Download Chemistry Resource Book

Chemistry
FE SEM - I/II
A.Y. 2020-21 (FE SEMESTER I/II)
Choice Based Credit Grading System with Holistic Student Development (CBCGS - H 2019)
Under TCET - Autonomy Scheme - 2019
Autonomous College Affiliated to University of Mumbai
Approved by All India Council for Technical Education(AICTE) and Government of Maharashtra
CHEMISTRY
Semester – I/II
Compiled, reviewed and edited by:
Mr. Rohitkumar Singh
Dr. Sunita Pachori
Dr. Kiran Sanap
Dr. Neha Mishra
TCET, Mumbai
STUDY MATERIAL FOR FE SEM-I / II
Under the guidance of Dr. B. K. Mishra, Principal
2nd Edition of revised syllabus, December 2020
© Thakur College of Engineering and Technology, Kandivali, Mumbai.
Published by:
Thakur College of Engineering and Technology
PREFACE
This Resource book is a structured and guided learning materials to be used effectively by
students to learn the contents of syllabus based on TCET Autonomy scheme. Top priority is
given to the students need and contents are presented lecture wise in order to help student keep
track of their learning after every lecture. The resource book emphasizes on the fundamental
definitions with a continuity of details easy to follow finally rendering to the practical
applications of the topic. The module is written with sequentially starting from prerequisite to
theoretical background eventually highlighting the formulae of the topics and covering all the
topics in detail. Numerical problems have been solved from each topic to make the
understanding of the subject simpler. Large number of short answer questions, long answer
questions and practice questions are included which will be helpful in preparing for
examinations.
This is a comprehensive (structured and guided) resource book covering the Chemistry syllabus
as per TCET Autonomy scheme. The details of content is presented module wise below:
Module-1: It primarily deals with Atomic structure, chemical bonding and co-ordination
chemistry and its co-relation. As part of atomic structure and its extension to chemical
bonding it includes Study of Schrodinger equation, Atomic Orbital and its Shapes in order to
understand the hybridization of atomic orbitals involving s, p and d orbitals including
Valence Bond theory (VBT), Concept of Molecular orbital theory, As part of co-ordination
chemistry it includes concept of complex compound formation and its application for
understanding and calculating hardness of water.
Module-2: It deals with methods of separation based on chromatographic techniques. It covers
various types of chromatography such as Thin Layer Chromatography, Gas Chromatography,
High Performance Liquid Chromatography, it also covers application of HPLC and GC in the
form of Chromatogram and TLC plate of various samples.
Module-3: This module covers basics of Green Chemistry its 12 principles. Design of Greener
route of synthesis over conventional route by making use of suitable catalyst/enzymes. It also
includes catalyst and its role in making the chemical process Green particularly by using
heterogeneous catalysts. It also deals with Catalyst design through artificial intelligence and
computer modelling
Module-4: This module includes types of Electrochemical Cell and its Electrode potential,
calculation of EMF by using Nernst Equation. It Covers Relation of free energy with EMF of
Cell, Numerical based on EMF and its feasibility prediction. The module also deals with
Corrosion and it’s Mechanism, factors affecting the rate of corrosion and ways and means to
reduce and control corrosion.
Module-5: This module primarily includes UV-Visible and IR Spectroscopy.
It covers
Interaction of electromagnetic radiation with matter including Beer-Lambert’s law and
numerical based on it. In later part it includes applications of spectroscopic techniques for
chemical structure elucidation including its functional group determination.
Module-6: It Covers Basics of Stereochemistry types of isomerism, study of optical activity and
elements of symmetry is also included. It covers concept
like
Enantiomerism,
Diastereoisomerism, mesomerism, racemic mixture, Representation by flying wedge, Fischer
and Newman projection is included in the module. Lastly Nomenclature of Stereo isomers like
D&L system, R-S Configuration , E-Z nomenclature etc.
This book will be used extensively for conducting day to day teaching learning process.
Sufficient number of Solved and exercise problems is included to develop the required
analytical thinking and problem solving skill.
At the end of each lecture it provides the learners an opportunity to check their learning from
lecture by using sections like takeaway from the lecture, self -assessment and self-evaluation.
General Guidelines for Learners:
1. Resource book is for structured and guided teaching learning process and
therefore learners are recommended to come with the same in every lecture.
2.
3.
4.
5.
6.
7.
While conducting teaching learning process this resource book will be extensively used.
Resource book is framed to improve the understanding of subject matter at depth and
therefore the learners are recommended to take up all the module contents, home
assignments and exercise seriously.
A separate notebook should be maintained for every subject.
Lectures should be attended regularly. In case of absence, topic done in the class should
be referred from the module before attending the next lecture.
Motivation, weightage and pre‐requisite in every chapter have been included in order to
maintain continuity and improve the understanding of the content to clarify topic
requirement from exam point of view.
For any other additional point related to the topic instructions will be given by the subject
teacher from time to time.
Subject Related Guidelines:
1.
The subject is pure as well as applied in nature and it requires thorough understanding of
2.
3.
4.
subject matter like knowledge of chemical reaction, process, mechanism and problem
solving skill.
Questions are expected from all modules and learners are instructed not to leave any
module in option.
Theory paper will be of 100 marks, practical examination will be of 25 marks and
continuous and systematic work during the term (Term work) will be of 25 Marks.
Importance should be given to term work and all internal assessment/continuous
assessment for improving overall percentage in examination.
Practice questions should be solved sincerely in order to enhance confidence level and to
excel in end semester examination.
Exam Specific Guidelines:
1.
All modules are equally important from examination point of view.
2.
3.
4.
5.
Wherever applicable chemical reaction, mechanism and neat labelled diagram must be
provided for writing effective answer.
Proper time management is required for completing the question paper within time
frame.
Read the question paper thoroughly first then choose the questions. Attempt the one that
you know the best first but do not change the internal sequence of the sub questions.
For further subject clarification/ doubt in the subject, learners can contact the subject
teacher.
Guidelines for Writing Quality Answer:
1.
Write content as per marks distribution.
2.
3.
4.
5.
Highlight the main points.
Write necessary content related to the point.
Draw neat and labelled diagrams wherever necessary.
While writing distinguishing points, write double the number of points as per the marks
given, excluding the example.
Guidelines for Solving Numerical:
Important steps should be written as they carry stepwise marks. The steps are as follows:
1.
Given data
2.
Diagrams (wherever applicable)
3.
Formula
4.
Substitution
5.
Calculation
6.
Answer with proper units
Choice Based Credit Grading Scheme with Holistic and Multidisciplinary Education - (CBCGSHME 2020)
TCET Autonomy Scheme (w.e.f. A.Y. 2020-21)
Course Name: Chemistry
Course Code: BSC102
Teaching Scheme (Program Specific)
Examination Scheme (Formative/ Summative)
Modes of Teaching / Learning / Weightage
Modes of Continuous Assessment / Evaluation
Practical/Oral Term Total
Theory
(25)
Work
(100)
(25)
Hours Per Week
Theory Tutorial Practical
Contact
Credits
Hours
IA
IP ESE
PR
TW
150
3
1
2
6
5
20
20
60
25
25
IA: In-Semester Assessment - Paper Duration - 1.5 Hours
IP: Innovative Practices
ESE: End Semester Examination - Paper Duration - 3 Hours
The weightage of marks for continuous evaluation of Term work/ Report: Formative (40%), Timely
completion of practical (40%) and Attendance/Learning Attitude (20%)
Prerequisite: 10+2 level of Science knowledge related to : intermediate of Science level knowledge of
Atomic Structure, chemical Bonding, Thermodynamics and Electrochemistry
Course Objective: The course is designed with the objectives of making them understand
microscopic chemistry at atomic and molecular level. It also emphasizes on latest methods of
instrumental analysis, spectroscopic techniques and principles of green chemistry used in
pharmaceutical industries and other chemical industries. The course provides an opportunity to
learn concept of electrochemistry and its application for controlling the rate of corrosion. It also
deals with stereo chemistry and determination of structure of organic molecules.
Course Outcomes: Upon completion of the course students will be able to:
Sr.
No.
CO1
CO2
CO3
CO4
Course Outcomes
RBT Levels
Analyse microscopic chemistry in terms of atomic and
molecular orbitals and intermolecular forces.
L1. L2, L3
Apply the knowledge of instrumental method of analysis for
analysis of various samples.
Understand and apply principles of catalysis and its application
in maintaining green matrix of reactions.
Understand electromagnetic spectrum used for exciting
different molecular energy levels in various spectroscopic
L1. L2, L3
L2, L3
L1. L2, L3
techniques
CO5
CO6
L2, L3
Understand and apply bulk properties and processes using
thermodynamic considerations
Understand the stereochemistry and determination of structures
of organic compounds
L2, L3
Detailed Syllabus:
Module
No.
1.0
Topics
Lectures
A. Atomic, Molecular Orbitals and Chemical Bonding
RBT Levels
L1, L2, L3
Schrodinger equation (No derivation), Atomic Orbital, Shapes of s, p d
and f Orbital, Hybridization of atomic orbitals involving s,p and d orbitals
Valence Bond theory (VBT), its application and drawback, Concept of
Molecular orbital theory, Molecular orbital Diagram (Energy level
09
diagrams) of diatomic molecules like H2, He2, Be2, C2, N2, O2, F2 and Ne2
(up to atomic no. 10), Application of Molecular Orbital Theory
B. Co-ordination chemistry and its application
Introduction to Coordination chemistry and coordination number,
Chelating agent, EDTA as chelating agent and its application for
estimating hardness of water, Hardness of water ( it’s source, unit of
measurement and Numerical based on hardness calculation) Estimation of
Hardness of water by Complexometric method, theory, Procedure,
calculation and numerical. Crystal field theory and the energy level
diagrams for transition metal ions (Tetrahedral and Octahedral geometry)
and their applications.
2.0
Instrumental Methods of Analysis
5
Introduction to Chromatography, Types of Chromatography (Adsorption
and partition chromatography), Thin Layer Chromatography (Theory,
Principle,
technique
and
applications),
Gas
Chromatography
–
(Introduction, theory, instrumentation., working) High Performance Liquid
Chromatography, – introduction, theory, instrumentation. Interpretation of
Gas/HPLC Chromatogram and TLC plate of various samples.
L1, L2, L3
3.0
A. Green Chemistry
7
L1, L2, L3
10
L1, L2, L3
Introduction to Green Chemistry, The 12 principles of Green Chemistry,
Prevention of waste, Atom Economy, Less hazardous Chemical synthesis,
Safer solvent and Auxiliary, Design for energy efficiency, use of
renewable feedstock, Reduction of derivatives, Catalysis, Design for
degradation, Real time analysis (Uses of electronic devices and sensors for
process control), inherently safer chemistry for accident prevention.
Design of Greener route of synthesis over conventional route, Numerical
based on calculation of Atom economy
B. Catalysis
Role of Catalyst in making the chemical process Green, Relevance and
examples,
Homogeneous
and
heterogeneous
catalysis,
Theory
of
Heterogeneous Catalysis (Adsorption Theory), Catalytic Converters, Acid
Base catalysis, Solid Acid Catalysis, Solid Base Catalysis, Transition metal
Catalysis, Metal and supported metal catalysis, Catalyst design through
artificial intelligence and computer modelling
4.0
Electrochemistry and its Application
A. Electrochemistry
Introduction, Concept of electrode potential, Concept of Electrochemical
cell, EMF of Cell, Cell potentials by Nernst equation, Relation of free
energy with EMF of Cell, Numerical based on EMF and its feasibility
prediction.
B. Corrosion:
Introduction of Corrosion, Fundamental reason, Mechanism of corrosion-i)
Electrochemical/Wet Corrosion mechanism a) Evolution of hydrogen gas
and b) Absorption of oxygen gas, ii)Direct Chemical/Dry/ Atmospheric
Corrosion a) Due to oxygen b) Due to other gases Factors affecting the rate
of corrosion, Types of corrosion-Galvanic cell corrosion, Concentration cell
corrosion (Differential aeration principle) Pitting Corrosion, Stress
corrosion, intergranular corrosion, waterline corrosion methods to minimize
the corrosion- I) Material selection and Proper design, II) Cathodic
protection i) Sacrificial anodic protection ii) Impressed current method
III)Anodic
protection,
IV)
Metallic
coating
i)
Anodic
coating
(Galvanization) ii) Cathodic coating (Tinning)
5.0
Spectroscopic techniques and applications
7
Electromagnetic radiation, electromagnetic spectrum, Interaction of
L1, L2, L3
electromagnetic radiation with matter, Beer-Lambert’s law ( mathematical
expression
and
derivation,
Numerical
expected),
UV
Visible
Spectrophotometer: Principle, Instrumentation (Single beam and double
beam spectrophotometer),
and Application. Vibrational Spectroscopy:
Principle, Instrumentation and Application. Identification of functional
group of compounds based on IR spectroscopy.
6.0
Stereochemistry
7
L1,L2, L3
Introduction of Stereochemistry, Optical Isomerism, Optical Activity,
Elements of symmetry, Specific Rotation, (Numerical Based on Specific
rotation) Chirality/Asymmetry, Optical isomerism in tartaric acid and 2,3
dihydroxy butanoic acid, Enantiomers, Molecules with two similar and
dissimilar chiral-centers, Diastereoisomers, meso structures, racemic
mixture. Representation by flying wedge, Fischer and Newman projection.
Nomenclature of Stereoisomers: D&L system, R-S Configuration
Geometrical Isomerism, E-Z nomenclature, Conformation analysis of
alkanes (ethane and n-butane); Relative stability with energy diagram
Books and References:
Reference/Text Books:
Sr.
No.
1
2
3
4
Name of Book
Author Name
Edition
Stereochemistry, conformation
and Mechanism
Physical Chemistry I & II
P.S Kalsi
10th
Dr. Hrishikesh
Chatterjee
Atkins
Skoog, Holler
Gauch
Vogel
2nd
Morrison Boyd
7th
Physical Chemistry
Principle of Instrumental
Analysis
Vogels Textbook of quantitative
chemical Analysis
Organic Chemistry
5
6
8th
7th
8th
List of Practical/ Experiments:
Practical
Number
Type of
Experiment
Practical/ Experiment Topic
Hrs.
Determination of Total hardness of water
by complexometric titration using EDTA
2
1
Basic
Experiments
Cognitive levels
of attainment as
per Bloom’s
Taxonomy
L1, L2, L3, L4
Determination of Permanent hardness of
water by complexometric titration using
EDTA
2
L1, L2, L3
Estimation Saponification value of an
lubricating oil.
2
L1, L2, L3
2
3
5
5
Advanced
Experiments
Basic
Experiments
8
9
Design
Experiments
10
11
L1, L2, L3, L4
Determination of surface tension using
Drop number method by Stalagmometer
2
L1, L2, L3
Separation of organic binary mixture
using Thin layer chromatography
2
L1, L2, L3
To determine λmax and Molar extinction
coefficient of given solution of KMnO4
using Colorimeter.
2
L1, L2, L3
To determine λmax and Molar extinction
coefficient of given solution of CuSO4 –
NH3 complex using Colorimeter.
2
L1, L2, L3
To determine the turbidity of given
unknown water sample.
2
L1, L2, L3
To study the effect of the change in
temperature or viscosity of a lubricating
oil by using a redwood viscometer no. 1.
2
L1, L2, L3
Removal of Hardness of water by using
Ion Exchange Colum
8
L1, L2, L3
Determination of Percentage of Iron in
Plain Carbon Steel.
6
7
2
Estimation of Acid value of used
lubricating oil.
4
Project
Based
ExperimentsGroup Activity)
(Students should
complete any one
project
Based
experiment from
the list or any
other project in
discussion
with
Faculty
inCharge)
2
Synthesis of a Meta dinitrobenzene (drug
intermediate).
Inorganic Preparation: Preparation of
Tetraamine coppe (II) Sulphate
Total
30
Table of Contents
Sr. No.
1
Name of Module
A. Atomic, Molecular Orbitals and Chemical Bonding
Page No.
1-42
Schrodinger equation (No derivation), Atomic Orbital, Shapes of
s, p d and f Orbital, Hybridization of atomic orbitals involving s,p
and d orbitals, Valence Bond theory (VBT), its application and
drawback, Concept of Molecular orbital theory, Molecular orbital
Diagram (Energy level diagrams) of diatomic molecules like H2,
He2, Be2, C2, N2, O2, F2 and Ne2 (up to atomic no. 10), Application
of Molecular Orbital Theory
B. Co-ordination chemistry and its application
Introduction to Coordination chemistry and coordination
number, Chelating agent, EDTA as chelating agent and its
application for estimating hardness of water, Hardness of water
(it’s source, unit of measurement and Numerical based on
hardness calculation), Estimation of Hardness of water by
Complexometric method, theory, Procedure, calculation and
numerical. Crystal field theory and the energy level diagrams for
transition metal ions (Tetrahedral and Octahedral geometry) and
their applications.
2
Instrumental Methods of Analysis
43-60
Introduction to Chromatography, Types of Chromatography
(Adsorption and partition chromatography), Thin Layer
Chromatography (Theory, Principle, technique and applications),
Gas Chromatography – (Introduction, theory, instrumentation.,
working) High Performance Liquid Chromatography, –
introduction, theory, instrumentation. Interpretation of
Gas/HPLC Chromatogram and TLC plate of various samples.
3
Green Chemistry and Catalysis
A. Green Chemistry
Introduction to Green Chemistry, The 12 principles of Green
Chemistry, Prevention of waste, Atom Economy, Less hazardous
Chemical synthesis, Safer solvent and Auxiliary, Design for
energy efficiency, use of renewable feedstock, Reduction of
derivatives, Catalysis, Design for degradation, Real time analysis
(Uses of electronic devices and sensors for process control),
inherently safer chemistry for accident prevention. Design of
Greener route of synthesis over conventional route, Numerical
based on calculation of Atom economy
B. Catalysis
Role of Catalyst in making the chemical process
Green, Relevance and examples, Homogeneous and
heterogeneous catalysis, Theory of Heterogeneous Catalysis
(Adsorption Theory), Catalytic Converters, Acid Base catalysis,
61-91
Solid Acid Catalysis, Solid Base Catalysis, Transition metal
Catalysis, Metal and supported metal catalysis, Catalyst design
through artificial intelligence and computer modelling.
4
Electrochemistry & Application (Corrosion)
A. Electrochemistry
92-133
Concept of electrode potential, Concept of Electrochemical cell,
EMF of Cell, Cell potentials by Nernst equation, Relation of free
energy with EMF of Cell, Numerical based on EMF and its
feasibility prediction.
B. Corrosion
Introduction of Corrosion, Fundamental reason, Mechanism of
corrosion-i) Electrochemical/Wet Corrosion mechanism a)
Evolution of hydrogen gas and b) Absorption of oxygen gas,
ii)Direct Chemical/Dry/ Atmospheric Corrosion a) Due to
oxygen b) Due to other gases Factors affecting the rate of
corrosion, Types of corrosion-Galvanic cell corrosion,
Concentration cell corrosion (Differential aeration principle)
Pitting Corrosion, Stress corrosion, intergranular corrosion,
waterline corrosion, Methods to minimize the corrosion- I)
Material selection and Proper design, II) Cathodic protection i)
Sacrificial anodic protection ii) Impressed current method
III)Anodic protection, IV) Metallic coating i) Anodic coating
(Galvanization) ii) Cathodic coating (Tinning)
5
Spectroscopic techniques and applications
134-165
Electromagnetic radiation, electromagnetic spectrum, Interaction
of electromagnetic radiation with matter, Beer-Lambert’s law
(mathematical expression and derivation, Numerical expected),
UV Visible Spectrophotometer: Principle, Instrumentation
(Single beam and double beam spectrophotometer), Application.
Vibrational
Spectroscopy:
Principle,
Instrumentation,
Application. Identification of functional group of compounds
based on IR spectroscopy.
6
Stereochemistry
Introduction of Stereochemistry, Optical Isomerism, Optical
Activity, Elements of symmetry, Specific Rotation, (Numerical
Based on Specific rotation), Chirality/Asymmetry, Optical
isomerism in tartaric acid and 2,3 dihydroxy butanoic acid,
Enantiomers, Molecules with two similar and dissimilar chiralcenters, Diastereoisomers, meso structures, racemic mixture.
Representation by flying wedge, Fischer and Newman projection.
Nomenclature of Stereoisomers: D&L system, R-S Configuration
Geometrical Isomerism, E-Z nomenclature, Conformation
analysis of alkanes (ethane and n-butane); Relative stability with
energy.
166-201
1
Atomic molecular and Chemical bonding
Module: 1A
Atomic, Molecular Orbitals and Chemical Bonding
Lecture : 1
1.1
Atomic, Molecular Orbitals and Chemical Bonding
1.2
Motivation:
Engineers use their knowledge of the structure of atoms to do everything from
developing new materials to exploiting the energy of nuclear reactions for electricity.
The matter and molecules that make up the world around us are formed mostly by
many different atoms bonding together — each having their own properties or
attributes. Knowledge of quantum theory open possibilities for the engineers to use
these techniques for a diverse range of technological applications, from building threedimensional nanomaterials and nanostructure to devices. Applications of
Coordination chemistry inspires engineers to apply their knowledge and skill to
develop the models which can assess the impurity level in water and design new
technology for its removal which can address the of critical need of society.
1.3
Historical background:
Quantum mechanics was developed independently in 1926 by Warner Heisenberg and
Erwin Schrodinger. Valence Shell Electron Pair Repulsion Theory was developed earlier
by Sidgwick and Powell in 1940 and it was further developed by Gillespie and Nyholm
in 1957.Valece Bond Theory was developed by Heitler and London in 1927 and further
improved by Pauling.
1.4
Syllabus:
Duration
(hour)
SelfStudy
(hours)
Introduction to atomic and molecular structure.
Schrodinger equation and terms involved in it. Atomic
Orbital
Shapes of s, p d and f Orbital, Hybridization of atomic
orbitals involving s, p and d orbitals
1
2
1
2
3
Valence Bond theory (VBT) application of VBT and
drawback,
1
2
4
Concept of Molecular Orbital Theory, Molecular orbital
Diagram
1
2
1
2
Lecture
No.
1
2
5
Contents
Energy level diagrams of diatomic molecules like H2, He2,
Be2, C2, N2, O2. Application of MOT
Chemistry Sem I/II
6
Introduction to Coordination chemistry and coordination
number, Chelating agent, EDTA as chelating agent and its
application for estimating hardness of water
1
2
7
Estimation of hardness of water, theory, procedure
1
2
8
Calculation and numerical.
1
2
9
Crystal field theory and the energy level diagrams for
transition metal ions (Octahedral geometry)
1
2
Energy level diagrams for transition metal ions
(tetrahedral geometry), and their applications.
1
2
10
1.5
Weightage: 18 Marks
1.6
Learning Objective:
•
•
•
•
•
•
1.7
2
Learners shall be able to understand the significance of Schrodinger equation.
Learners shall be able to understand the concept of hybridization and its
application.
Learners shall be able to draw the molecular orbital diagram of homoatomic
molecules.
Learners shall be able to explain the theory of EDTA and its application in
determination of hardness of water molecule.
Learners shall be able to apply the knowledge of Crystal Field theory in
calculating the crystal field stabilization energy
Learners shall be able to draw the energy level diagram of octahedral and
tetrahedral complex.
Prerequisite:
To understand the module knowledge of various model of atom is required such as
Rutherford model, Thomson model, Bohr’s model etc. Learner should know the laws
for filling the electrons in various orbitals which decides the type of hybridisation of
molecule.
1.8.1 Theoretical Background:
Quantum mechanical model of atom was established with the help of dual nature of
matter and Heisenberg’s uncertainty principle. Knowledge of inorganic chemistry helps
in better understanding of this chapter. Various theories put forward to understand the
Ψ = wavefunction
EDTA = Ethylene diaminetetraacetic acid
EBT = Eriochrome Black T
1.8.2 Key Definitions:
a) Wave Function: It is the mathematical function whose value depends upon the
coordinates of the electron in the atom and does not carry any physical meaning.
3
Atomic molecular and Chemical bonding
b) Degree of hardness: It is the total quantity of hardness causing salts present in water.
It is expressed in terms of CaCO3 equivalent quantity.
1.9
Course Content:
1.9.1
Introduction:
This module deals with the structure of atom and molecule. Schrodinger equation
which helps in understanding the nature of electron, orbit and orbital. As Valence
Bond theory suggest that covalent molecule is formed by the overlapping of atomic
orbitals in which the identity of atom retains even after bond formation. While on the
other side in Molecular Orbital Theory method valency electrons are considered to be
associated with all the nuclei in molecule. Crystal Field Theory emphasize the
formation of low spin and high spin complex based on distribution of electrons in d
orbitals Coordination chemistry focuses on its application in determining the hardness
of water sample by using EDTA method.
1.9.2
Simplest form of Schrodinger equation
Ĥψ=Eψ
Ψ = wavefunction of ē
Ε = energy of electron
Ĥ = Hamiltonian operator (mathematical operator) it is complex operator
Detailed Equation
𝒅𝟐 𝝍
𝒅𝒙𝟐
+
𝒅𝟐 𝝍
𝒅𝒚𝟐
+
𝒅𝟐 𝝋
𝒅𝒛𝟐
+
𝟖Л𝟐 𝒎
𝒉𝟐
(𝜺 – v) ψ = O
x.y.z are three space co-ordinates.
m is mass of electrons, φ is wavefunction
h is Planck constant
E is total energy
V is potential energy of electron
After solving this eqn we will get different values of φ ← quantum no. n, l, m
i) Energy level of electrons (Ĥφ – εψ)
Ψ2 = probability density of finding electron at given point.
More the magnitude of ψ2 at any point more will be the probability of finding the electron &
vice versa. Schrodinger equation tells the probability of finding the electron
1.9.3
Important features of Quantum Mechanical Model
1) The energy of electron in atom is quantized that means that a certain amount of energy is
associated with an electron present around the nucleus.
Chemistry Sem I/II
4
2) The electrons are present in quantized energy levels which are a direct result of wave like
properties of electrons and are permissible solutions of Schrodinger wave equation.
3) This gives the most probable region in an atom where the probability of finding the
electron is maximum. Therefore, this model is in accordance with the Heisenberg
uncertainty principle & does not specify the exact position & momentum of the electron.
But it talks about the most probable region i.e. orbitals.
4) An atomic orbital is described with the help of wave functions ψ for an electron in an
atom. Whenever an electron is described by a wave function, we say that electron occupies
that orbitals. Since many wave functions are possible for an electron there are many atomic
orbitals in an atom. In each orbital election has definite energy. Thus, all information about
the electron in an atom is provided by its orbital wave function.
5) The probability of finding the electron at a point within the atom is proportional no square
of the wave function i.e. (ψ)2 at that point. Therefore (ψ)2 is known as probability density
& is always positive.
6) The quantum mechanical model gives three constants, known as quantum numbers which
are required to specify the position & energy of electron in an atom.
1.9.4
•
•
•
•
•
Comparison of Atomic Orbital (AO) and Molecular Orbital (MO)
Atomic Orbital
Molecular Orbital
It contains a single nucleus i.e. it is • It contains more than one nucleus i.e. it is
monocentric.
polycentric.
Nucleus of the atom is fixed in space.
• Nuclei of the constituent atoms of the
molecule are fixed in space at their proper
relative.
They are named as s, p, d, and f.
• They are named as σ, Л, and δ.
It is derived for and from a particular • It is derived from the constituent atoms.
atom.
• The shape, size and energy of a M.O.
It has a definite shape, size and energy.
depends upon the shape, size and energy
of orbitals of constituent’s atoms.
1.9.5 Shapes of Atomic Orbitals
Electrons are not present in a fixed circular path. The orbital wave function is a
mathematical function of the coordinates of electron.
a) For 1s orbital, the wavefunction continuously decreases with the increase of distance.
b) For 2s-orbital, the wavefunction reaches a point where it becomes zero and then
negative. So, wavefunction is both positive and negative for 2s orbital, depending
upon the distance.
c) For 2p orbital, the wavefunction increases with distance, attains a maximum value and
then decreases.
5
Atomic molecular and Chemical bonding
1.9.6
Plot of wave function and distance
(a) 1s orbital
orbital
(b) 2s
(c) 2p orbital.
Fig:1.1
Variations of ψ2 with distance from the nucleus
(a) 1s orbital
(b) 2s-orbital
Fig: 1.2
d) Let’s check the take away from this lecture
e)
1) Any wave function can be written as a linear combination of -------a) Eigen vectors
b) eigen values
c) eigen functions
operators
d)
2) According to Valence Bond Theory, the direction of a bond which is formed due to
overlapping will be
a) In the same direction in which orbitals are concentrated
b) In the opposite direction in which orbitals are concentrated
c) Perpendicular to the direction in which orbitals are concentrated
d) None of the above
3) The probability density is the square root of wave function
a) square root of wave function
c) inverse of wave function
b) absolute value of wave function
d) absolute of square of wave function
Chemistry Sem I/II
6
Exercise
Q.1
Distinguish between atomic orbital and molecular orbital.
Q.2 Draw the plot of wave function and distance for 1s and 2s orbital
Questions/Problems for practice:
Q.3
Discuss Quantum Mechanical model of an Atom with the help of Schrodinger’s
Wave equation and its important features.
Q.4 Discuss the conditions which govern the combinations of atomic orbitals to form
molecular orbital.
Learning from this lecture: Learners will able to know the Schrodinger equation and its
significance in determining the probability of electron in an atom
Lecture: 2
Shapes of Orbital & Hybridization of atomic orbitals
1.9.7
Learning Objective:
Learners will be able to understand the Shapes of s, p d and f Orbital, Hybridization
of atomic orbitals involving s, p and d orbitals
1.9.8
Shapes of Orbital:
Shape of s orbital
a) s-orbitals are spherical in shape, so they are unidirectional. This is evident as value
of l for s orbital is 0 and even m value is zero. Therefore, they have only one
unidirectional orientation i.e. probability of finding the electron is same in all
directions from the nucleus.
Orbital Shape
Density of electron cloud
1s
Spherical Maximum at the nucleus
Node
No
node
Spherical Maximum at the nucleus and decrease at One
large distance
node
Size
Smaller
2s
Larger
ns-orbitals have (n-1) nodes there is no longer in s-subshell
7
Atomic molecular and Chemical bonding
Fig: 1.3 Shape of s Orbitals
p orbitals are dumb- bell shaped. For p orbitals(l=1), there are three possible orientations as
m = -1,0,+1 value. There are three p orbitals whose axes are mutually perpendicular which
are designated as px, py ,pz in each p subshells. p orbitals also increase in size and energy
with the increase in the principal quantum no. Number of radial nodes for p orbitals is given
by (n-20 formula. Thus, there is no radial node in 2p orbital
Orbital
2p
3p
4p
Radial nodes
0
1
2
Fig: 1.4 Shape of p- Orbital
Shape of d orbitals
d orbitals have dumb-bell shape except dz2 . There are five d orbitals correspond to l=2 and
therefore m values -l to +l including zero. So, m values come out to be 5 (i.e. -2, -1, 0, +1, +2).
The minimum value of principal quantum number associated with d orbitals has to be 3.
The five d orbitals are designated as dxy, dyz, dxz, dx2-y2,dz2
There are two types of nodes –
1) Radial nodes
2) Angular nodes
Chemistry Sem I/II
8
Radial nodes
orbital Radial node
There are to angular nodes in d-
3d
0
orbital
No. of angular nodes are given
Total no. of nodes=n-1
Radial nodes=n-l-1
Angular nodes=l
Nodal planes=l
4d
1
by=l
5d
2
Fig: 1.5 Shape of d-orbital:
It is oriented in 7 different ways and each orientation can hold 2 electrons. Therefore, f
orbitals together have 7 degenerates and hold 14 electrons. these have
corresponding magnetic quantum number m values in the set {−3, −2, −1, 0, +1, +2, +3}.\
Fig: 1.6 Shape of f orbitals
9
Atomic molecular and Chemical bonding
1.9.9 Hybridisation
The phenomenon of intermixing of the orbitals of slightly different energies so as to
redistribute their energies and to give new set of orbitals of equivalent energy & shape. The
new orbitals formed as a result of hybridization are called hybrid or hybridized orbitals. Many
types of hybridisation which depends on type of orbital involved such as sp, sp 2, sp3d, sp3
1.9.9.1 Features of Hybridisation
The number of hybridized orbitals formed is equal to the number of orbitals that get
hybridized. The hybridized orbitals are always equivalent in energy & shape. The hybrid
orbitals are directed in space in some preferred directions to have stable arrangements.
Therefore, the type of hybridization gives the geometry of molecule. The hybrid orbitals are
more effective in forming stable bonds than the pure atomic orbitals.
1.9.9.2 Prediction of hybridisation of central atom
The hybrid state(H) of central atom in covalent molecules or ionic species other than complex
ions can be predicated by following methods.
Locate the central atom.
i)
ii)
Find the number of surrounded atom or groups (indicate them as S)
Find the number of outers shell electron in the central atom in ground state
(indicate them as E)
Find the valency of the central atom in the given species. It is equal to the number
of monovalent groups like H, (OH), Cl, F, etc. or double the member of divalent
groups like O, S etc. directly linked to central atom (Indicated as V).
Find electric charge if species is ionic let it be represented by C.
Formula to calculate the no. of hybrid orbitals
iii)
iv)
v)
Find the value of H formula:
H= S+1/2 [E – V ± C]
+ sign with charge for anion & -ve sign with charge for cation. Find the value of H by using
formula:
Value of H
2
3
4
5
6
7
Hybrid state of central atom sp sp2 sp3 sp3d sp3d2 sp3d3
Examples
a) Identify the hybrid state of Boron atom in BF-4
H= S + ½ [E-V+C]
= 4+ ½ [3-4+1] = 4 ∴ B = sp3
b) Identify the hybrid state of S in SO2
H = S+1/2 [E-V± C]
Chemistry Sem I/II
10
= 2+ ½ [6-4+0]
= 03 = sp2
c) Identify the hybrid state of C in CH₄
H = S + ½ [E-V±C]
= 4+ ½ [4-4+0] = 4 = sp3
a) Identify the hybrid state of Sin SO42H = S + ½ [E-V±C]
= 4 + ½ [6-8+2] =4 ∴ S = sp3
1.9.9.3 Hybridization of atomic orbitals involving s, p and d orbitals
No. of
hybrid
orbital
2
3
Hybrid
state of
central
atom
sp
sp2
Number of
Number
surrounding of lone
atoms
pairs
Shape
Examples
2
3
0
0
C2H2,BeF2,CO2,NO2C2H4,BF3,AlCl3,SO3
1
0
1
2
0
1
2
3
0
linear
Trigonal
planor
bent
tetrahedral
Pyramidal
bent
Trigonal
bipyramidal
See saw
T-shape
linear
octahedral
4
sp3
5
sp3d
2
4
3
2
5
sp3d2
4
3
2
6
6
SO2,PbCl2
CH4,SO4-2,SiCl4
NH3,H3O+,PH3
H2O,NH2PCl5
SF4
ClF3
XeF2
SF6
Table:1.1
Lets’ check the take away from the lecture (MCQs)
1. Number of radial nodes present in 2p orbital is
a) 1
b) 3
c) 0
d) 5
2. The atoms in a molecule of acetylene adopt what kind of geometry?
a) linear
b) tetrahedral
c) octahedral d) trigonal planar
11
Atomic molecular and Chemical bonding
3. An octahedral complex is formed when central metal ion undergoes
hybridisation among ---------------------------orbitals
a) sp3
b) dsp2
c) sp3d
d) sp3d2
Exercise
Q.4
Define hybridisation. Write important features of hybridisation.
Q.5
Draw the geometry of p and d orbitals with proper justification.
Questions/Problems for practice:
Q.6
Try writing anodic half, cathodic half and overall cell reaction for 5 different set of
metal of your choice. You can refer Electrochemical/ Galvanic series for doing this job.
Geometry of ammonium ion is tetrahedral. Predict the most probable reason for the same.
Q.7What are possible types of hybridisation using s, p, d orbitals from hybridisation point of
view for deciding the geometry of molecule? Write example of each type of hybridisation.
Q.8 How do we predict the hybrid state of central metal ion? Write general formula to
calculate this by taking suitable example for this purpose.
Learning from this lecture: Learners will able to know the Shape of s, p, d and f orbitals
and their role in hybridisation for determination of geometry of molecule.
Lecture: 33
Valence Bond Theory (VBT), Application of VBT and Drawback
1.9.9.4 Learning Objective:
Learners will be able to understand the features of VBT its application in geometry of
molecule along with its drawbacks.
1.9.9.5 Valence Bond Theory (VBT)
This theory was developed by Heitler and London in 1927 and later extended by
Pauling and Slater. It is also called electron pair theory. Understanding VBT involves
the knowledge of atomic orbitals, electronic configuration of elements, the overlap
criteria
Postulates of Valence Bond Theory
1. Condition for bond formation: A covalent bond will be formed only when the halffilled atomic orbitals of one atom overlaps the half-filled atomic orbitals of another
atom
2. Strength of bond: The strength of bond depends upon the extent of overlapping.
Greater the overlapping of atomic orbitals, stronger would be the bond.
Chemistry Sem I/II
12
3. Energy of orbitals: In the formation of covalent bond, only those orbitals will overlap,
which have nearly the same energy.
4. Binding force: During the process of overlapping of atomic orbitals, the electron
density increases between the two nuclei. This causes the decrease in nuclear
repulsion and increase the attraction between the nuclei and the electrons, forming
strong covalent bond.
5. Stability of covalent bonds: When repulsion between two approaching atoms
becomes minimum, the energy of the system decreases to its minimum value and
stable covalent bond is formed.
6. Number of covalent bonds: The number of covalent bonds formed by an atom is
equal to the number of half-filled atomic orbitals present in an atom.
7. Geometry of molecule: Since each atomic orbital has particular direction n space, the
covalent bod formed by these atomic orbitals is also directional in nature.
8. Directional property: The overlap between the atomic orbitals can be positive,
negative or zero depending upon the characteristics of orbitals participating to
overlap.
a) The positive overlap involves the overlap of the lobes of same sign which leads to
attractive interactions.
b) Negative overlap involves the overlap of the lobes of opposite sign which leads to
repulsive interactions.
c) Zero overlap implies inability of any kind of interactions.
1.9.9.5 Interacting forces during the formation of covalent bonds:
1. Force of attraction between the nucleus of one atom and the electron of other atom and
vice versa.
2. Force of repulsion between the two nuclei of combining atoms, as the nuclei are
positively charged.
3. Force of repulsion between the two electrons of combining atoms, as the electrons are
negatively charged.
Types of overlapping of atomic orbitals:
i) s-s overlap: Example H2 molecule
Fig: 1.7
ii) s-p overlap
Example HCl molecule
13
Atomic molecular and Chemical bonding
Fig: 1.8
iii) p-p overlap
Example F2 molecule
Fig: 1.9
1.9.9.6 Application of Valence Bond Theory
1. Valence bond theory explains the concept of maximum overlap, resulting in the
formation of strong bond.
2. This theory is used to explain the covalent bond formation in many molecules.
3. For example, in the case of the F2 molecule, the F−F bond is formed by the overlap
of pz orbitals of the two F atoms, each containing an unpaired electron. Since the nature
of the overlapping orbitals are different in H2 and F2 molecules, the bond strength and
bond lengths differ between H2 and F2 molecules.
4. In an HF molecule the covalent bond is formed by the overlap of the 1s orbital of H
and the 2pz orbital of F, each containing an unpaired electron. Mutual sharing of
electrons between H and F results in a covalent bond in HF.
Drawbacks
The VB theory explains the formation, structures and magnetic behaviour of
coordination compounds, it suffers from the following shortcomings:
(i) It involves a number of assumptions
(ii) It does not give quantitative interpretation of magnetic data.
(iii) It does not explain the colour exhibited by coordination compounds.
(iv) It does not give a quantitative interpretation of the thermodynamic or kinetic
stabilities of coordination compounds.
(v) It does not make exact predictions regarding the tetrahedral and square planar
structures of 4-coordinate complexes.
(vi) It does not distinguish between weak and strong ligands.
Chemistry Sem I/II
14
Lets’ check the take away from the lecture (MCQ s)
1) s-s overlap present in
a) O2
b) H2
c) B2
d) N2
2) Nature of interacting forces present during the formation of covalent bond a) Force of attraction
b) force of repulsion c) Both of the above d) Vander waal’s force
3) Which type of overlap is present in HF molecule?
a) s-s overlap
b) s-p overlap
c) p-p overlap
d) none
Exercise
Q.1
1.State the applications of Valence bond theory.
Q.2. Identify the type of overlap present in O2 molecule
Questions/Problems for practice:
Q.3
What are the assumptions of Valence Bond Theory?
Q.4 How are interacting forces work between two combining atoms? Explain
Learning from this lecture: Learning from this lecture: Learners will able to understand
the type of overlap and nature of different interacting forces present in covalent
molecule
Lecture: 4
Concept of Molecular Orbital Theory, Molecular Orbital Diagram
1.9.9.7 Learning Objective:
Learners will be able to understand the Shapes of s, p d and f Orbital, Hybridization
of atomic orbitals involving s, p and d orbitals
1.9.9.8 Concept of Molecular Orbital Theory
MOT was put forward by F.Hund and R.S.Mulliken in 1932.This theory assumes that
in molecules the atomic orbitals lose their identity and the electrons present in
molecules are present in new orbitals called molecular orbitals which are not
associated with a particular atom belong to the molecule as a whole.
The salient features of this theory are
i) The electrons in a molecule are present in the various molecular orbitals as the
electrons of atoms are present in the various atomic orbitals.
ii) The atomic orbitals of comparable energies & proper symmetry combine to form
molecular orbitals
15
Atomic molecular and Chemical bonding
iii) While an electron in an atomic orbital is influenced by one nucleus, in a molecular
orbital it is influenced by two or more nuclei depending upon the number of atoms in
molecule. Thus an atomic orbital is monocentric while a molecular orbital is
polycentric.
iv) The no. of molecular orbitals formed is equal to the no. of combining atomic orbitals.
When two atomic orbitals combine, two molecular orbitals are formed. One is known
as bonding molecular orbital and other is called antibonding molecular orbital.
v) The bonding M.O has lower energy & has greater stability than the corresponding
antibonding molecular orbital.
vi) The electro probability distribution around a nucleus is given by an atomic orbital, the
electron probability distribution around a group of nuclei in a molecule is given by
molecular orbital.
vii) The molecular orbitals like atomic orbital are filled in accordance with Aufbau
principle obeying the Pauli’s exclusion Principle & Hunds rule
1.9.9.9 Linear combination of Atomic Orbitals LCAO method:
Molecular orbitals are formed by the combination of atomic orbitals of bonded atoms.
In wave mechanics atomic orbitals are expressed by wave functions(Ψs). These wave
functions are obtained as the solution of Schrodinger wave equation. The approximate
method to obtain the wave function for molecular orbitals is known as Linear
combination of Atomic Orbitals LCAO method.
Let us apply this theory to homonuclear diatomic molecules such as H 2
molecule. Consider the two atoms of hydrogen in the molecule is A and B. Each
hydrogen atom has one electron in 1s orbital in ground state. These atomic orbitals
may be represented by the wave functions ΨA and ΨB respectively. Then according to
LC AO method the molecular orbitals in the H2 molecule are given by linear
combination (addition or subtraction) of wave functions of individual atoms, i.e. ΨA
and ΨB as shown below.
ΨMO = Ψ A ± Ψ B
Ψb = Ψ A + Ψ B
Ψa = Ψ A - Ψ B
the molecular orbital Ψb formed by the addition overlap (constructive interference of
waves) of atomic orbital is called bonding molecular orbital (BMO) and the molecular
orbital Ψa formed by the subtractive overlap (destructive interference of waves) of
atomic orbital is called antibonding molecular orbitals is called antibonding molecular
orbital.(ABMO).
The probable probability density bonding molecular orbital is given by Ψb 2
(Ψ A + Ψ B )2 which means that shared electron density is higher than the sum of
electron densities of two separate orbital addition combination leads to increase in
electron density e between the two nuclei a and b similarly the probability density of
antibonding molecular orbital is given by Ψa2 or (Ψ A - Ψ B )2 which means that the
shared electron density is lower than the sum of electron densities of separate orbitals.
Chemistry Sem I/II
16
Thus subtraction combination leads to lowering of electron density in between the
nuclei.
1.10
Formation of Bonding and Antibonding MOs (Pictorial Representation):
Two atomic orbitals (say 1s) combine to give a pair of MOs – ψ – bonding and ψ* antibonding. The overlap of atomic orbitals along the axis joining the nuclei forms a σ
bond so these molecular orbitals are further called σ bonding MO and σ* antibonding
MO.
Fig: 1.10 Bonding and Anti bonding Molecular Orbital
1.10.1 Energy of Molecular Orbital
a)
Energy of all the molecular orbitals whether bonding or antibonding is not equal.
Their energy increase in the following order. No. of electrons >14
𝜎1𝑠 < 𝜎 ∗ 1𝑠 < 𝜎2𝑠 < 𝜎2 ∗ 𝑠 < [𝜋2𝑝𝑥 = 𝜋2𝑝𝑦] <𝜎2pz < [𝜋 ∗ 2𝑝𝑥 = 𝜋 ∗ 2𝑝𝑦] < 𝜎 ∗ 2pz
Above increasing energy order will be followed in molecular orbital energy level
diagram for homonuclear diatomic molecules like Li₂, Be₂ , B₂,C₂,N₂
If No. of electrons = or < 14 their energy increases in the following order
𝜎1𝑠 < 𝜎1 ∗ 𝑠 < 𝜎2𝑠 < 𝜎2 ∗ 𝑠 < 𝜎2pz < [𝜋2𝑝𝑥 = 𝜋2𝑝𝑦] < [𝜋 ∗ 2𝑝𝑥 = 𝜋 ∗ 2𝑝𝑦] < 𝜎 ∗ 2pz
Above increasing energy order will be followed in molecular orbital energy level
diagram for homonuclear diatomic molecules like O₂, F₂, Ne₂ etc.
17
Atomic molecular and Chemical bonding
Energy of 𝜋2𝑝𝑥 𝑎𝑛𝑑 𝜋2𝑝𝑦 molecular orbitals is equal, similarly energy of 𝜋 ∗ 2𝑝𝑥 = 𝜋 ∗
2𝑝𝑦 orbitals is equal.
The maximum number of electrons in 𝜎 molecular orbital is two and these two
electrons must be of opposite spin. (Paulis exclusion principle)
If there are two molecular orbitals of the same energy, pairing of electrons takes place
only after each orbitals of the same energy have one electron (Hund’s Rule)
Electrons are filled in these orbitals according to Aufbau’s principle. The molecular
orbitals of lowest energy are filled first.
After filling two electrons in 𝜎1𝑠 molecular orbital, filling of 𝜎 ∗ 1𝑠 starts. Similarly,
after filling two electrons in 𝜎2𝑠 molecular orbital, filling of 𝜎 ∗ 2𝑠 starts.
Fig: 1.11 Molecular Orbital of Hydrogen
1.10.2 M.O. Energy Level Diagram:
When two 1s orbitals combine, they form a ψ (or σ) MO and a ψ* (or σ*) M.O. Fig. 1.5
shows the energy levels of atomic orbitals and molecular orbitals. In this diagram, the
AOs are shown to the sides and the MOs formed are shown in the center. The two AOs
ψ 1 and ψ 2 are of identical energy and hence the MOs ψ and ψ* or σ and σ* contain
equal contribution from ψ 1 and ψ 2. The M.O. energy level diagram is symmetrical.
Comparison between Bonding MO and Antibonding MO
BMO
ABMO
1. It is obtained by addition of wave 1. It is obtained by subtraction of wave
functions of AOs.
functions of AOs.
Chemistry Sem I/II
18
2. It has lower energy than that of 2. It has higher energy than that of
combining AOs.
Combining AOs.
3. It is stable
3. It is unstable.
4. Electron density is concentrated in the 4. Electron
density
is
away
from
region between the nuclei.
internuclear region.
5. It results in bonding.
5. It results in nonbonding.
6. Nuclear repulsion is shielded
6. Nuclear repulsion is not shielded.
7. There is no node.
7. It has a node perpendicular to the bond
axis.
Lets’ check the take away from the lecture (MCQ s)
1) Energy of bonding orbital should be
a) Less
b) more
c) average
d) none
2) In which type of orbital nuclear repulsion is shielded?
a)
Atomic orbital b) bonding molecular orbital c) antibonding molecular orbital
d) both b and c
3) Name the type of orbital in which has node perpendicular to the bond axis
a) Atomic orbital
b) Antibonding Molecular Orbital
c) bonding Molecular Orbital
d) None
Exercise
Q.1
1.Describe the energy level of molecular orbital if total number of electrons are 14 or
less than 14.
Q2. Discuss any five important features of LCAO theory.
Questions/Problems for practice:
Q.3
Explain the term bonding and antibonding molecular orbitals.
Q.4Discuss any five important postulates of molecular orbital theory and with the help of
suitable example.
Q.5 What are the types of molecular orbitals formed after combining the atomic orbitals?
Learning from this lecture: Learners will able to know the concept of molecular orbital theory
and the order of filling the electrons in molecular orbitals.
19
Atomic molecular and Chemical bonding
Lecture: 5
Molecular Orbital Diagram
1.10.3 Learning Objective:
Learners will be able to draw to molecular orbital diagram of various
homomolecules and heteromolecules.
1.10.4 Hydrogen molecule
 It is formed by combination of two hydrogen atoms. Each hydrogen atom has one
electron in 1s orbital. Therefore, in all there are two electrons in hydrogen molecule
which are present in 𝜎1𝑠molecular orbital. So electronic configuration of hydrogen
molecule is
H₂: (𝜎1𝑠2)
The bond order of H ₂ molecule can be calculated as follows:
Bond order = Nb-Na/2 =2-0/2=1
 Two hydrogen atoms are bonded together by a single covalent bond. The bond
dissociation energy of hydrogen molecule has been found to be 438 kjmol-1 and bond
length equal to 74pm. Since no unpaired electron is present in hydrogen molecule,
therefore it is diamagnetic.
Fig: 1.12 Hydrogen molecule


