FOR 2020 - PG- PHYSICAL CHEMISTRY –I BY
Long live Scientist Newland
FOR 2020 - PG- PHYSICAL CHEMISTRY –I
Dr. C.SEBASTIAN, AP / CHEM. R.V.GOVT.ARTS COLLEGE,CHENGAIPATTU MOB: 9444040115 Page 2
1. Lithium Sodium Potassium
Rubidium Cesium Francium -2
They are known as Alkali metals-2( long)
2. Beryllium Magnesium Calcium
Strontium Barium Radium -2
They are known as Alkaline earths -2( long)
3. Boron belongs to third A group
Aluminium Gallium Indium Thalium -2
Are the members of Boron family -2
They form first group of p- block elements-2(long)
4.Carbon Silicon Germanium
Titanium Zirconium Hafnium – 2( long)
5. Phosphorous Arsenic Antimony Bismuth
Belong to fifth A Nitrogen family – 2
Sulphur Selenium Tellurium Polonium – 2
Belong to sixth A Oxygen family-2( long)
6. Fluorine Chlorine Bromine Iodine
Astatine are the five halogens-2
They do belong to seventh A group-2
They form salt with strong bases.-2( long)
7. Helium Neon Argon Krypton Xenon
Radon are the rare gases-2
They are known as noble gases -2
Which are basically inert nature-2( long)
8.Elements forming colour compounds
Scandium is the first member-2
Yttrium forms the first of second row -2( long)
9 Vanadium Niobium Tantalum
Chromium Molybdenum hard Tungsten-2
Which are Ferric Iron Cobalt Nickel – 2
Ruthenium Rhodium Palladium-2
Zinc Cadmium Mercury are their neighbours-2
They do belong to transition elements-2 ( long)
12.Elements following Lanthanum
Cerium Prasodium Neodymium-2
13.Promithyum Samarium Europium
Gadalonium Terbium Dysprocium-2
Luetecium are the rest of lanthanides-2. ( long)
14.Actinides are also fourteen numbers
Which are following element Actinium – 2
Thorium Protactinium Uranium -2
15.Neptunium Plutonium Americium
16. Rutherfordium Dubnium Seaborgium
Bohrium Hassium Meitnerium - 2
Darmastidium Roentgenium Copernicium - 2
17. Hydrogen resembles alkali metals
Halogens also same as hydrogen
So we are in need of your help in
Predicting the position of Hydrogen
Composed by: Dr. C.SEBASTIAN A NTONY SELVAN. ASST. PROF in CHEMISTRY,
R.V GOVT.ARTS COLLEGE, CHENGALPATTU ,
MOB: 9444040115 . OCT - 2017
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FOR 2020 - PG- PHYSICAL CHEMISTRY –I
Problem number refers to the problem number of on line test of this topic Page 158
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REGULATIONS (w.e.f. 2015-2016)
Inst.
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100
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Unit I: Chemical Kinetics - I
Effect of temperature on reaction rates - collision theory - molecular beams - collision cross sections - effectiveness of collisions - probability factors - potential energy surfaces – transition state theory - partition functions and activated complex. Eyring equation - estimation of free energy, enthalpy and entropy of activation and their significance.
Unit II: Chemical Kinetics - II
Reactions in solutions - effect of pressure, dielectric constant, ionic strength and salt effect - kinetic isotopic effects - linear free energy relationships-Hammett and Taft equations - Homogeneous catalysis - Acid base catalysis - mechanisms and Bronsted catalysis law.
Unit III: Group theory - I
Symmetry elements and operations. Concepts of groups, Sub groups, class, order, Abelian and Non-Abelian point groups. Products of symmetry operations and group multiplication table, point groups-identification and determination-reducible and irreducible representations-Direct product representation-orthogonality theorem and its consequences-character table – construction(NH3, H2O). Symmetry adapted linear combinations of atomic orbitals (water as example)
Unit IV: Group theory - II:
Hybrid orbital in non-linear molecules (CH4, XeF4, BF3, SF6 and NH3). Determination of representations of vibrational modes in non-linear molecules (H2O, CH4, XeF4, BF3, SF6 and NH3) Symmetry selection rules for infrared, Raman and electronic Spectra - mutual exclusion principle. Electronic Spectra of Ethylene and formaldehyde-Applications of group theory.
Unit V: Quantum Chemistry - I:-
Inadequacy of classical theory - black body radiation, photo electric effect - the Compton effect - Bohr's Quantum theory and subsequent developments -wave particle duality- de Broglie equation, Heisenberg uncertainty principle.
SEMI MICROW ANALYSIS – PROCEDURE – VIVA – Q &A
UNIT I
2. Collision Theory
3. Molecular Beams
10. Eyring Equation
11.- Estimation Of Free Energy, Enthalpy And Entropy Of Activation And Their Significance.
. EFFECT OF TEMPERATURE ON RATES OF A REACTION
The rate of the reaction depends on temperature of the system. When temperature increases the rate of the reaction increases. This is because when temperature increases number of collision between molecules increases. According to collision theory, the rate of the reaction depends on the number of collisions. Therefore the increase in temperature increases the rate constant of the reaction.
The relation between temperature and rate constant is given by
k = A
R – gas constant
This is known as Arhenius equation.
k = A ---------1
taking ‘ln’ on both sides
ln k = ln A - -----------2
If the experiment is carried out at different temperatures T1 and T2 with rate constants k1 and k2 , then
ln k1 = ln A - -------------3
ln k2 = ln A - -------------4
subtracting equation 3 from 4 , we get
ln k2 - ln k1 = - [ - ]
2.303 log ( = + [ ]
log ( = + [ ]
using this relation the rate constant at any desired temperature can be calculated.
Determination of Arhenius constant and Energy of activation:
Arhenius equation is k = A
Taking ‘ln’ on both sides
ln k = ln A -
changing ‘ln’ into ‘log’
Dividing by 2.303
= - ( )+ log A
This is of the form y = mx + c . Therefore The plot of ‘log k ‘vs’ ‘ gives a slope equal to
Problem :3 [Problem number refers to the problem number of on line test of this topic ] The activation energy of a reaction is 83.14 J. The rate constant is 1 s -1 at 10 K If the temperature is doubled what will be the rate constant.?
Solution:
log k2 – log ( = 0.22
log k2 – 2 = 0.22
= 162.98 s -1
Problem 4. Thermal decomposition of 2- nitro propane to yield propylene whose activation energy is 83.14 J. The frequency factor is 3 . Find its rate constant at 100 K if the frequeny factor is 10 units
Solution: k = A
= 9.048 s -1
Problem5. At what temperature the rate of reaction becomes 10 times for the reaction proceeding at 10 K with energy of activation 831.4 J
Solution:
= 12.99 K
Problem6 :. The rate constant of a reaction at 100 K was found to be ten times that at 10 K What will be the activation energy of a reaction?
Solution:
Ea = ?
= 10
= 2.12 J
Problem 7 Decomposition of benzene diazonium chloride is first order reaction. The plot of log k vs I/T gives a slope of 100. Find the energy of activation.
Solution:
slope =
=1.914KJ
Problem8.The activation energy and rate constant of certain reaction is 8314 J/mol and 100 units The change in rate constant when the temperature changes from 90 K to 100 K is [ = 0.3328]
Solution:
100 0.338
Problem 9. The decomposition of A and B have same pre-exponential factor. The rate constant at 100 K of A is same as for B at 200 K. If Ea for A is 50 J the Ea for B is
Solution:
= 100 J
Problem 10.: For a first order reaction the plot of log k vs gives
Problem 11.: Dissociation of HI gives the following graph Calculate the energy of activation. [ 3 marks]
Solution:
slope =
=1.914KJ
Problem 12.: The activation energy of a reaction is 230.3 Cal mol-1 and the value of rate constant at 100°C is 1000 . Calculate the frequency factor, A.
Solution: k = A
= 1000
= 1000
= 1319
13. Direction : This question contains two statements. Assertion and Reason and also has four alternatives only one of which is the correct answer. Select one of the codes
Code
a. Assertion is true ,Reason is true , Reason is the correct explanation of the Assertion
b. Assertion is true, Reason is true , Reason is not the correct explanation of the Assertion
c. Assertion is true, Reason is false
d. Assertion is false, Reason is true
Assertion: At high temperature reaction proceeds fast
Reason: Energy of activation is less at high temperature.
Solution:
Assertion is true reason is true, , reason is the correct explanation of the assertion
14. The activation energy and frequency factor of a reaction at 100 K was found to be 831.4 and 2 , its rate constant is [ = 4.5 ]
Solution:
k = 2
= 9
15. The rate constant of a reaction at 50 K was found to be doubled that at 40 K. Its energy of activation is . [ln (2) =0.693]
a. 152 KJ b. 1152 KJ
c. 1.52 KJ d. 45.5 KJ
Solution:
ln ( ) =
= 1152 .3J
16. From the following Arrhenius plot, the log of frequency factor of the reaction is
Solution:
log k = - ( ) + log A
intercept = log A
log A = 1000
17. From the following Arrhenius plot, the Arrhenius frequency factor of the reaction is
Solution:
log k = - ( ) + log A
A =
18. The Arrhenius plot, for reaction I and II are shown below
Ratio between Arrhenius factor of reaction II and I is
Solution:
Arrhenius equation in logarithmic form is log k = - ( ) + log A
intercept = log A
=
COLLISION THEORY
According to collision theory the rate of chemical reaction depends on the number of collisions.
