pM /RT) m (p N /RT)n
...
p lL p m p n
...
⎝ RT ⎠
B
(l+ m + n +
...
)
226
=
Kp
(RT) Äng
and ∴ Kp = Kc(RT)Δng
where Δng = total number of stoichiometric moles of gaseous products - total
number of stoichiometric moles of gaseous reactants
...
Case (i)
When Δng = 0, the total number of moles of gaseous products are equal to
the total number of moles of gaseous reactants
...
For example, 2H2O(g) + 2Cl2(g)
4HCl(g) + O2(g)
Δng = (4+1) – (2+2)
= 5–4=1
∴ Kp
Kp
= Kc (RT)1
= Kc RT
and Kp > Kc
Case (iii)
When Δng = –ve, the total number of moles of gaseous products are lesser
than the total number of moles of gaseous reactants
...
1
...
For the same reaction,
the dissociation equilibrium consists of the products in the place of reactants and
reactants in the place of products being written at the LHS and RHS of the
equilibrium sign respectively
...
For
example, consider the formation equilibrium reaction of SO3, from SO2 and O2
gases,
2SO2(g) + O2(g)
2SO3(g)
The equilibrium constant, Kc, is given by
Kc =
[SO3 ]2
dm 3 /mole
2
[SO 2 ] [O 2 ]
In the dissociation equilibrium reaction of SO3, the reactants become
products and vice versa
...
Usually, the equilibrium
constant of the dissociation equilibrium is the reciprocal of the equilibrium constant
of the formation equilibrium reaction
...
1
...
Let us assume ‘a’ and ‘b’ moles
of H2 and I2 gases being present in ‘V’ litres of the reaction vessel
...
Then, the equilibrium
concentrations in moles litre of H2, I2 and HI in the reaction mixture will be
(a-x)/V, (b-x)/V and 2x/V respectively
...
Kc =
=
[HI] 2
[H 2 ] [I 2 ]
(2x/V) 2
(a − x) (b − x)
×
V
V
4x 2
V2
= 2 ×
V
(a − x)(b − x)
4x 2
=
= Kp
(a − x)(b − x)
x is also known as the extent of reaction
...
07 moles of H2 and 9
...
38 moles of HI was formed at the equilibrium
...
07 moles
b = 9
...
69 moles
Kc =
2x = 13
...
38)2
(8
...
69)(9
...
69)
= 54
...
The equilibrium constant, Kc, value determines the direction and the extent
of the reaction that may proceed for maximum yield of the product for constant
conditions of temperature, pressure, initial concentrations of reactants or products
as applied on the reaction equilibrium
...
‘Q’ is defined as the ratio of product of initial concentrations
of products to the product of initial concentrations of reactants under nonequilibrium conditions
...
lL + mM
aA + bB
Let [A], [B], [L] and [M] be the actual concentrations present before the
occurrence of equilibrium
...
When Q is greater than Kc (Q>Kc) the reaction will proceed so as to reduce
Q
...
When Q is lesser than Kc (Q
reaction
...
230
Problem
Initially, 0
...
02 moles of HI gas
are mixed in a reaction vessel of constant volume at 300K
...
5 × 10-2]
...
5 × 10 − 2 at 300 K
[H 2 ]e [I 2 ]e
Under non equilibrium conditions
Q=
[HI]2
[H 2 ][I 2 ]
=
(2 × 10 − 2 ) 2
= 4 × 10 − 2
0
...
1
Thus Q > Kc
...
That is, concentrations of H2 and I2
should be increases, which is a reverse reaction
...
Dissociation equilibrium of PCl5
The dissociation equilibrium of PCl5 in gaseous state is written as
PCl5(g)
PCl3(g) + Cl2(g)
Let ‘a’ moles of PCl5 vapour be present in ‘V’ litres initially
...
V
V
[PCl3 ][Cl2 ]e
Kc =
[PCl5 ]e
=
x/V× x/V x 2
V
= 2×
(a − x)/V V (a − x)
231
Kc =
x2
(a − x)V
x is also known as the degree of dissociation which represents the fraction of
total moles of reactant dissociated
...
p Cl 2
p PCl 5
atm
x 2P
atm
(1 − x 2 )
If the degree of dissociation is small compared to unity, then (1–x) is
approximately equal to 1
...