It is formed by combination of two helium atoms. Each helium atom has two
electrons in 1s orbital.
Therefore, in all there are four electrons in helium molecule which are present in 𝜎1𝑠
and 𝜎 ∗ 1𝑠 molecular orbital. So electronic configuration of hydrogen molecule is
He₂: (𝜎1𝑠2), 𝜎1 ∗ 𝑠2
Chemistry Sem I/II
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The bond order of He₂ molecule can be calculated as follows:
Bond order = Nb-Na/2 =2-2/2=0

Since bond order of He₂ comes out to be zero, therefore this molecule is unstable and
does not exist. No unpaired electron is present in Helium molecule; therefore, it is
diamagnetic.
Lithium molecule


It is formed by combination of two lithium atoms. Each lithium atom has two electrons
in 1s orbital 1 electron in 2s orbital.
Therefore, in all there are six electrons in lithium molecule which are present in 𝜎1𝑠,
𝜎 ∗ 1𝑠 and 𝜎2𝑠 molecular orbital. So electronic configuration of lithium molecule is
Li₂: (𝜎1𝑠2), (𝜎1 ∗ 𝑠2), (𝜎2𝑠2)
The bond order of Li₂ molecule can be calculated as follows:
Bond order= Nb-Na/2 =4-2/2=1

Since bond order of Li₂ comes out to be one, therefore this molecule is stable. No
unpaired electron is present in lithium molecule, therefore it is diamagnetic.
21
Atomic molecular and Chemical bonding
Fig: 1.13 Lithium Molecule
Beryllium molecule
•
•
•
•
•
The electronic configuration of beryllium is 1s2,2s2 It is formed by combination of two
Beryllium atoms.
Each Beryllium atom has two electrons in 1s orbital and two electrons in 2s orbital
Therefore, in all there are eight electrons in Beryllium molecule which are present in
𝜎1𝑠 , 𝜎∗1𝑠, 𝜎2𝑠 , 𝜎∗2𝑠, molecular orbital.
So electronic configuration of Beryllium molecule is
Be ₂ : (𝜎1𝑠2), 𝜎1∗𝑠2, 𝜎2𝑠2 , 𝜎∗2s 2
The bond order of Be ₂ molecule can be calculated as follows:
Bond order=Nb-Na/2 =4-4/2=0
Since bond order of Be2 comes out to be zero, therefore this molecule is unstable and
does not exist. No unpaired electron is present in Beryllium molecule; therefore it is
diamagnetic.
•
Fig: 1.14 Beryllium molecule
Boron molecule
▪
▪
▪
The electronic configuration of boron is 1s2,2s2,2p1. Each boron atom has 5 electrons.
Thus, B ₂ molecule has 10 electrons which are to be accommodated in orbitals in order
of increasing energy.
The molecular orbital configuration of B ₂ molecule is:
(𝜎1𝑠)2, (𝜎1∗𝑠)2, (𝜎2𝑠 )2, (𝜎2∗𝑠) 2, (π2𝑝𝑥)1=π2𝑝𝑦)1 There are 6 bonding electrons and 4
anti-bonding electrons. Therefore, bond order may be calculated as:
Bond Order =1/2[Nb-Na/2] =6-4/2=1
Since 1-1electron is present in π2𝑝𝑥 and π2𝑝𝑦 hence it is paramagnetic molecule.
Chemistry Sem I/II
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Fig: 1.15 Boron molecule
Carbon molecule





The electronic configuration of oxygen is 1s2,2s2, 2p2. Each carbon atom has six
electrons. Thus, C ₂ molecule has 12 electrons which are to be accommodated in
orbitals in order of increasing energy. The molecular orbital configuration of C ₂
molecule is:
(𝜎1𝑠)2, (𝜎1∗𝑠)2, (𝜎2𝑠 )2, (𝜎2∗𝑠) 2,
(π2𝑝𝑥)2=π2𝑝𝑦)2 There are 8 bonding electrons and 4 anti-bonding electrons.
Therefore, bond order may be calculated as:
Bond Order =1/2[Nb-Na/2] =8-4/2=2
Since all electrons are paired, it is diamagnetic molecule.
23
Atomic molecular and Chemical bonding
Fig: 1.16 Carbon molecule
Nitrogen molecule






The electronic configuration of nitrogen is 1s2,2s2,2p3. Each nitrogen atom has seven
electrons. Thus, N ₂ molecule has 14 electrons which are to be accommodated in
orbitals in order of increasing energy. The molecular orbital configuration of N ₂
molecule is:
(𝜎1𝑠)2, (𝜎1 ∗ 𝑠)2, (𝜎2𝑠 )2, (𝜎2 ∗ 𝑠) 2, (𝜋2𝑝𝑥)2 = 𝜋2𝑝𝑦)2 ,(𝜎2pz) 2
There are 10 bonding electrons and 4 anti-bonding electrons. Therefore, bond order
may be calculated as:
Bond Order =1/2[Nb-Na/2] =10-4/2=3
The bond order suggests the bond energy is very high. The experimental value is 945
kj/mole.
Since all the electrons in nitrogen are paired, it is diamagnetic molecule.
Fig: 1.17 Nitrogen molecule
Chemistry Sem I/II
24
Oxygen molecule
▪
▪
▪
•
•
The electronic configuration of oxygen is 1s2,2s2,2p4. Each oxygen atom has eight
electrons.
Thus, O ₂ molecule has 16 electrons which are to be accommodated in orbitals in order
of increasing energy.
The molecular orbital configuration of O ₂ molecule is:
(𝜎1𝑠)2, (𝜎1∗𝑠) 2, (𝜎2𝑠 )2, (𝜎2∗𝑠) 2, (𝜎2pz) 2,
(π2𝑝𝑥)2=π2𝑝𝑦)2, (π2𝑝𝑥)1=π2𝑝𝑦)1 There are 10 bonding electrons and 6 anti-bonding
electrons. Therefore, bond order may be calculated as:
Bond Order =1/2[Nb-Na/2] =10-6/2=2
The bond order suggests the bond energy is high. The experimental value is 495
kj/mole and bond length121pm
Since it contains two unpaired electrons, it is paramagnetic molecule
Fig: 1.17 Oxygen molecule
Fluorine molecule