Rate = number of effective collision orientation facxtor
1 Number of collisions
Let us consider a system which contains NA molecules of ‘A’ and NB molecules of ‘B’
Consider a molecule A which moves with velocity ‘C’ relative to the molecules of B.
As the molecule A moves , it strikes all the molecules of B ,which are situated in the cylindrical volume traced by it.
Let’ σ’ be the diameter of the cylinder
Number of collisions of molecule A
=Volume swept out by molecule A × the number of molecules of B in that volume.
= V -----------------------1
Volume swept out by molecule A
V = π ( )2 × height of the cylinder ---------------3
height of the cylinder = The distance travelled by the molecule A in one second
= velocity × time
V = π ( )2 × -----------------4
From the kinetic theory of gases, average velocity = . Where k – Boltzmann constant
T – temperature, once all molecules are in motion we must use relative velocity.
=
=
Total number of collisions of all molecules of A = NA 2 ×
= NA -------3
= m and NA = = N Substituting in equation 3 we get
Total number of collisions =
2. Energy Factor
All collisions do not produce product. Only those collisions, with energy greater than or equal to threshold energy will lead to product formation. Such collisions are called effective collisions Consider a bi molecular reaction between molecule A and B
Number of effective collisions = Z
=
3. Probability Factor
The fraction of collisions that have the proper orientation is called the probability factor (P)
Therefore the rate constant k = P ×
= P ×
REVIEW:
According to collision theory the rate of chemical reaction depends on the number of collisions.
Rate = number of effective collision orientation facxtor
1 Number of collisions
2. Energy Factor
3. Probability Factor
The fraction of collisions that have the proper orientation is called the probability factor (P)
Therefore the rate constant k = P × 2
Problem 21[Problem number refers to the problem number of on line test of this topic]. Calculate the number of collisions per second in 1 cm 3 of homonuclear diatomic gas which contains Molecules / cm 3 at 100K and . The radius of the gas is 1.5 × 10 -5 cm . value at this temperature is 1 × 10 8 [ TRB]
Solution:
N =
σ A = 2 × 1.5 × 10 -5 cm [ diameter = 2 × radius ]
= 3 × 10 -5 cm
= 72
Problem22.Calculate the number of collisions per second between nitrogen and oxygen molecules in 1 cm 3 of an equi molar mixture of the gases which contain molecules / cm 3 . Collision diameter of nitrogen and oxygen are 3.0× 10 -8 cm 2.0 × 10 -8 cm value at this temperature is 1 × 10 8
Solution:
= 2 3.0× 10 -8 2.0 × 10 -8 1 × 10 8
= 48
Problem 23. Calculate the rate constant for the decomposition of HI at 100 K using collision theory assuming molecules are present in 1 cm3. Collision diameter of HI is 1 × 10 -5 cm given at 100K the factor (8×π kT ) = 2× 10 -10 The energy factor and probability factors are 1.5 and 0.2 respectively
Solution:
= 0.2 × × × 2× 10 -10 × 1.5
= 2.4 × 10 20
Limitation:
1. Collision theory yields good results with simple molecules like H2 , I2 etc. but for bigger molecules , the calculated value by collision theory, deviates from experimental values.
2. The orientation factor can not be simply calculated.
3. Collision theory treats molecule as hard spheres.
MOLECULAR BEAMS
Definition:
Molecular beam is produced by heating the substance to high temperature At this high temperature the substance becomes vapor. The vapor when passed into a low pressure chamber through a small orifice it comes out as molecular beam.
Application:
Molecular beams are used to fabricate thin film and artificial structure The formation of thin
film is called Molecular Beam Epitaxy (MBE)
Solar cells, Computer chips, semiconductor laser all are fabricated by MBE.
Method :
In this method, the substrate is placed on a substrate heater.
The substance to be deposited is heated to high temperature so that it becomes vapor.
The high pressure chamber is called ‘gun’.
From the gun the vapors are letting out through the orifice.
The vapors become a fine beam of particles while coming out of the orifice.
This beam is directed towards the substrate.
The molecules land on the surface of the substrate , condense and build up an ultra thin layer.
Depending upon the requirement different kinds of molecules can be shoot up.
For each kind of molecular beams we need different ‘gun’
26. Which is not true ?
a. Molecular Beam Epitaxy is formation of thin film on substrate
b. Molecular beam is produced by heating the substance to high temperature
c. Molecular beam is produced by cooling the substance below 0oC
d. Solar cells, Computer chips, semiconductor laser all are fabricated by MBE
COLLISION CROSS SECTION
The collisional cross section is defined as the area around a particle in which the center of another particle must be in order for a collision to occur.
1. The collisional cross section is an "effective area" that quantifies the likelihood of a scattering event when an incident species strikes a target species.
2. It is denoted by ‘ σ’  and measured in units of area.
3. It can be calculated using the following equation:
σ=π (+ )2
- Radius of atom A
- Radius of atom B
4. Assume that the two particles involved in the collision are the same in size and have the same radius.
σ=π (+ )2
5. Anytime the center of another particle is within this area, there will be parts of the two particles that will overlap, touch, and cause a collision.
Problems
Consider the following reaction: H+H→H2 The radius of hydrogen is 5.3×10−11m
What is the collisional cross section for this reaction?
Solution
Collisional Cross Section=π(2r)2
=3.53×10−20m2
1. H2+O2→H2O The radius of oxygen is 4.8 x 10-11 m. What is the collisional cross section for this reaction?
2. N2+O2→N2O. The radius of nitrogen is 5.6 x 10-11 m. If the distance between the two centers is 2.00 x 10-19 m, is there a collision between the two molecules?
3. F+F→F2  The center of the two atoms are 3.50 x 10-20 m apart from each other. How much closer do the centers have to be in order for a collision to occur?
Answers
1. The radius of H2 is 2(rH)=1.06 x 10-10 m. The radius of O2 is 2(rO)=9.6 x 10-11 m.
CollisionalCrossSection σ=π (+ )2
=1.28×10−19 m2
2. Yes, there is a collision. The radius of N2 is 2(rN)=1.12 x 10-10 m. The radius of O2 is 2(rO)=9.6 x 10-11 m.
CollisionalCrossSection σ=π (+ )2
=1.36×10−19 m2
1.36 x 10-11 m is the furthest the two molecules can be and still get a collision. Since 2.00 x 10-11 m is larger than that distance, the center of one molecule is not in the collisional cross section of the other molecule. Therefore, no collision occurs. The molecules are too far apart.
3.CollisionalCrossSection σ=π (+ )2
=2.22×10−20 m2
Because the molecules are 3.50 x 10-20 m apart, the center of one F is not in the Collisional Cross Section of the other F. They must be closer.
(3.50×10−20)−(2.22×10−20) =1.28×10−20
The molecules must be at least 1.28 x 10-20 m closer.
. EFFECTIVENESS OF COLLISIONS:
 Effective collision sare those that result in a chemical reaction. In order to produce an effective collision, reactant particles must possess some minimum amount of energy. This energy, used to initiate the reaction, is called the activation energy.
PROBABILITY FACTOR:
Translational energy = ½ mc2
Vibrational energy = ½ μ ( )2 + ½ k l2
Where c- linear velocity, I – moment of inertia, ω – angular velocity , , μ = reduced mass
L – distance from equilibrium position , k – force constant.
=
The probability that a molecule posses an energy greater than Ec ( collision energy) is given by
P = dEc
P = /
When the energy is composed only by two square terms then, put n= 2
P =
R = ZAB ×
= nA nB ( σAB ) 2 × (8πKT) ½ ×
Problem 1: What is the probability that at 100 K the energy of collision is 8.314K J in excess of average energy?
P =
=
=
Problem 2: Find the probability that a mplecule having 9 degrees of vibrational freedom has 8.31K J in excess of average energy at 100 K
P = /
= 2(9) + 2
. POTENTIAL ENERGY SURFACES:
The PES is the energy of a molecule as a function of the positions of its nuclei r. This energy of a system of two atoms depends on the distance between them.
At large distances the energy is zero, meaning “no interaction”.
The activation energy of chemical reactions can be calculated using the method of potential energy surfaces.
This method involves making a plot of energy as a function of various inter atomic distances in the complex that is formed.
If the reaction is , between two atoms A and B , then only one distance ( A-B) is involved and one could plot potential energy against this distance.
So the result would be two dimensional diagram
if linear A-B-C complexes are considered two distances ( A-B, B-C) are involved , and a three dimensional diagram is required.
1.It is a graphical function that shows a relationship between the energy and geometry.
2. energy is plotted on the vertical axis and geometric co – ordinates like bond distance valence angle etc are plotted on the horizontal axis
4. If the reaction is , between two atoms A and B , then only one distance ( A-B) is involved and one could plot potential energy against this distance. So the result would be two dimensional diagram
6. if linear A-B-C complexes are considered two distances ( A-B, B-C) are involved , and a three dimensional diagram is required.
7.
Consider the reaction between hydrogen atom and hydrogen molecule. The reaction is H + H - H – > H - H + H
The atom H and the molecule H - H come close to gether and give rise to anactivated complex H - H – H .
During the approach of the atom ,the potential energy of the system increases.
After this activated complex has been formed , the energy gradually decrease as H - H is formed.
Let as assume the distance between H and H is designated as r1, and the distance between H and H is as r2
the plot of energy against r1 and r2.
Diagram: H - H
r2 Q P r1
On the left- hand face of the diagram , the distance r2 is great , so we can assume H - H .Similarly , on the right- hand face of the diagram , there is a curve for the diatomic molecule H - H, the distance r1 now being sufficiently great so that H is far away.