∴ Kc =
x2
V
(or)
x á V But V á
xá
x 2 = Kc × V
1
P
1
P
where x is small, degree of dissociation varies inversely as the square root of
pressure (or) varies directly as the square root of volume of the system
...
2
...
For PCl5 dissociation equilibrium,
x 2P
Kp =
1− x2
P = total pressure = 1 atm
x = 0
...
2)2 (1
...
04
0
...
042 atm
2
1 − (0
...
04 0
...
5, P = ?
P=
=
10
...
042(1 − (0
...
042(1− 0
...
5)
0
...
042(0
...
126 atm
0
...
They are
concentration, temperature and pressure
...
Its presence merely hastens the approach of the
equilibrium
...
Let us consider the effects of changes
in temperature, concentration and pressure, on the equilibrium reactions and the
predictions of Le Chatelier’s principle
...
The concentrations of
reactant and product molecules are constant and remain the same as long as the
equilibrium conditions are maintained the same
...
Since the system possesses equilibrium concentrations
of reactants and products, the excess amount of NO react in the reverse direction
to produce back the reactants and this results in the increase in concentrations of
N2 and O2
...
Concentration of NO is
raised in the reaction mixture
...
Effect of change of temperature
A chemical equilibrium actually involves two opposing reactions
...
If the forward reaction in a chemical equilibrium is endothermic
(accompanied by absorption of heat) then the reverse reaction is exothermic
(accompanied by evolution of heat)
...
0 kJ/mole
In this equilibrium, the reaction of the product formation (NO2) is endothermic
in nature and therefore, the reverse reaction of reactant formation (N2O4) should
be exothermic
...
According to Le Chatelier’s principle, the equilibrium
will shift in the direction which tends to undo the effect of heat
...
Therefore, generally, when the temperature is raised in a
chemical equilibrium, among the forward and reverse reactions, the more
endothermic reaction will be favoured
...
e
...
Effect of change of pressure
If a system in equilibrium consists of reactants and products in gaseous
state, then the concentrations of all components can be altered by changing the
total pressure of the system
...
According to Le Chatlier’s principle, the change can be
counteracted by shifting the equilibrium towards decreasing the moles of products
...
In case of a gas phase equilibrium which is accompanied by decrease in
number of moles of products formed, the effect of pressure can be considered as
follows,
2NH3(g)
N2(g) + 3H2(g)
If the pressure is increased then the volume will decrease proportionately
...
, favours the formation reaction of NH3
...
Thus at higher
pressures, the yield of ammonia will be more
...
Ammonia is commercially
produced in industries from the gaseous elements nitrogen and hydrogen in air by
means of Haber’s process
...
Fe
N2(g) + 3H2(g)
2NH3(g)
ΔH0f = –22
...
Therefore, nearly 300-500 atm pressure is applied on 3:1 mole ratio of H2:N2
gas mixture in the reaction chamber for maximum yield of ammonia
...
By Le Chatlier’s principle, increase in temperature
favours decomposition reaction of ammonia
...
Hence an optimum temperature
close to 500°C-550°C is maintained
...
In order to maintain the
equilibrium conditions, steam is passed to remove away the ammonia as and
when it is formed so that the equilibrium remains shifted towards the product
side
...
Contact Process
This process involves the equilibrium reaction of oxidation of SO2 gas by
gaseous oxygen in air to manufacture large quantities of SO3 gas
...
By Le Chatlier’s principle, when large pressure is applied, forward
reaction is favoured
...
SO3 production
is an exothermic reaction
...
However, lowering of temperature prolongs the time of attainment
of equilibrium
...
The most widely used catalyst for SO3 production is porous vanadium
pentoxide (V2O5)
...
Only dry and
pure SO2 and O2 gases are used over the catalyst
...
SO3 is the anhydride of H2SO4
...
236
SELF EVALUATION
A
...
2
...
c) none
d) K1 = 1/K22
c) K2 = (K1)2
b) K1 = 1/K2
B is K1
In the reversible reaction 2HI
H2 + I2, Kp is
a) greater than Kc
b) less than Kc
c) Equal to Kc
d) Zero
4
...