The electronic configuration of Fluorine is 1s2,2s2,2p5. Each Fluorine atom has 9
electrons. Thus, F ₂ molecule has 18 electrons which are to be accommodated in orbitals
in order of increasing energy. The molecular orbital configuration of F ₂ molecule is:
(𝜎1𝑠)2, (𝜎1∗𝑠)2, (𝜎2𝑠 )2, (𝜎2∗𝑠) 2, (𝜎2pz) 2,
(π2𝑝𝑥)2=π2𝑝𝑦)2, (π2𝑝𝑥)2=π2𝑝𝑦)2There are 10 bonding electrons and 8 anti-bonding
electrons. Therefore bond order may be calculated as:
Bond Order =1/2[Nb-Na/2] =10-8/2=1
Since it contains no unpaired electron, it is diamagnetic molecule.
25
Atomic molecular and Chemical bonding
Fig: 1.18 Fluorine molecule
Lets’ check the take away from the lecture (MCQ s)
1) Bond order present in O2molecule will be
a) 4
b) 2
c) 3
2)
d) 1
Diamagnetic and paramagnetic behaviour of molecule will be decided on
a)
b)
c)
d)
No. of unpaired electrons
No. of paired electrons
No. of paired and unpaired electrons
None
Exercise:
Q.1Bond order in lithium molecule is one. Explain on the basis of Molecular Orbital Theory.
Q.2 Discuss the criteria to predict the magnetic behavior of molecule.
Questions for practice:
Q.3 Draw the molecular orbital diagram of following molecules and comment on molecular
electronic configuration, bond Order, stability and magnetic characteristics.
i) N2
ii) F2
Q.4 Distinguish between bonding and antibonding molecular orbitals.
Learning from this lecture: Learners will able to draw molecular orbital diagram of
homoatomic molecules and able to predict stability, bond order and magnetic behaviour of
molecules.
Chemistry Sem I/II
26
Module: 1B
Lecture: 6
Introduction to Coordination chemistry and Coordination number, Chelating agent, EDTA
as chelating agent and its application for estimating hardness of water.
1.11
Learning Objective:
Learners will be able to understand the concept of Coordination chemistry and
Coordination number, Chelating agent, EDTA as chelating agent and its application
for estimating hardness of water.
1.11.1
Introduction to Coordination chemistry
Compounds formed due to combination of two or more simple stable salts which
retain their identity in solid as well as in dissolved state are called coordination
compounds.
Coordination compound is one which contains a central atom or ion surrounded by a
number of oppositely charged or neutral molecules. In a coordinate compound there
are coordinate linkages between the central metal atom or ion & opp. Charged ions or
molecules formed by donation of lone pair of ē by these ions or neutral molecules to
the central atom.
E.g.: Ni (CO)4 Ni is linked to 4 CO molecules.
Types of Complexes:
a) Cationic complexes: - Complex ion carries the charge [ Co (NH3)6] +2\
[Ni (NH3)6] +2
b) Anionic complexes: - [Fe (CN)6]-3, [Ag (CN)2]c) Neutral complex: - [Ni (CO)4], [Co Cl2(NH3)3]
Ligand is charged or neutral group or molecule that coordinates to a central atom or ion in a
coordination compound. It may be neutral, negative or positively charged entity.
E.g. Monodentate Ligand having single donor atom, e.g.NH3, H2O, NH3 etc.
Bidentate Ligand can bind through two donor atoms: H2N-CH2 -CH2 -NH2 (en)
Polydentate ligand: Ligands having several donor atoms e.g. Hexadentate EDTA
1.11.2
Coordination number:
It is the number ligands attached directly to the central metal atom or ion by coordinate
bonds in the given complex or coordinate ion.
E.g.
i)
ii)
ii)
iii)
Ion
[Ag (NH3)2] +
[Fe (CN)6]-3
Ni (Co)4
[Cr (H2O)6] +3
Ligand
NH3
CNCO
H2O
Co. No.
2
6
4
6
27
Atomic molecular and Chemical bonding
1.11.3
Chelating agent:
Chelates are complex cyclic compounds in which the substituents are linked to the
central metal atom by main & sec. valence bonds. A chelating agent is a substance whose
molecules can form several bonds to a single metal ion. In other words, a chelating
agent is a multidentate ligand. An example of a simple chelating agent is
ethylenediamine.
Chelation is a type of bonding of ions and molecules to metal ions. It involves the
formation or presence of two or more separate coordinate bonds between a polydentate
(multiple bonded) ligand and a single central atom. These ligands are called chelates,
chelators, chelating agents,
Fig: 1.19 Chelating agent
Many essential biological chemicals are chelates. Chelates play important roles in oxygen
transport and in photosynthesis. Furthermore, many biological catalysts (enzymes) are
chelates. In addition to their significance in living organisms, chelates are also economically
important, both as products in themselves and as agents in the production of other chemicals.
A chelate is a chemical compound composed of a metal ion and a chelating agent. A chelating
agent is a substance whose molecules can form several bonds to a single metal ion. In other
words, a chelating agent is a multidentate ligand. An example of a simple chelating agent is
ethylenediamine. A single molecule of ethylenediamine can form two bonds to a transitionmetal ion such as nickel (II), Ni2+. The bonds form between the metal ion and the nitrogen
atoms of ethylenediamine. The nickel (II) ion can form six such bonds, so a maximum of three
ethylenediamine molecules can be attached to one Ni2+ ion. Ni NH2 NH2 CH2 CH2 OH2 OH2
OH2 OH2 2+ chelate with one ethylenediamine ligand 2+ Ni NH2 NH2 NH2 NH2 CH2 CH2 CH2
CH2 OH2 OH2 chelate with two ethylenediamine ligands 2+
1.11.4 ETDA as chelating agent & its application for estimating hardness of water
EDTA is a multipurpose chelating agent. It can form four or six bonds with a metal ion, and
it forms chelates with both transition-metal ions and main-group ions. EDTA is frequently
used in soaps and detergents, because it forms a complex with calcium and magnesium ions.
These ions are in hard water and interfere with the cleaning action of soaps and detergents.
The EDTA binds to them, appropriating them and preventing their interference. In the
calcium complex, [Ca(EDTA)]2– , EDTA is a tetradentate ligand, and chelation involves the
Chemistry Sem I/II
28
two nitrogen atoms and two oxygen atoms in separate carboxyl (-COO‾) groups. EDTA is
also used extensively as a stabilizing agent in the food industry.
Fig: 1.20 Structure of EDTA
Lets’ check the take away from the lecture (MCQ s)
1) Chelate compounds are
a) Linear
b) cyclic
c) non cyclic
d) none
2) Ethylenediamine is ------------------ ligand
a) Monodentate b) bidentate
c) tridentate
d) hexadentate
3) EDTA has tendency to form compounds which is
a) Simple
b) complex
c) coordinate
d) all of
the above
Exercise:
1. Define coordination number . Find the coordination number of Nickel in nickel tetra
carbonyl.
2. Write a short note on chelating agent.
Questions for practice:
1) Define Coordination number. Explain with suitable examples.
2) EDTA as chelating agent is used in determination of hardness of water. Justify this
statement.
Learning from this lecture: Learners will able to understand the concept of coordination
number and its application for understanding the chelating concept.
Lecture: 7
Estimation of hardness of water, theory, Procedure, calculation and numerical
1.11.5 Learning Objective:
Learners will be able to understand the estimation of hardness of water by
complexometric titration (EDTA method).
29
Atomic molecular and Chemical bonding
1.11.6 Estimation of hardness of water
Theory of EDTA titration method
Theory of EDTA:
It is called complexometric titration, as EDTA forms complex ions with hardness causing Ca+2
and Mg+2 ions present in water. It is a titration of water sample, buffered at pH 10 against a
standard solution of Ethylene Diamine Tetra Acetic Acid whose structure is
HOOC – H2C
CH2- COOH
N–– CH2 –– CH2 –– N
HOOC – H2C
CH2– COOH
Disodium salt of EDTA
NaOOC – H2C
CH2- COOH
N–– CH2 –– CH2 –– N
HOOC – H2C
CH2–COONa
Complex structure of M-EDTA
Disodium EDTA reacts quickly with hardness causing metal ions to form cyclic co-ordinate
complex (chelate). pH of 10 is required for this reaction; otherwise it won’t go to completion.
For determination of end point, alcoholic solution of blue coloured dye Eriochrome Black -T
is added to water sample at the beginning. On this addition initially, Ca2+, Mg2+etc. form wine
red unstable complexes with the dye.
Mn++ EBT
[M – EBT]
On the addition of EDTA, these complexes are replaced by stable complexes. At the end point,
wine red complex disappears, and the original blue colour of the dye appears
M – EBT + EDTA
M – EDTA + EBT
∴the schematic reactions are
(i)
M++ + EBT →[ M-EBT]
(ii)
[M-EBT] + EDTA →[ M-EDTA] + EBT
(iii)
M++ + EDTA →[ M-EDTA]
Chemistry Sem I/II
30
1.11.7 Procedure
(1) Preparation of standard Hard water: Dissolve 1g of CaCO3in minimum quantity of dil. HCl. Boil it to dryness to expel excess of
acid & CO2dissolve the residue in distilled water to make 1 lit solution. It is standard hard
water. It contains 1 mg CaCO3 equivalent hardness.
(2) Preparation of EDTA Solution:Dissolve 4.0 g EDTA solid crystals and 0.1 g MgCl2in one liter distilled water.
(3) Preparation of indicator: Dissolve 0.5 g of Eriochrome Black-T in 100 ml of alcohol.
(4) Preparation of buffer solution: Add 67.5 g of NH4Cl to 570 ml of conc. ammonia solution and then dilute with distilled water
to 1 ml.
Titration 1:Standardization of EDTA: Mix 50 ml of standard hard water sample with 15 ml of buffer
solution and add to this mixture 4 or 5 drops of indicator. Titrate it against EDTA solution till
wine- red colour changes to blue. Let the volume of EDTA be V1ml.
Titration 2:Determination of total hardness: Mix 50 ml of hard water sample with 15 ml of buffer
solution and add to this 4 or 5 drops of indicator. Titrate it against EDTA solution till wine
red colour changes to blue. Let the volume of EDTA be V2ml.
Titration 3:
Estimation of permanent hardness
Take 250 ml of water sample in large beaker. Boil it till the volume is reduced to 50 ml. Filter
and wash the precipitate with distilled water, Collect the filtrate and washings in a 250 ml
measuring flask with distilled water and titrate 50 ml of it with EDTA solution. Let the volume
used be V3ml.
1.11.5.2 Calculations:50 ml of standard hard water = V1ml of EDTA solution.
∴50 x 1 mg of CaCO3 = V1ml of EDTA
∴1 ml of EDTA = 50 / V1mg of CaCO3
Note:- 50 ml of hard water = V2ml of EDTA = mg of CaCO3
∴Total hardness of water = 1000 V2/ V1mg / L
= 1000 V2 /V1ppm
Now 50 ml of boiled water = V3ml of EDTA
= V3/V1X50 mg of CaCO3
∴1000 ml of boiled water = 1000 V3/V1mg of CaCO3
∴Permanent Hardness = 1000 V3/V1ppm
31
Atomic molecular and Chemical bonding
& Temporary hardness = [Total – Permanent]
= 1000 x [V2/V1-V3/V1] ppm
= 1000(V2-V3)/V1ppm
1.11.8 Numerical
Eg.1) 0.28 gm of CaCO3 was dissolved in HCl and the solution made up to 1000 ml with
distilled water. 100 ml of the solution required 28 ml of EDTA solution for titration. 100
ml of hard water sample required 33 ml of EDTA and after boiling and filtering
required 10 ml of EDTA solution. Calculate each type of hardness of water.
Soln.
Strength of standard hard water
water
= 0.28gm CaCO3 / 1000 ml of distilled
= 28 mg in 1000 ml water
= .28 mg/ml
Now, 100 ml of SHW required º28 ml EDTA solution
So, 28 ml EDTA solution
º
1 ml EDTA solution
hardness
º
Now, 100 ml of unknown hard water sample required
28 mg CaCO3 equivalent hardness
(28/28)
º
mg
CaCO3 equivalent
33 ml EDTA solution
i.e.100 ml of hard water contains
=
(33 x 28/28) mg CaCO3 equivalent
So,total hardness per litre of sample
º
(33 x 1000)/100=330 mg/L
Total hardness
º
330 ppm
Now, 100 ml of boiled water sample required
º
10ml EDTA solution
i.e.100 ml of boiled water contains
=
(10 x 28/28) mg CaCO3 equivalent
So, permanent hardness per litre of sample
º
(10 x 1000/100=100 mg/L
Permanent hardness
º
100 ppm
Temporary hardness º
º
330-100=230 mg/L
º
230 ppm
Total hardness- Permanent hardness
2) 0.5 gm of CaCO3 was dissolved in HCl and the solution made up to 500 ml with distilled
water. 50 ml of the solution required 48 ml of EDTA solution for titration. 50 ml of hard
water sample required 15 ml of EDTA and after boiling and filtering required 10 ml of
EDTA solution. Calculate temporary hardness of water.
Soln.
Strength of standard hard water
= 0.5 gm CaCO3 / 500 ml of distilled water
= 500 mg in 500 ml water
= 1 mg/ml
Chemistry Sem I/II
32
Now, 50 ml of SHW required
º
48 ml EDTA solution
So, 48 ml EDTA solution
º
1 ml EDTA solution
50 mg CaCO3 equivalent hardness
º (50/48) mg CaCO3 equivalent hardness
Now, 50 ml of water sample required
Hardness of sample
ml
º
º
15 ml EDTA solution
(15 x 50/48) mg CaCO3 equivalent hardness for 50
So, hardness per litre of sample
º
(10 x 50/48) x 1000/50 mg/L
Permanent hardness
º
208.33 ppm
Temporary hardness º
º
312.5 – 208.33
º
104.17 ppm
Total hardness- Permanent hardness
3.) 50ml of standard hard water containing 1 mg pure CaCO3 per ml, consumed 20ml EDTA
solution.50ml water sample consumed 30ml EDTA solution using EBT indicator.
Calculate hardness of water sample.
Soln.
Strength of standard hard water
= 1 mg CaCO3 / ml
Thus 50 ml of standard hard water
= 1 x 50 mg CaCO3 equivalent
50 ml of standard hard water
= 20 ml EDTA
= 50 mg CaCO3equi.
1 ml EDTA = 50/ 20 mg CaCO3
Therefore, 50 ml water sample
1000 ml water sample
= 30 ml EDTA Soln.
= 1000 x 30
50
Because, 1 ml EDTA solution
600 ml EDTA solution
= 50/20 mg CaCO3
= 600 x 50 /20 mg CaCo3 equivalent
= 1500 mg CaCO3 equivalent
Hardness of water sample
= 1500 mg of CaCO3.
4) 15 gm of CaCO3 was dissolved in HCl and the solution made up to 1000 ml with distilled
water. 20 ml of the solution required 25 ml of EDTA solution for titration. 100 ml of hard
water sample required 18 ml of EDTA and the same sample after boiling and filtering
required 12 ml of EDTA solution. Calculate each type of hardness of water.
Soln.
Strength of standard hard water
water
= 15 gm CaCO3 / 1000 ml of distilled
= 15000 mg in 1000 ml water
= 15 mg/ml
33
Atomic molecular and Chemical bonding
Now, 20 ml of SHW required
º25 ml EDTA solution
So, 25 ml EDTA solution
hardness
º
20x15=300 mg CaCO3 equivalent
1 ml EDTA solution
hardness
º
(300/25) mg CaCO3 equivalent
=12 mg CaCO3 equivalent hardness
Now, 100 ml of unknown hard water sample required
º
18 ml EDTA solution
i.e.100 ml of hard water contains
=
(18x12) mg CaCO3 equivalent
So, total hardness per litre of sample
º
2160 mg CaCO3 equivalent
Total hardness
º
2160 ppm
Now, 100 ml of boiled water sample required
º
12ml EDTA solution
i.e.100 ml of boiled water contains
=
(2x12) mg CaCO3 equivalent
So, permanent hardness per litre of sample
=144 mg CaCO3 equivalent
Permanent hardness
º
Temporary hardness º
º
2160-1440=720 mg/L
º
720 ppm
1440 ppm
Total hardness- Permanent hardness
5)1.2 gm of CaCO3 was dissolved in HCl and the solution made up to 1000 ml with distilled
water. 50 ml of the solution required 32 ml of EDTA solution for titration. 50 ml of hard
water sample required 14 ml of EDTA and after boiling and filtering required 8.5 ml of
EDTA solution. Calculate temporary, permanent and total hardness of water.
Soln.
Strength of standard hard water
water
= 1.2 gm CaCO3 / 1000 ml of distilled
= 120 mg in 1000 ml water
= 0.12 mg/ml
Now, 50 ml of SHW required º32 ml EDTA solution
So, 32 ml EDTA solution
hardness
º
50x0.12=6 mg CaCO3 equivalent
1 ml EDTA solution
º (6/32) mg CaCO3 equivalent hardness
= 0.15 mg CaCO3 equivalent hardness
Now, 100 ml of unknown hard water sample required
º 14 ml EDTA solution
i.e.100 ml of hard water contains
= (14x0.15) mg CaCO3 equivalent
So, total hardness per litre of sample
º 21 mg CaCO3 equivalent
Total hardness
º 21 ppm
Now, 100 ml of boiled water sample required
º 8.5ml EDTA solution
Chemistry Sem I/II
34
i.e.100 ml of boiled water contains
= (8.5x0.15) mg CaCO3 equivalent
So, permanent hardness per litre of sample
= 12.75mg CaCO3 equivalent
Permanent hardness
º
Temporary hardness º
º
21-12.75=8.25 mg/L
º
8.25 ppm
12.75 ppm
Total hardness- Permanent hardness
Lets’ check the take away from the lecture (MCQ
1) For Complexometric titration pH range required to be maintained is--------a) 6-8
b) 10-12
c) 7
d) at any Ph
2) What is used as an indicator in complexometric titration
a) EDTA
B) Murexide
c) Starch
d) EBT
Exercise:
1. Explain theory of EDTA method
2. Name the indicator used in EDTA titration. What is the colour change at the end point of
titration?
Questions for practice
3. How is hardness of water determined by EDTA method? Write the necessary calculation
1.
1.0 gm of CaCO3 was dissolved in HCl and the solution made up to 1000 ml with distilled
water. 50 ml of the solution required 45 ml of EDTA solution for titration. 50 ml of hard
water sample required 28 ml of EDTA and after boiling and filtering required 15 ml of
EDTA solution. Calculate the hardness of water.
2)
0.45 gm of CaCO3 was dissolved in HCl and the solution made up to 500 ml with distilled
water. 50 ml of the solution required 50 ml of EDTA solution for titration. 50 ml of hard
water sample required 18 ml of EDTA and after boiling and filtering required 10 ml of
EDTA solution. Calculate each type of hardness of water.
3)
1.5gm of CaCO3 was dissolved in HCl and the solution made up to 1000 ml with distilled
water. 50 ml of the solution required 32 ml of EDTA solution for titration. 50 ml of hard
water sample required 16 ml of EDTA and after boiling and filtering required 9 ml of
EDTA solution. Calculate temporary, permanent and total hardness of water.
4)
0.30 gm of CaCO3 was dissolved in HCl and the solution made up to 1000 ml with
distilled water. 100 ml of the solution required 30 ml of EDTA solution for titration. 100
ml of hard water sample required 35 ml of EDTA and after boiling and filtering required
12 ml of EDTA solution. Calculate temporary hardness of water.
5)
0.25 gm of CaCO3 was dissolved in HCl and the solution made up to 250 ml with
35
Atomic molecular and Chemical bonding
distilled water. 50 ml of the solution required 20 ml of EDTA solution for titration. 50
ml of hard water sample required 18 ml of EDTA and after boiling and filtering required
10 ml of EDTA solution. Calculate temporary
Learning from this lecture: Learners will able to understand the method/Process of EDTA
titration and its mathematical expression for calculating all types of hardness.
Lecture: 8
Crystal Field Theory
1.12
Learning Objective:
Learners will be able to understand the features of Crystal Field Theory and it’s
application for calculating CFSE of Octahedral complexes.
1.12.1 Postulates of Crystal Field Theory
i) This theory considered the metal-ligand bond is purely ionic.
ii) Anionic ligands are treated as negative point charges and neutral ligands as point
dipoles.
iii) Interaction between metal atom and ligand is purely electrostatic. There is no
intermixing of atomic orbitals.
iv) The arrangement of ligand around the central metal ion in such a way that repulsion
should be minimum.
v) The d orbitals in the metal atom in the free state are degenerate i.e. they have the same
energy.
vi) In complex formation d orbitals split into different energy level under the influence of
approaching ions towards the transition metal ion. E.g. orbitals (dz2 & dx2-y2) have
their lobes along the axes. (Doubly degenerate set) t2g (dxy, dyz, dxy) have their lobes
lying between the axes. (triply degenerate set)
Fig: 1.21 Energy Level Diagram
Chemistry Sem I/II
36
1.12.2 Crystal field splitting of d-orbitals in different geometries
Crystal field splitting of d-orbitals in Octahedral complexes: 1. In this type 6 Ligands participate in forming a complex
2. There is attraction between +ve metal ion & -vely charged ligands which holds the
ligand to metal in complex
3. Electrostatic repulsion between l.p. ligands & ē in d orbitals of metal.
4. As a result of M-L interactions the energy of fine d orbitals is nullified & give high
lying equal energy levels dx2-y2 & dz2 & three low lying equal energy levels (dxy, dyz
& dxz)
The energy differentiate between these 2 sets of orbitals is called crystal field splitting (∆ or 10
Da)
Distribution of electrons in Octahedral Complex
i)
Distribution of d ē of central metal cation in t2g & e.g. orbitals in Octahedral complexes
ii) For d1, d2 & d3 -------- ē → t2g
iii) For d8, d9 & d10 ------- first 6 ē go to t2g while the remaining two (in d8 ion), three (in d9
ion) & four (in d10) occupy e.g. level
iv) For d4, d5, d6 & d7 distribution of d ē depends upon whether ē are in weaker or stronger
ligand field.
Weak field ligands
I-<Br-<S-2<Cl-<No-3<F-<OH-<EtOH<oxalate<H2O<EDTA<NH3<en<NO2-<CN-<CO
(strong field ligand)
Distribution of d ē of central metal cation in t2g & eg orbitals in Octahedral complexes
i)
ii)
For d1, d2 & d3 -------- ē will enter in t2g orbital singly according to Hunds rule.
For d4 ions, the fourth electron could either enter the tDig level with pairing or avoid
pairing by occupying eg level.
The two options are:
a) If ∆ₒ (CFSE) < (Pairing energy), the fourth electron enters one of the eg orbitals giving
configuration t₂g 3 eg 1 .
Here ligands are weak field ligands and form high spin complex.
If ∆ₒ (CFSE) > (Pairing energy) , the fourth electron enters t₂g orbital with configuration
t₂g 4eg 0. Here ligands are strong field ligands and form low spin complex.
b) For d4, d5, d6 & d7 distribution of d ē depends upon whether ē are in weaker or
stronger ligand field.
d4 to d7 coordination entities are more stable for strong field as compared to weak
field cases.
c) For d8, d9 & d10 the first 6 ē go to t2g while the remaining two (in d8), three (in d9 ) &
four (in d10) occupy eg level.
37
Atomic molecular and Chemical bonding
1.12.3 CFSE and Electron Arrangement in Octahedral Complexes
Number of d
electrons
d4
Arrangement in
Weak ligand field
3
t g eg
1
CFSE
∆ₒ
-0.6
2
d5
3
t g eg
4
t g eg
2
0
5
t g eg
2
-0.4
6
t g eg
2
-0.8
6
t g eg
2
-1.2
6
t g eg
2
0
-2.0
6
0
-2.4
6
1
-1.8
6
2
-1.2
6
3
-0.6
6
4
0
t g eg
t g eg
t g eg
2
3
-0.6
2
d10
5
t g eg
2
2
d9
0
2
2
d8
4
t g eg
2
2
d7
CFSE
∆ₒ
-1.6
2
2
d6
Arrangement in
strong ligand field
t g eg
2
4
0
t g eg
2
1.12.4 CRYSTAL FIELD STABILISATION ENERGY (CFSE)
The electron occupying t2g orbital stabilize the orbital by 0.4∆ₒand the electron present
in eg orbital destabilize the complex by 0.6∆ₒ.
i)
The average of the energies of electron present in these orbitals is referred to as
Crystal Field Stabilisation Energy.
∆ₒ(CFSE)=[(-4* ne-(t₂g)+(+6* ne-(eg )]Dq
Where ne- (t₂g) and ne- (eg ) are number of electrons in t₂g and eg set of orbitals
respectively
and Dq is unit of CFSE.
ii)
The energy of two eg orbitals will increase by (3/5) ∆ₒand that of three t2g
orbitals will be decrease by (2/5) ∆ₒ
iii)
CFSE depends upon geometry of complex, size of metal atom, oxidation state
of metal ion and field produced by ligand.
Lets’ check the take away from the lecture (MCQ s)
1. The CFSE for a high spin d4 octahedral complex is
a) -0.6 ∆o
b) -1.8 ∆o
c) -1.6∆o+P
d) -1.2∆o +P
2. The value of the spin only magnetic moment for one of the following
configurations is 2.84 BM. The correct one is
a) d4 in strong field ligand
b) d4 in weak field ligand
c) d3 in weak as
well as strong field ligand
d) d5 in strong field ligand
Chemistry Sem I/II
38
Exercise:
1.Discuss the distribution of d4,d7 electrons in octahedral complex in weak and strong
field ligand.
2. What are the essential requirements for formation of low spin and high spin complex.
Questions for practice:
1. Discuss the features of Crystal Field Theory.
2. Define Crystal Field Stabilisation Energy.
3. What do you understand by 10Dq in an octahedral complex?
Learning from this lecture: Learners will able to know the features of Crystal Field Theory
theory and its application to calculate the CFSE for Octahedral complexes.
Lecture: 9
1.13
Learning Objective:
Learners will be able to understand the applications of crystal field theory in
calculating CFSE for tetrahedral complexes.
1.13.1 Splitting of d orbitals in tetrahedral crystal field
Fig: 1.22 Splitting of d orbitals in tetrahedral complexes
CFSE and electronic arrangement in Tetrahedral Complexes
No. of d electron
Arrangement of electrons
Tetrahedral
CFSE
d1
eg1 t2g0
-0.6
d2
eg2t2g0
-1.2
39
Atomic molecular and Chemical bonding
d3
eg2t2g1
-1.2+0.4=-0.8
d4
eg2t2g2
-1.2+0.8=-0.4
d5
eg2t2g3
-1.2+1.2=0
d6
eg3t2g3
-1.8+1.2=-0.6
d7
eg4t2g3
-2.4+1.2=-1.2
d8
eg4t2g4
-2.4+1.6=-0.8
d9
eg4t2g5
-2.4+2.0=-0.4
xd10
eg4t2g6
-2.4+2.4=0
Applications of Crystal Field Theory
•
This theory helps in calculating the Crystal Field Stabilization Energy of octahedral
and tetrahedral complexes.
•
The model can be used to understand, interpret and predict the magnetic behavior and
some structure of coordination complexes.
•
This theory helps in calculating hydration and lattice energies of molecules.
•
CFT predict the electronic spectra of molecules.
Lets’ check the take away from the lecture (MCQ s)
1) Crystal field theory is applicable in
a)
b)
c)
d)
To understand, interpret and predict the magnetic behavior
To predict the electronic spectra of molecules
In calculating the Crystal Field Stabilization Energy
All of the above
Exercise:
1. Calculate the CFSE of d5 configuration of element in tetrahedral complexes.
Questions for practice:
1. Explain the stability of tetrahedral complexes on the basis of Crystal field theory.
Learning from this lecture: Learners will be able to calculate the crystal field stabilisation
energy of tetrahedral complexes.
Chemistry Sem I/II
40
1.14 Conclusion:
Solution of Schrodinger equation is a wave function which gives us information about the
particle's behaviour in time and space. The Valence bond theory and Molecular Orbital theory
have their own unique approach to determine bonding and thus the overall structure of a
molecule. MO theory involves the production of molecular orbitals from the combination of
atomic orbitals in a molecule. VB theory utilizes the overlap of molecular orbitals. The
overlaps are where the bonds are formed and these bonds are localized. The theory of orbital
hybridization can be used for more complex molecules. Knowledge of CFT helps in
calculating CFSE which leads to stability of molecule. EDTA theory is applied in predicting
the hardness of water sample which helps in designing the water treatment plant.
1.15 Add to knowledge: (Content beyond syllabus)
Crystal field theory has been used to describe various spectroscopies of transition
metal coordination complexes, in particular optical spectra (colours). CFT successfully
accounts for some magnetic properties colours, hydration enthalpies, and spinel structures of
transition metal complexes. Several epidemiological investigations have demonstrated the
relation between risk for cardiovascular disease, growth retardation, reproductive failure, and
other health problems and hardness of drinking water or its content of magnesium and
calcium.
Set of Questions for FA/CE/IA/ESE
1. What do you understand by Hybridization?
2.What is the difference between atomic orbital and molecular orbital?
3. 0.25 gm of CaCO3 was dissolved in HCl and the solution made up to 250 ml with distilled
water. 50 ml of the solution required 20 ml of EDTA solution for titration. 50 ml of hard water
sample required 18 ml of EDTA hardness and 50 ml of same sample after boiling and filtering
required 10 ml of EDTA solution. Calculate temporary of water.
4. Draw the molecular orbital diagram and write molecular electronic configuration. Also
Calculate Bond Order and comment on its stability and magnetic characteristics.
i) O2
ii) N2
iii) He2
iv) F2
5. Explain formation of following on the basis of Hybridization.
a) BF3
References:
i)
ii)
iii)
b) IF7
c) CH4
d) PF5
Inorganic Chemistry by Shriver and Atkins
Inorganic Chemistry by James E. Huheey
A textbook of Engineering Chemistry – Jain and Jain
Self-assessment1. What do you understand by atomic orbital and molecular orbital? (level 1)
2. Discuss the characteristics of hybridization. (level 2)
41
Atomic molecular and Chemical bonding
3) Draw the molecular orbital diagram of Nitrogen and with the help of MOT predict its
bond order and magnetic behavior. (level 3)
4) Explain the principle of EDTA. How it will be helpful for calculating the hardness of
water sample? (level 4)
Chemistry Sem I/II
42
Self-evaluation
Name of
Student
Class
Roll No.
Subject
Module No.
S.No
Tick
Your choice
1.
2.
3.
4.
5.
Do you understand the concept of atomic
orbital?
o
Yes
o
o
No
Yes
o
No
Will you able to describe the Molecular
Orbital theory?
o
Yes
o
No
Are you able to understand the Crystal Field
Theory for octahedral and tetrahedral
complexes?
o
Yes
o
No
Do you understand module ?
o
Yes, Completely.
o
Partialy.
o
No, Not at all.
Do you understand different types of
hybridisaton?
43
Instrumental methods of Analysis
Module:02
Instrumental methods of analysis
Lecture : 10
Instrumental methods of analysis
2.1
2.1.1 Motivation: The purpose of this module is to provide a better understanding on
components separation from complex mixtures and measurement techniques and its
application. The important step in any chemical process or bioprocessing industry is a need
of purifying the product from complex mixture. Chromatography is a technique which
separate a mixture with great precision. These techniques have wide range of applications in
organic chemistry which include separation, preparation, purification and identification of
organic compounds. By learning this technique, we can easily predict the pollutants such as
pesticides and carcinogens in air, water and food supplies.
2.1.2 Historical Background:- The Russian botanist Mikhail Tswett coined the term
chromatography in 1906.The first analytical use of chromatography was described by James
and Martin in 1952, for the use of gas chromatography for the analysis of fatty acids mixtures
2.1.3 Syllabus:
Lecture
No.
Duration
(hour)
Contents
Types
(hours)
10
Introduction
to
chromatography
of
1
2
11
Thin Layer Chromatography (Theory, Principle,
technique and applications),
1
2
12
Gas Chromatography –
instrumentation., working)
1
2
13
High Performance Liquid Chromatography,
introduction, theory, instrumentation
–
1
2
Interpretation of Gas/HPLC Chromatogram and TLC
plate of various samples.
1
2
14
chromatography,
Self-Study
(Introduction,
theory,
2.1.4 Weightage: Marks 10 marks
2.1.5 Learning Objective:
•
Learners shall be able to understand the principle of chromatography
44
Chemistry Sem I/II
•
Learners shall be able to understand the types of chromatography and its
application.
•
Learners shall be able to draw the instrumentation of various types of
chromatography.
•
Learners shall be able to explain the theory of TLC, GC and HPLC.
•
Learners shall be able to apply the knowledge of chromatography in
interpretation of Gas/HPLC Chromatogram
2.1.6 Prerequisite: To understand the module chromatography learners should know the
concept of dynamic equilibrium between the concentrations of a solute in a system of two
coexisting phases.
2. 1.7 Theoretical Background: Chromatography techniques based on the concept of phases
and helps in separation, identification and purification of components from complex
mixture. It emphasizes the use of two mutually immiscible phases, stationary phase and
mobile phase.
2.1.8. Key Definitions:
a) Chromatography: It is an important biophysical technique that enables the separation,
identification and purification of the components of a mixture for qualitative and
quantitative analysis.
b)
Chromatogram: It is a visible record (such as a graph) showing the result of separating
the components of a mixture by chromatography.
c) Stationary phase: It is a phase that is fixed in place, either packed in a column or on a
fixed planar surface. This phase is always composed of a “solid” phase or “a layer of
a liquid adsorbed on the surface solid support”.
d) Mobile phase: The mobile phase carries the mixture of the compounds to be separated
along with it and moved through stationary phase. This phase is always composed of
“liquid” or a “gaseous component.”
e) Distribution ratio: It is defined as the ratio of concentration of a solute in two phases
(stationary and mobile) at equilibrium.
f) Retention time: The retention time is the time between injection of a sample and the
appearance of a solute peak at the detector.
2.1.9 Course Content:
Introduction and types of chromatography
2.1.10 Learning Objective:
Learners will be able to understand the principle of chromatography and its types which
works on different principle and helps in understanding its application in various fields
45
Instrumental methods of Analysis
2.1.11 Introduction:
Chromatography is based on the principle where molecules in mixture applied onto the
surface or into the solid, and fluid stationary phase (stable phase) is separating from each
other while moving with the aid of a mobile phase.
.A wide range of chromatographic procedures makes use of differences in size, binding
affinities, charge, and other properties to separate materials. It is a powerful separation tool
that is used in all branches of science and is often the only means of separating components
from complex mixtures.
The factors effective on this separation process include molecular characteristics related to
adsorption (liquid-solid), partition (liquid-solid), and affinity or differences among their
molecular weights. Because of these differences, some components of the mixture stay longer
in the stationary phase, and they move slowly in the chromatography system, while others
pass rapidly into the mobile phase, and leave the system faster.
These components thus form the basis of the chromatography technique.
2.1.12 Types of Chromatography
Chromatographic techniques can be classified on the basis of nature of mobile phase or the
stationary phase or the mechanism of separation.
i)
Based on the nature of mobile phase, chromatography is of two types:
a) Liquid chromatography
ii)
b) Gas chromatography
Based on the nature of stationary phase, these chromatography techniques are
further divided into various types:
a) Liquid chromatography-four types i) Liquid-Liquid Chromatography (LLC) ii)
Liquid-Solid Chromatography (LSC) iii) Ion Exchange Chromatography (IEC) iv) Size
Exclusion Chromatography (SEC)
b) Gas chromatography-i) Gas Liquid Chromatography (GLC) ii) Gas Solid
Chromatography (GSC)
iii)
Based on the mechanism of separation or the interacting phenomenon between the
solute molecules and the stationary phase, the various chromatographic
techniques have also been classified into the following types:
a) Adsorption chromatography b) Partition chromatography
c)Ion exchange chromatography d) Gas permeation e) Electrophoresis
iv)
Based on the nature of support used for holding the stationary phase, the
chromatography is broadly classified as:
a) Paper
chromatography
chromatography
b)
Thin
layer
chromatography
c)
Column
46
Chemistry Sem I/II
Type
Adsorption
chromatography
Partition
chromatography
Fig. 2.1 Different types of chromatography
Stationary phase
Mobile phase
Mechanism
Solid adsorbent
Liquid or gas
The
selective
adsorption of the
components of the
mixture occurs on
the surface of the
solid.
Thin film of liquid Liquid or gas
The separation of
formed on the
different
surface of solid
components
is
inert support
based upon their
selective partition
between
the
moving phase and
a thin film held on
an inert solid .
a) Let’s check the take away from this lecture
b)
1.
In chromatography which of the following can the mobile phase be made of
a) Solid or liquid
b) gas only
2.
c) liquid or gas
d) liquid only
Based on the nature of support used for holding the stationary phase, the
chromatography is broadly classified in
47
Instrumental methods of Analysis
a) 2
b) 3
c) 4
d) 5
Exercise:
Q.1 Distinguish between adsorption and partition chromatography\
Q.2How are chromatographic techniques classified? Explain on the basis of different criteria.
Questions for practice
Q.3 What is chromatography? Describe the principle of chromatographic separation.
Q.4 Define and explain the following terms: a) Stationary phase b) Mobile phase.
Learning from this lecture : Learners will able to know the term chromatography and its types
which serves the purpose of understanding the concept.
Lecture:11
2.2 Theory, principle, technique and applications of Thin Layer Chromatography
2.2.2 Learning Objective:
Learners will be able to understand the theory, principle, technique and application of Thin
layer chromatography.
2.2.3 Theory of TLC
The essential requirements for TLC are a suitable adsorbent, glass plates, a suitable device to
apply a thin adsorbent layer, a means of holding the plates and a tank to run the plates.
In TLC, a solid phase, the adsorbent, is coated onto a solid support (thin sheet of glass, plastic
and aluminium) as a thin layer (about 0.25mm thick). The mixture to be separated is dissolved
in a solvent and the resulting solution is spotted on to the thin layer plate near the bottom. A
solvent mixture of solvents, called eluatant is allowed to flow up the plate by capillary action.
The mobile phase rises up the TLC plate by capillary action and the components dissolve in
the solvent and move up to the TLC plate. A competition is set up between the silica gel plate
and the development solvent for the spotted material. The very polar silica gel tries to hold
the spot in its original place and the solvent tries to move the spot along with it as it travels
up the plate. The outcome depends upon a balance among three polarities that of the plate,
the development solvent and the spot material.
2.2.4 Principle:
Thin Layer Chromatography (TLC) is a solid-liquid technique in which the two phases are a
solid (stationary phase) and a liquid (moving phase).
48
Chemistry Sem I/II
The principle of TLC is based on principle of separation. The separation depends on the
relative affinity of compounds towards stationary and mobile phase. The compounds under
the influence of the mobile phase (driven by capillary action) travel over the surface of the
stationary phase. During this movement, the compounds with higher affinity to stationary
phase travel slowly while the other travels faster. Thus separation of components in the
mixture is achieved. Once the separation occurs, the individual components are visualised as
spots at respective level of travel on the plate. Their nature or character are identified by
means of suitable detection techniques.
2.2.5 Technique:
Fig. 2.2 Thin Layer Chromatography
In this technique following steps are involved:
i) Preparation of thin layer plate
ii) Application of material on the plate
iii) Development of the plate
iv)
Visualisation
v)
Calculation of Rf values
i) Preparation of thin layer plate
The plates are usually prepared using glass plate or plastic plates of the size 20 x 20 cm or 5 x
20 cm. The thin layer thus spread is then dried in an oven at 373K to activate the adsorbent
plate. The layer of adsorbent is about 0.15 mm to 0.2 mm thick. Silica gel is the most widely
used adsorbent. The prepared thin layer on glass is called a chromaplate. Chromaplates are
prepared by applying a uniform layer of silica or alumina in the form of thin aqueous paste.
The solid adsorbent is mixed with a small amount of binder such as calcium sulphate which
helps in adherence of adsorbent on to the plate. The adsorbent (stationary phase) is finely
divided and made into a slurry in water and spread over a glass plate with a spreader.
ii) Application of material on the plate
The material is dissolved in an appropriate solvent and deposited on the thin layer with the
help of capillary. The capillary is first dipped in the sample solution, the solution goes up the
capillary. The sample is applied on the plate by touching the capillary to the adsorbent surface
49
Instrumental methods of Analysis
at about 2 cm from the edge. Position of spot is marked with a pencil so that the Rf values may
be calculated after separation.
iii) Development of the plate
When the spot is dry, the chromaplate is placed in a jar containing small amount of solvent.
The jar must be saturated with solvent vapour before the plate is placed in it. The dry plate is
next placed vertically in a tank in such a way that the lower edge of the plate dips to a depth
of about 1 cm in the solvent, used as mobile phase, placed in the tank. The mobile phase rises
on the plate due to capillary action and dissolves the solutes in the spot. There is a competition
for the solute between the adsorbent which tries to retain the solutes, and the solvent which
tries to dissolve the solutes and take them up during the rise of the solvent on the plate. This
leads to the separation of the components of the mixture sample.
iv) Visualisation
After the run, the plate is taken out and dried without heating. Usually, a current of dry air is
passed over the plate surface. The separated components are located and identified by their
Rf values. If the components of the mixture are coloured they can be detected by simple visual
examination. If they are colourless, they are located by
a) Keeping the plate in a closed dry tank containing crystals of iodine. The spot shows up
brown spots.
b) Irradiation with UV light to locate the substances which give fluorescence on exposure
to UV.
v) Calculation of Rf values
Rf value (Retention factor) is defined as the ratio of the distance travelled by the component
at its point of maximum concentration to the distance travelled by the solvent.
Retention factor=Distance moved by component
Distance moved by solvent front
The Rf is characteristic of a substance for a given solvent.
2.2.6 Applications
(i)
The method is used in the laboratory for a quick check on whether a reaction is
complete or not.
(ii) TLC often serves in the identification of plant extracts, drugs and adulterants in food
products.
(iii) It is a more versatile technique compared to paper chromatography.
(iv) TLC is also faster and has a better reproducibility than paper chromatography.
Separation of metals with similar chemical properties can be easily carried out by TLC e.g.,
Nickel, Cobalt, Manganese and Zine can be separated
50
Chemistry Sem I/II
a) Let’s check the take away from this lecture
1)Thin layer chromatography is used in
a) Biochemical analysis
b) Identification of compounds
c) In pharmaceutical industry
d) All of the above
2) The size of a thin layer of adsorbent is about --------a) 0.1 mm
b) 0.2 mm
c) 0.3 mm
d) 0.4 mm
Exercise:
1. How Thin layer plate can be formed?
2. State the applications of Thin layer chromatography.
Questions for practice:
1) State the principle of Thin Layer Chromatography.
2) Discuss the process of Thin Layer Chromatography (TLC).
Learning from this lecture: Learners will able to know the principle, theory and application of
thin layer chromatography
Lecture:12
2.3 Gas Chromatography – (Introduction, theory, instrumentation., working)
2.3.1 Learning Objective:
Learners will be able to understand the theory, instrumentation and working of Gas
chromatography.
2.3.2 Introduction
Gas chromatography was discovered by John Porter Martin as a separation technique to
separate compounds. GC is a term used to describe the group of analytical separation
technique used to analyse volatile substances in the gas phase. Gas liquid chromatography is
most commonly used method to separate organic compounds. The combination of gas
chromatography and mass spectrometry is important tool in the identification of molecules.
2.3.3 Theory of GC
51
Instrumental methods of Analysis
In gas chromatography, the components of a sample are dissolved in a solvent and vaporized
in order to separate the analytes by distributing the sample between two phases: a stationary
phase and a mobile phase. The mobile phase is a chemically inert gas that serves to carry the
molecules of the analyte through the heated column. Gas chromatography is one of the sole
forms of chromatography that does not utilize the mobile phase for interacting with the
analyte. The stationary phase is either a solid adsorbant, termed gas-solid chromatography
(GSC), or a liquid on an inert support, termed gas-liquid chromatography (GLC).
2.3.4 Instrumentation
Fig. 2.3 Gas Ghromatography
Typical gas chromatograph consists of an injection port, a column, carrier gas flow control
equipment, ovens and heaters for maintaining temperatures of the injection port and the
column, an integrator chart recorder and a detector.
Gas – Liquid Chromatography (GLC)
In GLC, the liquid stationary phase is adsorbed on to a solid inert packing or immobilized on
the capillary tubing walls. The column is considered packed with inert small spherical inert
supports. The liquid phase adsorbs on to the surface of these beads in a thin layer.