The course of the reaction is considered as the transition from the point P to the point Q on the potential energy surface. The point R corresponds to the system H - H - H
Calculation of activation energy:
2. The activation energy of chemical reactions can be calculated using the method of potential energy surfaces.
Imagine that the atom H is removed to infinite, there remain the diatomic molecule H -H and the coloumbic and exchange energies for this are designated as A and . Similarly if H is removed to infinity a diatomic molecule H - H is left whose coloumbic and exchange energies are B and .The removal of H leaves the diatomic species H - H , the energies for which are called C and . According to London’s treatment the energy of the system is shown as below.
A+ C+
B+
The energy is given by E = A+B+C+ ½ [a – )2 + ( - )2 + ( - )2 ]
Experimentally it has been found out that the activation energy of the reaction is 8.8 kcal/mole.
TRANSITION STATE THEORY
Absolute Reaction Rate theory(ARR theory)/ Activation Complex theory/
According to this theory, when molecules collide with each other, they form an activated complex which is in transition state This complex either converted into products or changes back into reactants. The complex AB* is in equilibrium with the reactants Hence the name Activated Complex Theory( ACT) OR Transition State Theory(TST).
Thermodynamic formulation of Activated Complex theory:
Consider the reaction A+ B → Products
According to this theory,
AB* is the activated complex
K* - equilibrium constant,
k – rate constant
Since the complex AB* is in equilibrium with the reactants
K* =
Rate of the reaction = k K* -----------------2
Rate constant =
From quantum mechanics , k =
Rate constant = K* --------------4
The relation between free energy change( G) and equilibrium constant ( K*) is given by
G = - RT ln K*
Substituting in the above equation we get
ln K* =
Rate constant =
.This is known as Eyring ’s equation
The expression for rate constant on the basis of Activated Complex theory , involves only thermodynamic factors like enthalpy change and entropy change, which are absolute in nature and hence the name Absolute Reaction Rate Theory
PARTITION FUNCTIONS AND ACTIVATED COMPLEX.
Partition function ( q) is defined as the quantity, which indicates how the gas molecules of an assembly are distributed or partitioned among the various energy levels. If the energy of i th level is i and gi is the degeneracy, then partition function (q) is given by
q =
The partition function, q is the sum of all possible energy states. 
. EYRING’S EQUATION.
Rate constant =
The expression for rate constant on the basis of Activated Complex theory , involves only thermodynamic factors like enthalpy change and entropy change, which are absolute in nature and hence the name Absolute Reaction Rate Theory.
ESTIMATION OF FREE ENERGY, ENTHALPY AND ENTROPY OF ACTIVATION:
1.FREE ENERGY OF ACTIVATION:
k =
rearranging, =
= ln ( )
= RT ln ( )
Using this relation , the free energy of activation can be calculated
ENTHALPY OF ACTIVATION
dividing by T , =
rearranging = - ( ) + ln (+
A plot of ln ( ) versus gives a straight line with a slope equal to
Slope =
ENTROPY OF ACTIVATION
Using this relation S can be calculated.
Problem -1 : Reaction between methyl iodide and sodium ethoxide was carried out at 100 K. ln ( ) vs 1/T gave a slope of 10 units and intercept of 20.31. Find the enthalpy of activation and entropy of activation.
slope =
4.Salt Effect
7. Taft Equations
8. Homogeneous Catalysis
10. Bronsted Catalysis Law.
REACTIONS IN SOLUTIONS
Rate of reaction in solution is lower than that of gaseous reaction because in solution the molecules are bound with solvent and hence they can not move as like gaseous molecules. Therefore number of collisions will be lesser than in gaseous reactions which results lower rate.
EFFECT OF PRESSURE( VANT HOFF EQUATION)
The rate of the reaction is directly proportional to the hydrostatic pressure of the system. The relation between pressure and rate constant can be derived as follows.
The free energy and rate constant is related as G = - RT lnK
Differentiating with respect to pressure, at constant temperature,
= - RT -----------------------1
The relation between change in free energy and change in pressure is given by
G =H + PV
Differentiating with respect to pressure at constant temperature, we get
T = V --------------------------2
rearranging d (ln k ) = - dp
integrating on both sides = - dp
ln k = - p + c
when p = 0, k = k 0 the above equation becomes,
ln k0 = 0 + c
substituting the value of c , ln k = - p + ln k0
ln k = + ln k0 - p
ln k - ln k0 = - p
ln ( ) = - p
This is of the form y = - mx Therefore plot of ln ( ) versus p will give straight line passing through the origin.
slope at any pressure is equal to -
Problem 1. The rate constant os a reaction in solution was found to be 250 and 25 at a pressure of zero and 23 atmosphere respectively. Find the volume of activation.
P = 0, k0 = 250 T = 10 K, P = 23 atm k = 25 V = ?
ln ( ) = - p
V = - ln ( )
= 8.314
Problem 2. What is the Pressure of the reaction whose rate constant is 10 times greater than at zero atmosphere pressure the volume of activation at 10 K being 8.314
K0 = 10 k , V = 8.314, T = 10 K , P = ?
ln ( ) = - p
+ p = ln ( )
p = ln ( )
= 23 atm
Problem 3. At what temperature volume of activation will be 2.303 if the rate constant at 83.1 atm is 10 times greater that that at zero atm.pressure.
K0 = 10 k , V = 2.303, P = 83.1, T = ?
ln ( ) = - p
+ p = ln ( )
T = ln ( )
= 10 K atm [ log 10 = 1 ]
Problem 4. What is the rate constant of the reaction at 10 atm if the rate constant at zero atmosphere pressure is 20 . the volume of activation at 10 K is 8.314
k0 = 20, T = 10 K, P = 10 atm k = ? V = 8.314
ln ( ) = - p
ln k – ln k0 = - p
EFFECT OF OF SOLVENT(DIELECTRIC CONSTANT)
The ability of substance to store electrical energy in an electric field is known as dielectric constant(DC).
DC =
Double sphere model:
Consider an ionic reaction, involving two ions A and B with charges z1e and z2e in solution in which the solvent has dielectric constant . Let the ions are initially at infinite distance from each other. In activated state, they are considered to be intact forming a double sphere and hence the model is called double sphere model.. The distance between the two ions in the activated state is d AB. The situation is represented as below.
d AB
initial state activated complex
According to Absolute Reaction Rate theory, the rate constant (k) is related to free energy by the equation
k = e –
The free energy of activation for N molecules is given by
G = G electrostatic +G non-electrostatic
i.e G = G e .s +G n. e .s.
Substituting the value of G,
k = e –
= e – e –
To find G e .s
G e .s = work done to move the ions from ∞ to a distance of ‘d’
= force × displacement
= dx [ force =
= +
Therefore electrostatic free energy for N molecules is G electrostatic = N
Substituting in 1 ,we get
k = k0
= k0 [ = k ]
ln k = - ( ) + ln k0
ln k = ln k0- ( )
This shows that the rate constant of reactions in solution inversely related with the dielectric constant of the solvent used. A graph drawn between ln k and 1/ is a straight line with a negative slope equal to . From the slope, the distance d AB can be calculated.
Problem 1. In the double sphere model, the graph drawn between log k and 1/ of an uni-uni valent electrolyte gives a slope equal to . Find the distance between the ions.
Slope =
= 1 pm [ for uni- uni valent electrolyte Za = Zb = 1 ]
[ I pico meter = meter]
Problem In the double sphere model, the graph drawn between log k and 1/ of an uni-uni valent electrolyte is given Find the distance between the ions if =0.05
Solution:
Slope =
= 0.1 pm
Problem The dielectric constant of CH3COOH, CH3OH and H2O are 6, 33 and 78 respectively. The rate of reaction in these solvents will be ?
Solution:
[ BRONSTED-BJERRUM equation , primary kinetic salt effect]
Ionic Strength is the measure of electrical intensity, due to the presence of ions in the solution. It is given by = ½
The effect of ionic strength on the rate constant of a reaction involving ionic species is called primary salt effect or kinetic salt effect.
Consider a bimolecular reaction between A and B in solution giving an activated complex [X] * which decomposes into product.
A+ B ↔ [X]* Product
Rate of the reaction in terms of activity coefficient, ( f) is given by
k = k0 [,f- activity coefficient, ]
Taking log on both sides log k = log k0 + log f A + log f B – log f X ------------4
The relation between activity (f) and ionic strength is given by Debye Huckel limiting law as log f = - Q z2 ,
substituting in 4
log k = log k0 - Q - Q + Q ( zA + zB ) 2
= log k0 - Q [ + - ( zA + zB ) 2 ]
= log k0 - Q [ + - zA 2 - zB 2 - 2 z A zB ]
= log k0 - Q [ - 2 z A zB ]
= log k0 + 2 Q z A zB
For aqueous solution, Q= 0.51, substituting
log k = log k0 + 2(0.51) z A zB
= log k0 + 1.02 z A zB
log k - log k0 = 1.02 z A zB
log = 1.02 z A zB ---------------------------------5
.
1. If one of the reactant is a neutral molecule then z A zB = 0 and the rate constant will be independent of ionic strength. Example: base- catalysed hydrolysis of ethyl acetate.
CH3 COOC2H5 + OH – z A zB = 0
2.For ions of similar charges z A zB is positive , a positive slope will be obtained.