For the homogeneous gas reaction at 600 K
4NH3 + 5O2
4NO(g) + 6H2O(g)
(g)
(g)
the equilibrium constant Kc has the unit
a) (mol dm-3)-1
b) (mol dm-3) c) (mol dm-3)10
d) (mol dm-3)-9
6
...
0
dm3 vessel in which it partially dissociates at high temperature
...
0 mole of ammonia remains
...
An equilibrium reaction is endothermic if K1 and K2 are the equilibrium
constants at T1 and T2 temperatures respectively and if T2 is greater than T1
then
a) K1 is less than K2
b) K1 is greater than K2
c) K1 is equal to K2
d) None
[Ans: 1-a; 2-b; 3-c; 4-d, 5-b, 6-a and 7-a]
237
B
...
Dissociation of PCl5 decreases in presence of increase in Cl2 why?
9
...
11
...
13
...
CO(g) + H2O(g)
N2O4(g)
CO2(g) + H2(g)
2 NO2(g)
State Le Chatelier’s principle
...
Calculate Δng, for the following reactions
i) H2(g) + I2(g)
2HI(g)
ii) 2H2O(g) + 2Cl2(g)
4HCl(g) + O(g)
C
...
Derive the relation Kp = Kc (RT)Δng for a general chemical equilibrium
reaction
...
State Le Chatelier’s principle
...
N2g) + O2(g)
2NO(g)
17
...
D
...
The equilibrium constant Kc for A(g)
B(g) is 2
...
The rate constant
of the forward reaction is 0
...
Calculate the rate constant of the
reverse reaction
...
0 sec-1]
19
...
Total pressure of the reaction mixture is 60 atm
...
[Ans: 10 atm, 20 atm, 30 atm]
20
...
What will be the degree of dissociation
assuming only a small extent of 1 mole of PCl5 has dissociated?
[Ans: x = 1
...
At temperature T1, the equilibrium constant of reaction is K1
...
Predict whether the equilibrium is
endothermic or exothermic
...
At 35°C, the value of Kp for the equilibrium reaction N2O4
2NO2 is
0
...
2382 atm
...
5768]
23
...
75 × 10-6 at 790°C
...
Hint: Kp = Kc(RT)Δng
[Ans: Kp = 3
...
For the equilibrium 2SO3(g)
constant is 4
...
At equilibrium, if the concentrations of SO3
and SO2 are 0
...
15M respectively
...
[Ans: 0
...
Hydrogen iodide is injected into a container at 458°C
...
At equilibrium, concentration of HI is found to
be 0
...
04 × 10-2M, at 458°C
...
[Ans: Kc = 2
...
Dissociation equilibrium constant of HI is 2
...
At
equilibrium, concentrations of HI and I2 are 0
...
15M respectively
...
[Ans: [H2] = 1
...
The equilibrium constant for the reaction 2SO3(g)
2SO2(g) + O2(g) is
0
...
Calculate the equilibrium constant for the reaction
2SO3(g) at the same temperature
...
67 mol-1 dm3]
28
...
[Ans: 10
...
How much PCl5 must be added to one litre volume reaction vessel at
250°C in order to obtain a concentration of 0
...
0414 mol dm-3 at 250°C
...
341 moles PCl5]
30
...
0 atm is 1
...
Calculate equilibrium constant in molar concentration
(Kc) at same temperature and pressure
...
aA + bB
K=
•
lL + mM
[L]l [M]m k f
=
[A]a [B]b k r
Equilibrium constants in terms of molar concentration (Kc) and partial
pressures (Kp) are related as Kp = Kc(RT)Δng for gaseous reactants and
products in equilibrium
...
•
•
•
The reaction quotient (Q) of non equilibrium concentrations to equilibrium
concentrations of reactants and products is related to equilibrium constant
(K) of an equilibrium reaction as: when Q
...
of moles dissociated
Degree of dissociation ‘x’= ________________________
total no of moles present initially
Application of Le Chatelier’s principle to explain effect of pressure,
temperature and concentration on the equilibrium reactions
...
ii) For endothermic equilibrium, increase in
temperature favours product formation
...
References
1
...
Glasstone, Mac Millan, India Limited
...
Physical chemistry by G
...
240