A separation brought about by an exchange between a mobile gas phase and a liquid
stationary phase held in a column is called gas – liquid chromatography. The column is
generally made of glass or stainless steel; depending upon the chemical reactivity of the
substances to be separated. Columns may be 1 to 20 meters long and are usually coiled. The
internal diameter of the column is generally about 4 mm, the column is packed with an inert
support material such as celite. Alumina and micro glass beads have also been used. The
liquid that is immobilized or held on the column is usually a hydrocarbon of high molecular
weight, e.g., squalene. Polar liquids such as polyethylene glycol or mixtures of ethers, esters
and hydrocarbons have also been used. The liquid thus immobilized on the column acts as
the stationary phase.
The mobile is a gas which is selected for its chemical inertness. Helium, nitrogen, argon and
hydrogen are the most commonly used gases. Hydrogen, unless unavoidable, is not used as
52
Chemistry Sem I/II
it is explosive. The gas used as the mobile phase is called a carrier gas. The column is operated
at a known constant temperature which may vary from room temperature to 623 K and this
depends upon the volatility of the liquid phase.
2.3.5 Working
To separate the compounds in gas-liquid chromatography, a solution sample that contains
organic compounds of interest is injected into the sample port where it will be vaporized. The
vaporized samples that are injected are then carried by an inert gas, which is often used by
helium or nitrogen. This inert gas goes through a glass column packed with silica that is coated
with a liquid. Materials that are less soluble in the liquid will increase the result faster than
the material with greater solubility.
The carrier gas emerging from the column passes into a detector which gives an electric signal.
The signal response is proportional to the concentration of the substance present in the carrier
gas. Consider a mixture containing three components A, B and C is to be separated by GLC.
A very small quantity, 0.2 µ1 (micro dm3) is injected into the front end of the column through
a rubber septum by means of a syringe. The heating unit or the oven vaporizes the liquid and
the vapor mixes with the carrier gas and the mixture sweeps through the column at a definite
rate. Each component of the mixture in its passage through the column is retarded by the
stationary liquid phase. The degree of retardation of a component depends upon the partition
coefficient of that component in the liquid phase, i.e., the tendency of the component to
dissolve in the liquid.
2.3.6 Detectors:
Estimation of the separated components in chromatography is a necessary and important
step. In general, there are two types of detectors. Integral detectors and Differential detectors.
Integral detectors provide a measure of the total quantity of solute which has emerged up to
a given instant. Differential detectors give a measure of the instantaneous concentration of the
component. The following detectors are in common use.
(i) Flame Ionization Detector (FID)
(ii) Thermal Conductivity Detector (TCD)
(iii) Electron Capture Detector (ECD)
(iv) Flame Photometer Detector (FPD)
2.3.7 Applications of Gas Chromatography
Gas chromatography, both GLC and GSC, is a very useful tool for the analysis of small
samples of mixtures of gases or volatile liquids. Therefore, gas-liquid chromatography is used
more often than gas-solid chromatography. The applications given below will show the
versatile nature of gas chromatography.
1. Separation of benzene (b.p. 353.1K) and cyclohexane (b.p.353.8 K). This separation is
virtually impossible by fractional distillation. By GLC, the separation of the two can be
accomplished in a few minutes.
2. By using molecular sieves, gas-solid chromatography has been used to separate a
mixture of H2, CO2, CO, O2, CH4, C2H2, C2H4, and C2H6.
3. Automobile exhaust gases, a main pollution hazard, have been analyzed by GLC.
53
Instrumental methods of Analysis
4. Volatile substance such as human breath, environmental air and urine have been
analyzed by GLC. By using 300m capillary columns with methyl silicone as the
stationary phase 150 substances have been obtained in the human breath. Gas
chromatography is thus increasingly being used in biomedical analysis.
5. Flavors and aromas of flowers and foods are the result of a combination of hundreds of
organic compounds in trace amounts. These have been separated by GLC.
6. The high degree of resolution of GLC allows purity of samples to be checked.
7. GLC has also been used in the separation of radioactive products.
8. Gas-liquid chromatography has been used to separate samples from less than a
microgram to 100 g. Traces of components of the order of 10-15 g have been detected.
9. Gas chromatography has also been used to study reaction mechanisms. For instance, the
dehydration of the two isomers of butanol has been studied.
10. Let’s check the take away from this lecture
1) Basic principle involved in gas solid chromatography is
a) Exclusion
c) ion exchange
b) Adsorption
d) absorption
2) Material used in construction of capillary tube in gas chromatography
a) Glass
c) stainless steel
b) Metal
d) fused silica
Exercise:
1. Discuss the applications of gas chromatography.
2. Classify the types of gas chromatography.
Questions for practice:
1)Discuss the principle of Gas chromatography.
2) How do we inject sample in column of Gas chromatography?
3) Discuss in brief, the principle of separation of the components of sample in Gas
Chromatography.
Learning from this lecture: Learners will able to know the principle, types and working of
Gas Chromatography.
54
Chemistry Sem I/II
11. Lecture :13
2.4 High Performance Liquid Chromatography, – Introduction, theory,
instrumentation
2.4.1 Leaning Objective:
Learners will be able to understand the theory, instrumentation and working of High
Performance Liquid Chromatography.
2.4.2 Theory:
High performance (pressure) liquid chromatography is an improvement of the classical
method of liquid or partition chromatography. In liquid chromatography, a solute (or solutes)
is partitioned between two immiscible liquids. In other words, both the mobile phase and the
stationary phase are liquids. One of the liquids is immobilized or fixed on the surface of a
solid and this fixed liquid acts as the stationary phase.
High Performance Liquid Chromatography is the chromatography in which liquid is the
mobile phase and very finely divided particles comprises the stationary phase. To increase
the efficiency or to obtain increased flow rates, the liquid is pressurized to several hundred
pounds per square inch. HPLC is useful in separation of complex mixtures and more versatile
than gas chromatography.
2.4.3 Instrumentation
The equipment used in high performance liquid chromatography consists of the following.
a) Solvent reservoir
b) Pumping systems
c) Sample injection system
d) Analytical column
e) Detector
f) Data control and display
55
Instrumental methods of Analysis
Fig: 2.4 Flow diagram of HPLC
High Performance Liquid Chromatography (HPLC)
a) Solvent Reservoir:
The solvent (mobile phase) reservoir system consists of one or more glass or stainless-steel
vessels which can contain 1 to 2 dm3 of solvent. Dissolved oxygen and nitrogen in the solvent
are removed by degassers. These gases form bubbles in the analytical column and the
resolution of peaks is affected. Two types of elution techniques are used. If a single solvent
is used, the separation is called isocratic elution. Separation efficiency can be increased by
using two or more solvents. Such an elution is called gradient elution. This type of elution
shortens the time of separation without seriously affecting the resolution of peaks. If
gradient elution is used, the solvents from the reservoirs are mixed in the mixing vessel and
pumped into a precolumn by means of a high-pressure pump.
b) Pumping system:
In HPLC, the columns used are quite narrow and are packed with particles of small sizes.
Hence there is high resistance to flow of solvent. Therefore, high pressures are required to
achieve constant flow rate. The most popular type I the pump which can provide accurately
controlled rates of flow of solvents (up to 20ml/min.) at pressures up to 300-400
atmospheres.
c)
Sample Injection System:
Sample can be introduced into the column either by a syringe injection into the solvent
stream or by a sample loop from which it is swept into the column by the solvent. The
experimental sample is injected into the column by means of a self-sealing Teflon or
neoprene septum. Sometimes, the sample can also be introduced, by stopping momentarily
the flow of solvent, removing the cap on the column and injecting the sample on to the
column.
d)
Column:
The analytical column consists of a stainless tube about 15 to 30 cm in length. These straight
columns can withstand very high pressure of up to 15-50cm length and are corrosion
resistant. The selection of the column packing depends on the chemical nature of the sample
components and the mobile phase to be used. The column is generally surrounded by a
water jacket to maintain a constant temperature. The sample gets separated on the column
due to partition between the mobile phase and stationary phase and the components register
on a detector.
Column packing material:
In general, three types of particles are used for column packing in HPLC.
Porous polymeric materials: These are based on styrene-divinyl benzene copolymers.
Porous layer beads: These are porous supports where a thin, porous, active layer is coated
onto a solid core such as impervious glass beads. The thickness of the porous layer is 1-3µm.
56
Chemistry Sem I/II
Totally porous silica particles: In these packings the silica particles, which are generally 510µm in diameter, form the packing material.
e)
Detectors:
The function of detector is to monitor the composition of mobile phase as it is comes out from
the column. Two types of detector are generally employed:
Bulk property detectors: These monitor the difference in some physical property of solute in
the mobile phase as compared to pure mobile phase. E.g. Refractive index, Conductivity
detectors
Solute property detectors: These detectors respond to the physical and chemical property of
solute such as fluorescence and electrochemical detectors.
f)
Recorders:
The signals from the detectors are recorded as deviation from the base line. The peak position
along the curve, relative to the starting point, denotes the particular component.
2.4.4 Application of HPLC
High performance liquid chromatography is an efficient, higher selective method of
separation. Small sample sizes can be separated. The only condition is that a suitable
immiscible solvent pair must be available. Generally, the more polar of the two is made the
stationary phase. In HPLC, separation is carried out a room temperature. Therefore, thermally
unstable substances, which cannot be separated by GLC, can be separated. The method is also
applicable for the separation of inorganic ions.
c)
d) Let’s check the take away from this lecture
1)
a)
b)
c)
d)
Which of the following is not correct statement about HPLC
It requires high pressure for separation of species
There is no need to vaporise the samples
It is performed in columns
It has high sensitivity
2)
a)
b)
c)
d)
In Normal phase HPLC there is
Non polar solvent / polar column
Polar solvent
/ non polar column
Non polar solvent /non polar column
All of the above
Exercise:
1) Explain the principle of HPLC.
2) HPLC is mainly useful in pharmaceutical industry. Justify this statement.
Questions for Practice:
1)
What are the important basic components of HPLC.
57
Instrumental methods of Analysis
2)
Draw a Schematic diagram of High performance Liquid Chromatography (HPLC)
unit and describe briefly functions of each part.
3)
Write a note on Applications of HPLC.
Learning from this lecture: Learners will able to know the principle, theory and
instrumentation of High performance Liquid chromatography
12.
Lecture :14
13. Interpretation of Gas/HPLC Chromatogram and TLC plate of various samples
Learning objective: Learners will be able to understand the interpretation of GC/HPLC
Chromatogram and TLC plates of various samples.
A chromatogram is a representation of the separation that has (chromatographically)
occurred in HPLC system. A series of peaks rising from a baseline is drawn on a time axis.
Each peak represents the detector for a different compound. The chromatogram.is a graph
that monitors the signal in the detector over time, As chemicals are detected by the
instrument, the signal increases, and the chromatogram displays the peak. Each peak in the
chromatogram indicates the presence of a chemical in sample.
Chromatogram generated by Gas Chromatography
While the instrument run, the computer generated a graph from signal called chromatogram
X axis shows retention time
Y axis show the intensity of the signal
Fig: 2.5 Chromatogram of GC
A chromatogram is a pictorial record of the detector response as a function of elution
volume or retention time.
58
Chemistry Sem I/II
It consists of a series of peaks, symmetrical in shape, representing the elution of individual
analytes.
It is a two dimensional plot with the ordinate axis giving concentration in terms of detector
response (AU) and abscissa represents the time(t).
The base line represents any time period during which only mobile phase is passing through
the detector.
14. Fig: 2.6 Interpretation of TLC plate of various samples
15.
A high Rf -value indicated that the compound has travelled far up the plate and is less
polar, while a lower Rf value indicates that the compound has not travelled far, and is
more polar.
e) Let’s check the take away from this lecture
1) High Rf value indicates that compound is
i)
Less polar ii) more polar iii) neutral iv) none
2) Detection of compound is identify by
i)
Symbol ii) peaks iii) value iv) all of the above
Exercise:
1.What are the basic features required for interpretation pf HPLC chromatogram?
Questions for practice:
1) What is the significance of Rf value?
2) Define chromatogram.
Learning from this lecture: Learners will able to apply the knowledge of Gas/HPLC and
TLC for interpretation of chromatogram
59
Instrumental methods of Analysis
Conclusion:
We conclude that chromatography is an effective way of determining the components
present in a compound. In TLC chromatography, certain properties of the analyte or the
mobile phase can be determined such as its polarity and its solubility towards the eluate or
the solvent. HPLC method play a critical role in analysis of pharmaceutical products. In GLC
compounds are separated according to their partition coefficients. Hence Chromatography is
accepted as an extremely sensitive, and effective separation method.
Add to knowledge: (Content beyond syllabus)
Chromatography is used for quality analyses and checker in the food industry, by identifying
and separating, analysing additives, vitamins, preservatives, proteins, and amino
acids. Chromatography like HPLC is used in DNA fingerprinting and bioinformatics. The
Environmental Protection Agency makes the method of chromatography to test drinking
water and to monitor air quality. Pharmaceutical industries use this method both to prepare
huge quantities of extremely pure materials, and also to analyze the purified compounds for
trace contaminants. The other applications of chromatography especially HPLC is used in
Protein Separation like Insulin purification, Plasma Fractionation and Enzyme Purification.
These separation techniques like chromatography gain importance in different kinds of
companies, different departments like Fuel Industry, biotechnology, biochemical processes,
and forensic science.
Set of Questions for FA/CE/IA/ESE
1.
2.
3.
4.
What is the principle of chromatography?
What is the difference between adsorption and partition chromatography?
Discuss Application of Thin Layer Chromatography (TLC) in different fields.
Discuss in brief, the principle of separation of the components of sample in Gas
Chromatography.
5.
Draw a Schematic diagram of High performance Liquid Chromatography (HPLC) unit
and describe briefly functions of each part.
References:
1. Willard H H ,Merritt 11,Dean J A, Settle FA. Instrumental Methods of Analysis
2. Sharma B K Instrumental Methods of Chemical Analysis
Self assessment1) Define Chromatography. State its principle and types. (level 1)
2) Explain the principle and technique of TLC to separate the components of mixture.
(level 2)
3) Explain the various components in High Performance Liquid Chromatography. (level
3)
4) How will you do interpretation of Gas and HPLC chromatogram? (level 4)
60
Chemistry Sem I/II
Self-evaluation
Name of
Student
Class
Roll No.
Subject
Module No.
S.No
Tick
Your choice
1.
2.
3.
4.
5.
Do you understand
chromatography?
the
concept
of
o
Yes
o
No
Do you understand different types of
chromatography?
o
Yes
o
No
Will you able to describe the principle of Gas
Chromatography?
o
Yes
o
No
Are you able to understand the working of
High-Performance Liquid Chromatography?
o
Yes
o
No
Do you understand module ?
o
Yes, Completely.
o
Partialy.
o
No, Not at all.
Green Chemistry and Catalysis
61
Module 3
Green Chemistry and Catalysis
Lecture: 15
3.1Motivation:
The term green chemistry was coined by Paul Anastas in 1991. However, it has been
suggested that the concept was originated by Trevor Kletz in his 1978 paper in Chemistry and
Industry where he proposed that chemists should seek alternative processes to those
involving more dangerous substances and conditions. Also in Chemistry and Industry the
solvent free green chemistry version of “The Hajos-Parrish Cyclisation” has been highlighted
in 1996 by Professors Andrew B. Holmes and G. Richard Stephenson.
Currently green chemistry has got lot of attention. This chapter imparts knowledge to the
students which they can use in minimizing or preventing environmental deterioration and
prevail safe working condition. The use of catalyst alone is capable of making the process
green to a greater extent and hence we will look into catalyst , catalysis and use of catalyst for
designing greener chemical processes.
3.2 Syllabus:
Module
Contents
Duration
(Hrs)
Self
Study
(Hrs)
15
16
17
18
19
Introduction:
Need and Concept of Green chemistry: What is Green
chemistry?
12 principles of Green Chemistry. Prevention of waste,
Design safer chemical and Products, Design less hazardous
Chemical synthesis
Synthesis of Indigo by Conventional and Greener Route, Use
of renewable feedstock, Synthesis of Adipic acid by
conventional and greener route, Use of catalyst , Maximize
Atom economy and Numerical based on it.
Avoid Chemical Derivatives, Synthesis of Ibuprofen by
conventional and Greener route
The use of auxiliary substances (solvents, separation agents,
etc.) should be made unnecessary whenever possible and,
when used, innocuous, Increase Energy efficiency, Design
chemicals and products to degrade after use, Analyze in real
time to prevent pollution, Minimize the potential for
accidents
Role of Catalyst in making the chemical process Green,
1
2
1
3
1
2
1
2
1
2
Chemistry Sem – I/II
62
Relevance and examples, Homogeneous and heterogeneous
catalysis,
Theory of Heterogeneous Catalysis (Adsorption Theory)
20
Catalytic Converters, Acid Base catalysis,
1
2
21
Solid Acid Catalysis, Solid Base Catalysis, Transition metal
1
2
Catalysis, Metal and supported metal catalysis, Catalyst
design through artificial intelligence and computer
modelling
3.3 Learning Objective:
• Learners will be able to understand goals and principles of green chemistry
• Learners will be able to understand the applications of green chemistry in designing
new molecule with inherent safety.
• Learners will be able to understand concept and practices of atom economy.
• Learner will be able to understand the role of catalyst in design and development of
Green processes.
• Learner will be able to understand the concept of Catalytic converter and its uses.
• Learner will be able to understand the use of heterogeneous catalyst for hazard
reduction.
3.4Theoretical Background:
Green Chemistry is a chemical philosophy encouraging the design of products and processes
that reduce recycle or eliminate the use and generation of hazardous substance by finding
creative ways to minimize the human and environmental impact without stifling scientific
progress. In 1990 the Pollution Prevention Act was passed in the United States. This act
helped create a modus operandi for dealing with pollution in an original and innovative way.
It aims to avoid problems before they happen.
3.5Key Definitions:
1.
Green Chemistry: It is a chemical philosophy encouraging the design of products and
processes that reduce recycle or eliminate the use and generation of hazardous substance by
finding creative ways to minimize the human and environmental impact without shifting
scientific progress.
2.
Atom economy: It may be defined as the ratio of formula weight of target molecule to formula
weight of all the starting materials and the reagents.
3.
Catalyst: It is defined as a substance which alters the rate of chemical reaction, while itself
remaining chemically unchanged at the end of the reaction.
4.
Catalysis: The phenomenon of altering the velocity of a chemical reaction by the presence of
catalyst is known as Catalysis.
Introduction and Twelve Principles of Green Chemistry
Learning Objective: In this lecture student will able to learn about Green Chemistry Principle.
Course Content:
Introduction:
Green Chemistry and Catalysis
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
63
Green chemistry, also called sustainable chemistry, It is a chemical philosophy encouraging
the design of products and processes that reduce, recycle or eliminate the use and generation
of hazardous substances by finding creative ways to minimize the human and environmental
impact without stifling scientific progress.
3.6 Twelve Principle of Green Chemistry
Prevention Try not to make waste, then you do not have to clean it up.
Atom economy The final product should aim to contain all the atoms used in the process.
Less hazardous chemical synthesis Wherever it is possible, production methods should be
designed to make substances that are less toxic to people or the environment.
Designing safer chemicals Chemical products should be designed to do their job with
minimum harm to people or the environment.
Safer solvents When making materials try not to use solvents or other unnecessary
chemicals. If they are needed then they should not be harmful to the environment in any way.
Design for energy efficiency The energy needed to carry out a reaction should be minimized
to reduce environmental and economic impact. If possible, processes should be carried
out at ambient temperatures and pressures.
Use of renewable feedstocks A raw material should be renewable wherever possible.
Reduce derivatives Try not to have too many steps in the reaction because this means more
reagents are needed and more waste is made.
Catalysis Reactions that are catalysed are more efficient than uncatalysed reactions.
Design for degradation When chemical products are finished with, they should break down
into substances that are not toxic and do not stay in the environment.
Real-time analysis for pollution prevention Methods need to be developed so that harmful
products are detected before they are made.
Inherently safer chemistry for accident prevention Substances used in a chemical process
should be chosen to minimise the risk of chemical accidents, including explosions and fire.
3.7 Basic Principles of Green Chemistry:
The basic principles of Green Chemistry are as follows:
1.
Prevent waste:
It is better to design chemical synthesis to prevent waste rather than leaving no waste to treat
or clean up. In most of the cases, the cost involved in the treatment and disposal of wastes
adds to the overall cost of the production. The unreacted starting materials also form part of
the waste. The waste if discharged in the atmosphere, sea or land not only causes pollution
but also requires expenditure for cleaning up. Hence it is truly said, “Prevention is better than
Cure “.
2.
Design safer chemicals and products:
Fully effective chemical products should be designed with little or no toxicity. It is very
important that the synthesized chemcial should be safe to use. When any medicinal
formulations are to be put in market, they are put first on trials to check their toxic effects. If
found to be toxic then alternatives methods are prepared keeping in mind the function of the
medicine with reduced toxicity. In many insecticides like DDT, gamaxane which are found to
be toxic to humans, the use of these restricted and biological pesticides are more in use.
For example, unsafe drug thalidomide was used for reducing the effects of nausea and
vomiting during pregnancy. As the children born to women taking thalidomide suffered birth
defects, the use of the drug was banned.
3.
Design less hazardous chemical synthesis:
Chemistry Sem – I/II
64
Wherever practicable chemical syntheses should be designed to use and generate substances
with little or no toxicity to humans and the environment. One of the most important principle
of green chemistry is to prevent or at least minimize the formation of hazardous product
which may be toxic and or enviornmentally harmful. In case hazardous products are formed,
their effects on the workers must be minimized.
The starting material selected should be least toxic. Thus, e.g. like Pyridine or beta-naphthyl
amine, benzene, Aniline and other aromatic compound being known to be carcinogenic
should be avoided as starting material. A very well known example fitting to this principle of
green chemistry is synthesis of indigo from tryptophan rather than Aniline.
The reactions with toxic intermediates or reagents or products should not be followed instead
alternative pathways should be used for synthesis
Let’s check the take away from this lecture
1)
2)
Chemistry is applicable in the field of –
a) Medicines
b)dyes
The objective of green chemistry is –
a) to minimize environmental pollution
c) to use safer chemicals
c) insecticides
d)all of the above
b) to design harmless chemical processes
d) all of the above
Exercise:
Q.1 What is green chemistry? List the principles of green chemistry
Q.2
Prevention of waste is an important principle of green chemistry. Explain
Questions/problems for practice:
Q.3
State 12 Principle of Green Chemistry
Learning from the lecture :
Student will be able to define Green Chemistry its Twelve Principles and key terms
involved in it with an appropriate example.
Lecture 16
Numerical on atom economy, synthesis of the adipic acid, carbaryl, Ibuprofen
Learning objective: In this lecture students will able to understand about atom economy and
its calculation
Synthesis of Indigo: [Based on R2,Chapter 23 , page 793]
Indigo is the dye which is used to colour blue Jeans. The conventional route makes use of
aniline as a starting material. The principle involved in this synthesis is use of non hazardous
chemical synthesis i.e. Green chemistry recommends the design of synthesis to use and
generate substances with little or no toxicity to humans and the environment.
This commercial process is facing the problems of
i)
Use of highly toxic aniline.
ii)
Generation of considerable amount of waste salts, thereby causing disposal problems.
The conventional route for the production of indigo using hazardous aniline is given below
Green Chemistry and Catalysis
NH2
Cl
65
O
O
HN
OH
OH
OH
NH2
chloroacetic acid
aniline
NH
Na
(phenylamino)acetic acid
1H-indol-3-ol
Air
NH
O
O
NH
Indigo
Fig. 4.1 Conventional Route to Synthesize Indigo
The greener route makes use of the reaction in which the side chain of tryptophan is removed
enzymatically to give indole. It can be dehydroxylated enzymatically and then oxidized with
oxygen to indigo.
OH
NH
O
OH
Tryptophanase
NH
Naphthalene
Deoxygenase
OH
NH2
N
H
1H-indole
2,3-dihydro-1H-indole-2,3-diol
L-Tryptophan
Air,O2
O
H
N
N
H
O
Indigo
4.
Fig.4.2 Greener Route to Synthesis Indigo
Use renewable feedstock:
Raw materials and feedstock should be renewable rather than depleting whenever technically
and economically practicable. Renewable feedstock are often made from agricultural products
or are the wastes of other processes; depleting feedstock are made from fossil fuels.
For e.g. A new method is developed to prepare adipic acid from glucose obtained from
cellulose. This is green process because it replaces benzene as a starting material for the
production of the same product.
Synthesis of Adipic acid: [Based on 2, Chapter 23, page 789]
For the manufacture of adipic acid initially the substrate used was benzene which is
carcinogenic. The continuous use of benzene affects the human health.
The traditional synthetic pathway for the manufacture of adipic acid is given below:
Chemistry Sem – I/II
66
O
Ni/Al2O3
CO,O2
370-800 Psi
+
120-140 Psi
cyclohexane
benzene
OH
cyclohexanol
cyclohexanone
Cu, NH4VO3
HNO3
O
HO
OH
O
Adipic acid
Fig. 4.3 Synthesis of Adipic Acid
In the new synthetic pathway, traditionally used benzene is substituted by a new substrate
glucose. Being non-toxic substance glucose is absolutely safe to use as a substrate. The
alternative greener pathway is given below
O
OH
OH
HO
O
OH
E-Coli
O
OH
OH
6-(hydroxymethyl)tetrahydr
o-2H-pyran-2,3,4,5-tetrol
or D-Glucose
E-Coli
O
HO
OH
OH
O
OH
4,5-dihydroxy-3-oxocyclohe
x-1-ene-1-carboxylic acid
or 3-Dehydroshikimate
(2E,4E)-hexa-2,4-dienedioic acid or
Cis, Cis-Muconic Acid
Pt, H2 , 50 Psi
O
HO
OH
O
hexanedioic acid
or Adipic Acid
Fig. 4.4 Synthesis of Adipic Acid by Greener Route
5.
Use catalysts, not stoichiometric reagents:
Minimize waste by using catalytic reactions. Catalysts are used in small amounts and can
carry out a single reaction many times. They are preferable to stoichiometric reagents, which
are used in excess and work only once. Catalysts are selective in their action in that the degree
of reaction that takes place is controlled. Catalytic reactions are faster and hence require less
energy. In recent years, many processes are developed which use non-toxic, recoverable
catalyst & biocatalyst.
Application of the use of catalyst is as follows:
Green Chemistry and Catalysis
67
i) Hydrogenation of olefins in presence of nickel catalyst gives much better yields.
Ni
H3C
+
CH2
2+
H2
H3C
prop-1-ene
CH3
propane
Synthesis of Carbaryl
Carbaryl is a wide-spectrum carbamate insecticide which controls over 100 species of insects
on citrus, fruit, cotton, forests, lawns, nuts, ornamentals, shade trees, and other crops, as well
as on poultry, livestock and pets. It is also used as a molluscicide and an acaricide. Carbaryl
works whether it is ingested into the stomach of the pest or absorbed through direct contact.
The chemical name for carbaryl is 1- naphthol N-methylcarbamate. Carbaryl is formulated as
a solid which varies from colorless to white to gray, depending on the purity of the
compound. The crystals are odorless. This chemical is stable to heat, light and acids under
storage conditions. It is non-corrosive to metals, packaging materials, or application
equipment. It is found in all types of formulations including baits, dusts, wettable powder,
granules, oil, molassas, aqueous dispersions and suspensions.
O
OH
O
O
+
naphthalen-1-ol
Cl
Cl
NaOH
Cl
carbonyl dichloride
naphthalen-1-yl carbonochloridate
H3C
NH2
O
CH3
O
NH
Carbaryl
Fig.4.5 Synthesis of Carbaryl
Greener Way to Synthesize Carbaryl
O
CH3
OH
O
NH CH3
+
H3C
naphthalen-1-ol
Kat
NH
O
N,N'-dimethylurea
Carbaryl
Fig 4.6 Synthesis of Carbaryl by Greener Technique
NH
Chemistry Sem – I/II
6.
1)
2)
68
Maximize atom economy:
Design chemical synthesis so that the final product contains the maximum proportion of the
starting materials.
The percentage yield for any reaction is calculated by % yield =
Actual yield of the product
x 100
Theoretical yield of the product
From the calculations it is clear that if one mole of a starting material produces one mole of the
product then the yield is 100%. Such a synthesis even though is 100% is not considered to be a
green synthesis as it may generate significant amount of waste or byproducts which is not
visible in the above calculation. A reaction or a synthesis is considered to be green if there is
maximum incorporation of the starting materials or reagents in the final product. The
percentage atom utilization, which is determined by the following equation
% Atom economy = MW of desired product x 100
MW of all the reactants
Consider the addition reaction of bromine to propene (Bromination of propene)
H3C CH =CH2 + Br2
H3CCHBrCH2Br
Propene 1,2 – dibromopropane. In the above reaction all the elements of the reactants are
incorporated into the final product. So this reaction is 100% atom economical reaction.
For the synthesis of any product various chemical reactions can be evaluated and for the
different reaction atom economy can be calculated & thus the process giving maximum atom
economy can be selected.
The process with 90% atom economy yield are considered as excellent & those with 20% yield
is considered as poor.
Calculate the percentage atom economy for the following reaction with respect to allyl
chloride Propene + Chlorine ( Cl2) → Allyl chloride + Hydrochloric acid
Solution –
Reaction given is –
Propene + Chlorine ( Cl2) → Allyl chloride + Hydrochloric acid
42
71
76.5
36.5
% Atom economy =
(Molecular weight of the product / Total molecular weight of the reactants) x 100
= ( 76.5 / 42 + 71) x 100
= ( 76.5 / 113) x 100
= 67.7%
% Atom economy = 67.7%
Calculate the percentage atom economy for the following reaction with respect to all
maleic anhydride
Butene + 3 molecules of oxygen → Maleic anhydride + Water
56
96
98
Solution –
Reaction given is –
Butene + 3 molecules of oxygen → Maleic anhydride + Water
56
96
98
% Atom economy =
(Molecular weight of the product / Total molecular weight of the reactants) x 100
= ( 98 / 56+96) x 100
Green Chemistry and Catalysis
3)
4)
5)
69
= ( 98/152) x 100
= 64.5%
% Atom economy = 64.5%
Calculate the percentage atom economy for the following reaction with respect to
Acetanilide.
C6H5NH2 + (CH3CO)2O→ C6H5NHCOCH3 + CH3COOH
Solution –
Reaction given is –
C6H5NH2 + + (CH3CO)2O → C6H5NHCOCH3 + CH3COOH
93
102
135
% Atom economy =
(Molecular weight of the product / Total molecular weight of the reactants) x 100
= ( 135 / 93+ 102) x 100
= ( 135 / 195) x 100
= 69.23%
% Atom economy = 69.23%
Calculate the percentage atom economy for the following reaction with respect to chloro
benzene
C6H6 + Cl2 → C6H5Cl + HCl
Solution –
Reaction given is –
C6H6 + Cl2 → C6H5Cl + HCl
78
71
112.5
% Atom economy =
(Molecular weight of the product / Total molecular weight of the reactants) x 100
= ( 112.5 / 78+ 71) x 100
= ( 112.5 / 149) x 100
= 75.5%
% Atom economy = 75.5%
Calculate the percentage atom economy for the following reaction.
C6H6 + CH3Cl → C6H5CH3 + HCl
Solution –
Reaction given is –
C6H6 + CH3Cl → C6H5CH3 + HCl
78
50.5
92
% Atom economy =
(Molecular weight of the product / Total molecular weight of the reactants) x 100
= ( 92 / 78 + 50.5) x 100
= ( 92 / 128.5) x 100
= % Atom economy = 71.59%
Chemistry Sem – I/II
70
Let’s check the take away from this lecture
3)
4)
5)
The starting material for synthesis of indigo by conventional route is
(a) benzene
(b) styrene
(c) aniline
(d) L – Tryptophan
Maleic anhydride can be prepared by oxidation of –
(a) benzene
(b) butane
(c) butane
(d) all of the above
The starting material for synthesis of indigo by greener route is –
(a) benzene
(b)styrene
(c) aniline
(d) L – Tryptophan
Exercise:
Q.4 Calculate the percentage atom economy for the following reaction.
CH3CH=CH2+H2→CH3CH=CH2+KBr+H2O
Ans- 100%
Q.5 Give the synthesis of adipic acid by the conventional as well as the greener route.
Q.6 Give the synthesis of Carbaryl by the conventional route.
Questions/problems for practice:
Q.7
Calculate the percentage atom economy for the following reaction
CH3CH2CH2Br+KOH→CH3CH2CH3
Ans- 27%
Q.8 Calculate the percentage atom economy for the following reaction.
C6H6+4.5O2→C4H2O3+2CO2+2H2O
Ans- 44.1%
Q.9 Calculate the percentage atom economy for the following reaction.
CH3CH2CH=CH2+3O2→C4H2O3+3H2O
Ans64.5%
Learning from the lecture
Student should be able to Calculate Atom economy, Also would be able to write synthesis of
adipic acid, indigo, carbaryl by using conventional technique as well as greener route
Lecture 17
Synthesis of Ibuprofen
Learning objective: In this lecture students will able to understand synthesis of Indigo
7.
Avoid chemical derivatives:
Avoid using blocking or protecting groups or any temporary modifications if possible. A
commonly used technique in organic synthesis is the use of protecting or blocking group. This
type of approach in organic synthesis increases number of steps and yield of desired product.
This can be better understood by synthesis of ibuprofen by conventional as well as greener
route. Conventional route for Ibuprofen synthesis completes in six steps whereas Greener
route developed for it requires merely three steps which increase the yield of Ibuprofen.
Green Chemistry and Catalysis
71
O
CH3
CH3COO)2O
H3C
Al Cl3
H3C
H3C
(2-methylpropyl)benzene
or Iso-Butyl Benzene
O
CH3
1-[4-(2-methylpropyl)phenyl]ethanone
Or 4-Acetyl iso butyl benzene
OH
ClCH2COOC2H5 , C2H5oNa
CH3
CH3
O
O
H3C
CH3
O
3-[4-(2-methylpropyl)phenyl]butanoic acid
or Ibuprofen
H3C
CH3
CH3
N
ethyl 3-methyl-3-[4-(2-methylpropyl)phenyl]oxirane-2-carboxylate
H+/H20
CH3
H3C
H+/H20
O
CH3
2-[4-(2-methylpropyl)phenyl]propanenitrile
OH
CH3
N
P205
NH2OH
CH3
H3C
CH3
2-[4-(2-methylpropyl)phenyl]propanal
H3C
CH3
(1E)-N-hydroxy-2-[4-(2-methylpropyl)phenyl]propan-1-imine
Fig.4.7 Synthesis of Ibuprofen by Conventional method
O
OH
(CH3COO)2O
CH3
CH3
propan-2-ylbenzene
or iso-butyl
benzene
CH3
H2
HF
H3C
Catalyst
H3C
H3C
CH3
CH3
1-[4-(propan-2-yl)phenyl]ethanone
1-[4-(propan-2-yl)phenyl]ethanol
Co, Pd
O
OH
CH3
H3C
CH3
2-[4-(propan-2-yl)phenyl]propanoic acid
Or IBUPROFEN
Fig. 4.8 Synthesis of Ibuprofen by Greener Method
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Let’s check the take away from this lecture
6)
An example of a green reagent is –
a) dimethyl carbonate
b) styrene
c) benzene
d) xylene
Exercise:
Q.10 Explain conventional and green route of production of Ibuprofen. By this reaction which
principle of green chemistry is shown.
Questions/problems for practice:
Q.11 Explain Greener Route to Synthesize Ibuprofen
Learning from the lecture:
Student should be able understand green chemistry principle and apply in synthesis of ibuprofen
Lecture 18
Green Solvents (ionic liquid supercritical CO2 ), Products from natural materials
Learning objective: In this lecture students will able to understand the application of green
solvent and its application
8.
The use of auxiliary substances (solvents, separation agents, etc.) should be made
unnecessary whenever possible and, when used, innocuous.
A number of solvents like methylene chloride, chloroform, carbon tetrachloride, benzene and
other aromatic hydrocarbons which are suspected human carcinogens have been used in
many reactions due to their excellent solvent propeties. The solvent selected for a particular
reaction should not cause any enviornmental pollution and health hazard.
For example, chlorofluro carbons (CFC) which have been used as cleaning agents, blowing
agents and as refrigerants are found to be responsible for depleting the ozone layer having
disastrous effect on human survival. If a solvent is necessary, water is a good medium as well
as certain eco-friendly solvents that do not contribute to smog formation or destroy the ozone.
Solvents such as acetone, benzene, ether being highly inflamable should be avoided. For dry
cleaning the fabrics, the toxic solvents like perchloro ethylene (tetrachloro ethylene) was used
which is replaced by liquid CO2.
9.
Increase energy efficiency:
In any chemical synthesis, the requirement of energy should be kept to a minimum. The
reactions in which the reaction mixture has to be heated to reflux for completing the reaction,
the time required for completion of the reaction should be minimum, so that less amount of
energy is required. Even the use of a catalyst lowers the requirement of energy of a reaction. If
the final product obtained is impure, purification is carried out by distillation, recrystallisation
or ultrafiltration. As it requires energy, the process should be designed in such a way that
there is no need for seperation or purification.
The recently found substitutes are microwave radiations and ultrasound. Thus energy
efficiency can be increased by proper heat transfer and minimum wastage of energy during
the process.
Green Chemistry and Catalysis
10.
73
Design chemicals and products to degrade after use:
Design chemical products to break down to innocuous substances after use so that they do not
accumulate in the environment.
The problem of non–biodegradability is generally associated with pesticides, plastics and
other organic molecules. Most of the pesticides are organo- halogen based compounds which
generally tend to bioaccumulate in plants and animals. For example, pesticide DDT, because
its residue remains in soil for many years for causing pollution. The alternative to this is use of
biological insecticides.
The chemical with functional group which are susceptible to hydrolysis, photolysis have been
used to ensure that products will undergo biodegradation. Care should be taken that the
biodegradation products should not be toxic.
11.
Analyze in real time to prevent pollution:
Analytical methodologies and technology have been developed to allow the prevention and
minimization of the generation of hazardous substances in chemical processes. Using various
techniques, a chemical process can be monitored for generation of hazardous by products and
side reactions. These procedures can prevent any accident which may occur in chemical
plants.
12.
Minimize the potential for accidents:
Design chemicals and their form to minimize the potential for chemical accidents including
explosions, fires, and releases to the environment. At times it is possible to increase accident
potential in an attempt to minimize the generation of waste in order to prevent pollution. An
attempt of recycling the solvents from a process usually increases the potential for a chemical
accident.
Let’s check the take away from this lecture
1)
2)
3)
DDT is used as –
a) herbicide
b) pesticide
c) germicide
Green chemistry utilizes _______ principles
a) 8
b) 12
c) 3
Example of carcinogenic chemical is –
a) beta naphthyl amine b) acetone
c) DMC
d) all of the above
d) 6
d) critical carbondioxide
Exercise:
Q.1 Write Short Note on Super critical liquid
Questions/problems for practice:
Q. 2 Write application of Super critical liquid
Learning from the lecture
Student should be able to understand green solvent and also products occurring from nature
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Module 3B
Catalysis and Green Chemistry
LECTURE 19
Role of Catalyst in making the chemical process Green, Relevance and examples, Homogeneous
and heterogeneous catalysis, Theory of Heterogeneous Catalysis (Adsorption Theory)
Learning objective: In this lecture students will able to understand the Catalyst and its
classification, mechanism of working of catalyst
3.8 Introduction: J.J. Berzelius reviewed a no. of observations where the rate of a reaction was increased by the
presence of some substances that remain unchanged at the end of the process.
Eg. a) Decomposition of alk. H2O2 presence of certain metals. Such substances which help to loosen
the bonds which holds the atoms in the reacting molecule together called as catalyst and the
phenomena is named as catalysis.
Substances, which can alter rate of chemical reactions and themselves, remain chemically and
quantitatively unchanged after the reactions are catalysts.
3.9 Types of catalysts
Homogenous Catalysts
Heterogeneous Catalysts
When the reactants and When reactants & catalyst are in
catalyst are in same phase
different phases
e.g. Lead chamber process
e.g. Contact process
SO2 (g)+ O2 (g) ¾NO(g)
¾¾
® SO3
( )
SO 2 ( g ) + O 2 ( g ) ⎯⎯⎯
→ SO 3
CH 3 COOCH 3 ( l ) + H 2 O ( l )
Vegetable oil. (l) + H2(g)
Pt s
( )
⎯⎯⎯
→ vegetable ghee(s)
Ni s
HCl ( l )
⎯⎯⎯→ CH 3 COOH + CH 3 OH
3.9.1
Role of Catalyst in making the chemical process Green