S2O3 2- + I - z A zB = 2,
3. For ions of un similar charges z A zB is negative, a negative slope will be obtained.
H+ + Br - + H2O2 z A zB = -1,
SALT EFFECT [ SECONDARY SALT EFFECT]
In the case of weak electrolyte, addition of salt causes change in ionic strength of the medium and also change the concentration of catalytic species. This increases the rate of the reaction. This is known as secondary salt effect.
KINETIC ISOTOPIC EFFECTS
The kinetic isotope effect (KIE) is the change in the reaction rate of a chemical reaction when one of the atoms in the reactants is replaced by one of its isotopes.
It is the ratio of  rate constants  for the reactions involving the light (kL) and the heavy (kH) isotopically substituted reactants:
{\displaystyle KIE={\frac {k_{L}}{k_{H}}}}This change in rate results from heavier  isotopologues  having a lower velocity and an increased stability from the higher  dissociation energies  when compared to the compounds containing lighter isotopes.
Primary kinetic isotope effects
When a bond to the isotopically labeled atom is being broken in the rate-determining step of a reaction it is called primary kinetic isotope effect.
Secondary kinetic isotope effects
A secondary kinetic isotope effect is observed when no bond to the isotopically substituted atom in the reactant is broken or formed in the rate-determining step.
types of kinetic isotope effect experiments involving C-H bond functionalization:
A) KIE determined from absolute rates of two parallel reactions
In this experiment, the rate constants for the normal substrate and its isotopically labeled analogue are determined independently, and the KIE is obtained as a ratio of the two.
Uses:
The study of kinetic isotope effects can help the elucidation of the  reaction mechanism  of certain chemical reaction
INFLUENCE OF SUBSTITUENTS ON REACTION RATE
(HAMEET EQUATION)
According to Hammet, the rate constant of aromatic compounds having substituent at meta or para position is related to the value for unsubstituted compound in terms of two parameters ρ and σ
log k = log k0 +
where k is the rate constant for the substituted compound, k0 is the rate constant for the parent ( unsubstituted) compound, σ – substituent constant and ρ – reaction constant.
Similarly, the equilibrium constant of aromatic compounds having substituent at meta or para position is related to the value for unsubstituted compound in terms of two parameters ρ and σ
log K = log K0 + σρ
where K and K0 are the equilibrium constant for substituted and the parent ( unsubstituted) compound. σ ,is substituent constant while ρ is reaction constant.
Equations 1 and 2 are called Hammett equations.
Significance of σ and ρ
σ is substituent constant ρ is reaction constant
A value of unity is chosen for ρ, for the ionisation of benzoic acid in aqueous solution. Reactions with positive ρ values are accelerated by electron withdrawing group
with negative ρ values are retarded by electron withdrawing group
Hammet equations are linear free energy relations.
G = G 0 = constant
Substituent constants
The   chemical equilibrium  for which both the substituent constant and the reaction constant are arbitrarily set to 1 for : the  ionization  of  benzoic acid  or  benzene carboxylic acid  (R and R' both H) in water at 25 °C.
Having obtained a value for K0, a series of equilibrium constants (K) are now determined based on the same process,
1.with variation of the para substituent—for instance,  p-hydroxybenzoic acid  (R=OH, R'=H) or p -aminobenzoic acid  (R=NH2, R'=H). These values, combined in the Hammett equation with K0 and remembering that ρ = 1, give the para substituent constants 
2.. Repeating the process with meta-substituents afford the meta substituent constants. This treatment does not include  ortho-substituents , which would introduce  steric effects .
Rho value
With knowledge of substituent constants it is now possible to obtain reaction constants
the alkaline  hydrolysis  of  ethyl benzoate  (R=R'=H) in a water/ethanol mixture at 30 °C. Measurement of the  reaction rate  k0 combined with that of many substituted ethyl benzoates ultimately result in a reaction constant of +2.498.]
The reaction constant, or sensitivity constant, ρ, describes the susceptibility of the reaction to substituents, compared to the ionization of benzoic acid.
It is equivalent to the slope of the Hammett plot.
Information on the reaction and the associated mechanism can be obtained based on the value obtained for ρ.
If the value of:
1. ρ>1, the reaction is more sensitive to substituents than benzoic acid and negative charge is built during the reaction (or positive charge is lost).
2. 0<ρ<1, the reaction is less sensitive to substituents than benzoic acid and negative charge is built (or positive charge is lost).
3. ρ=0, no sensitivity to substituents, and no charge is built or lost.
4. ρ<0, the reaction builds positive charge (or loses negative charge).
These relations can be exploited to elucidate the mechanism of a reaction.
As the value of ρ is related to the charge during the rate determining step, mechanisms can be devised based on this information.
a Hammett plot can be constructed to determine the value of ρ.
If one of these mechanisms involves the formation of charge, this can be verified based on the ρ value.
Conversely, if the Hammett plot shows that no charge is developed, i.e. a zero slope, the mechanism involving the building of charge can be discarded.
Hammett plots may not always be perfectly l
HAMETT PLOT
log = log + σρ
if ‘H’ is replaced by’ ‘D then the above equation becomes
log = log + σρ
log = - σρ
a plot of log versus ‘σ’ is called Hamett plot
Nonlinearity
However, nonlinearity emerges in the Hammett plot when a substituent affects the rate of reaction or changes the  rate-determining step  or  reaction mechanism  of the reaction.
For the reason of the former case, new sigma constants have been introduced to accommodate the deviation from linearity otherwise seen resulting from the effect of the substituent. σ+ takes into account positive charge buildup occurring in the transition state of the reaction.
Therefore, an electron donating group (EDG) will accelerate the rate of the reaction by resonance stabilization and will give the following sigma plot with a negative rho value.
σ- is designated in the case where negative charge buildup in the transition state occurs, and the rate of the reaction is consequently accelerated by electron withdrawing groups (EWG). The EWG withdraws electron density by resonance and effectively stabilizes the negative charge that is generated. The corresponding plot will show a positive rho value.
Problem The rate constant of alkaline hydrolysis of m-chloro ethyl Salicylate was found to be units. Calculate the rate constant of hydrolysis of ethyl Salicylate if substituent constant for m-chloro is 3. The reaction constant for the hydrolysis was 4
Solution:
log k0 = 8 – 12
= -4
k0 =
Problem The rate constant of alkaline hydrolysis of ethyl Salicylate was found to be units. Calculate the rate constant of hydrolysis of m-chloro ethyl Salicylate if substituent constant for m-chloro is 3. The reaction constant for the hydrolysis was 4.0
Solution:
log k = log () +3(4)
log k0 = 8 – 12
= -4
k0 =
Problem The ratio between rate constant of alkaline hydrolysis of ethyl Salicylate and m-chloro ethyl Salicylate was found to be 2 . Calculate the substituent constant for m-chloro ethyl Salicylate if the reaction constant for the hydrolysis was 0.4
Solution:
log ( ) = σ(0.4)
= 5
Problem The ratio between rate constant of alkaline hydrolysis of ethyl Salicylate and m-chloro ethyl Salicylate was found to be 2 . Calculate the reaction constant for the hydrolysis reaction The σ m-cl = 0.2
Solution:
log ( ) = (0.2) ρ
2 = 0.2 ρ
= 10
Problem . From the following Calculate the reaction constant for the hydrolysis reaction The σ m- NO2 = 2.0
Solution
log ( ) = 2σ
1 =
Problem The rate constant of alkaline hydrolysis of m-chloro ethyl benzoate and p- methyl ethyl benzoate was found to be and units. Calculate the reaction constant if substituent constant for m-chloro and p- methyl are 0.6 and 0.4 respectively.
Solution:
ρ = ?
subtracting 2 from 1
reaction constant ρ = 10
Problem The rate constant of alkaline hydrolysis of m-chloro ethyl benzoate and p- methyl ethyl benzoate was found to be and units The substituent constant for m-chloro and p- methyl are 0.6 and 0.4 respectively. Calculate the rate constant of hydrolysis of ethyl benzoate
Solution:
ρ = ?
subtracting 2 from 1
substituting in 1 we get, log () = log k0 + 60
8 = log k0 + 60
log k0 = 8- 60
6 = log k0 +40
log k0 = 6- 20
[ PLEASE CHEQUE THE ANSWER]
Problem. The rate constant of alkaline hydrolysis of m-chloro ethyl benzoate and p- methyl ethyl benzoate was found to be and units. The substituent constant for m-chloro and p- methyl are 0.6 and 0.4 respectively. Calculate the rate constant of hydrolysis of p – nitro ethyl benzoate if σ p-nit = -0.4
Solution:
subtracting 2 from 1
substituting in 1 we get, log () = log k0 + 60
8 = log k0 + 60
log k0 = 8- 60
log k p- nit = log k0 + σ p- nit ρ
= - 52 + (- 0.4 )×(10)
log k = log k0 + σ *ρ *
where k is the rate constant for a particular member of reaction series and k0 is the value for parent compound.
 While the Hammett equation accounts for how  field ,  inductive , and  resonance  effects influence reaction rates, the Taft equation also describes the  steric effects  of a  substituent .
The Taft equation is written as: log ( ) = σ *ρ * +
{\displaystyle \log \left({\frac {k_{s}}{k_{{\ce {CH3}}}}}\right)=\rho ^{*}\sigma ^{*}+\delta E_{s}}where log ( ) {\displaystyle \log {\frac {k_{s}}{k_{{\ce {CH3}}}}}} is the ratio of the  rate  of the substituted reaction compared to the reference reaction,
σ* is the polar substituent constant that describes the field and inductive effects of the substituent,
Es is the steric substituent constant,
ρ* is the sensitivity factor for the reaction to  polar effects ,
and δ is the sensitivity factor for the reaction to steric effects.
Rearranging equation 2 we get
ln K - ln K0 = σρ
ln ( ) = σρ
This is of the form y = mx. So the plot of ln ( ) versus σ gives a straight line passing through origin. This plot is known as Hammett plot.
Hammet Equations Are Linear Free Energy Relations.
The relation between equilibrium constant and change in free energy is given by VantHoff isotherm which is G = - RT ln K
Rearranging ln K =
similarly ln K0 =
Substituting in Hammett equation , ln K = ln K0 + σρ we get
= + σρ
G = G0 –RT σρ
Dividing by ρ, = - RTσ
. For a second series of homologous reactions , having a reaction constant ρ’
G * = G0 * – RT σρ’
Dividing by ρ’, = - RTσ
G = G 0
= constant
Thus a linear relationship between the free energies of activation for one homologous series of reactions and those for another.
Limitations:
2. . It is applicable only for meta and para substituents.
THE TAFT EQUATION:
TAFT proposed an equation for ortho substituted derivative .
 While the Hammett equation accounts for how  field ,  inductive , and  resonance  effects influence reaction rates, the Taft equation describes the  steric effects  of a  substituent .
The Taft equation is written as:
{\displaystyle \log \left({\frac {k_{s}}{k_{{\ce {CH3}}}}}\right)=\rho ^{*}\sigma ^{*}+\delta E_{s}} log k = log k0 + σ *ρ * + δ Es 
σ* is the polar substituent constant
ρ* is the sensitivity factor for the reaction to  polar effects ,
Es is the steric substituent constant,
δ is the sensitivity factor for the reaction to steric effects.
Polar substituent constants, σ*
To determine σ* Taft studied the  hydrolysis  of  methyl   esters  (RCOOMe).
The hydrolysis of esters can occur through either  acid and base catalyzed mechanisms , both of which proceed through a  tetrahedral   intermediate .
In the base catalyzed mechanism the reactant goes from a neutral species to negatively charged intermediate in the  rate determining (slow) step , while in the acid catalyzed mechanism a positively charged reactant goes to a positively charged intermediate.
Due to the similar tetrahedral intermediates, Taft proposed that under identical conditions any steric factors should be the same for the two mechanisms and therefore would not influence the ratio of the rates.
However, because of the difference in charge buildup in the rate determining steps it was proposed that polar effects would only influence the reaction rate of the base catalyzed reaction since a new charge was formed.
He defined the polar substituent constant σ* as:
{\displaystyle \sigma ^{*}=\left({\frac {1}{2.48\rho ^{*}}}\right){\Bigg [}\log \left({\frac {k_{s}}{k_{{\ce {CH3}}}}}\right)_{B}-\log \left({\frac {k_{s}}{k_{{\ce {CH3}}}}}\right)_{A}{\Bigg ]}} σ* = log - log ]
where log(ks/kCH3)B is the ratio of the rate of the base catalyzed reaction compared to the reference reaction,
log(ks/kCH3)A is ratio of a rate of the acid catalyzed reaction compared to the reference reaction, and ρ* is a reaction constant that describes the sensitivity of the reaction series.
For the definition reaction series, ρ* was set to 1 and R = methyl was defined as the reference reaction (σ* = zero). The factor of 1/2.48 is included to make σ* similar in magnitude to the  Hammett σ values .
Steric substituent constants, Es]
Although the acid catalyzed and base catalyzed hydrolysis of esters gives  transition states  for the rate determining steps that have differing  charge densities , their structures differ only by two  hydrogen  atoms.
Taft assumed that steric effects would influence both reaction mechanisms equally. Due to this, the steric substituent constant Es was determined from solely the acid catalyzed reaction, as this would not include polar effects. Es was defined as:
{\displaystyle E_{s}={\frac {1}{\delta }}\log \left({\frac {k_{s}}{k_{{\ce {CH3}}}}}\right)}
where ks is the rate of the studied reaction and {\displaystyle {\ce {{\mathit {k}}_{CH3}}}} is the rate of the reference reaction (R = methyl). δ is a reaction constant that describes the susceptibility of a reaction series to steric effects. For the definition reaction series δ was set to 1 and Es for the reference reaction was set to zero. This equation is combined with the equation for σ* to give the full Taft equation.
Sensitivity factors]
Polar sensitivity factor, ρ*
Similar to  ρ values  for Hammett plots, the polar sensitivity factor ρ* for Taft plots will describe the susceptibility of a reaction series to polar effects.
The polar sensitivity factor ρ* can be obtained by plotting the ratio of the measured reaction rates (ks) compared to the reference reaction ({\displaystyle {\ce {{\mathit {k}}_{CH3}}}}) versus the σ* values for the substituents. This plot will give a straight line with a  slope  equal to ρ*. Similar to the Hammett ρ value:
· If ρ* > 1, the reaction accumulates negative charge in the transition state and is accelerated by  electron withdrawing groups .
· If 1 > ρ* > 0, negative charge is built up and the reaction is mildly sensitive to polar effects.
· If ρ* = 0, the reaction is not influenced by polar effects.
· If 0 > ρ* > −1, positive charge is built up and the reaction is mildly sensitive to polar effects.
· If −1 > ρ*, the reaction accumulates positive charge and is accelerated by  electron donating groups .
Steric sensitivity factor, δ
Similar to the polar sensitivity factor, the steric sensitivity factor δ for a new reaction series will describe to what magnitude the reaction rate is influenced by steric effects. When a reaction series is not significantly influenced by polar effects, the Taft equation reduces to:
{\displaystyle \log \left({\frac {k_{s}}{k_{{\ce {CH3}}}}}\right)=\delta E_{s}}
A plot of the ratio of the rates versus the Es value for the substituent will give a straight line with a slope equal to δ. Similarly to the Hammett ρ value, the magnitude of δ will reflect to what extent a reaction is influenced by steric effects:
A very steep slope will correspond to high steric sensitivity, while a shallow slope will correspond to little to no sensitivity.
Since Es values are large and negative for bulkier substituents, it follows that:
If δ is positive, increasing steric bulk decreases the reaction rate and steric effects are greater in the transition state.
If δ is negative, increasing steric bulk increases the reaction rate and steric effects are lessened in the transition state.
. Taft plots in QSAR
 have used Taft plots in studies of polar effects in the  aminolysis  of  β-lactams .
Problem 1. The ratio between rate constant of hydrolysis of ethyl propionate and that of chloro ethyl acetate in alkaline medium and acidic medium were found to be 8.5 and 3.5 respectively. Find σ*
σ* = log - log ]
HOMOGENEOUS CATALYSIS
In catalytic reactions, if the catalyst and the reactants are of same phase , the reactions are called homogeneous catalysis.
For example, in the conversion of sucrose in the presence of mineral acid, the catalyst mineral acid as well as the reactant sucrose , both are in liquid medium
Mechanism:
Consider a reaction between A and B in the presence of catalyst C. This takes place in two steps.
1. Formation of an intermediate complex between one of the substrates and the catalyst
2. Decomposition of intermediate it may react with other reactant to give product
A + C X
X + B P
Applying Steady State Approximation to X
Rate of formation of X = k1 [A][ C]
Rate of disappearance of X = k-1 [X] + k2 [X][B]
According to steady state principle
Rate of formation of X = Rate of disappearance of X
k1 [A][ C ] = k-1 [X] + k2 [X][B]
= [X] { k-1 + k2 [ B] }
If K-1 >> k2 X is called Arhenius intermediate
If K2 >> k1 X is called Vant Hoff intermediate
HOMOGENEOUS CATALYSIS
HETEROGENEOUS CATALYSIS
catalyst and the reactants are different phase.
2.
Stoichiometry of catalyst can not be well defined
3.
Distillation, solvent extraction etc should be used
Mere by filtration we can separate catalyst from product
4.
5.
Example: conversion of sucrose in the presence of mineral acid,
Synthesis of NH3 by Haber process
ACID – BASE CATALYSIS
Acid base catalysis include, reactions in solution, which are catalysed by acids or bases or both. These are classified in to two types
1.Specific acid catalysis( Hydronium ion catalysis)
A reaction which is catalysed by H+ ions but not by other Bronsted acids, is called specific acid catalysts.
Examples:1. inversion of sugar,
2. keto enol transformation,
This involves two steps
Eequilibrium between the substrate and H+ followed by the rate determining step to form product.
S + H + S H+
The equilibrium constant K =
Rate of the reaction = k2 K [ S] [ H+]
Here rate of the reaction depends [ H+] .
2.General acid catalysis
A reaction that is catalysed by any Bronsted acid is called general acid catalysis.
Examples
This involves two steps.
Equilibrium between the substrate and H+ followed by the rate determining transfer of proton to the base A.
S + H + S H+
SH+ + A- Product + HA
Rate of the reaction = k2 [ SH+] [A] --------------1
The equilibrium constant K =
Rate of the reaction = k2 K [ S] [ H+] [A] --------------2
The ionization reaction of the acid is given by
HA H+ + A –
= k2 K [ S] KHA [ HA]
= K’ [ S] [ HA] Where K’ = k2 K KHA
Here rate of the reaction depends [ HA] .
3.Specific base catalysis
Specific base or hydroxide ion catalysis refers to reactions catalysed by only hydroxide ions.
Examples
.Mechanism:
Proton abstraction from the substrate by a base B establishes a rapid equilibrium and the conjugate base of the substrate reacts with R in a slow step to give rise to products.