A substance, which decreases the rate of chemical reaction, is called negative catalyst. For
example, in self decomposition of hydrogen peroxide
H 3 PO 4 or glycerol
H 2 O 2 ⎯⎯⎯⎯⎯⎯
→ H 2O + O 2
negative catalysts

Sometimes, product formed in a chemical reaction can act as positive or negative catalyst.
Such reactions are called as autocatalysis and auto poisoned reactions, respectively.
For example, in oxidation of oxalic acid by permanganate
® K2SO4 + MnSO4 + H2O + CO2
KMnO4 + H2SO4 + H2C2O4 ¾¾
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75
Product MnSO4 acts as a catalyst and rate of the reaction with time will be as shown

When one reaction catalyse an other reaction than the phenomenon is called induced
catalysis. For example, reduction of HgCl2 to Hg2Cl2 by oxalic acid is very slow while that of
KMnO4 is fast,
but mixture of HgCl2 and KMnO4 is reduced rapidly by oxalic acid as
reduction of KMnO4 catalyse the reduction of HgCl2.

Promotors are the substances that increase activity of a catalyst and catalytic poisons
(inhibitors)
decrease it. Promotors and catalytic poisons are not catalysts. A promotor increases
the number of sites on the surface whereas poisoning is due to preferential adsorption of poison
on the surface of catalyst.
For example, (i) Haber’s process
(ii)
Fe Catalyst
N 2 + 3H 2 ¾Mo
¾¾¾
® 2NH 3
Promotor
ZnO Catalyst
CO + 2H 2 ⎯⎯⎯⎯⎯
→ CH 3 OH
Cr2 O 3 Promotor
(iii) Rosenmund reaction
Pd Catalyst
RCOCl + H 2 ⎯⎯⎯⎯⎯
→ RCHO + HCl
BaSO 4 Poison
Definitions: a) Catalyst: - It is defined as a substance which alters the rate of chemical reaction, while itself
remaining chemically unchanged at the end of the reaction.
b) Catalysis: - The phenomenon of altering the velocity of a chemical reaction by the presence
of catalyst is known as Catalysis.
Eg.: - i) 2MnO4- + 6H+ + COOH
2Mn+2 + 10CO2 + H2O
1
COOH
ii) C2H5OH
Cn CH3CHO
Ethanol
Ethanal
3.9.2 Classification of Catalysis: Catalysed reactions are broadly classified into the following two groups based on mechanism of
reactions and phases
(1) Homogeneous Catalysis: - In these reactions reactants and catalyst are in the same phase and
the reactions involve molecules, ions or free nodicals as intermediates.
(a) In gas phase: i) Oxidation of SO2 to SO3 by the oxides of nitrogen (as in lead chamber process for the
manufacture of sulphuric acid)
2SO2 (9) + O2 (9) NO (9) 2SO3 (9)
ii) Decomposition of ozone in presence of nitric oxide
2O3(9) NO (9) 3O2(9)
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(b) In liquid phase: i) Hydrolysis of esters
CH3COOC2H5 + H2O H+ CH3COOH + C2H5OH
(c) Homogeneous Catalysis: - It refers to catalytic reactions in which catalyst and reactants are
in same phase
i) Inversion of cane sugar
C12H22O11 () + H2O () H+ C6H12O6 () + C6H12O6 ()
Glucose
Fructose
(2) Heterogeneous Catalysis: - In these reactions the reactants and catalysts are present in different
phase & the reactions proceed at an interface between the two discrete phases.
Example: a) Heterogeneous catalysis with solid reactants
2KClO3 (s) MnO2 (s) 2KCl (s) + 3O2 (9)
Decomposition of potassium chlorate in presence of manganese dioxide as a catalyst.
b) With gaseous reactants Synthesis of ammonia by Haber’s process
N2(9) + 3H2(9) Fe(s) 2NH3 (9)
c) With liquid reactants
C6H6(l) + CH3COCl Alll3(s) C6H5COCH3 + HCl
Benzene Acetyl
Acetophenone
Chloride
3.9.3Types of Catalyst
•
Positive Catalyst
•
Negative Catalyst
Features of Catalyst
1. Catalyst not consumed during the reaction.
2. Catalyst is specific in nature.
3.9.4 Adsorption theory of Heterogeneous catalysts
Old theory
Increase in concentration of reactants on the surface increases the rate of reaction.
Heat
released in adsorption is utilized in increasing rate of a reaction further by Arhenius
theory also.
Modern theory
Following steps are proposed
(I)
(II)
Diffusion of reactants to surface of catalyst
Adsorption of reactant molecules on surface.
Green Chemistry and Catalysis
(III)
(IV)
(V)
Formation of an intermoderate on catalysts surface and reaction occur
Desorption of products from surface and it is available again for new reactants
Diffusion of products from catalyst surface far away.

This theory cannot explain action of promoters and poisons.
77
Let us take example of addition of H2 gas to C2H4 in presence of Ni catalyst

This theory is extended to explanation to functions of promotors and poisons
recently.
Promotors occupy interstitial voids as a result; surface area for adsorption increases
but
inhibitors (poisons) decreases surface area due to preferential adsorption as shown
and rate of
reaction decreases.
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3.9.5 Important features of solid catalysts
1)
Activity
Activity of a catalyst depends on strength of chemisorption. Adsorption of
reactants
should be strong enough to become active but not much strong that they stuck and
surface is
not available further. It has been found that for hydrogenation
reaction, the catalytic
activity increases from Group 5 to Group 11 of d block
metals with maximum activity
being shown by groups 7-9 elements of
theperiodic table.
Pt
2H 2 + O 2 ⎯⎯
→ 2H 2 O
2)
Selectivity
Ability of a catalyst to direct a reaction to yield a particular product is called
as
selectivity of the catalyst. For example, here reactants are same in all
the reactions but products
are different as catalysts are different.
CO + H 2 ¾Ni¾
® CH 4 + H 2O
Cu
CO + H 2 ⎯⎯
→ HCHO
Cu ZnO – Cu 2 O 3
CO + H 2 ⎯⎯⎯⎯⎯→
CH 3 OH
3)
Physical state of a catalyst may change during the reaction. Finely divided state of
catalyst at optimum temperature is more efficient for the reaction due to more adsorption.
4)
In some cases rate of reaction depends on amount of catalyst also and in some rare
cases catalysts initiate the reaction also. For example, H2 and O2 don’t react at room
temperature but reaction starts and occurs rapidly in presence of Pt black.

Zeolites are 3-D alumino silicate and having honeycomb like structure. They are
good shape
selective catalysts in which reaction depends on pore structure of catalyst & size of
reactants & products. They are used in crackling of hydrocarbons & isomerization.
Alcohols ¾ZSM–5
¾¾® Gasoline(petrol)
ZSM − 5 = Na x ( AlO 2 ) x ( SiO 2 ) 96– x  .16 H 2 O
0 < x < 27
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79
Pictoral representation of Mechanism of Heterogeneous Catalysis
i) Adsorption Theory: - According to the adsorption theory, for a general heterogeneous
catalytic reaction such as –
A (9) + B (9) Solid Catalyst C (9) + D (9)
The following steps are involved:
Step-1: - Adsorption of reactant molecules on the Catalytic surface by weak Vander Waals forces
(physical adsorption) or by partial chemical bonds (chemisorption).
Step-2: - Formation of an unstable and intermediate activated complex, A…. B from the adsorbed
reactant species adjacent to one another.
Step-3: - Decomposition of the unstable activated complex to form the products.
Steps-4: - Desorption of the stable products formed to release the fresh catalytic surface for a fresh
cycle of the above steps.
Step-1: -
Adsorption of reactant species on catalyst surface
Step-2: -
Formation of an unstable activated complex
Step-3: -
Catalyst Surface
Decomposition of the activated complex & formation of product.
Step-4: -
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Catalyst Surface
Desorption of stable product species & release of the free catalyst surface for another cycle of the
alone steps
Depending on the nature of the reactants, the mechanism may slightly
Mechanism
Step -1: - Absorption of H2 molecule on the Catalyst Surface due to residual Valence bond of Ni
atom
Catalyst Surface
Step-2: - Breaking of H-H bond which are held to Catalyst surface
Catalyst Surface
Step-3: - Chemisorbed H atoms are attached to CzHy molecule by partial Chemical bond
Catalyst Surface
Step-4: - Unstable complex is decomposed to yield the product and catalyst surface is released
Green Chemistry and Catalysis
81
Catalyst Surface
Pictorial presentation of mechanism of the hydrogenation of ethene to ethane in presence of Ni as a
Catalyst
Let’s check the take away from this lecture
1.
When a catalyst is added to a system the
(a) equilibrium concentrations are increased.
(b) equilibrium concentrations are unchanged.
(c) the rate of forward reaction is increased and that of backward reaction is decreased.
(d) value of equilibrium constant is decreased.
2.
Catalytic poison act by
(a) coagulating the catalyst.
(b) getting adsorbed on the active centers on the surface of catalyst.
(c) chemical combination with any one of the reactants.
(d) none of the above
3. Following are the events taking place to explain adsorption theory of catalysis
I: desorption
II: diffusion of the reactants long the surface
III: adsorption of the reactants
IV formation of the activated surface complex
These events are taking place in the following order:
(a) I, II, III, IV
(b) II, III, IV,I
(c) III, IV, I, II
(d) IV, III, II, I
Exercise:
Q.10 Give Classification of Catalysis.
Questions/problems for practice:
Q.11 What is catalysts? Explain role of catalyst in greener technique
Q.12Describe Theory of Adsorption Catalysis and its mechanism
Learning from the lecture:
Student should be able understand catalyst and its application
Lecture 20
Catalytic Convertor, Acid Base Catalysis
Learning objective: In this lecture students will able to understand the application of Catalyst
3.10 Catalytic Converter
A Catalytic converter is an exhaust a miss control device that reduces toxic gases & pollutants in
exhaust gas from an internal combustion engine into less toxic pollutants by catalyzing a redox
reaction. They are usually used with internal combustion engine fueled gasoline or diesel.
Applications: - Catalytic converters are commonly applied to exhaust system in automobiles, they
are also used on electric generator, mining equipment, trucks, buses, locomotives and motorcycles.
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Placement of Catalytic Convertors: - Catalytic convertors require a temperature of 800o F (426o C) to
efficiently convert harmful exhaust gases into gases, such as CO2 & water vapor.
Eg: - In running cars, the temperature inside the engine becomes high. At this temperature nitrogen
& oxygen gases react to form nitric oxide
N2(g) + O2 (9)
2NO (9)
NO (9) further combines with O2 to form NO2. NO2 & other gases which are also emitted by the
automobiles like CO & imburnt hydrocarbons are major causes of air pollution.
To overcome this difficulty, most of the cars are fitted with catalytic convertors. The catalyst
employed are PT or PD. Sometimes transition metal oxides such as CuO or Cr2O3 are also used.
An efficient catalyst converter serves the following purpose.
i)
It oxidizes CO & unburnt hydrocarbons to CO2 & H2O.
ii)
It reduces NO & NO2 to N2 & O2.
Fig5.1 Catalytic Converters for an Automobile
Let’s check the take away from this lecture
1. Which type of metals form effective catalysts?
(a) Alkali metals
(b) Transition metals
(c) Alkaline earth metals
(d) Radioactive metals
2. Which statement is not correct for a catalyst?
(a) It enhances the rate of reaction in both directions.
(b) It changes enthalpy of reaction.
(c) It reduces activation energy of reaction.
(d) It is specific in nature.
3.
Efficiency of a catalyst depends on its
(a) particle size
(b) solubility
(c) molecular weight (d) none of these
Green Chemistry and Catalysis
83
Exercise:
Q.14Describe Catalytic Convertor
Questions/problems for practice:
Q.15 Describe Catalytic Convertor with Diagrammatic representation
Learning from the lecture:
Student should be able understand catalyst and its application
Lecture 21
Solid Acid Catalysis, Solid Base Catalysis, Transition metal Catalysis, Metal and supported metal
catalysis, Catalyst design through artificial intelligence and computer modelling
Learning objective: In this lecture students will able to understand the application of Catalyst
3.11 Uses of Solid Acids as Catalyst
1. Major source of waste in the (fine) chemicals industry is derived from the widespread use of
liquid mineral acids (HF, H2SO4) and a variety of Lewis acids. They cannot easily be recycled
and generally end up, via a hydrolytic work-up, as waste streams containing large amounts
of inorganic salts.
2. Their widespread replacement by recyclable solid acids would afford a dramatic reduction
in waste. Solid acids, such as zeolites, acidic clays and related materials, have many
advantages in this respect.
3. They are often truly catalytic and can easily be separated from liquid reaction mixtures,
obviating the need for hydrolytic work-up, and recycled. Moreover, solid acids are noncorrosive and easier (safer) to handle than mineral acids such as H2SO4 or HF.
4. Solid acid catalysts are, in principle, applicable to a plethora of acid-promoted processes in
organic synthesis. These include various electrophilic aromatic substitutions, e.g. nitrations,
and Friedel-Crafts alkylations and acylations, and numerous rearrangement reactions such
as the Beckmann and Fries rearrangements
5. A prominent example is Friedel-Crafts acylation, a widely applied reaction in the fine
chemicals industry. In contrast to the corresponding alkylations, which are truly catalytic
processes, Friedel-Crafts acylations generally require more than one equivalent of, for
example, AlCl3 or BF3. This is due to the strong complexation of the Lewis acid by the ketone
product. Zeolite beta is employed as a catalyst, in fixed-bed operation, for the acetylation of
anisole with acetic anhydride, to give p-methoxyacetophenone. The original process used
acetyl chloride in combination with 1.1 equivalents of AlCl3 in a chlorinated hydrocarbon
solvent, and generated 4.5 kg of aqueous effluent, containing AlCl3, HCl, solvent residues
and acetic acid, per kg of product.
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6. The catalytic alternative, in stark contrast, avoids the production of HCl in both the acylation
and in the synthesis of acetyl chloride. It generates 0.035 kg of aqueous effluent, i.e. more
than 100 times less, consisting of 99% water, 0.8% acetic acid and < 0.2% other organics, and
requires no solvent.
7. Furthermore, a product of higher purity is obtained, in higher yield (>95% vs. 85– 95%), the
catalyst is recyclable and the number of unit operations is reduced from twelve to two.
Hence, the Rhodia process is not only environmentally superior to the traditional process, it
has more favorable economics. This is an important conclusion; green, catalytic chemistry, in
addition to having obvious environmental benefits, is also economically more attractive.
Catalytic cracking is normally performed over solid acid catalysts such as aluminosilicates
and zeolites.ZSM-5 zeolites are the most promising in the catalytic cracking of bio-oil
•
•
The traditional catalyst hydrogen fluoride, an extremely corrosive, hazardous and
toxic chemical used in the production of linear alkyl benzenes, has been successfully
replaced by a solid acid catalyst, viz. fluoride silica-alumina catalyst, which does not
require special material of construction (of the container), involves lower operating
costs
Microencapsulated Lewis acids have replaced traditional corrosive monomeric Lewis
acids in the reactions like Michael Addition, Friedel Crafts reaction
Uses of Solid Bases as Catalyst
1. The replacement of conventional bases, such as NaOH, KOH and NaOMe, by
recyclable solid bases, in a variety of organic reactions, is also a focus of recent
attention. For example, synthetic hydrotalcite clays, otherwise known as layered
double hydroxides (LDHs) and having the general formula Mg8-xAlx(OH)16(CO3)x/2 ·
nH2O, are hydrated aluminum-magnesium hydroxides possessing a lamellar
structure in which the excess positive charge is compensated by carbonate anions in
the interlamellar space.
Green Chemistry and Catalysis
85
2. Calcination transforms hydrotalcites, via dehydroxylation and decarbonation, into
strongly basic mixed magnesium-aluminum oxides, that are useful recyclable
catalysts for, inter alia, aldol , Knoevenagel and Claisen-Schmidt condensations.
3. Another approach to designing recyclable solid bases is to attach organic bases to the
surface of, e.g. mesoporous silicas. For example, aminopropyl-silica, resulting from
reaction of 3-aminopropyl(trimethoxy)silane with pendant silanol groups, was an
active catalyst for Knoevenagel condensations .
4. A stronger solid base was obtained by functionalisation of mesoporous MCM-41
with the guanidine base, 1,5,7-triazabicyclo-[4,4,0]dec-5-ene (TBD), using a surface
glycidylation technique followed by reaction with TBD.
5. The resulting material was an active catalyst for Knoevenagel condensations,
Michael additions and Robinson annulations.
6. A solid base like Na/ Al2O3 which has large surface area is very active for the
isomerization of olefins. For example, the isomerization of 1-pentene attains its
equilibrium in 60 min even at 30°C, and that of 1-butene in only 0.6 min at 25°C.
7. The isomerization of trans-crotonnitrile to cis-isomer was investigated by using solid
base catalysts such as, Al2O3 MgO, CaO, Na2CO3 and NaOH supported on silica gel
and solid organic compounds.
8. Alkylation of aromatics with olefins was also catalyzed by solid bases such as , Na/
Al2O3 NaH, K/graphite etc.
9. In contrast to the areas of heterogeneous oxidation catalysis and solid acid catalysis,
the use of solid base catalysis in liquid phase reactions has not met the same level of
breakthrough. The industrial applications of basic catalysts are in the alkylation of
phenol, side chain alkylation and isomerisation reactions
Example 1:- Alkylation of Phenol
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Example 2:- Alkylation of side chain
Transition Metal Catalyst
• Presently, organometallic catalysts play major role in homogeneous catalysis.
• Organometallic complex consist of a central transition metal ion bonded to organic
ligands such as R2C=CR2, RCO, R3P, R3N, CO etc.
• Catalysis occurs through dissociation of ligands followed by co-ordination of
reactant molecule to the metal ion.
• The transition metal ions react through exchange of d electrons.
• Organometallic complexes usually have octahedral or tetrahedral geometry.
Reactions catalyzed by organometallic complexes include hydrogenation,
hydroformylation, carbonylation and decarbonylation, hydrocarbon rearrangement,
partial oxidations etc.
Example :The Habers Process combines Hydrogen and nitrogen to make ammonia using an Iron
Catalyst
N2(g) + 3H2(g)
2NH3(g)
Example:• Carbon monoxide and hydrogen have been known to be used for the manufacture of
methyl alcohol. Also, the first product to be manufactured by the hydroformylation
of propene is butyraldehyde
Carbonylation of 1-( 4~isobutylphenyl)ethanol gives ibuprofen
Phase Transfer Catalyst
• Most of the pharmaceuticals or agricultural chemicals (insecticides, herbicides, plant
growth regulators) are the result of organic synthesis.
Green Chemistry and Catalysis
•
•
•
•
•
•
87
Most of the syntheses require a number of steps in which additional reagents,
solvents and catalysts are used. In addition to the syntheses of the desired products,
some waste material (by-products) is generated, the disposal of which causes
problems and also environmental pollution.
One of the most general and efficient methodologies that takes care of the above
problems is to use a phase-transfer catalyst (PTC)
Difficulties are often encountered in organic synthesis if the organic compound is
soluble in organic solvent and the reagent in water. In such cases, the two reactants
will react very slowly and the reaction proceeds only at the interface where these two
solutions are in contact.
The rate of the reaction can, of course, be slightly increased by stirring the reaction
mixture and by using aprotic polar solvents, which solvate the cations so that the
anions are free. Such solvents (like dimethylsulfoxide, dimethylformamide) are
expensive and their removal is difficult. Also the use of strong bases (which are
necessary for the reactions like Wittig etc.) create other problems and many side
reactions take place.
These problems can be overcome by using a catalyst, which is soluble in water as
well as in the organic solvent. Such catalysts are known as phase-transfer catalysts
(PTC).
The PTC reaction, in fact, is a methodology for accelerating the reaction between
water insoluble organic compounds and water soluble reactants (reagent). The basic
function of PTC is to transfer the anion (from the reagent) from the aqueous phase to
the organic phase.
Examples:• Williamson's Ether Synthesis The PTC technique provides a simple method for conducting
Williamson ether synthesis. Use of excess alcohol or alkyl halide, lower temperature and
larger alcohol (e.g. CSH1PH) give higher yields of ethers
•
Crown Ethers This is a group of cyclic polyethers which are used as phase transfer catalysts.
These have been used for esterifications, saponifications, anhydride formation, oxidations,
aromatic substitution reactions, elimination reactions, displacement reactions, generation of
carbenes, alkylations etc. Some of the examples are as follows:
3.12 Catalyst design through artificial intelligence and computer modelling
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AI is penetrating various fields of science and technologies, for example, it has been used for
image/speech recognition, highly personalized online services and strategic board games. Despite
its great success, there are important challenges in the deployment of an intelligent framework for
designing new catalytic materials. Arguably, this is due to the complex nature of catalytic processes
and the stringent requirements of ideal catalysts; for example, that they are highly efficient,
environmentally benign, stable under working conditions and made of earth-abundant elements. In
practice, the discovery of viable catalysts in industrial processes largely relies on a trial-and-error
approach with chemical intuition or luck. Recent advances in computing infrastructures and
electronic structure methods have popularized a high-throughput computational approach to
materials design. However, it requires high-level user interaction and is practically intractable due
to the immense chemical and chemistry space.
Let’s check the take away from this lecture
1.
2.
In which pH range, enzymes are highly active
(a) 2 to 6
(b) 5 to 7
(c) 7 to 9
(d) 10 to 14
Which enzyme catalyst is responsible for conversion of milk into curd?
(a) pepsin
(b) zymase
(c) lactobacilli
(d) urease
Exercise
Q.1
What is Phase Transfer Catalyst
Q.2 Define Solid Acid and Solid Base Catalyst
Questions/Problems for practice:
Q.3
Describe in brief Transition Metal Catalyst.
Learning Outcome In this lecture student will be able to understands the application of catalysts
3.12 Conclusion
Though the tenets of green chemistry might seem simple to implement, improvements can still be
made in a large number of chemical processes. A lot of the chemical products we all utilise come
from processes that still fail to meet a number of these principles; plenty of these products are
still derived from chemicals from crude oil, and many still produce large amounts of waste.
There are, of course, challenges involved in meeting some of the principles in a large number of
processes, but it can also drive new research and the discovery of new chemistry. It is to be
hoped that, in the coming years, many more processes will be adapted with these principles in
mind.
3.13 Add to Knowledge
Green Chemistry and Catalysis
89
Increasing demand for fuels and chemicals, driven by factors including over-population, the
threat of global warming and the scarcity of fossil resources, strains our resource system and
necessitates the development of sustainable and innovative strategies for the chemical industry.
Our society is currently experiencing constraints imposed by our resource system, which drives
industry to increase its overall efficiency by improving existing processes or finding new uses
for waste. Food supply chain waste emerged as a resource with a significant potential to be
employed as a raw material for the production of fuels and chemicals given the abundant
volumes globally generated, its contained diversity of functionalised chemical components and
the opportunity to be utilised for higher value applications. The present manuscript is aimed to
provide a general overview of the current and most innovative uses of food supply chain waste,
providing a range of worldwide case-studies from around the globe. These studies will focus on
examples illustrating the use of citrus peel, waste cooking oil and cashew shell nut liquid in
countries such as China, the UK, Tanzania, Spain, Greece or Morocco. This work emphasises
2nd generation food waste valorisation and re-use strategies for the production of higher value
and marketable products rather than conventional food waste processing (incineration for
energy recovery, feed or composting) while highlighting issues linked to the use of food waste
as a sustainable raw material. The influence of food regulations on food supply chain waste
valorisation will also be addressed as well as our society's behavior towards food supply chain
waste. “There was no ways of dealing with it that have not been known for thousands of years. These
ways are essentially four: dumping it, burning it, converting it into something that can be used again,
and minimizing the volume of material goods – future garbage – that is produced in the first place.”
William Rathje on waste (1945–2012) – Director of the Tucson Garbage project
3.14.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
Set of Questions for FA/CE/IA/ESE
Define Green chemistry.
What are the principles of green chemistry?
Why there is a requirement of following Green Chemistry concept in today’s context?
What is the significance of practicing green chemistry concept?
Name one Green Solvent.
Mention any two advantages of following Green Chemistry concept.
What is the significance of understanding concept of green chemistry?
Explain the basic ideas involved in the field of green chemistry w.r.t three research oriented
examples.
What are the industrial applications of green chemistry?
State and explain the 12 principles of green chemistry?
Give the synthesis of indigo by the conventional as well as the greener route.
Explain the use of supercritical carbon dioxide as solvent?
Explain supercritical fluid system.
4.13. References :
R1 :
Engineering Chemistry
R2 :
Engineering Chemistry
R3 :
Green Chemistry
R4 :
Green Chemistry
By Jain & Jain
By S.S. Dara
By V.K. Ahluwalia
By Rashmi Sanghi and M M Srivastava
Applied Chemistry – Sem I
90
Self Assesment
Q.1 List the principles of green chemistry.
[Level 1]
Q.2 Giving conventional & green chemistry route of production of adipic acid. Highlight
the green chemistry principles addressed in this case .
[Level 2]
Q.3 Prevention of waste is an important principle of green chemistry. Explain
[Level 3]
Q.4 Why it is essential to have green process?
[Level 4]
Q.5 Write a short note on supercritical fluids.
[Level 5]
Module 4: Green Chemistry
91
Self-evaluation
Name of
Student
Class
Roll No.
Subject
Module No.
S.No
Tick
Your choice
1.
2.
3.
4.
5.
Do you understand the concept of catalysis
o
Yes
o
No
o
Yes
o
No
Will you be able to calculate the atom
economy of any given reaction?
o
Yes
o
No
Are you able to understand theory of
catalysis
o
Yes
o
No
Do you understand module ?
o
Yes, Completely.
o
Partialy.
o
No, Not at all.
Will you be able to define Green Chemistry?
Chemistry Sem-I/II
92
Module:04
Electrochemistry and its Application (Corrosion)
Lecture: 22
4.1 Motivation:
Concept of electrochemistry is key in understanding the concept of corrosion and its prevention and
minimization. Corrosion is the disintegration of an engineered material into its constituent atoms
due to chemical and electrochemical reactions with its surroundings. This chapter helps us in
understanding the different types of corrosion and various factors affecting the rate of Corrosion. It
includes the different measures which need to be taken for control of corrosion.
4.2 Syllabus:
Duration
Lecture
no.
Content
22
Introduction, Concept of electrode potential, Concept of
Electrochemical cell, EMF of Cell, Cell potentials by Nernst
equation, Relation of free energy with EMF of Cell,
Numerical based on EMF and its feasibility prediction
1
2
23
Introduction: Types of Corrosion (I) Dry or Chemical
Corrosion i) Due to O2 ii) due to other gases.
1
2
24
Wet or Electrochemical Corrosion: ‐ Mechanism i) Evolution
of H2 ii) Absorption of O2.
1
2
25
Factors affecting the rate of corrosion
1
2
26
Types of Corrosion – Galvanic, Concentration cell corrosion.
1
2
27
Types of Corrosion –
1
2
(Hr)
Self-Study
(Hrs)
Intergranular, Stress corrosion and Pitting corrosion
28
Methods to decrease the rate of Corrosion, proper
designing, use of pure metal, alloys
1
2
29
Cathodic protection
1
2
30
Anodic Protection
1
2
31
Metallic coatings, Anodic Coating and Cathodic Coating
1
2
Module 4 : Electrochemistry and its Application (Corrosion)
93
4.3. Weightage: 16-18 Marks
4.4 Learning Objective:
•
Learner should know the key concept of Electrochemistry.
•
Learners shall be able to illustrate the concept of corrosion and its various types.
•
Learners shall be able to recall galvanic and emf series
•
Learners shall be able to explain the mechanism of corrosion.
•
Learners shall be able to identify the corrosion type.
•
Learners shall be able to list various factors affecting on rate of corrosion.
•
Learners shall be able to summarize the concept, types and factors affecting on rate of
corrosion.
•
Learners should able to explain the methods to decrease the rate of corrosion.
•
Learners should able to compare cathodic and anodic coatings
4.5 Theoretical Background:
One has to have very good knowledge of electrochemistry for understanding the basic
reason of corrosion. Corrosion is the disintegration of an engineered material into its constituent
atoms due to chemical and electrochemical reactions with its surroundings. Formation of an oxide
of iron due to oxidation of the iron atoms in solid solution is a well-known example of
electrochemical corrosion, commonly known as rusting. This type of damage typically produces
oxide(s) and/or salt(s) of the original metal. In other words, corrosion is the wearing away of metals
due to a chemical reaction.
Many structural alloys corrode merely from exposure to moisture in the air, but the process can be
strongly affected by exposure to certain substances Corrosion can be concentrated locally to form a
pit or crack or it can extend across a wide area more or less uniformly corroding the surface.
Because corrosion is a diffusion-controlled process, it occurs on exposed surfaces. As a result,
methods to reduce the activity of the exposed surface, such as passivation and chromateconversion, can increase a material's corrosion resistance. However, some corrosion mechanisms are
less visible and less predictable.
4.6 Key Definitions:
1.
Electrochemistry is the area of chemistry which is concerned with interconversion chemical
and electrical energy.
2.
Electrodes are the surfaces on which oxidation and reduction half reactions take place.
Electrodes may or may not participate in the reactions. The electrodes which do not take part
in reactions are inert electrodes.
Chemistry Sem-I/II
94
3.
Cathode is an electrode at which the reduction takes place. At this electrode the species
undergoing reduction gains electrons.
4.
Anode is an electrode at which oxidation takes place. At this electrode, the species undergoing
oxidation loses electrons.
5.
Corrosion: It is defined as the deterioration or destruction of metal (alloys) by an unwanted
chemical or electrochemical reaction with its environment starting at its surface.
6.
Hydrogen Embrittlement: Action of hydrogen on metals at low temperatures is called
hydrogen Embrittlement.
7.
Passivity: Passivity or passivation is the phenomenon in which a metal or an alloy exhibits a
much higher corrosion resistance than expected from its position in the electrochemical series.
8.
Oxidation Potential: It is the potential developed between the electrode and its ions in the
solution due to oxidation reaction at equilibrium at given temperature.
9.
Reduction Potential: It is the potential developed between an electrode and its ions in the
solution due to reduction reactions at equilibrium at given temperature.
10.
Metal Cladding: It is the process by which a dense homogeneous layer of coating metal is
bonded firmly and permanently to the base metal on one or both sides.
11.
Galvanizing: It is the process of coating iron or steel sheets with a thin coat of zinc to prevent
them from rusting.
12.
Tinning: It is the process of coating iron or steel sheets with a thin coat of tin to prevent them
from rusting.
4.7 Course Content
4.7.1. Galvanic/ Voltaic/Electrochemical Cell
In galvanic or voltaic cells, electricity is generated through the use of spontaneous chemical
reactions.
A galvanic (or voltaic) cell is made of two half cells. Each half cell consists of a metal strip immersed
in the solution of its own ions of known concentration. For example, a strip of zinc metal immersed
in 1 M aqueous solution of zinc ions forms an half-cell.
It follows that two metal plates and the solutions of their ions with known concentrations are
required for the construction of a galvanic (voltaic) cell. Two half cells are constructed by immersing
the two metal plates in solutions of their respective ions placed in separate containers. The two half
cells so constructed are combined together to form the galvanic cell. The metal plates called
electrodes are connected through voltmeter by a conducting wire for transfer of electrons between
them. To complete the circuit the two solutions are connected by conducting medium through
which cations and anions move from one compartment to the other. This requirement is fulfilled by
a salt bridge.
Module 4 : Electrochemistry and its Application (Corrosion)
95
Electrochemical cell set-up of Cu/Ag system
Salt Bridge
In a galvanic cell, the two solutions are connected by a salt bridge. It is an U tube containing a
saturated solution of an inert electrolyte such as KCl or NH4NO3 and 5 % agar solution. The ions of
electrolyte do not react with the ions of electrode solutions or the electrodes.
Salt bridge is prepared by filling a U tube with hot saturated solution of the salt and agar agar
solution allowing it to cool. The cooled solution sets into a gel which does not come out on inverting
the tube. The salt bridge is kept dipped in distilled water when not in use.
Functions of Salt Bridge
Chemistry Sem-I/II
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The salt bridge serves the following functions:
• It provides an electrical contact between two solutions and thereby completes the electrical
circuit.
• It prevents mixing of two solutions.
• It maintains electrical neutrality in both the solutions by transfer of ions.
Electrode Potential:
A galvanic cell is composed of two half cells, each consisting of electronic (metal plates) and
electrolytic (solution of ions) conductors in contact. At the surface of separation of solid metal and
the solution, there exists difference of electrical potential. This potential difference established due
to electrode half reaction occurring at the electrode surface, is the electrode potential. The potential
is associated with each of the half reaction, be it oxidation or reduction. The potential associated
with oxidation reaction is oxidation potential while that associated with reduction gives the
reduction potential. The overall cell potential, also called electromotive force (emf), is made of the
contributions from each of the electrodes.
In other words, the cell potential is algebraic sum of the electrode potentials,
Ecell =Eoxi (anode) + Ered (cathode)
where Eoxi is the oxidation potential of anode (-) and Ered is the reduction potential of cathode (+).
When galvanic cell operates, electrons are generated at the anode. These electrons move through
external circuit to the cathode. The cell potential is the force that pushes electrons away from anode
(-) and pulls them toward cathode where they are consumed.
Dependence of cell potential on concentration( Nernst Equation)
The standard cell potential tells us whether or not the reactants in their standard states form the
products in their standard states spontaneously. To predict the spontaneity of reactions for anything
other than standard concentration conditions we need to know how voltage of galvanic cell varies
with concentration. Dependence of cell voltage on concentrations is given by Nernst equation.
For any general reaction,
aA + bB -------------------cC + dD
The cell voltage is given by
Ecell = E0cell - RTnF1n [C]c [D]d/[A]a [B]b
= E0cell - 2.303RTnF log10[C]c [D]d/[A]a [B]b
Module 4 : Electrochemistry and its Application (Corrosion)
97
where n = moles of electrons used in the reaction, F = Faraday = 96500 C, T = temperature in kelvin,
R = gas constant = 8.314 J K-1mol-1
Let’s check the take away from this lecture
1)
Electrochemical cell converts
a)
Chemical energy into Electrical Energy
b)
Electrical Energy into Chemical Energy
c)
Both
d)
None
2)
What are the uses of salt bridge in an electrochemical cell
a)
Completes electrical circuits
b)
It maintains electrical neutrality
c)
it prevents mixing of two solutions
d)
All of the above
3)
Electrochemical Cell consist of
a)
Anode half-cell
c)
Cathode half-Cell
b)
Both
d)
None
4)
EMF of cell is dependent on
a)
Concentration of electrolytes
b)
Set of Metal being used as an electrode
c)
Temperature
d)
All of the above.
5)
Free Energy is
a)
Capacity to do the work
c)
Energy available to do useful work
b)
Energy available freely
d)
None of the above
Chemistry Sem-I/II
98
6)
Actual EMF od Cell is calculated using
a)
Nernst Equation
c)
Both
b)
Arrhenius Equation
d)
None
Exercise
Q.1
Discuss the construction and working of an electrochemical cell for harnessing electrical
energy.
Q.2
why it is important to have salt bridge in an electrochemical cell?
Questions/Problems for practice:
Q.3
Try writing anodic half, cathodic half and overall cell reaction for 5 different set of metal of
your choice. You can refer Electrochemical/ Galvanic series for doing this job.
Q.4
Discuss the usefulness of Free energy for predicting spontaneity of any physical and
chemical Process.
Q.5
What do you mean by Electrode potential. How do we practically determine it?
Questions/Problems for practice:
Q.6
Numerical Based on calculation of EMF and Feasibility.
Learning from this lecture: Learners will be able to understand construction and working of an
Electrochemical cell. They will be also able to calculate standard EMF, Actual EMF of Cell, Free
Energy and its application of deciding feasibility of any physical and chemical process.
Lecture: 23
4.7.2 Corrosion and its types
Learning objective:
In this lecture learners will able to understand corrosion and it types.
4.7.2.1 Introduction to Corrosion
Corrosion is the process of destruction of metals by the chemical or electrochemical attack of
environment, starting at surface. The losses incurred due to corrosion are enormous. The machines,
equipment’s, various metallic structures and metallic parts get damaged due to corrosion. Thus it
results in loss of materials as well as financial loss and the loss regarding high cost of fabrication of
equipment’s / machinery / metallic structures. The severity of the problem can be understood from
loss to the tune of Rs. 1, 00,000 crores per year, all over the world.
Module 4 : Electrochemistry and its Application (Corrosion)
99
The corrosion of metals occurs by chemical reaction of environment is slow process but the
corrosion of metals with cell formation is faster.
Most of the metals occur in the compound state (except gold, platinum) i.e. in the form of metal
compounds like oxides, sulphides, carbonates, silicates, chlorides, hydroxides. Etc. The metal
compounds are thermodynamically at lower level of energy. i.e. high chemical stability.
In the process of manufacturing of metals from their compounds (metallurgy) we add heat or
electrical energy and chemical energy and metal so formed is at higher state of energy. There is
natural tendency to take metal to lower state of energy by attack of certain corroding medium and
again form certain metal compounds.
Thus the phenomenon of destruction of metal by action of corroding medium is natural and
obvious.
Definition: Corrosion is an unwanted and undesirable deterioration of a metallic substance by
chemical or electrochemical transformation caused by the environment, starting at its surface. It is
also called as weathering of metals. The destructive effects of corrosion are 1) Metal loses its efficiency and its useful properties get altered.
2) Maintenance cost and cost of material increases while production rate decreases.
3) The purity of products gets affected.
Because of these destructive effects of corrosion, it becomes essential to understand the
mechanism of corrosion, which is discussed in this module.
4.7.2.2 Typess of Corrosion
There are two types of corrosion depending on whether metal is in environment of dry air or moist
air. Some time metal can be immersed in liquid/solution partially or fully.
I] Dry / Direct Chemical Corrosion / Atmospheric corrosion.
II] Wet/ Immersed / Electrochemical Corrosion
Chemistry Sem-I/II
100
I] Dry / Direct Chemical Corrosion / Atmospheric corrosion.
This type of corrosion occurs due to direct attack of atmospheric gases like oxygen, carbon dioxide,
sulphur dioxide, halogen, hydrogen sulphide, nitrogen, hot flowing liquid or anhydrous inorganic
liquid with metal surfaces. The metal forms corresponding compounds such as oxides, carbonates,
sulphides, halides or sulphates.
Rate of this type of corrosion depends upon:
(i) Temperature (ii) Chemical affinity between metal and gas
(ii) Nature of oxide film on surface of metal, (iv) Moisture in air
It can be classified into:
a. Corrosion due to oxygen – Leading to formation of oxides.
b. Corrosion due to other gases – Leading to formation of other compounds.
c. Corrosion due to other corrosive liquids - caused by molten liquids
4.7.2.3 Corrosion due to oxygen (Oxidation corrosion):
Among the corroding gases Oxygen is largest in amount in atmosphere and it is present
everywhere. Oxygen in absence of moisture directly attack metal surface at low as well as high
temperature. At low temperature Na, K, Li, Be, Ca, Mg etc. are rapidly oxidized. At high temp all
metals except Ag, Au, Pt are oxidized.
After reacting with metal surface it will form the oxide film on metal surface. Usually more active
metal get corroded faster than less active metals. For example alkali metals and alkaline earth
metals get oxidized even at low temperatures as compared to other metals which are less active.
Mechanism of chemical reaction due to oxidation
2M
(Metal)
2M
n+
(Metal ion)
+
)
(Electrons
n/2 O2 + 2ne
Net reaction =
-
2ne
-
2M + n/2 O2
Metal + Oxygen
2-
nO
M2On
Metal Oxide
Module 4 : Electrochemistry and its Application (Corrosion)
101
The extent of corrosion depends upon the types of oxide films formed by metals. Fe, Cr, Al, Zn
forms there corresponding oxide such as Fe2O3, Cr2O3, Al2O3, ZnO by reacting with oxygen. The
oxide films formed are classified in to three categories depending on the nature of oxide film. The
nature of oxide film plays an important role in the further corrosion to continue or not. Oxide films
occurs in following forms.
(i) Stable (a) porous (b) non-porous.
(ii) Unstable
i)
(iii) Volatile.
Stable oxide film: It normally get adhered to the underlying parent metal surface and hence
provide a defensive layer on surface, thereby reducing rate of further corrosion. The extent of
further corrosion depends on its type.
a) Porous oxide layer: The name itself suggests it has pores or cracks. In such case
atmospheric oxygen (from air) has access to the underlying surface of metal, through the
pores or cracks of the layer. These is get clear by the example of alkali metals Li, Na, K
or alkaline earth metal as Ca and Mg etc. These metals after forming oxides such as
Na2O, K2O, MgO, etc. Where volume of oxide formed is less than, volume of reacted
metal. Thus, oxide layer is in less quantity and therefore it can not cover the entire
surface of metal block and leads to further destruction.
b. Non-porous oxide layer: The rate of corrosion reduces considerably and in most cases it
stops completely, if the nature of film is non porous in nature. Metals like Al, Cr etc.
forms their oxides such as Al2O3, Cr2O3 etc. whose volume of oxide film is greater than
the volume of metal. Hence, oxide film after its formation covers the entire metal surface
and not giving any chance for further attack of oxygen. Thus, on metal surface protective
layer of oxide is developed which becomes passive and stops further corrosion.
Chemistry Sem-I/II
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(ii) Unstable oxide film: They get decomposed on metal surfaces back to metal & oxygen.
Consequently, oxidation corrosion is not possible in such a case. Thus Pt, Au and Ag don’t go
under oxidation corrosion.
Metal Oxide
Metal + Oxygen
(iii) Volatile oxide film: They are vaporized from metal surfaces as soon as they are formed,
leaving behind the underlying metal surfaces for further attack of oxygen. For e.g. The film formed
by Molybdenum metal is a volatile film. Thus, in these cases corrosion continues till the metal is
available.
Metal + Oxygen
Metal Oxide
Module 4 : Electrochemistry and its Application (Corrosion)
103
Corrosion due to other gases:
Gases like CO2, H2S, Cl2, SO2, F2 etc. react on the metal surface resulting in formation of a film
corresponding compound. The extent of corrosion is mainly governed by chemical affinity of metal
and attacking gases. The rate of further corrosion depend on the volume of the film of corrosion
product and the volume of the metal. In case of attack of other gases e.g. Cl2, the nature of product
differs from metal to metal. E.g. Cl2 react with silver to give highly protective film of silver chloride
2Ag + Cl2
2AgCl (Protective)
Sn + 2Cl2
SnCl4 (Volatile)
whereas same gas reacts with tin to form corresponding chloride which is volatile in nature. Here,
the film of silver chloride is protective (non porous), hence rate of further corrosion reduces, while
that of stannic chloride is volatile in nature, hence rate of corrosion is enhanced. Thus it is obvious
that the corrosion of silver due to attack of chlorine stops on surface of metal whereas that of tin
continues till the tin metal gets completely vanished.
(b) Hydrogen embrittlement: It is caused by gases like hydrogen & hydrogen sulphide. At high
temp hydrogen undergoes dissociation to give nascent hydrogen. In atomic state, hydrogen
H2