SH + B S - + BH+
SH + S - + BH+
Rate of the reaction = k2 [R] --------------2
B + H2O BH+ + OH –
Kb =
Concentration of H2O can be neglected . therefore the above equation becomes
Kb =
Rearranging
[ B] =
= k2 [R] [SH+] ×
Here rate of the reaction depends [ OH- ] .
4.General base catalysis.
Mechanism:
Slow abstraction of proton from the substrate by the base followed by the rapid reaction of the conjugate base S- with R to give product.
SH + B S - + BH+
S- + BH+ Product + B
Example: muta rotation of glucose
Enolisation of ketone
Equilibrium leads to complex formation between S and HA. This is followed by the slow transfer of a proton to a base B
S + H A S.HA(complex)
Rate of the reaction = K’[ S][ HA] [B]
The solvent being amphiprotic, it can play the role of a base if the catalyst is HA
And it can play the role of acid if the catalyst is B
s.no
CATEGORY
K’[ S][ HA] [B]
BRONSTED CATALYSIS LAW( Catalytic activity and acid strength)
There exists a correlation between the effectiveness of the catalyst and its strength as an acid or base . This strength is a measure of the case with which the catalyst transfers a proton to or from a water molecule. The dissociation constant of acid is represented as
HA + H2O H3O+ + A -
The relation between the catalytic constant ka and the dissociation constant Ka is given by
Catalytic constant (dissociation constant )α
ka Ga Ka
ka = Ga Ka ------------------1
Wher G and ‘ are costants
Derivation:
dividing by ρ , we get
log k = log k0 + σ -------------------2
where k is the rate constant for the substituted compound, k0 is the rate constant for the parent ( unsubstituted) compound, σ – substituent constant and ρ – reaction constant.
Similarly, for equilibrium constant
dividing by ρ’ , we get
log K = log K0 + σ -------------4
Subtracting 4 from 2
log k - log K = Constant
log - log = Constant
= C
= C
k =
For base catalyst, kb = Gb Kb -----------------2
The relation between catalytic constant of a base and the acid strength of conjugated acid is Kb = G’b ( ) ------------3
equations 1,2 and 3 are known as Bronsted relations.
On investigation of general base catalysed decomposition of nitramide ( NH2NO2) in to nitrous oxide and water.
Verification:
log kb = log Gb - β log Ka
log kb = log Gb + β pKa
a plot of log kb versus pKa should be linear with a slope equal to β and intercept log Gb
Problem : The general base catalyzed hydrolysis of ethyl dichloroacetate was investigated using variety of bases at 25 0 C. The plot of log Kb versus pKa gives a slope of 0. 5 and an intercept of -6.5 . Calculate β and Gb
From Bronsted Catalysis law
2. Concepts of groups,
9. Group multiplication table
12. Direct product representation
14. Character table
15. Construction of -character table NH3 and H2O
16. Symmetry adapted linear combinations of atomic orbitals (water as example)
GROUP THEORY
Symmetry operation and Symmetry elements:
Symmetry operation is the movement of the molecule about a point or an axis or a plane in a such a way that the resulting configuration of the molecule is indistinguishable from the original.
Symmetry elements are the point , axis or plane about which a symmetry operation is carried out in order to generate an equivalent orientation.
There are five types of symmetry elements. They are
1. Identity operation:
1.. Identity operation:
1. Any operation which retains the orientation of an object is called an identity operation
2. denoted by E
3. It is an operation of doing nothing or rotation through 360 o.
4. This element is present in all groups.
5. If a sequence of symmetry operations, brings the molecule back, to its original configuration, the net operation is called identity operation.
For example, two successive rotations about the C2 axis in H2O are equivalent to identity.
C2 .C2 = C22
2.. Centre of symmetry:
1. It is a point inside the molecule, from which lines drawn to opposite directions will meet similar points at exactly the same distance. Examples:
2. A molecule can not have more than one centre of symmetry.
Examples:
O ==== C =====O, H ---C === C ----H, Trans--F-N==N-F, XeF4, SF6, N2O4
Examples for compound having centre of symmetry
a. Carbon di oxide
d. N2O4
e. F2N2
A molecule can not have more than onecentre of symmetry
2. Planar symmetry( reflection):
1. It is a symmetry element , that acts as a mirror, and makes one half of the molecule, the mirror image of the other half. For example
2. The three types of reflection planes are
There are three types of reflection plane
1. Those planes which are lying along the direction of the principal axis are known as vertical planes. (σv)
2.Those planes which are perpendicular to the principal axis and along the bondsare known as horizontal planes (σ h)
3.Those planes which are perpendicular to the principal axis and bisect the bond angles are known as dihedral planes (σ d)
Horizontal plane (σ h) : The plane perpendicular to principal axis of the molecule
Example: in BCl3 molecule C3 axis is the principal axis.
It passes through boron and is perpendicular to molecular plane
Therefore the molecular plane is σ h
Vertical plane (σ v) : Plane containing the principal axis
Water molecule has two vertical planes
Dihedral plane (σ d) : Plane bisecting two C2 axes
Example : staggered conformation of ethane
3. Linear molecules having a centre of symmetry ( H2, CO2 ) have
an infinite number of planes of symmetry passing through the length of the molecule
and an additional plane of symmetry perpendicular to these planes and passing through
the centre
one bisects the H-O-H angle
the second cuts every atom into hemispheres
5. NH3 contains 3 planes of symmetry
6. BF3 contains
one plane of symmetry in the plane of the molecule
and 3 other planes of symmetry each of which passes through one of the B-F bonds and
bisects the F-B-F angle.
7.Square planar molecules like XeF4, PtCl4, has 5 planes of symmetry
8. Tetrahedral molecules like CH4, CCl4, Ni (CO)4 has 6 planes of symmetry.
9. A cube has 9 planes of symmetry
10. SF6 has 9 planes of symmetry
Molecule
3 . Proper axis of symmetry:
1. It is a simple rotation of a molecule, about an axis, through an angle of , .
2. It is denoted by Cn, where n’ being 2,3,4… For example
C2 axis means rotation about = 180 and
C3 axis means rotation about = 120.
3. Water molecule has one C2 axis of symmetry.
4. BF3 has three C2 axis of symmetry and one C3 axis of symmetry
5. If a molecule contains more than one axis of symmetry, the axis having highest value of ‘n’
is called principal axis. For example BF3 contains C2 and C3 axis of symmetry. Here C3 is
principal axis of symmetry.
, , , , ---- are the Operations generated by the axis of symmetry:
If an object has ‘Cn’ axis = E,
= E,
= E
E .Improper axis of symmetry( Sn):
This is a compound operation combining a rotation (Cn) with a reflection through a plane perpendicular to the Cn axis σh.(Cn followed by σh)
σCn=Sn
Molecules showing S3
PCl5, SO3, BCl3 – all trigonal planar and trigonal bipyramidal ( with same atoms) molecules
Molecules showing S4 are PtCl4 SF6
Molecules showing S6
Benzene
Operations generated by S3 [ Note: we must do until we get E]
[ example BCl3 : D3h]
S3 = σ C3
= = = E E EEE = E
In general in Sn
For example
S4 = σ C4
S No
identity
E
CONCEPT OF GROUP
A set of elements, with an operator , is called group if it obeys the following conditions.
1. Law of Closure :The combination of two elements in the set results an element, which is also the element of the set. For example consider a set A = { a, b, c, d} with operator * then if a* b = c , the set obeys law of closure.
1. Law of Identity :There should be an element in the set, which on operation with other elements , leave them unchanged. This element is called identity.
For example consider a set A = { a, b, c,d} with operator * then
If a* b = b then ‘a’ is called identity
1. Law of Inverse : Every element in the set ,should have an inverse, which is also an element of the set. For example consider a set A = { a, b,c,d} with operator * then
If ‘a’ is the identity element and a* b = a then ‘b’ is called inverse of ‘a’
1. Associative law :Every element of the set , must obey the associative law .
For example consider a set A = { a, b,c,d} with operator * then
(a* b ) * c = a* ( b * c)’
Properties of a group:
1. Law of Identity
The identity element of a group is unique. i.e, a group has only one identity element
1. Law of Inverse
1. Right cancellation law
If a, b and c are the elements of a group with operator * and if
a*b = c * b then a = c
1. Left cancellation law
If a, b and c are the elements of a group with operator * and if
a*b = a * c then b = c
1. Inverse of inverse law
If ‘a’ is an element of a group with a -1 as its inverse then
(a –1 )-1 = a
SUB GROUP:
The subset of the elements of a group obeying the following rules are said to be sub group of the main group
1. It must obey the closure and associative laws
2. It must contain the identity element
3. = integer
For example consider a set A = { 1,-1,i,-i}} with operator * then
{1,-1}, {1, i}, {1, -i} are called sub groups of A
Problem. Find the sub groups of the set A = { 1,-1,i,-i}} with complex multiplication operator( *)
Solution:
{1,-1}, {1, i}, {1, -i} are the sub groups of A
Problem 1. Find the sub groups of C2v point group
C2v point group containing the following elements { E, C2v, σv, σv,}
O ( h) = 4
There are two sub groups.
s.no
order
elements
C2
Problem 2. Find the sub groups of D2h = { E, 3C2, 3σv, i }
O ( h) = 8
∴ Order of sub group must be 1 or 2 or 4
The sub groups are
D2
CLASS
A set of conjugate elements in a group which satisfy the following condition ( similarity transformation) is called a class.
If A , B, X are the elements of the group, then if X-1AX = B we said A and B belongs to same class .