⎯⎯
→ H +H
is highly reactive. It diffuses into metals & gets collected into cavities present in the metal. In
cavities, they combine to form hydrogen molecule.
H + H → H2
When pressure of hydrogen increases beyond the limit, it causes blisters & cracks metal making it
weak.
Atomic hydrogen diffused in metal combine with C, S, O, N which are present as impurities in
metals.
e. g.
C + 4H →
C H4
Fe + H 2 S → FeS + 2 H .
Let’s check the take away from this lecture
1)
Corrosion is –
Chemistry Sem-I/II
104
a)
deterioration of a metal due to its reaction with the environment.
b)
destruction of a metal due to unwanted reactions
c)
an electrochemical process in which there is flow of current between anode and cathode.
d)
all of the above.
2)
Galvanic series comprises of –
a)
both metals and non metals
c)
both metals and alloys
b)
metals , non metals and alloys
d)
none of the above
3)
Metals react with oxygen present in the atmosphere forming-
a)
oxides
c)
sulphates
b)
sulphides
d)
nitrates
Exercise
Q.1
Define corrosion. Explain losses due to corrosion. Give the classification of corrosion
according to environment of surrounding.
Q.2
What is dry corrosion? Explain with the example how nature of oxidized product affects the
rate of corrosion.
Questions/Problems for practice:
Q.3
Why silver, gold and platinum do not undergo oxidation corrosion?
Learning from this lecture: Learners will be able to state various gases which causes the corrosion.
Lecture: 24
4.7.3 Electrochemical corrosion (Immersed / Wet corrosion)
Learning objective:
Module 4 : Electrochemistry and its Application (Corrosion)
105
In this lecture learners will able to understand the mechanism of corrosion.
4.7.3.1 Electrochemical corrosion:
An electrochemical reaction is the one which results in the transfer of electrons. It is observed that
when two different metals are present in contact with each other through a common liquid, one of
the metals undergo corrosion which can be explained by electrochemical theory. Metal having
higher oxidation potential acts as anode & with lower oxidation potential acts as cathode. Anode
undergoes corrosion and cathode is protected. It is called electrochemical corrosion. Wet corrosion
is more common than dry corrosion.
Theory: This types of corrosion occur under following conditions, (i) Where conducting liquid is in
contact with metal, or (ii) Two dissimilar metal or alloys are either immersed or dipped partially in
the solution.
These types of conditions are always suitable for formation of separate anodic and cathodic areas
between which electric current can flow through the conducting solution.
At anode, the liberation of electron takes place, i.e. Oxidation reaction. Consequently at anode, the
destruction of metal occurs, either by dissolving it as metal ions or by converting it into a
compound such as its oxide.
At cathode, the absorption of electron takes place. i.e. reduction reaction. These reactions do not
affect the part of metal acting as a cathode, since most of metal cannot be further reduced. Thus
corrosion occurs always at anode.
Formation of metal ion (Mn+) at anode causes dissolution of metal and free electrons are liberated.
These free electrons are consumed in cathodic reactions with either by—
a) With evolution of hydrogen.
b) With absorption of oxygen.
a] With evolution of hydrogen:
This type of mechanism is generally observed when environment around metal is acidic. As
environment is acidic there can be availability of H+ ion around metal surface. In these cases, final
product of reaction will be hydrogen. It means there will be evolution of hydrogen. This type of
mechanism is observed in situations where there will be transport of corrosive liquid through
metallic pipe. If the environment is acidic there will be availability of H+ in large quantity. Generally
metal is having tendency to lose electron (oxidation), therefore there is large anodic part in this
mechanism (Fig). On the same metal surface, there can be some parts (smaller) where this reaction
is not taking place; it can be either because of impurity or any other reason. In smaller part as no
reaction (oxidation) taking place, therefore there will be less charge density compare to other parts
(larger). Whatever electrons are generated at larger part, they all will move towards the smaller
Chemistry Sem-I/II
106
part. This small area where electrons are accumulating it will act as small cathodic area. Around this
smaller part there is availability of H+ ions (acidic environment), these H+ ions will also attract to
this smaller part. So at smaller part electrons will absorb by the hydrogen and get converted to H2.
H+ H+
2 H + + 2 e - H 2( g )
Fe
Fe2+ + 2e-
Small cathodic
area
Electrochemical Reactions:
At anode,
Fe → Fe2+ + 2e−
(Oxidation / Corrosion) At Anode,
2 H + + 2e − → H 2  (Reduction reaction at Cathode)
Overall reaction,
Fe + 2 H + → Fe2+ + H 2 
i.
Iron tank acts as anode undergoes corrosion as Fe atoms pass into acidic solution as Fe ++
ions.
ii.
Free electrons which are accumulated at cathode are taken up by hydrogen ions to from H 2
gas. All the metals above H 2 in the electro chemical series get dissolved in acidic solution
with simultaneous evolution of hydrogen.
In this type of corrosion anode has a large area and cathode has a smaller area.
b] Absorption of oxygen:
This type of corrosion is observed when surrounding environment is alkaline or neutral. In this type
of corrosion oxygen is adsorbed by the system, initially it will get converted in to oxide then finally
to hydroxide. Normally metal have tendency to lose electron to get converted to cations. For e.g.
Fe
→ Fe2+ + 2e−
Module 4 : Electrochemistry and its Application (Corrosion)
107
The electron which are released in above reaction it will absorb O2 and get converted to oxides.
Suppose, if we have taken any metal, if it is exposed to atmosphere, oxide layer will form on it. If
this oxide layer gets broken, because of vibration or any other defects, cracks can be developed. The
cracked area (smaller part) will behave as anode. In anodic area metal will get oxidise and release
electrons. As this part is smaller, therefore there will be smaller anodic area and the remaining area
(total metal surface rather than crack) will acts as the cathode. Electron will flow from anode to
cathode through iron metal and these electron are taken up by the dissolved oxygen in the presence
of water and as a result the reaction taking place at cathode is
½ O2 + H2O + 2e
-
-
2OH
The ferrous ions at the anode and hydroxyl ion at the cathode diffuse towards the anode and then
they combine to form ferrous hydroxide.
At anode,
2+
-
Fe + 2OH
Fe(OH)2
(Brown rust)
In the presence of sufficient amount of dissolved oxygen, ferrous hydroxide is further oxidized to
ferric hydroxide.
4 Fe(OH)2 + O2 + 2H2O
4Fe(OH)3
Ferric hydroxide is almost insoluble in water and precipitates as yellow rust having composition
Fe2O3.H2O and forms a protective layer which decreases corrosion rate.
1
O2 + 2e- +
2
H2O
e
small
2OH
oxide film
e
cathode
large
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(c) Liquid metal corrosion: It is due to chemical action of moving liquid over solid metal surface. It
involves : (i) The slow dissolution of solid metal into flowing liquid metal at high temperature.
(ii) The penetration of liquid metal into solid metal by capillary action.
Comparison:
Chemical corrosion
(Atmospheric corrosion / Dry)
Electrochemical corrosion
(Immersed/Wet corrosion )
It occurs in dry condition.
It occurs in presence of aqueous solution or
electrolytes.
The direct chemical attack of the metal by It occurs through a large number of galvanic cells
environment.
It can be explained by absorption mechanism
It can be explained by electrochemical reaction
Corrosion products accumulate at the same Corrosion products generally accumulate at the
spot where corrosion starts. Hence, further cathodic area. Hence, further corrosion occurred
corrosion is prevented and it is a slow process.
and it is rapid process.
It occurs on both
heterogeneous surfaces.
homogeneous
and It occurs in heterogeneous surfaces.
Corrosion is uniform.
It is not uniform.
Rate of corrosion is governed by nature of Rate of corrosion is governed by cathodic product
corrosion product
Let’s check the take away from this lecture
1) The rate of corrosion is high at –
a) high temperature and high O2 availability
c) low temperature and high O2 availability
b) high temperature and low O2 availability
d) low temperature and low O2 availability
2) Electrochemical series comprises of –
a)
both metals and non metals
c)
both metals and alloys
b)
metals , non metals and alloys
d)
none of the above
Exercise:
Q.1
Explain wet corrosion in acidic medium with a schematic diagram and mechanism.
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Questions/problems for practice:
Q. 2
Explain wet corrosion in neutral medium with a schematic diagram and mechanism.
Learning from this lecture:
Learners will able to understand the mechanism of corrosion with the help of different theories of
corrosion. They will also able to learn how corrosion is takes place on a piece of metal if it is
subjected to dry atmospheric condition or wet atmospheric condition.
Lecture : 25
4.7.4 Factors affecting the rate of corrosion
Learning objective:
In this lecture, learners will able to understand the factors affecting on rate of corrosion.
There are two things involved in corrosion, metal and environment and hence we need to discuss factors
related to metal as well as environment which affects the rate of corrosion.
1) Factors Related to metal:
i) Nature of metal: Nature of metal depends upon position in galvanic series and potential
difference.
a) Position of metal in galvanic series: This is major factor of corrosion of metals. If two dissimilar
metals are in corroding environment, the metal having higher electrode potential and position in
the galvanic series undergoes corrosion,. i.e. It acts as anode.
b) Potential difference: Amongst the two metals in contact, greater the difference in electrode
potential, higher is the rate of corrosion.
ii) Oxidation potential: The extent of corrosion depends upon position of metal in electrochemical
& galvanic series. The rate of corrosion depends upon difference in the oxidation potential
(potential developed between the electrode and its ions in the solution due to oxidation reaction at
equilibrium at given temperature) & greater the difference in oxidation potential, faster is the
corrosion of anodic metal.
iii) H2 Over voltage: All metals above H 2 in electrochemical series(the series developed by
arranging the electrode potentials) liberate H 2 when immersed in acidic solution some metals
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above hydrogen do not liberate H 2 from acid solution. It can be explained by hydrogen over
voltage.
Over potential of H 2 is difference between the potential of electrode at which evolution of H 2 gas
is observed & theoretical value of potential at which H 2 gas evolution takes place.
iv) Relative areas of anode & cathode: If two dissimilar metal are in contact, one forming anode
and while another forming cathode, then the corrosion of anodic metal is directly proportional to
the ratio of area occupied by the cathode and anode.
Therefore, Corrosion at anode = Area of cathodic part/ Area of anodic part
If anodic area is very small as compared to cathodic area, corrosion occurs. The reason is current
density at a smaller anodic area is much greater and the demand for electron by the cathodic area is
more.
v) Purity of metal: If metals are impure, then impurities present in them cause heterogeneity which
gives rise to small electro chemical cells at the sites where metal & impurities are exposed to
environment, & thus corrosion starts affecting the entire metal. Example: Zinc metal, if with the
impurities of Fe or Pb, undergoes corrosion at the sites where the impurities are exposed because of
the formation of local small electrochemical cell. Thus the more the percentage of impurity, higher
is the corrosion of zinc metal.
vi) Physical state of metal: Small is the grain size of metal, greater will be the rate of corrosion. It is
also influenced by the orientation of the crystals at the metal surface. Even the pure metal, the areas
under stress tend to be anodic on which corrosion takes place.
vii) Nature of oxide film: If a protective, non-porous oxide film formed is strongly adhered to a
metal surface then it will protect a metal surface. But if the film is porous, loosely adhered, then
oxygen will diffuse through it and bring about further corrosion.
viii) Solubility of corrosion products: If product formed is soluble in the corroding medium, the
rate of corrosion will be faster. If corrosion product forms an insoluble, dense & adherent layer on
metal surface then rate of corrosion decreases.
2) Factors related to environment:
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i) Temperature: The rate of the reaction is accelerated with rise in temperature. Rate of dry
corrosion increases with increase in temperature but rate of wet corrosion is increased slightly with
rise in temperature.
ii) Presence of moisture: Corrosion increases rapidly in presence of moisture. It is due to gases like
O2 , CO2 , SO2 etc. which get dissolved in presence of moisture giving conducting solution certain
metals like Al, Zn, Mn, Cr, Fe may be corroded in presence of moisture even in absence of oxygen.
In presence of moisture, rate of corrosion increases due to presence of chemically active substances
in atmosphere which increase conductivity.
iii) Effect of pH of moisture: Acidic mediums are more corrosive than neutral or alkaline mediums.
All the metals have a particular pH value at which it has highest corrosion resistance, below &
above that value it corrodes faster.
iv) Nature of ions present in the medium: Anions like silicates inhibit corrosion by formation of
insoluble reaction products. Anions like chloride ions , destroy protective and passive film surface
& expose metal surface for further corrosion.
v) Conductance of corrosion medium : Corrosion of underground structures depend on conducting
medium. Conductance of dry, sandy soils is lesser & therefore if leads to slow corrosion of
underground pipelines. Conductance of clay and mineralized soils leads to higher corrosion.
Let’s check the take away from this lecture
1)
Corrosion at anode is directly proportional to –
i) Cathodic area
area
2)
ii) anodic area
iii) anodic area/cathodic area
iv) anodic area/cathodic
Which of the following factor affect the rate of corrosioni) Nature of metal
ii) Temperature
iii) pH
iv) all of the above
Exercise:
Q.1
Explain how are the following factors influence the rate of corrosion
(1) Position of metal in galvanic series. (2) Relative areas of anode and cathode.
Questions/problems for practice:
Q.2
Explain the effect of pH, Purity of metal and nature of corrosion product on rate of corrosion.
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Learning from the lecture: Learners will able to know what are the different factors affecting on the
rate of corrosion which will act as prerequisite
Lecture : 26
4.7.5 Types of Corrosion – Galvanic and Concentration cell corrosion
Learning objective:
In this lecture learners will able to know the different types of corrosion.
Different Types of Corrosion
A) Galvanic or Bimetallic Corrosion: [Based on R1,15th Edition, Ch 7, pg 334-335]
It is wet type corrosion. When two dissimilar metals are in electrical contact with each other and are
exposed to an electrolyte, a potential difference is created between two dissimilar metals. This
potential difference produces electron flow between them. The less noble metal will dissolve and act
as anode while more noble metal will act as cathode. This type of corrosion is called galvanic
corrosion.
In the above figure two dissimilar metals, zinc and copper plates are in electrical contact with each
other and are immersed into a solution of an electrolyte. The more electropositive metal zinc placed
higher in electrochemical series acts as anode and is attacked and gets dissolved. Copper lower in
electrochemical series acts as cathode. Zinc undergoes oxidation releasing electrons and forms
metals ions. These released electrons are taken up by ions causing reduction at the cathode. The emf
is generated due to oxidation reduction reaction causing dissolution of metal at anode causing
corrosion. The flow of electrons will be from zinc to copper. The corrosion will be more if the emf
generated is large.
The reaction at the anode will be : Zn →Zn2+ + 2e-
In the galvanic corrosion, cathodic metal is always protected from the corrosion attack. Then extent
of corrosion depends on corrosive environment as well as the difference in the electrode potential of
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two contacting metals i.e. their position in the galvanic series. Further apart the metal from each
other more is the intensity of corrosion. In acidic medium the corrosion takes place by the hydrogen
evolution process while in neutral or slightly alkaline solution corrosion takes place by oxygen
absorption method. Galvanic corrosion depends on the relative areas of anode and cathode. The
electron current flows from the anodic metal zinc which undergoes corrosion to the cathodic metal
copper. Thus corrosion occurs at anodic metal while the cathodic part is protected.
The various examples of galvanic corrosion are
1)
Lead-antimony solder around copper wire (copper being lower in series act as cathode)
2)
Steel screws in marine hardware, made of brass (iron higher up than brass in electrochemical
series gets corroded)
3)
Steel pipe connected to copper plumbing (iron in steel higher in electrochemical series
becomes anode and gets corroded)
B) Concentration Cell Corrosion: [Based on R1,15th Edition, Ch 7, pg 335-336]
This type of corrosion is due to electrochemical attack on the metal surface, exposed to varying
aeration or varying electrolyte concentration. Thus concentration cell corrosion occurs either due to
varying ionic concentration of electrolyte in contact with a metal or due to difference in aeration of
air or oxygen over the metal surface. Due to this a potential difference is developed between the two
areas resulting in corrosion. This may be the result of local difference in metal-ion concentration,
caused by local temperature differences or inadequate agitation or slow diffusion of metal ions
produced by corrosion.
Differential aeration corrosion is the most common and important type of concentration cell
corrosion. This type of corrosion occurs when one part of the metal is exposed to a different air
concentration from the other part of the metal. This develops a difference in potential between
differently aerated areas. It has been found experimentally that poor oxygenated parts act as anode
and highly oxygenated areas act as cathode. Thus, a flow of electron from anode to cathode takes
place due to differential aeration and is called differential current.
Differential aeration corrosion occurs when metals are partially immersed in a solution just below
the waterline. A metal (zinc) is partially immersed in a dilute solution of a neutral salt (say NaCl)
and the solution is not agitated properly. The parts above and closely adjacent to the waterline are
strongly aerated and act as cathode. The parts immersed to greater depth show a smaller oxygen
concentration i.e. poorly aerated and act as anode.
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Differential aeration corrosion of a metal immersed partially in solution
Hence, a difference of potential is developed which causes flow of current between the two
differentially aerated areas of the same metal.
Zinc will dissolve at the anodic areas.
Zn → Zn2+ + 2e- (oxidation)
Oxygen will take up electrons at the cathodic areas to form hydroxyl ions.
½ O2 + H2O + 2e- → 2 OH- (reduction)
Circuit gets completed by flow of OH- ions through the electrolyte and flow of electrons from anode
to cathode through metal.
In a similar way, corrosion of iron by water drops (or salt solution) can be easily explained. Areas
covered by droplets, having no access of oxygen become anodic with respect to the other areas,
which are freely exposed to air.
From the above it is clear that oxygen concentration cell increases corrosion but it occurs where the
oxygen concentration is lower.
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Formation of an oxygen concentration cell on metal under a drop of water/ salt solution.
Let’s check the take away from this lecture
1) Metals acting as anode are –
a) the ones placed higher in the galvanic series
c)
both of the above
b) the ones placed lower in the galvanic series
d)
none of the above
a) the ones placed higher in the galvanic series
c)
both of the above
b) the ones placed lower in the galvanic series
d)
none of the above
2) Metals acting as cathode are –
Exercise:
Q.1
Explain concentration cell corrosion with the help of a suitable example.
Q.2
Explain bimetallic corrosion.
Questions/problems for practice:
Q. 3
Explain differential aeration corrosion with the help of suitable example.
Learning from the lecture: In this lecture learners will be able to understand differential cell corrosion
and bimetallic corrosion.
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Lecture : 27
4.7.6: Types of corrosion Intergrannular, stress and pitting corrosion
Learning objective: In this lecture learners will able to interpret corrosion type on any specific object.
C) Intergranular corrosion :
This type of corrosion occurs along grain boundaries under following conditions, a) The material is
highly sensitive to corrosive attack and, b) corrosive liquid possesses a selective character of
attacking only at the grain boundaries, but leaving the grain centers (interiors) untouched or only
slightly attacked. This type of corrosion is due to the fact that the grain boundaries contain material,
which shows electrode potential, more anodic than that of grain center in the particular corroding
medium. This may be due to precipitation of certain compounds at grain boundaries, Due to such
disproportioned precipitation; a solid solution is aggregated at center, adjacent to boundary. The
solid solution thus formed is anodic with respect to the grain centers as well as to the precipitated
compound. Hence it gets attacked preferentially in the corrosive environment.
As grain boundaries contain material which has high electrode potential it becomes anode and
grain center becomes cathode & corrosion occurs at grain.
Intergranular corrosion is encountered in alloys causing loss of strength of the alloy &
disintegration at grain boundaries. e. g. stainless steel ( 18 % Cr, 8 % Ni ) containing more than 0.1 %
carbon when heated for welding & cooled slowly, accelerates the reaction between carbon &
chromium. To form chromium carbide at grain boundaries & results in the formation of galvanic
cells. Precipitation of chromium of chromium carbide at grain boundaries reduces concentration of
chromium at grain boundary area & will become anodic. Grain center being rich in chromium act as
cathodic & corrosion takes place as grain boundaries.
D) Stress Corrosion (Stress Cracking):
It is similar to intergranular corrosion. Stress corrosion is the combined effect of tensile stresses &
corrosive environment on a metal. It is characterized by highly localized attack which occurs even
when overall corrosion is negligible. For stress corrosion to occur the essential conditions are: a)
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Presence of tensile stress and specific corrosive environment. The specific substances acts as
corrosive agents and are selective for particular material in particular environment. This type of
corrosion is observed in fabricated articles of certain alloys (like high zinc brasses and nickel
brasses) due to the presence of stresses caused by heavy working like rolling, drawing or
insufficient annealing. On a metal stress results from poor design, riveting, residual cold working,
welding, bending, pressing etc. The part of metal under stress becomes anodic with respect to
unstressed portion and undergoes corrosion in a corrosive environment. The pure metals are
relatively immune to stress corrosion. Stress corrosion involves in a localized electrochemical
corrosion, forming anodic areas with respect to the more cathodic areas at the metal surface.
Presence of stress produces strain, which results in localized zones of higher electrode potential.
These chemically active sites are attacked, even by mild corrosive environment. This results in the
formation of crack, which grows and propagates in the plant. The attack of corrosion continues till
failure of structure occurs.
Well known examples of stress corrosion are
1. ‘Season cracking’ occurs in sold drawn brasses especially in presence of moisture & traces of
ammonia or amines. This tendency can be eliminated by a low temperature annealing.
2. Caustic embitterment occurs in mild steel when exposed to alkaline solution at high temp.
Specific corrosive environments for different metals.
Metal / alloy
Corrosive environment.
(1)
Mild steel
Soln. of NaOH
(2)
Low Carbon steel
Soln. of nitrates of Ca & Na. Strong
(3)
Nickel
Soln. of NaOH & KOH.
(4)
Brass
Soln. of Ammonia vapour & ammonical solutions.
Stress corrosion can be minimized by :
a) Suitable treatment to reduce internal stresses.
b) Suitable heat treatment to prevent heterogeneity.
c) Removal of critical environment.
d) Selecting a better resistant material.
E) Pitting Corrosion:
It results when small particles of dust, dirt or scale are deposited on metal surfaces. The portion
covered by dust is not aerated The concentration cells are set up in which covered portion acts as
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anode and exposed surface acts as cathode In presence of moisture containing dissolved oxygen,
corrosion starts beneath the dust forming area & forms a small shallow depression in metal surface
called pits. After pit is formed, rate of corrosion increases as anodic area is small compared to a
large cathodic area. Cathodic area induces small anodic area to corrode faster by electrons released
from anodic area resulting in the enlargement of pit.
Pitting corrosion occurs because of
a) Metal surfaces being not homogeneous (surface roughness)
b) Scratches due to friction or sliding under heavy load.
c) Corrosion products are insoluble.
d) Films are not uniformly perfect.
e) Severe action of some chemicals or solutions.
This corrosion is highly dangerous & causes damage and destruction of pipes, tubes and reaction
vessels. It can be avoided by use of pure & homogeneous metal with a highly polished metal
surface.
Pitting corrosion can be controlled or minimized by a) Finishing metal surfaces properly. b) By
maintaining metal coating intact. c) by using lubricants to control friction. d) By avoiding unwanted
contact with corroding solutions/dust/ impurities.
untouched or only slightly attacked.
Let’s check the take away from this lecture
1) For intergrannular corrosion to occur, criteria is a) The material is highly sensitive to corrosive attack
c)
both of the above
b) corrosive liquid of attacking only at the grain boundaries
d)
none of the above
d)
both b and c
2) Pitting corrosion is which type of corrosion –
a) Dry
b)
Wet
c) Electrochemical
Exercise:
Q.1
Explain the term Inter-granular corrosion.
Q.2
Explain stress corrosion.
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Questions/problems for practice:
Q. 3
Write short note on pitting corrosion and intergrannular corrosion
Learning from the above lecture: Learners will able to know various other types of corrosion such as
stress, intergrannular and pitting corrosion.
Lecture : 28
4.7.8: Methods to decrease rate of Corrosion, Selection of Materials and Design
Learning objective:
In this lecture learners will able to understand various methods used for decreasing the rate of
corrosion, specifically selection of materials and Design
Methods to decrease rate of Corrosion:
Corrosion is one of the major reasons of increasing maintenance cost in most of the industries,
therefore either prevention or subsequent control of corrosion becomes more important. The
methods of corrosion control vary from condition to condition depending on the type of corrosion.
Therefore it makes very difficult to apply any single method to solve the problem. The methods to
prevent the corrosion are as follows:
1) Material Selection & Design
2) Cathodic protection i) Sacrificial anodic protection, ii) Impressed current method
3) Anodic protection method
4) Application of protective coatings
a) Metallic coatings.
b) Nonmetallic coatings.
1) Material Selection and Design:
a] Right type of material is necessary.
b] Though noble elements are more resistant to corrosion, can’t be used on account of its cost. High
purity metals may reduce corrosion but its corrosion resistance depends on nature of corrosive
environment.  Alloying with suitable elements can increase corrosion resistance of metals &
their strength.
c] Proper heat treatment like annealing reduces internal stresses & reduce stress corrosion.
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d] To prevent galvanic corrosion contact of two dissimilar metals should be avoided. If their use is
essential then two metals should be selected that their electrode potential are as close as possible
to each other. Anodic material should be larger than cathodic area.
e] Metals used should not come in contact with porous material. Otherwise porous material
absorbs moisture & holds liquid which act as electrolytic conductors.
f] Sharp bends, corners & recesses should be avoided by use of joints & weldings. Welded joints
are preferred over riveted or bolted lap joints to prevent stress corrosion. Crevices and
obstructions should be avoided in tanks and pipelines to avoid formation of concentration cell.
The equipment should be kept free from dust and dirt. It should be supported on lags instead of
on large blocks to allow free circulation of air & prevent formation of stagnant pool or damp
areas.
Using Pure Metal
If the metal used to manufacture machine parts is 100% pure, the corrosion resistance is more as
compared to impure metal. This is because, impurity cause heterogeneity. Thus use of pure metal or
purifying the metals prevents the corrosion. This method is useful only if corrosion proceeds by
electrochemical mechanism.
Using Metal Alloys
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Corrosion resistance of metal can be improved by alloying the metal, if it gives the homogenous
product. Example: iron can be alloyed with chromium and carbon to give steel, which has better
corrosion resistance. Here, chromium metal has a tendency to form an oxide which gets adhered to
the surface. Also, this oxide film if broken gets healed immediately. Thus steel containing 13%
chromium are normally used to make cutlery, surgical instruments, springs, etc. while higher
percentage of chromium such as up to 25% are used to prepare turbine brackets, heat resisting parts
etc. Brass is another example. Individually, copper or zinc are less resistant to corrosion, but alloy
of these two metals possesses better resistance to the corrosion.
Let’s check the take away from this lecture
1) From the following which one of the method is not used for corrosion control?
i) Using impure metal
ii) using pure metal iii) using metal alloys
iv) none of the above
Exercise:
Q.1
How proper designing of metal article can minimize the corrosion?
Questions/problems for practice:
Q. 2 Explain the following methods of corrosion prevention. I) Using pure metals ii) Using metal
alloys.
Learning from the lecture:
Learners will able to explain the different methods of corrosion control. They will also learn that
which type of material should be selected to avoid corrosion.
Lecture : 29
4.7.9: Cathodic protection
Learning objective:
In this lecture learners will able to understand how cathodic protection works for the preventing
corrosion.
Cathodic protection
It is a method of protecting metallic object from corrosion by making them completely cathodic and
no area of metallic object is allowed to act as anode. It is achieved by supplying electrons to the
metal structure to be protected.
There are two types of cathode protection.
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a) By using galvanic or sacrificial anode or Auxiliary anode method
b) By Using impressed current method
a) By using galvanic or sacrificial anode: In this method, more active metals and their alloys is
connected to metal structure to be protected. These results in the corrosion of the piece of
metal connected, thereby saving the base metal. Since the more active metal sacrifices itself,
by undergoing corrosion and saving the base metal, the method is named as sacrificial
anode or auxiliary anode method. More active metal supply electrons to the metal structure
by dissolution of sacrificial anode. When this piece of more active metal gets corroded
completely, it can be simply replaced by fresh new piece. The metals normally used are Mg,
Zn or Al. The method is used to protect buried steel pipelines, industrial water tanks which
are normally embedded under the soil. Applications of this method are seen to protect
cables or iron pipelines, by connecting them to Mg-Blocks; and in case of marine structures,
ships are protected by using Zn-plates as sacrificial anode. Even boiler, water tanks are
protected by using Zn metal. It requires low installation cost, minimum maintenance & short
term protection.
Cathodic protection using sacrificial anode.
b) By Using impressed current method
In this method, a current is applied in the opposite direction to that of corrosion current; thereby
nullifying the effect of later one on base metal, i.e. converting the base metal, to cathode from an
anode. Such an impressed current can be obtained by using dc source such as battery or dry
cells along with an insoluble anode such as platinum, stainless steel, graphite etc. The anode is
normally embedded underground, To this with the help of dc source the impressed current is
applied, and whole of this assembly is connected to the metallic structure to be protected. The
connections are carried out by using wires. The insoluble anodes are kept inside the back-fill
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which is made up of gypsum or any such material, which can help in increasing the electrode
contact with the soil. Single anode can be used, if the area of metallic structure to be protected is
small, or many anodes connected in series can be used if the area of metallic structure to be
protected is wider, i.e. Long pipeline. Disadvantage of this method is that, due to the application
of impressed current, the anode deteriorates and hence, it has to be replaced from time to time.
The method is used for protecting marine structures like tanks, pipelines condensers etc. This
protection is suited for large structures & for long-term operations
Fig. Cathodic protection using impressed current.
Let’s check the take away from this lecture
1)
Cathodic protection is applicable to -
a)
Akali metals
c)
alkaline earth metals
b)
transition metals
d)
all metals
2) Cathodic protection can be achieved by –
a)
use of sacrificial anode
c) use of impressed current method
b)
use of auxillary anode method
d) all of the above
Exercise
Q.1
What is cathodic protection? Discuss Sacrificial anode method for the corrosion control.
Chemistry Sem-I/II
Q.2
124
Write a short note on cathodic protection by impressed current method.
Questions/Problems for practice:
Q.3
With a neat labeled diagram explain sacrificial anode method to control corrosion..
Learning from this lecture ‘cathodic protection’:
Learners will able to know the methods of cathodic protection using which corrosion can be
prevented for metallic pipes. They will also able to know that how cathodic protection plays the role
for inhibiting the corrosion.
Lecture : 30
4.7.10 :Anodic protection:
Learning Objective:
In this lecture, learners will able to illustrate the mechanism of anodic coatings.
Anodic protection:
Metals and alloys, which become passive can’t be offered cathodic protection. The corrosion rate
can be slowed by use of anodic current. The principle of anodic protection is the formation of a
protection oxide film, which occurs on application of an appropriate strength of anodic current on a
metal specimen in a suitable oxidizing environment. The metal gets passivated & reduces the
dissolution of metals.
Anodic protection system for tanks
Potentiostat is a device to maintain the metal at constant potential w. r. t. reference electrodes. A
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Potentiostat has three terminals, one connected to the structure that is to be protected, the other to
the auxiliary platinum electrode and third to reference calomel electrode. The metal specimen to be
protected is kept in a oxidizing atmosphere & acts as anode Potentiostat maintains a constant
potential between specimen & reference electrode.
This method requires small current & operation cost is low. The method is applicable in extremely
corrosive environments.
Comparison:
Cathodic protection
Anodic protection
This method is applicable to all metals.
Useful for weak
environments.
or
moderate
The method applicable to only those
metals, which show active-passive
behaviors.
corrosive More corrosive environments can be
handled.
Installation cost is lower.
Installation cost is higher.
Operating cost is higher.
Operating cost is lower.
Standard & well established method.
Better corrosion protection can
achieved with fewer electrodes.
be
Applied current can’t give any indication of Applied current gives an indication of
corrosion.
corrosion rate.
Feasibility cannot be predicted by laboratory Feasibility and design can be predicted
experiments.
by laboratory experiments.
It can provide steady, consistent protection for It may not provide steady protection;
long duration.
but if system goes out of control, rate of
corrosion increase suddenly.
Let’s check the take away from this lecture
1)
By anodic protection is more suitable for which kind of environment?
i) Less corrosive
ii) more corrosive
iii) both i) and ii)
2)
iv)none of the above
Anodic protection is possible fori) All metals
ii) all non-metals
iii) Metal showing active passive character
iv) none of the above
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Exercise
Q.1
Write a short note on anodic protection.
Questions/Problems for practice:
Q.2
Distinguish between anodic and cathodic coatings.
Learning from the Lecture ‘Anodic protection’ :
Student will be able to learn the different methods available for the control of corrosion. They will
also able to know the effective use of electrochemical series for the application of corrosion control
by anodic and cathodic protection.
Lecture : 31
4.7.11: Metallic coatings, hot dipping, galvanizing, tinning, metal cladding, metal spraying.
Learning objective:
In this lecture learners will able to know various methods of application of metallic coatings for
prevention of corrosion.
Applications of protective coatings
This method is most easy and effective of all the methods discussed earlier in the chapter, for the
simple reasons, that it can provide a continuous barrier between the metal surface and the corrosive
environment. The problem of localized corrosion starts only when the protective coat gets broken,
due to friction or by any other wear and tear of the surface.
There are mainly two types of protective coatings a) Metallic Coatings, b) Non metallic coatings
Metallic coatings: [Based on R1,15th Edition, Ch 7, pg 349-352]
They are used for the prevention of corrosion of metals. Metallic coatings can be divided into
A) Anodic coatings (coating with more active metal): If the active metals whose electrode
potentials are higher than that of the base metal form anodic coatings. e.g. zinc, aluminum
cadmium on iron or steel. If any portions of coatings are broken, a galvanic cell is formed. In
case of galvanized steel, zinc being anodic to iron will undergo corrosion; iron being cathodic
will be protected. Anodic coatings will be good under all conditions & protect base metal from
corrosion irrespective of whether coating is porous or non-porous. Only drawback is they lack
luster and reflectivity.
Module 4 : Electrochemistry and its Application (Corrosion)
127
B) Cathodic coatings: More noble metal having lower electrode potential than base metal is used.
They protect as they have higher corrosion resistance than base metal. If coatings are punctured,
coating metal becomes cathode and exposed base metal acts as an anode.
Therefore a galvanic cell is set up and localised corrosion occurs at exposed part Anodic coating
takes the care of corrosion and cathodic coating gives better luster & reflectivity.
Anodic coating
Cathodic coating
Coating metal is higher placed in EMF series Coating metal is lower placed in EMF series
than base metal.
than base metal.
Base metal remains safe as long as coating Base metal undergoes fast corrosion, if coating
metal present on base metal, if coating ruptured, to produce pits/cracks.
ruptured.
More preferred coating.
Coating is less preferred.
Methods of applying metallic coatings Following are the different methods used for applying metallic coatingsA) Hot dipping: It is used to apply coating of low melting metals like Zn, tin, cadmium, aluminium.
It includes 1) Pickling operation: to clean surface with acid.
2) Fluxing: to facilitate wetting.
3) Dipping: To dip the article in molten bath of controlled composition.
4) Wiping: To clean the dipped pieces to regulate thickness & uniformity of coatings.
i) Galvanizing: Coating of zinc on iron or steel is called galvanizing. It is applied on sheets rods,
wires, pipes, buckets, nails, tanks etc. Galvanized iron vessels can’t be used for storing foods due to
formation of poisonous products by action of food on zinc. In such cases tin coatings are preferred.
ii)Tinning: Coating of tin on iron or steel is called tinning. It is resistant against atmosphere. Being
nontoxic, it is used for coating of steel, copper and brass sheets used for manufacturing containers
for storing foodstuffs, ghee, oil etc. Tin coatings are also used for production of copper cables
against attack of sulphur present in rubber insulation.
Chemistry Sem-I/II
128
Comparison :
Tinning
Galvanizing
It is the process of covering iron with thin It is process of covering mild steel with thin
layer of zinc to prevent it from rusting.
layer of tin to prevent it from rusting.
Zn protects iron from rusting, as it is more Tin protects base metal from corrosion, as it is
electropositive than iron & does not allow iron less electropositive than iron, therefore it is
to pass into solution.
resistant to chemical attack.
It is kind of anodic coating
It is kind of cathodic coating.
Zn continues to protect metal by galvanic cell Tin protects metal till coating is perfect any
action, even if coating is broken.
break in coating cause rapid corrosion.
Galvanized ware can’t be used for storing Tin coated utensils can be used for storing
foodstuffs, which are acidic in nature as Zn food stuffs as it is more resistant to organic
dissolves in dilute acids forming poisonous Zn acids avoids food poisoning
compounds.
B) Metal cladding: In this process a dense and homogeneous layer of metal is permanently bonded
to base metal on one or both sides. The protective layer of covering metal is called cladding. It is
done by arranging thin sheets of coating metal over base metal sheet in the form of sandwich.
Cladding is done by (i) Hot rolling (ii) casting. The method is used in aircraft industry.
C) Metal spraying: In the molten state the coating metal is sprayed from a spraying gun on rough
surface of base metal. The advantages of this method are (I) greater speed of working (ii) Ease of
application (iii) applicability to large surfaces. The strength of adhesion coatings are lesser than that
of hot dipping or electroplating method
Let’s check the take away from this lecture
1) Galvanizing is the process of coating iron with a thin coat of –
a)
Zn
c)
Al
b)
P
d)
S
2) Tinning is the process of coating iron with a thin coat of –
Module 4 : Electrochemistry and its Application (Corrosion)
a)
Zn
c)
Al
b)
P
d)
Sn
129
Exercise
Q.1 What is metallic coating? Distinguish between anodic and cathodic coatings.
Q.2
What are the methods of Metallic coating? Describe the metal cladding with appropriate
diagram.
Questions/Problems for practice:
Q.3
Write a note on: Metallic coating. Explain galvanizing.
Learning from this lecture ‘Metallic coatings’:
Learners will able to know how metallic coatings are also effective for corrosion
4.8: Conclusion
The study of corrosion shows the impact of corrosion on the deterioration of a metal. Corrosion of
materials leading to the degradation of their physical properties is of great concern. It plays an
important role in energy production, transportation, biomedical engineering, water distribution and
sewage, electronics and nanotechnology. Remedial action based on a better and widespread
understanding of the corrosion phenomenon could reduce significantly the financial burden of
corrosion to the nation.
Add to Knowledge:
Key differences between erosion and corrosion: While these words appear to be similar sounding,
but they are totally different in terms of the many circumstances. Corrosion generally refers to the
destruction of materials through chemical reactions. As for erosion, it is basically the physical
phenomenon involving the movement of small fragments of rocks or the topsoil due to the
influence of natural forces like gravity, water, wind, etc.
Interestingly, one similar fact about corrosion and erosion is that they both happen due to certain
external actions on a surface. While these are very important natural processes that we see and hear
about in our everyday lives let us look at the distinct characteristics of each.
Chemistry Sem-I/II
130
Set of Questions for FA/CE/IA/ESE
Q. 1)
Explain wet corrosion in acidic and neutral medium with a schematic diagram and
mechanism.
Q. 2)
Define corrosion. Explain losses due to corrosion. Give the classification of corrosion
according to environment of surrounding.
Q. 3)
Discuss the various factors affecting the rate of corrosion.
Q. 4)
What is corrosion? Discuss the corrosion caused due to combination of metals of different
electrode potential.
Q. 5)
Discuss the concentration cell theory of corrosion.
Q. 6)
Explain concentration cell corrosion with the help of a suitable example.
Q. 7)
How do the following factors influence the corrosion rate?
(i) Position of metals in galvanic series.
(ii) Formation of oxygen concentration cell.
(iii) Temperature.
(iv) Relative areas of cathodic and anodic parts
Q. 8)
How are the following factors responsible for rate of corrosion:
a) Relative area of Anode & Cathode
b) pH
c) Over voltage
d) Nature of surface film.
e) Position of metal in galvanic series
Q. 9) Define corrosion. List the various type of corrosion.
Q.10) Gold does not corrode due to oxidation.
Q. 11) Define Corrosion. Explain Stress corrosion with appropriate diagram and example.
Q.12) Write a note on any two of the following factors
i) Passivity
iii) Effect of PH
Q.13) Explain the direct Chemical Corrosion.
Q.14) Discuss the corrosion due to combination of metals of different electrode potential.
Module 4 : Electrochemistry and its Application (Corrosion)
Q.15)
Why
silver,
gold
and
platinum
do
131
not
undergo
oxidation
corrosion?
Q. 16)
Explain differential aeration corrosion with the help of suitable example.
Q. 17)
What is electrochemical corrosion? With the help of suitable diagram and electrode
reactions, explain electrochemical mechanism of rusting of iron in neutral, aqueous
medium.
Q.18)
What is oxidation corrosion? Explain why pure aluminium metal exhibits good corrosion
resistance in atmospheric oxygen.
Q. 19)
What is dry corrosion? Explain with the example how nature of oxidized product affects
the rate of corrosion.
Q.20) Distinguish between anodic and cathodic protection.
Q. 21) Write a note on: Metallic coating. Explain galvanizing.
Q. 22) Distinguish between galvanizing and tinning.
Q. 23) What is metallic coating? Distinguish between anodic and cathodic coatings.
Q. 24) Explain why zinc coating gives better protection for iron than tin.
Q. 25) Explain the advantages of Galvanizing over Tinning.
Q.26) What is metal Cladding?
Q.27) With a neat labeled diagram explain sacrificial anode method to control corrosion.
Q.28) What are the methods of Metallic coating? Describe the metal cladding with appropriate
diagram.
Q.29) What is cathodic protection? Discuss Sacrificial anode method for the corrosion control.
Q.30) Give the Anode & cathode Reactions involved in the corrosion of steel pipe connected with
copper plumbing.
Q.31) What is cathodic protection? Describe impressed current method of corrosion control.
References:
1)
Engineering Chemistry By Jain & Jain.
2)
Engineering Chemistry By S.S. Dara.
3)
Engineering Chemistry By B. K. Sharma
Chemistry Sem-I/II
132
Self-assessment
Q.1)
Define corrosion. With the help of suitable diagram and electrode reactions, explain
electrochemical mechanism of rusting of iron in neutral, aqueous medium.
Q. 2)
What are the different types of electrochemical corrosion? Explain differential aeration
corrosion with neat labeled diagram.
Q. 3)
How corrosion of the metal is influenced by the pH of the medium and relative areas of
anode and cathode.
Q.4) What is oxidation corrosion? Explain with the example how nature of oxidized product affects
the rate of corrosion.
Q.5) Galvanized steel is not used for storing food stuffs. Explain?
Module 4 : Electrochemistry and its Application (Corrosion)
133
Self-evaluation
Name of
Student
Class
Roll No.
Subject
Module No.
S.No
Tick
Your choice
1.
2.
3.
4.
5.
Do you understand the various types of corrosion?
o
Yes
o
No
o
Yes
o
No
Do you know the various factors affecting on the
rate of corrosion?
o
Yes
o
No
Do you understand
contolled?
o
Yes
o
No
o
Yes, Completely.
o
Partialy.
o
No, Not at all.
Do you understand
corrosion?
the
concentration
how corrosion
Do you understand module ?
can
cell
be
Chemistry Sem – I /II
134
Module 5
Spectroscopic techniques and applications
Lecture : 32
5.1. Motivation:
This module will help the student to know the various spectroscopic techniques available for
structure elucidation of organic molecules. They will able to make right choice of techniques
for analysis of particular type of organic molecules.
5.2. Syllabus:
Lecture
No
Content
Duration
SelfStudy
32
Electromagnetic radiation, electromagnetic spectrum,
1 Lecture
2 hours
33
Absorption of Light, Interaction of electromagnetic
radiation with matter, Beer-Lambert’s law statement
1 Lecture
2 hours
34
Beer-Lambert’s law (mathematical
derivation and Numericals
35
Instrumentation for Colorimeter & UV Visible
Spectrophotometer (components of spectrophotometer)
36
Working of Single beam and double beam 1 Lectures
spectrophotometer and Application of UV Visible
Spectroscopy.
2 hours
37
Vibrational Spectroscopy: Introduction, The infrared
absorption process (Principle), Types of Molecular
absorption.
1 Lectures
2 hours
38
Instrumentation of Vibrational Spectroscopy, of 1 Lectures
functional group Identification of compounds based on
Vibrational Spectroscopy and Application of Vibrational
Spectroscopy
2 hours
expression
and 1 Lectures
1 Lecture
2 hours
2 hours
5.3. Weightage: 16-18 Marks
5.4. Learning Objectives: Students should be able to1. To understand the relationship electromagnetic radiation and absorption of light by
molecule.
2. To apply the knowledge of Beers- Lamberts law for measuring the absorbance of light by
molecule.
3. To study the principle of UV-VIS spectroscopy.
Module 5: Spectroscopic methods and applications
135
4. To know the instrumentation and applications of UV-VIS spectroscopy.
5. To study the principle of Vibrational spectroscopy.
6. To apply the knowledge of Vibrational spectroscopy for prediction of functional group
present in the organic molecule.
5.5 Theoretical Background:
The various kinds of methods are being used by chemists in order to reveal the structure of
molecules with as many details as possible. E.g. size, shape, bond length, bond angle etc.
These methods are belonging to two categories
1] Chemical methods 2] Physical methods
Physical methods have certain advantages over chemical methods
i) Amount of sample required is very small (few mg)
ii) Sample can be recovered in almost all spectroscopic methods to the extent of 100%.
iii) These methods are very quick.
iv) Gives more detailed structural information about the molecule than chemical methods.
Although certain limitations are also there
i) All of them involve use of costly instruments.
ii) Trained persons are required for operations of the instruments.
iii) Experts required for interpretation of data.
Among the available physical methods, the characterisation of pure as well as mixture of
substances is achieved with spectroscopic methods. Spectroscopy techniques are widely
applied in almost all impurities and workplaces in research, production and quality control
where chemistry is involved.
5.6. Abbreviations:
UV-VIS = Ultraviolet & Visible
IR
= Infrared
5.7. Formulae:
1)
A = €bc
A is the absorbance,
“b” is the path length in cm,
€ is molar absorptivity in L/[(mole)(cm)]
“c” is the concentration of the (sample) in mol/L
5.8. Key Definitions/Laws:
Lambert’s law:
When a ray of monochromatic light passes through the absorbing medium, its intensity
decreases exponentially as the length of absorbing medium (b) is increases.
Or We can say that absorbance is directly proportional to path length of medium. i.e. A ∝ b
Beer’s Law:
When a ray of monochromatic light passes through the absorbing medium, its intensity
decreases exponentially as the concentration of solution (c) is increases. Or We can say that
absorbance is directly proportional to concentration of solution i.e.
A∝c
Combined statement and derivation of Lambert- Beers Law:
Chemistry Sem – I /II
136
When a monochromatic light is passed through a solution containing the absorbing substance,
the decrease in the intensity of light with path length is proportional to the concentration of
the solution and the intensity of light.
Or For the given system and thickness of the medium the absorption of medium is directly
proportional to the concentration of an absorbing species.
A = €bc
5.9. Course Content:
5.9.1: Electromagnetic Radiations
The interaction of radiation with matter is the subject of science called spectroscopy. In
spectroscopy, the molecule is exposed to some kind of radiations and the responses given by
the molecules are recorded –generally in the form of graph- called as ‘spectrum’. A given
molecule when exposed to radiations, it absorbs some part of it and get excited and goes to
higher energy. The amount and type of wavelength of radiations absorbed by the molecule in
order to reach the excited states depends upon the structural features of the molecules.
Therefore, by studying the type of radiations absorbed, it is possible to predict what kind of
structural features are present in the molecule. The absorptions of different types of radiations
such as UV-Visible, Infrared (IR) region, microwave, Radio Waves produce different kinds of
excitation in the molecule. Each excitation will provide some important information about the
structure.
Fig 5.1: Electromagnetic spectrum.
The range of electromagnetic radiations extends considerably beyond the visible region.
Above figure 5.1 shows that gamma rays and x rays have very short wavelengths, while
ultraviolet, visible, and infrared and radio waves have progressively longer wavelengths. For
colorimetry and spectrophotometry, the visible region and adjacent ultraviolet region are of
major importance. Spectrophotometry is mainly concerned with the following regions of
spectrum a] Ultraviolet 180 nm-400 nm. B] Visible 400-800 nm and c] infrared 800 - 1500 nm.
Type of energy transition taking place in each region of the electromagnetic spectrum has been
given in the following table 5.1. Thus, we can say that if the frequency of transition is in the
range 1020-1024 Hz, then we may call this transition as gamma ray transition. In this region of
spectrum nuclear transition takes place.
Table 5.1: Type of transition in different region of electromagnetic spectrum.
Type of Radiation Frequency Range (Hz)
Wavelength Range
Type of Transition
gamma-rays
1020-1024
<1 pm
nuclear
X-rays
1017-1020
1 nm-1 pm
inner electron
ultraviolet
1015-1017
400 nm-1 nm
outer electron
Module 5: Spectroscopic methods and applications
137
visible
4-7.5 x 1014
750 nm-400 nm
outer electron
infrared
1013-1014
25 µm-2.5 µm
molecular vibrations
microwaves
3x1011-1013
1 mm-25 µm
molecular rotations,
electron spin flips
radio waves
<3x1011
>1 mm
nuclear spin flips
5.9.1.2: Nature of electromagnetic radiations:
Electromagnetic radiations are a form of energy that is transmitted through space at enormous
velocities. Electromagnetic radiations exhibit dual nature- particle and wave form. In studying
spectroscopy, our emphasis is more on wave form of the radiations. All radiations when
propagate in waveform produce electric field and magnetic field. The direction of propagation
of these two fields are mutually perpendicular to each other. Waves are having properties
such as wavelength, frequency, velocity and amplitude.
Fig 5.2: Nature of electromagnetic radiations
Wavelength (): The distance between two successive crests and troughs is known as
wavelength. The unit for measurement of wavelength is nanometers (nm), Angstrom (Å) or
micron (µ).
1 nm = 10-9 m,
1 Å = 10-10 m,
1 µ = 10-6 m
Frequency (Ѵ): The frequency of a wave is the number of wavelengths that pass by a point
each second. Or The no of waves per unit time is called frequency of radiations (ν). Frequency
has unit of second inverse (s-1). Frequency is also expressing in terms of Hertz (Hz). One hertz
= one vibration or one oscillation per second. Frequency and wavelength are related by
equation. Ѵ = c/ , C = velocity of light (3 x 108 m/s).
Wavenumber (Ṽ): The number of waves per unit length (cm) is known as wavenumber. The
wavenumber is reciprocal of . i.e. Ṽ = 1/ . It has unit of m-1 or cm-1. These all three wave
properties. i.e. Wavelength, frequency and wavenumber are correlated by the equation Ѵ =
c/  = c Ṽ.
Amplitude: The amplitude of a transverse wave is half the distance between a crest and
trough. As the distance between crests and troughs increases, the amplitude of a transverse
wave increases.
Chemistry Sem – I /II
138
Electromagnetic radiations possess certain amount of energy. The energy of this unit radiation
is called the photon and can be given as. E = hѴ = hc/. E = Energy of photons in ergs, h =
Planck’s constant = 6.625 x 10-27 erg. Sec.
Let’s check the take away from this lecture
1) Spectroscopy is the study of
a) Study of atomic levels
c) Study of interaction of radiation with matter
b) Study of photons and matter
d) Study of molecular levels
2) The energy of Gamma rays is ___________X rays
a) Higher
c) Similar
b) Lower
d) none of the above
3) Which of the following property/ies radio waves possesses?
a) Longer wavelength
c) lower energy
b) Lower frequency
d) all of the above
Exercise
Q.1
What is electromagnetic spectrum? What are the types of transitions takes place in
different region of electromagnetic spectrum?
Questions/Problems for practice:
Q.2
What are the properties of electromagnetic radiations? Explain each of them in detail.
Questions/Problems for practice:
Learning from this lecture: Learners will be able to state various properties of electromagnetic
radiations
Lecture: 33
5.9.1.3: Absorption of Light
Learning objective: In this lecture learners will able to understand the basis of spectroscopy.
Ground and Excited State
When a chemical absorbs light, it goes from a low energy state (ground state) to a higher
energy state (excited state)
Energy required of photon to give
this transition:
∆E = E1 – E0
∆
E = hѴ
Fig 5.3: Excitation process
Figure 5.3 depicts that excitation process is quantized. The electromagnetic radiation that is
absorbed has energy exactly equal to the energy difference between the excited state and
ground state. i.e. Only photons with energies exactly equal to the energy difference between
Module 5: Spectroscopic methods and applications
139
the two electrons states will be absorbed. Since different chemicals have different electron
shells which are filled, they will each absorb their own particular type of light.
5.9.2: Interaction of electromagnetic radiations with matter
In spectroscopic methods, the molecules of matter are exposed to radiations.
What happens when radiations strike the molecules?
When radiations strike the molecules, the molecule absorbs parts of it. The wavelength
of frequency of radiations absorbed depends upon the structural features of the
molecules. As a result of absorption of energy, molecule undergoes excitation. i.e. it
raised to higher energy states. The type of excitations produced depends upon the
energy of radiation employed.
Molecular Energy Levels
Molecules can have the following types of energy
a) Kinetic (due to motion)
b) Electronic (PE and KE of electrons)
c) Vibrational (Oscillation of atoms in bonds) d) Rotational
All Except KE are quantized.
Emolecule = Erotational + Evibrational + Eelectronic
Fig.5.4: Molecular energy level diagram
Rotational excitations: First the molecule rotates about various axes, the energy of rotation being
definite at definite energy levels, so the molecule may absorb radiations and be raised to
higher rotational energy in a rotational transition. Here the energy required is minimum.
Vibrational Excitations: Second the atom or group of atoms within a molecule vibrates relative
to each other and energy of these vibration occur at definite quantized levels. The molecule
then absorb a discrete amount of energy and be raised to higher vibrational energy levels in a
vibrational transitions. As energy of infrared transitions is higher than microwaves, so along
with vibrational excitations rotational excitation also takes place.
Electronic Excitations: If radiations from UV-Visible regions are used, it brings about electronic
excitations. As energy of these radiations is higher than microwaves and IR, along with
electronic excitations, vibrational and rotational excitations also takes place.
Since each of these internal energy transitions are quantized, they will occur only at definite
wavelength corresponding to energy hv equal to the quantized jump in the internal energy.
Chemistry Sem – I /II
140
There are however many different energy levels for each type of transitions and several
wavelengths may be absorbed
The energy levels of these transition processes are in the following order: electronic >
vibrational > rotational. Each of these transitions differs by an order of magnitude.
Rotational transitions occur at lower energies (longer wavelengths) and this energy is
insufficient and cannot cause vibrational and electronic transitions but vibrational (near infrared) and electronic transitions (ultraviolet and visible region of the electromagnetic spectrum)
require higher energies.
5.9.2: What is UV/Vis Spectroscopy?
Beside chemical analysis, the characterization of pure as well as mixtures of substances is
achieved with physical methods. Among other techniques such as, determination of melting
point, refractive index and density, optical spectroscopy in the ultraviolet and visible light
region is widely applied in almost all industries and workplaces in research, production, and
quality control etc. where chemistry is involved. Uv/Vis spectroscopy is based on the
absorption of light by a sample. Depending upon the amount of light and its wavelength
absorbed by a sample, valuable information can be obtained, such as purity of the sample,
concentration etc.
Moreover, the amount of absorbed light is related to the amount of sample and thus
quantitative analysis is also possible by Uv/Vis spectroscopy. We already discussed that
spectroscopy is nothing but interaction of radiations with matter.
5.9.2.1: Absorption of light as Analytical Tool:
Light absorption can be used in analytical chemistry for characterisation and quantitative
determination of substances. UV-Vis spectroscopy is a technique based on the absorption of
light by an unknown substance/sample. Here the sample is illuminated with electromagnetic
rays of various wavelengths in the UV-VIS region. Depending on the substance, part of light
is absorbed and part of light is transmitted. Transmitted light is recorded as a function of
wavelength by a suitable detector, providing the sample UV-Vis spectrum. As a result,
because each substance absorbs light in different ways, a unique and specific relationship