For example consider the C2v point group
The symmetry elements of C2v are {E, C2,σv, σv’ }
The similarity transformation of E is
E-1E E= E-1E = E
C2-1E C2= C2-1 C2 = C2 C2 = E [ E C2 = C2, Each element is its inverse ∴ C2 -1 = C2 ]
σv -1 E σv = σv -1 σv = σv σv = E
Since the similarity transformation of E with all the operations generates E, E forms a separate class.
The similarity transformation of C2 is
E-1 C2 E= E-1 C2 = EC2 = C2 [ the inverse of E is E]
C2-1 C2 C2= C2-1 E = C2 E = C2 [C2 C2 = E, Each element is its inverse ∴ C2 -1 = C2 ]
σv -1 C2 σv = σv -1 σ ‘v = σv σ ‘v = C2
Since the similarity transformation of C2 with all the operations generates C2, C2 forms a separate class.
The similarity transformation of σv is
E-1 σv E= E-1 σv= σv [ the inverse of E is E ]
C2-1 σv C2= C2-1 σ ‘v = C2 σ ‘v = σv [, Each element is its inverse ∴ C2 -1 = C2 ]
σv -1 σv σv = σv -1 E = σv E = σ ‘v
Since the similarity transformation of σv with all the operations generates σv , σv forms a separate class.
The similarity transformation of σ ‘v is
E-1 σ ‘v E= E-1 σ ‘v= σ ‘v [ the inverse of E is E ]
C2-1 σ ‘v C2= C2-1 σ‘v = C2 σ v = σ ‘v [, Each element is its inverse ∴ C2 -1 = C2 ]
σv -1 σ’v σv = σv -1 E = σ ‘v E = σ ‘v
Since the similarity transformation of σ ‘v with all the operations generates σ’v , σ ‘v forms a separate class.
Thus the four elements of the C2v point group form four different classes.
{ E}, {C2}, { σ ‘v }, { σ ‘v }
Find the various classes of NH3 molecule
Ammonia posses C3v point group.
The symmetry elements of C3v are { E,C3, C3 2 ,σ v σ ‘v σ ‘’v }
E forms a class by itself
C3 and C32 form a class.
σ v σ ‘v σ ‘’v - all the three σ v form one class.
Thus there are three classes of elements in C3v point group. or NH3 They are { E, } { C3, C32, } { σv, σv’ σ’’v }.
ORDER
Order of a group :
Order of a group is defined as the total number of symmetry elements present in a group. It is denoted by ‘h’. For example
The symmetry elements in H2O are [E, C2, σv, σv’]
∴ order ( h) = 4
The symmetry elements in NH3 are [E, C3, C32, σv1 σv2 , σv3 ]. ∴ order ( h ) = 6
ABELIAN AND NON-ABELIAN
Abelian group:
A group is said to be abelian group , if it obeys commutative property. . For example consider a group A = { a, b,c,d} with operator * then if
a* b = b*a then the group A is called abelian group.
Non - Abelian group:
A group is said to be non- abelian group , if it does not obey commutative property. . For example consider a group A = { a, b,c,d} with operator * then if
a* b ≠ b*a then the group A is called non - abelian group.
POINT GROUPS.
.
1.Cn point group: A molecule possesses the proper axis of symmetry Cn and the identity element belongs to Cn point group Here C stands for cyclic and ‘n’ refers to the order of the axis. Cn = { E, Cn },
C2 = { E, C2 }- trans -1,2 dibromo cyclo propane, 1,1 dibromo cyclo propane
1,2- dichloro allene
C3 = { E, C3 C3 2 } - 1,1,1- trichloro ethane
2.Cnv point group: A molecule which contains E, Cn, and n v as symmetry elements is denoted by Cnv point group.
Cnv = { E, Cn , n σv}
C2v = { E, C2 , 2 σv} : H2O
C3v = { E, C3 , C32 , σv, σ’v , σ’’v, } : NH3
3.Cnh point group: If E, Cn , i and h are the elements of symmetry, then the molecule belongs to Cnh
Cnh = { E, Cn , σh, , i }
C\2h = { E, C2 , σh, , i } : Trans – N2F2(dinitrogen difluoride)
, trans 1,2 – dichloro ethylene
4.Cs point group:
This group contains identity element and a reflection plane. Here C stands for cyclic and ‘s’ refers to S1 axis.
Example: HOCl
5.Dn point group:
Molecules having Cn principal axis , and’ n’ perpendicular C2 axis belong to dihedral point group.
The group is denoted by Dn , if it has Cn axis and ‘n’ perpendicular C2 axis.
Dn = { E, Cn , n C2 }
D3 = { E, C3 , 3 C2 } : [Co (en)3] +
6.Dnd point group:
A molecule , which has Cn axis, ‘n’ perpendicular C2 axis and ‘n’ dihedral planes is denoted by Dnd group.
Dnd = { E, Cn , n C2 , n σd}
D2d = { E, C2 , 2 C2, 2 σd} : Biphenyl
D3d = { E, C3 , 3 C2, nσd} : Chair form of cyclo hexane: D3d
7..Dnh point group
A group which contains the principal Cn axis, ‘n’ perpendicular C2 axis, centre of inversion, and horizontal plane of symmetry comes under Dnh point group.
Dnh = { E, Cn , n C2 , i, σh}
D2h = { E, C2 , 2 C2, i, , σh} : Naphthalene , p-dichloro benzene
D3h = { E, 2C3 , 3 C2 , σh} : 1,3,5 – tri bromo benzene
8.Tetrahedral: it is denoted by Td
Td = { E, 4 C3, 3 C2, 3 S4, 6σd }
Example: silicon tetraflouride, methane, Ni (CO)4
9.Octahedral : Oh
Oh = { E, 4 C3, 6C2, 3 C4, 3C2’ 3 S4, 4S6, 3σh, 6σ d, i}
Example: SF6
Example: B12 H122- and B12 Cl 122-
11 Sn point group(‘ n’ is only even) The S2 point group has the identity element and inversion centre. S2 = { E, i }
Example for S2 : staggered form of meso tartaric acid, 1,4 – dibromo- 3,5 - dichlor
cyclohexane(chair form)
GROUP MULTIPLICATION TABLE
All combinations of symmetry elements are represented by a multiplication table called group multiplication table.
In this table all the elements of the group are written at the top row and in the extreme left of the column. The point group symbol is shown on the left top column. The body of the table contains the binary products. AB means operation B followed by A.
Transformation of co- ordinate in symmetry operations:
On symmetry operations the co – ordinates of the points are transformed as follows
E ( x, y, z) = (x, y, z) [ no change]
C2 (x, y, z) = ( - x, - y, z) [ only change in x and y ] [ z- axis ]
v (x, y, z) = (x, - y, z) [ only change in y ] [ xz plane ]
’v (x, y, z) = ( - x, y, z) [ only change in x ] [ yz plane]
Product of symmetry operations:
Product of two symmetry elements A and B is given by AB. Here B is the first operation to be carried out.
∴ E C2(z) = C2 (z) [ use of (x,y,z) method is not applicable for NH3 since the angle is not 180)
For example The multiplication table for C2v point group is constructed as follows.
1 .E E ( x,y,z) = E ( x, y, z)
= ( x, y, z)
= ( -x, - y, z)
= ( x, - y, z)
= (- x, y, z)
5. C2 E(x,y,z) = C2 (x, y, z)
= ( - x, - y, z )
= ( x, y, z )
= ( - x, y, z )
= ( x, -y, z )
= ( x, - y, z )
= ( - x, y, z )
= ( x, y, z )
= ( - x, -y, z )
= ( - x, y, z )
14. ’v C2 (x,y,z) = ’v (- x, - y, z)
= ( x, -y, z )
= ( - x, -y, z )
∴ ’v C2 = C2
16. ’v ’v (x,y,z) = ’v ( -x, y, z)
= ( x, y, z )
C3v point group contains the following elements
C3v = { E, C3 , C32 , σv1, σv2 , σv3, }
C3v
E
C3
C32
σv1
σv2 ,
σv3,
E
E
C3
C32
σv1
Assigning point groups
The steps in assigning point group for molecules involve the identification of
1. The presence of an axis of symmetry
1. This axis as the principal axis or not
1. The existence of subsidiary axes
1. The existence of h
1. The presence of nv ‘s
1. If there is no principal axis, then look for C1,Ci,Cs,S4 point groups.
Yes – No response flow chart to find out point group:
Is there an axis of symmetry
No Yes
No Yes No Yes
perpendicular to Cn
Cs Td Oh
No Yes No Yes
Is there an Sn is there a h is there a h
Ci
Ci Sn Cnh Dnh
No Yes No Yes
Cn Cnv Dn Dnd
( 6C6 )
no yes
( σ h ) No
( i )no yes yes
( n σ v ) ( σ h ) ( n perpendicular C 2 axis )C1 Ci Cs
No yes
( σ h )
Identification of point groups:
2. It is unique.
4. No h
5. No v
2. H2O
2. The C2 axis is the principal axis
3. No nC2’s perpendicular to Cn
4. It has no h
5. It has twov’s
Therefore the point group is C2v
The symmetry elements of C2v are { E, C2,σv, σv’ }
Other examples for C2v are cyclohexane (boat),1,1-dichloroethene,
3. NH3
2. The C3 axis is the principal axis
3. No nC2’s perpendicular to Cn
4. It has no h
5. It has three v’s
Therefore the point group is C3v
The symmetry elements of C3v are { E, C2,σv, σv’ }
Other examples for C3v are
4. N2F2
2. The C2 axis is the principal axis
3. No nC2’s perpendicular to Cn
4. It has h
Other examples for C2h are trans 1,2 – dichloroethene,
C3h : H3BO3
5. BF3
1. It has three C2 axis
2. It has a C3 axis which is highest order and unique.
3. There are 3C2’s perpendicular to C3
4. There is a h
Hence the point group is D3h
Other examples for C3h are PF5, ethane (eclipsed)
C2H4 : D2h, C6H6 : D6h
6. CH2 = C= CH2(allene)
2. The C2 axis is the principal axis
3. There are two more C2 axis perpendicular to the principal axis(C2)
4. It has no h
5. It has two d’s
Therefore the point group is D2d
Other examples : ethane(staggered), cyclohexane( chair) : D3d ferrocene : D5d,
7. CH4
2. It has also 3 C2 axis
3. No C4 axis
8. SF6
1. It has a 3 C4, 4C3, and 6C2 axes
2. It has 3h
9. HCl
2. Infinite number of v plane
3. No C2 axis
The point group is C∞v
All linear molecules without centre of symmetry belong to this class
10. H2
2. Infinite number of C2 axis perpendicular to C∞ axis
3. There is a σh
Order of a point group :
Order of a group is defined as the total number of symmetry elements present in a group. It is denoted by ‘h’. For example
The symmetry elements in H2O are [E, C2, σv, σv’] ∴ order = 4
The symmetry elements in NH3 are [E, C3, C32, σv, σv’ σ’’v]. ∴ order = 6
Class of a point group:
Symmetry elements having same type of operation constitute a class
For example There are 3 symmetry classes for NH3 { E, } { C3, C32, } { σv, σv’ σ’’v }.
H2O belongs to C2V point group. Symmetry operations C2V = {E,C2, xy , yz }
Number of classes =4 [ {E }, {C2 }, { xy , yz }]
SIMILARITY TRANSFORMATION:
The transformation of matrix A in to another matrix B by C – 1 AC = B is known as Similarity transformation.
The matrices A and B are said to be conjugate each other.
Matrix C is called modal matrix.( transformation matrix)
Diagonalisation is the process of finding a matrix C such that C – 1 AC = D where D is a diagonal matrix.
Diagonal matrix is one in which aij = 0 for all ‘i’ and ‘j’ and i ≠j
For example D = is a diagonalised matrix
Using Similarity transformation we can diagonalise a matrix.
REDUCIBLE AND IRREDUCIBLE REPRESENTATIONS.
If the matrix can be reduced to diagonalised matrix, by similarity transformation, then it is called RR.
If it is not further possible to reduce the matrix, by similarity transformation, then it is
called IR
If the matrix can be reduced to diagonal matrix, then it is called RR.
Irreducible Representation(IRR)
If it is not possible to reduce the matrices, to diagonalised form, the representation is called IRR
The number of irreducible representation (IRR)
= [ ∑ ( number of symmetry element) (RR) ( its character from table) ]
Where ‘h’ is the order of the group. This is known as reduction formula.
THE STANDARD REDUCTION FORMULA
The number of times (ni) the ith representation occurring in the reducible representation is given by
n(τi) = × )
n(τi) – Frequency of occurrence of ith IR (τ1, τ2 etc)
h - order of the point group
– order of the class under R operation
) – character of the RR in that class
) - character of the IR in that class
1.Find the number of times A1 appearing in C2v point group.
C2V
E
C2
xy
yz
A1
1
1
1
1
RR
5
3
-1
1
Solution:
Order of the group (h) = total number of symmetry elements
= 4
= 2
C3V
Order of the group (h) = total number of symmetry elements
= 6
= 2
= 0
= 1
= 2 (A1 ) + 0 (A2 ) + 1 E
= 2 A1 + E
C2V
Order of the group (h) = total number of symmetry elements
= 4
Number of A1 = [ (1)(4)(1) + (1)(0)(1) + (1)(2)(1) + (1)(2)(1) ] [using the table]
= 2
Number of A2 = [ (1)(4)(1) + (1)(0)(1) + (1)(2)(-1) + (1)(2)(-1) ] [using the table]
= 0
Number of B1 = [ (1)(4)(1) + (1)(0)(-1) + (1)(2)(1) + (1)(2)(-1) ] [using the table]
= 1
Number of B2 = [ (1)(4)(1) + (1)(0)(-1) + (1)(2)(-1) + (1)(2)(1) ] [using the table]
= 1
= 2(A1 ) + 0 ( A2 ) + 1 (B1 ) + 1 (B2)
= 2A1+ B1 + B2
Reducible representation(RR)
If the matrix can be reduced to diagonalised matrix, by similarity transformation, then it is called RR.
The number of times (ni) the ith representation occurring in the reducible representation is given by
ni =
Number of IRR = [ ∑ ( number of symmetry element)(RR)( its character from table) ]
Where ‘h’ is the order of the group. This is known as reduction formula.
1.Find the number of times A1 appearing in C2v point group.
C2V
E
C2
xy
yz
A1
1
1
1
1
Total(RR)
5
3
-1
1
Solution:
Order of the group (h) = total number of symmetry elements
= 1 +1 +1+1
= 2
C3V
Order of the group (h) = total number of symmetry elements
= 1 + 2 + 3 = 6
Number of IRR = [ ∑ ( number from character set)(RR)( number from character table) ]
Number of A1 = [ (1)(4)(1) + (2)(1)(1) + (3)(2)(1) ]
= 2
= 0
= 1
= 2 (A1 ) + 0 (A2 ) + 1 E
= 2 A1 + E
Irreducible Representation(IRR)
If it is not possible to reduce the matrices, to diagonalised form, the representation is called IRR
1. Convert the following representation into Irreducible representation( IRR):
C2V
Order of the group (h) = total number of symmetry elements
= 1 + 1+1+1
= 4
Number of IRR = [ ∑ ( number from character set)(RR)( number from character table) ]
Number of A1 = [ (1)(4)(1) + (1)(0)(1) + (1)(2)(1) + (1)(2)(1) ] [using the table]
= 2
Number of A2 = [ (1)(4)(1) + (1)(0)(1) + (1)(2)(-1) + (1)(2)(-1) ] [using the table]
= 0
Number of B1 = [ (1)(4)(1) + (1)(0)(-1) + (1)(2)(1) + (1)(2)(-1) ] [using the table]
= 1
Number of B2 = [ (1)(4)(1) + (1)(0)(-1) + (1)(2)(-1) + (1)(2)(1) ] [using the table]
= 1
= 2(A1 ) + 0 ( A2 ) + 1 (B1 ) + 1 (B2)
= 2A1+ B1 + B2
DIRECT PRODUCT REPRESENTATION
GREAT ORTHOGONALITY THEOREM(GOT)
It states “ the summation of the product of elements of ‘m’ th row and ‘n’ th column of IRRi and the elements of ‘m’’ th row and ‘n’’ th column of IRRj for a particular symmetry operation R is equal to δijδmnδm’n’
Mathematically
Consequences of GOT
1. Number rule: The number of IRR is equal to number of symmetry classes.
For example
In NH3 there are 3 symmetry classes for NH3 { E, } { C3, C32, } { σv, σv’ σ’’v }.
∴Number of IRR = 3
1. Dimension rule.
The sum of squares of dimensions of IRR is equal to the order of the group
∑ li 2 = number of symmetry classes
For example for NH3 molecule l12 + l22 + l3 2 = 6
1. Character of Identity operation rule:
The sum of squares of characters of identity operation (E) in the IRR is equal to the order of the group.
∑ χ(E) 2 = h.
χ(E1) 2 + χ(E) 2 + χ(E) 2 = 6
1. Adjacent IRR character rule:
The characters of two different IRR satisfy the following.
∑ g χ(E1) χ(E1) = 0 --------------1
∑ g χ(E1) χ(E2) = h ---------------2
where g denotes the number of symmetry operation of each element.
For example for NH3 molecule ( g ( E) =1, g(C3) =2, g ( σv) =3
Reduction formula.
Construction of character table for C2v and C3v point groups.
CHARACTER TABLE
A character table lists the characters of all the symmetry classes for all the IRRs.
It contains four sections:
Section –I:
1. In the top row Schoenflies symbol for the point group is given.
1. In the first column Mulliken symbol for the different IRR are given
1. Depending on the dimensions the IRR are denoted as
One Dimension
T
when character for the rotation about the principal axis is +1 Symbol A is used,
1. character for the rotation about the principal axis is -1 B .
1. For a molecule having centre of symmetry,
symmetric - the subscript ‘g’
antisymmetric. the subscript ‘u’
symmetric - the subscript ‘1’
antisymmetric. the subscript ‘2’
symmetric - ‘’’
antisymmetric. ‘ ‘’’
Section II:
It gives the characters of all the symmetry operations of different IRR.
Section III :
2. It gives x,y,z represents Transition dipole selection rule
3. rotation operators Rx, Ry Rz represents spin orbit coupling
4. brackets indicate degeneracy
3. others represent ‘d’ orbital
4. all are operators for Raman spectra selection rule
The character table of water which belongs to C2v point group is
Section -III
Section -II
Section -III
Section -III
Orthogonality Theorem And Its Consequences [Great Orthogonality Theorem(GOT)
It states “ the summation of the product of elements of ‘m’ th row and ‘n’ th column of IRRi and the elements of ‘m’’ th row and ‘n’’ th column of IRRj for a particular symmetry operation R is equal to δijδmnδm&rsq