exists between substance & its UV-Vis spectrum. The spectrum can be used to identify or
quantify the substance.
Fig. 5.5: Absorbance and transmittance of light.
Fig 5.5: depicts that light passing through sample solution is partially absorbed by compounds
and remaining light is transmitted.
UV-Vis spectroscopy is applied to organic molecules, inorganic ions or complexes in
solutions. By UV-Vis spectroscopy, concentration of analyte in solution can be determined by
measuring absorbance at specific wavelength.
1: Laws of light absorbance:
Module 5: Spectroscopic methods and applications
141
When a light (monochromatic or heterochromatic) is incident upon homogenous medium,
part of incident light is reflected, some of the part is absorbed medium and remainder is
transmitted.
I0 = Ia+ It + Ir
--------- (i)
I0 = Incident light,
Ia = absorbed light It = transmitted light, Ir = reflected light
If a comparison or same cell is used, the value of reflected light (Ir) which is very small can be
eliminated for air glass interferences. Under this condition above eqation becomes
I0 = Ia+ It
Lambert’s law and beer’s law are known as the basic law of photochemistry or
spectrophotometry.
a) Lambert’s law:
When a ray of monochromatic light passes through the absorbing medium, its intensity
decreases exponentially as the length of absorbing medium (b) is increases.
Or We can say that absorbance is directly proportional to path length of medium. i.e. A ∝ b
b) Beer’s Law:
When a ray of monochromatic light passes through the absorbing medium, its intensity
decreases exponentially as the concentration of solution (c) is increases. Or We can say that
absorbance is directly proportional to concentration of solution i.e.
A∝c
c) Combined statement and derivation of Lambert- Beers Law:
When a monochromatic light is passed through a solution containing the absorbing substance,
the decrease in the intensity of light with path length is proportional to the concentration of
the solution and the intensity of light.
Or For the given system and thickness of the medium the absorption of medium is directly
proportional to the concentration of an absorbing species.
A = €bc
A is the absorbance,
€ is molar absorptivity in L/[(mole)(cm)]
“b” is the path length in cm,
“c” is the concentration of the (sample) in mol/L
Importance of Molar absorptivity or Molar Extinction coefficient:
i) Molar extinction coefficient (€) is a measurement of how strongly a chemical species absorbs
light at given wavelength.
ii) It is an intrinsic property of chemical species, i.e. it depends upon chemical composition
and structure of material, it independent on concentration.
Let’s check the take away from this lecture
1) Molecules can have the following types of energy
a) Kinetic (due to motion)
d) Rotational
b) Vibrational (Oscillation of atoms in bonds) e) All of the above
c) Electronic (PE and KE of electrons)
2) At what condition electronic excitation will takes place from ground state to excited
state?
a) Incident light have higher frequencies
c) ΔE > hv
b) ΔE = hv
d) ΔE < hv
3) Wavelength of maximum absorbance (λ max) is depends on concentration-
Chemistry Sem – I /II
142
a) This is stated by Lambert's law
b) above statement is true
c) above statement is false
d) This is stated by beer's law
4) Beers law states that
a) Absorbance is directly proportional to concentration
b) λMax is depend on concentration
c) Absorbance is directly proportional to path length
d) Absorbance is inversly proportional to concentration
Exercise
Q.1
Which laws governs the absorbance of light? State those laws.
Q.2
What is the basis of spectroscopic studies?
Questions/Problems for practice:
Q.3
If molecule will be irradiated with UV-VIS light, what are the transitions will takes
place at molecular energy levels? Explain each of them in details with the help of suitable
energy level diagram.
Learning from this lecture: Learners will be able to state different laws of light absorbance.
Lecture : 34
5.9.2.2 Derivation of Lambert- Beers Law :
Learning objective: In this lecture learners will able to understand about how to derive Beer
Lamberts law.
Derivation,
So, I can write here, -dI ∝ I (Intensity)
(The decrease in intensity is directly proportional to intensity of light)
Now, if we say about concentration
So –dI ∝ c (concentartion)
 - dI ∝ b
𝑑𝐼
∝ cI
𝑑𝑥
𝑑𝐼
− 𝑑𝑥
= kcI
−
Rearranging the equation
– dI/I = kcdx
Integrating on both the sides where limits are I =I 0 when b = o and I =I when path length x =
b
𝐼
𝑏
𝑑𝐼
∫
= −𝑘𝑐 ∫ 𝑑𝑥
𝐼0 𝐼
0
After solving it,
[ln 𝐼]𝐼𝐼0 = −𝑘𝑐[𝑥]𝑏0
𝑙𝑛𝐼 − 𝑙𝑛𝐼0 = −𝑘𝑐[𝑏 − 0]
Module 5: Spectroscopic methods and applications
143
𝑙𝑛𝐼
= −𝑘𝑐𝑏
𝐼0
𝑙𝑛𝐼0
= 𝑘𝑐𝑏
𝐼
𝐼0
2.303 log10 = 𝑘𝑐𝑏
𝐼
2.303
𝐼0
log10
𝑏𝑐
𝐼
𝑘
1
𝐼0
= log10
2.303 𝑏𝑐
𝐼
𝑘=
𝑘
As 𝜖 = 2.303
1
∈ = 𝑏𝑐 log10
∈ 𝑏𝑐 = log10
𝐼0
𝐼
𝐼0
𝐼
As 𝐴 = 𝜖𝑏𝑐
𝐴 = log10
𝐼0
𝐼
Now transmittance,
𝑇 = 𝐼/𝐼0
log 𝑇 = log10 𝐼/𝐼0
= − log10
𝐼0
𝐼
log 𝑇 = −𝐴
A = −log 𝑇
𝑇 = 10−𝐴
𝑇 = 10−𝜖𝑏𝑐
5.9.2.3: Numerical:
1) A ferrous ammonium sulfate solution of unknown concentration is placed into a
spectrophotometer. A student finds that the solution’s absorbance is 0.920. The molar
absorptivity of ferrous ammonium sulfate is 7.35 Lit mol-1 cm -1 and the path length of the
solution container is 5.00 cm. What is the concentration of solution?
Solution:
Data:
Absorbance (A) = 0.920
Molar Absorptivity (€) = 7.35 Lit mol-1cm-1
Pathlength (b) = 5 cm
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According to Lambert Beers Law: A = €bc
𝐴
0.920
 c = €𝑏 = 7.35 𝑥 5
c = 0.025 mol/lit
 We can say that concentration of FAS = 0.025 Mol/Lit
2) A solution of KMnO4 of concentration 2.8 x10-5 M shows an absorbance of 1.1 at its
wavelength of maximum absorption. Calculate the concentration of an unknown KMnO4
solution which shows an absorbance of 0.81 in the same cuvette.
Solution:
Data: Concentration (C1) = 2.8 x 10-5 M
Abs (A1) = 1.1
Abs (A2) = 0.81
To find out, concentration of unknown KMnO4 (C2) =?
According to Lambert Beers Law, A = €bc
As same cuvette is used for both known and unknown sample of KMnO 4
 b1 = b2 and €1 = €2
Thus,
𝐴1
𝐴2
=
€1b1c1
€2b2c2
𝑐1
= 𝑐2
Substituting the values of A1, A2 and C1
Thus,
1.1
=
0.81
2.8 x 10−5
c2
C2 = 2.06 x 10-5
Thus the concentration of unknown KMnO4 solution is = 2.06 x 10-5 M.
3) When a 0.005 M solution is placed in 4 cm path length cell. It shows an absorbance of 0.25.
What will be the absorbance of solution if it is placed in 1 cm path length cell?
Solution:
Data:
Concentration (C1) = 0.005 M
Pathlength (b1) = 4 cm
Absorbance (A1) = 0.25
Path length (b2) = 1 cm
To find out, Absorbance (A2) = ?
According to Lambert Beers Law: A1 = €1b1c1
0.25 = €1 x 4 x 0.005
€ = 12.5 Lit/mol/cm
Now when b2 = 1 cm
A2 = €2b2c2
= 12.5 x 1 x 0.005
A2 = 0.00625
Thus, absorbance at path length for 1 cm is = 0.0625.
4) A solution of concentration 1 x 10 -2 M is placed in 3 cm path length cell shows an
absorbance of 0.6. What will be the absorbance of solution if it is placed in 2 cm path length
cell?
Solution:
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Data:
Concentration (C1) = 0.01 M
Pathlength (b1) = 3 cm
Absorbance (A1) = 0.6
Pathlength (b2) = 2 cm
To find out, Absorbance (A2) = ?
According to Lambert Beers Law: A1 = €1b1c1
0.6 = €1 x 0.01 x 3
€ = 20.0 Lit/mol/cm
Now when b2 = 2 cm
A2 = €2b2c2
= 20 x 2 x 0.01
A2 = 0.4
 Solution will show absorbance of 0.4 when path length for 2 cm.
Let’s check the take away from this lecture
1) According to Beer Lamberts law ‘A’ is directly proportional to
a)
both b & c
b) only b
c) only C
d) None of the above
2) Unit of Molar absorptivity coefficient is?
a) L/(mole)
c) L/(mole)(cm)
b) Mol/Lit
d) (mole)(cm)/ L
3) Absorbance of 0.02 M KMnO4 solution is 0.7 in 1 cm path length cell, its transmittance
in 5 cm path length cell using same instrument will be__________
a) zero
b) increase
c) decrease
d) same
4) Lamberts law states that
a) Absorbance is directly proportional to concentration
b) λMax is depend on concentration
c) Absorbance is directly proportional to path length
d) Absorbance is inversely proportional to concentration
Exercise
Q.1
A solution of concentration 2 x 10-2 M is placed in 4 cm path length cell shows an
absorbance of 0.8. What will be the absorbance of solution if it is placed in 2 cm path length cell?
Q.2
A copper sulfate solution of unknown concentration is placed into a spectrophotometer.
A student finds that the solution’s absorbance is 0.750. The molar absorptivity of copper sulfate
is 7.35 Lit mol-1 cm -1 and the path length of the solution container is 10.00 cm. What is the
concentration of solution?
Questions/Problems for practice:
Q.3 When a 0.007 M solution is placed in 5 cm path length cell. It shows an absorbance of 0.35.
What will be the absorbance of solution if it is placed in 1 cm path length cell?
Learning from this lecture: Learners will be able to solve different numerical on Beer
Lamberts law.
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Lecture : 35
5.9.3 Instrumentation:
Learning objective: In this lecture learners will able to understand the working of single
beam colorimeter.
1) Single beam colorimeter
The photometric colorimeter is discussed on the basis of principle, construction and working
as follows
Principle: When a beam of monochromatic light is allowed to pass through a coloured solution
at constant path length, the absorbance (optical density) varies directly with the concentration
of the absorbing material.
Construction: It is consisting of polychromatic radiations source. The light radiation is
allowed to pass through collimating lens and an adjustable slit. After this slit there is an
arrangement to insert suitable colored filter to get definite wavelength. There is socket to
insert the cell or cuvette (sample holder). Behind this photocell (detector) is placed which is
connected to recorder through amplifier and which can be directly read in terms of absorbance
or percent transmittance. The basic components of a single beam photometric colorimetric are:
a) A source of light such as tungsten filament lamp with conclave reflector and
colliminating lens.
b) An adjustable diaphramgs or slits
c) A Coloured glass filter
d) A cuvette such as glass tube for holding solution/solvent
e) A detector such as photovoltaic cells
f) A reorder such as galvanometer.
Working: The cuvette is filled with distilled water and inserted in the sample cell. Once the
light source will be on, the light radiation will pass through collimating lens, slit and filter.
Finally, the light passes through the distilled water placed in a cell and then fall on
photovoltaic cells (detector). Thus, the photoelectric current is generated depending on the
intensity of radiations falling on the photocell. For the distilled water recorder is adjusted to
100% transmission. In next cycle, distilled water is taken out, from the cell, rinsed and filled
with solution of given substance of known concentration and absorbance & transmittance
noted.
Fig 5.6: Schematic representation of single beam colorimeter
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2. Components used in spectrophotometers
The spectrophotometer is used in both visible and ultraviolet region. Spectrophotometers
have the following components as shown in following block diagram Fig. 5.7.
a) Source: A stable continuous source of radiant energy covering the region of spectrum
in which instrument is designed to work.
b) Filter or Monochromater: both filter and monochromater allow the light of required
wavelength to pass through but absorb the light of another wavelength.
c) Sample Cells: All instruments must contain a transparent container, called as cuvette
for the sample and solvent.
d) Detector: It is used for measuring the radiant energy transmitted through the sample
to a electrical signal. In some instruments detectors are associated with readout system
for the presentation of detector response.
Fig 5.7. : Block diagram of basic components of spectrophotometer
Let us discuss each component in brief
a) Radiation sources- the radiation sources must fulfill the following requirements
i) it must be stable
ii) it must produce sufficient intensity of radiation so that the transmitted intensity can
be detected at the end of optical path.
iii) it must supply continuous spectrum of radiation over the entire wavelength in
which it is used.
In the visible region (400-800 nm) a tungsten lamp is most common source of radiation
and it is most widely used. Its construction is similar to the household lamp. It contains
if a piece of tungsten wire which is heated in controlled atmosphere. To avoid fluctuation
in reading, it is necessary that tungsten lamp should provide constant radiant energy, this
is achieved by employing a constant power supply to the lamp
In UV region (200-400 nm) hydrogen lamp, deuterium lamp, xenon discharge lamp and
Mercury arc lamps are used as a source of radiant energy.
b) Filters or monochromators: A radiation source is generally emitting a continuous spectra.
i.e. it gives all possible wavelength of the region. However a narrow band of wavelength
is always necessary for colorimetric and spectrophotometric analysis. Therefore it is
necessary to have the device which will select a narrow band of wavelength of the
continuous spectra. For this purpose filters or monochromators are used.
c) Filters: a light filter is device that allows light of required wavelength to pass but
absorbs light of other wavelength wholly or partially. Thus a suitable filter can select the
desired wavelength band that is a particular filter may be used for specific analysis. There
are two types of filter viz
i) Absorption filter
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ii) Interference filters
Absorption filters: These filters work by selective absorption of unwanted wavelengths. i.e. this filters limit
radiation by absorbing certain portion off the light spectrum as per given in the following
figure 5.8.
Fig. 5.8: Absorption filter
Interference filters: These filters generally provide somewhat narrower bandwidth
(as low as 10 nm) than do absorption type filters. These filters work under interference
phenomenon at desired wavelengths, thus neglecting unwanted radiation bye selective
reflection. When a ray of light is incident upon it, a part of light reflects back whereas
remaining light is transmitted.
d) Monochromator: - monochromator is a device which converts polychromatic radiation
to monochromatic radiation usually a prism or diffraction grating are used as
monochromators. The function of prism or grating is to disperse heterochromatic
radiation into its component wavelength. Following figure 5.9 depicts the conversion of
polychromatic light into monochromatic light.
Fig 5.9. : Prism monochromator
e) Sample cell or cuvettes: Sample cell holding the sample (usually solution) should be
transparent to the wavelength region being recorded. The cell
used is called as cuvette. The cell or cuvette maybe rectangular
or cylinder equal in shape as shown in following figure 5.10.
The cuvette made up of glass is used for visible region and that
made up of quartz is used for ultraviolet region.
a variety of sample sales are available for the visible and
ultraviolet wavelength regions. The choice of sample is based
upon following factors.
i) the path length, shape and size
ii) the relative expense
iii) The transmission characteristics of
desired wavelength.
Fig.5.10: Sample cuvette
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f) Detector:- A detector is a device which converts radiant energy in to electrical energy.
Generally, three types of detectors are used to detect radiations. These are Photovoltaic
cells, Phototubes and Photomultiplier tubes.
Let’s check the take away from this lecture
1) Usually ______________________are used as monochromators.
a) Prism
c) none of these
b) Diffraction grating
d) both Prism and Diffraction grating
2) In UV region (200-400 nm) hydrogen lamp, deuterium lamp, xenon discharge lamp and
Mercury arc lamps are used as a source of radiant energy.
a) True but only mercury arc lamp is used.
c) False
b) True but only hydrogen lamp is used.
d) completely true
3) Cuvette should be made up of ____________to be used in UV region.
a) Glass
b) quartz
c) metal
d) non metal
4) A detector is a device which converts radiant energy in to____________.
a) mechanical energy
c) electrical energy
b) Chemical energy
d) none of these
Exercise
Q.1
Give the principle, construction and working of single beam colorimeter?
Questions/Problems for practice:
Q.2
Enlist the basic components used in spectrophotometer and explain each of them in
detail?
Learning from this lecture: Learners will be able to know about various components of
colorimeter and spectrophotometer.
Lecture: 36
5.9.3.1: Single beam spectrophotometer:
Learning objective: In this lecture learners will able to understand the difference between
single beam spectrophotometer and double beam spectrophotometer.
Single beam spectrophotometer:
Working:
In single beam spectrometer as shown in Fig. 5.11: A beam of light is generated by source, it
passes through entrance slit and then goes to monochromator, which allows selected narrow
band of light to reach sample. Thereafter the transmitted beam reaches detector. The amount
of light absorbed by material is measured by knowing the intensity of light reaching the
detector. This intensity is compared with the intensity of light reaching the detector without
placing the sample. This type of instrument is called single beam spectrophotometer. The
observations for sample and reference samples are taken at different times, so this setup has
space for errors as the intensity of beams received from sample and reference are measured at
different times and compared afterwards. This may give rise to difference in signal due to
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variation of different parameters which may give rise to inaccurate readings. To overcome
this challenge a double beam design is used, which compares the intensities of two beams in
real time and considered to be more reliable and accurate.
Fig. 5.11: Single beam spectrophotometer
5.9.3.2: Double Beam Spectrophotometer: These are the modern, general purpose ultraviolet
visible spectrophotometer. A double beam spectrometer has same components as that of
single beam instrument. The only difference is that little bit more optics is involved in it as
shown in Fig. 5.12. In this setup the incident beam is divided into two equal beams with the
help of chopper mirrors. A chopper mirror is a circular disc having reflecting, blocking and
transmitting areas, each part covering one third part of disc. This disc is rotated at known
speed and is synchronized with the similar disc at the exit point of beam. As the disc rotates
it allows the beam to travel in the straight line for one third of the rotation time, allowing beam
to strike the sample. Where as in other one third time the beam strikes the reflecting surface
and attains the alternate path to reach the reference sample. Both the beams reach detector at
different times. The detector is programmed to receive the signals of two beams separately at
different times. These two signals received by detector are processed digitally by two different
channels and compared for final output. In this way the measurements of sample and
reference are observed in real time and errors are removed to the great extent. This instrument
is very useful for quantitative analysis in which the total spectrum is required in the range
between about 200 -800 nm.
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Fig 5.12: schematic diagram of double beam spectrophotometer.
Differences between colorimeter and spectrophotometer.
Sr.
No
1
2
3
4
5
6
7
8
Colorimeter
Spectrophotometer
It uses filters
It uses narrow band of wavelengths
It is used only in the visible region
It operates at the wavelengths of the
filter available.
It uses photovoltaic cell as detector, its
sensitivity is less.
It gives less accurate results
It is less expensive
It is less informative, since broad band
of wavelengths used
It uses monochromators
It uses single wavelength
It is used in ultraviolet and visible region.
It operates at any wavelength simply by
rotation of prism or grating.
It uses photomultiplier tube as a detector,
its sensitivity is high.
It gives more accurate results
It is expensive
It is much informative, since narrow band
of wavelength used.
5.9.4:Applications
1) Quantitative Analysis:
The colorimetric and spectrophotometric quantitative analysis is basically due to beers law.
In short, the analysis is based upon the absorption of light radiation by the absorbing materials
in the solution and the amount of light radiation absorbed is directly proportional to the
concentration of absorbing species in the solution. We will discuss this application in
following cases.
i) Unknown concentration by method 1:
This is done by construction of calibration curve as per beers law. The curve is prepared by
preparing a series of solution of known concentration. The colorimeter is set at a wavelength
( max) where the absorbance of these standard solution is maximum. After this absorbance
of each of this standard solution is determined by referring the solvent in which the solution
is prepared as a blank for absorbing 100 % transmittance or zero absorbance. The absorbance
are then plotted against the concentration of solution. A best straight line is obtained. The
curve thus constructed is called calibration curve. The slope of line is 𝜖𝑏. The nature of plot is
given in Fig. 5.13
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Fig 5.13: Calibration curve plot
Solution of unknown concentration is treated in same way and its absorbance is determined
using same cuvette and wavelength. Knowing the absorbance of unknown, its concentration
is determined with the help of above calibration curve.
ii) Alternatively, the unknown concentration can be determined without constructing the
calibration curve. From the Beer’s Lambert law, we can say that
For known concentration: A1 = €1b1c1
For unknown concentration: A2 = €2b2c2
Since absorbing species are same in both the cases, therefore ε1 = ε2 and cuvette used are also
identical therefore b1 = b2. Under this condition the ratio of above equation is. Thus,
𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝐾𝑛𝑜𝑤𝑛 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑈𝑛𝑘𝑛𝑜𝑤𝑛 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛
=
𝑐1
𝑐2
=
𝐴1
𝐴2
𝐴𝑏𝑠𝑜𝑟𝑏𝑎𝑛𝑐𝑒 𝑜𝑓 𝑘𝑛𝑜𝑤𝑛 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝐴𝑏𝑠𝑜𝑟𝑏𝑎𝑛𝑐𝑒 𝑜𝑓 𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝐶1 𝑥 𝐴2
 C2 =
𝐴1
Thus, knowing the concentration of known solution (C1) and the measured value of
absorbances of known (A1) and unknown (A2) solution. The concentration of unknown
solution (C2) can be determined very easily.
Let’s check the take away from this lecture
1) In spectrophotometer wavelength selector is used as _____________.
a) filter
c) none of these
b) monochromators
d) both filter and monochromators
2)_________________ is used in ultraviolet and visible region.
a) Colorimeter
c) both
b) Spectrophotometer
d) none of these
3) Which of the following statement is true about spectrophotometer
a) It gives more accurate results
b) It is expensive
c) It is much informative, since narrow band of wavelength used.
d) It is used only in the visible region
4) Which of the following statement is true about colorimeter
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a) It uses filters
b) It uses narrow band of wavelengths
c)It is used in ultraviolet and visible region.
d) It is less expensive
Exercise
Q.1
Give the principle, construction and working of single beam spectrophotometer?
Q.2
How one can find out the concentration of unknown solution by constructing
calibration curve, explain in detail.
Questions/Problems for practice:
Q.3
Give the principle, construction and working of double beam spectrophotometer?
Q.4
What are the major differences between colorimeter and spectrophotometer.
Learning from this lecture: Learners will be able to explain differences between colorimeter
and spectrophotometer.
Lecture : 37
5.9.5: Infrared Spectroscopy
Learning objective: In this lecture learners will able to know about Vibrational spectroscopy.
5.9.5.1: Introduction:
Almost most of the organic compounds having covalent bonds absorbs various frequencies
of electromagnetic radiations in the infrared region of spectrum. This region lies at
wavelength longer than those associated with visible light, which range from approximately
400-800 nm but lies at wavelength shorter than those associated with microwaves, which are
longer than 1 mm. For chemical purposes, we are interested in vibrational portion of infrared
region. It includes radiation wavelength () between 2.5 µm and 16 µm. In previous part of
this chapter, we already that energy of molecule can be resolved in to at least four components.
i) Translation
ii) Rotational
iii) Vibrational
iv) electronic
In earlier part we have also discussed that near UV & Visible radiation (200-800 nm) which
are energetic enough to bring about excitation in the electronic energy levels in the molecule.
Now we will consider radiations which have longer wavelength (IR radiation) and are capable
of affecting both the vibrational and rotational energy levels in the molecules. The whole IR
region is subdivided in to three parts as shown in following table 5.2.
Table 5.2: Different region of IR and wavelength
Region
Wavelengths
Wavenumber
1] Near IR
0.8 -2.5 µ
12500 – 4000 cm-1
2] IR
2.5 -16 µ
4000 – 625 cm-1
a) Functional group
2.5 -7.7 µ
4000 – 1300 cm-1
b) Finger print
7.7 -11 µ
1300 – 909 cm-1
c) Aromatic Region
11 -16 µ
909 – 625 cm-1
3] Far IR
16 -200 µ
625 – 50 cm-1
The infrared absorption process:
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Atoms in molecule and molecule as a whole is never stationary, they continuously rotate,
vibrate and also moves from one point to another point even at 0 K. when the kinetic energy
of entire molecule is zero, the atomic nuclei vibrate about the bonds which connects them.
The atoms in the molecule vibrate in number of ways. Each vibration required definite amount
of energy. i.e. molecule has no of vibrational energy levels, each of which is quantized, if a
molecule is exposed to IR radiations and as a result get excited to higher vibrational energy
levels. A molecule absorbs only selected frequencies (energies) of infrared radiations. The
absorption of infrared radiations corresponds to energy changes of the order of 8-40 kJ/Mole.
Radiation in this energy range corresponds to the range encompassing the stretching and
bending vibrational frequencies of the bonds in most covalent bonds. In the absorption
process, those frequencies of infrared radiations that match the natural vibrational frequencies
of the molecule are absorbed, and the energy absorbed serves to increase the amplitude of the
vibrational motions of the bonds in the molecule. Note however that not all bonds in a
molecule are capable of absorbing infrared energy, even if the frequency of the radiations
exactly matches that the bond motion. Only those bonds that have a dipole moment that
changes as function of time are capable of absorbing infrared radiations.
The type of IR wavelengths absorbed by the molecule depends on types of atoms and chemical
bonds in the molecule. When these absorptions are recorded, we get IR spectrum. IR spectrum
consists of number of peaks, each peak is due to particular vibrational excitation in the
molecule. The position of absorption peak can be specified in terms of frequency (Ѵ) or the
wavelength () or the wavenumber (Ṽ) of the IR radiation absorbed. The main reason for
chemists to prefer wavenumber as unit is that they are directly proportional to energy (a
higher wavenumber corresponds to higher energy). Energy ∝ Ѵ ∝ Ṽ.
5.9.5.2: Types of Molecular Vibration:
We know that covalent bond between atoms are not rigid elastic. A molecule can be said to
resemble system of balls of different masses (corresponding to the atoms) connected with each
other by springs (corresponding to chemical bonds) of varying strength. These imaginations
help in explaining interaction between IR radiations and organic molecules. A molecule is
constantly vibrating, its bond stretch, contract or bend with respect to each other. The number
of different ways in which a molecule can vibrate are known as fundamental modes of
vibration. This no. depends on total number of atoms present in the molecule and geometry
of the molecule. There are two types of vibration in the molecule.
a) Stretching Vibration
b) Bending Vibrations
a) Stretching Vibrations: These are characterized by change of internuclear distance. Thus, the
distance between two atoms increases or decreases, but atoms remain along the same bond
axis. During stretching the relative distance between the atoms in a bond will either increase
or decrease but relative bond angle will not change as seen in following figure 5.14. These
methods of vibration are further divided in to a) Symmetrical stretching b) Unsymmetrical
stretching
H
H
C
C
H
H
Fig. 5.14: Symmetrical and Unsymmetrical stretching
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Bending vibrations: These are characterized by change of angle between two covalent bonds,
due to change in the position of atoms relative to original bond axis. These modes of vibration
are further divided into four types.
a) Scissoring
b) Rocking
c) Wagging
d) Twisting
The first two vibrations, scissoring and rocking are occurred in a plane while the last two i.e.
wagging and twisting occur out of plane. Bending vibrations generally require less energy
than stretching vibrations.
Fig 5.16: Different type of bending vibrations
The number of fundamental vibrations for a molecule can be calculated from the number of
atoms in the molecule. The molecule containing N number of atoms will have 3N degrees of
freedom. Depending upon the geometry of the molecule the fundamental modes of vibrations
are calculated as follows.
i) For linear molecules, containing N atoms there are 3N-5 possible fundamental modes of
vibrations.
ii) For nonlinear molecules: containing N atoms there are 3N-6 possible fundamental modes
of vibrations. Following table shows fundamental modes of vibrations for different molecules.
Molecule
NO
CO2
H2O
NH3
CH4
C6H6
Atoms
(N)
2
3
3
4
5
12
Geometry of the Molecule
Linear (3N-5)
Linear (3N-5)
Non Linear (3N-6)
Non Linear (3N-6)
Non Linear (3N-6)
Non Linear (3N-6)
Fundamental modes
of vibrations
1
4
3
6
9
30
Even though a molecule can vibrate in number of ways, each and every vibration does not
absorb IR. The absorption of IR radiations takes place only if following two conditions are full
filled.
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i) When the frequency of the vibration of a bond and frequency of IR radiation used for
excitation match, then only energy of IR radiation is absorbed.
i.e Frequency of vibration of bond = Frequency of IR radiation
ii) Only those vibrations which results in change in dipole moment of the molecule absorb IR
radiations, such vibrations are said to IR active.
Let’s check the take away from this lecture
1) IR region falls in which of the following region?
a) 12500 – 4000 cm-1
c) 20000- 12500 cm-1
-1
b) 4000 – 625 cm
d) none of these
2)For a molecule to be IR active ____________
a) It must have dipole moment in it.
b) or there should be change in dipole moment during IR absorption.
c) both a) and b) are correct
d) only a) is correct
3) For CO molecule no of possible fundamental vibrations are__________
a) 1
b) 5
c) 4
d) 2
4) For BF3 molecule no of possible fundamental vibrations are__________
a) 1
b) 5
c) 4
d) 6
Exercise
Q.1
Give the principle of IR spectroscopy.
Q.2
What is fundamental modes of Vibrations? Calculate it for i) Methane ii) Ethane
Questions/Problems for practice:
Q.3
What is molecular vibrations? Explain stretching and bending vibrations in details.
Learning from this lecture: Learners will be able to understand the principle of IR
spectroscopy.
Lecture : 38
5.9.5.3: Infrared Spectroscopy instrumentation
Learning objective: In this lecture learner will able to tell us about instrumentation and
applications of IR spectroscopy.
Instrumentation:
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Fig. 5.17: infrared Spectrometer
The instrument that determines absorption spectrum for a compound is called an infrared
spectrometer or more precisely a spectrophotometer. Above figure 5.17 schematically
illustrates the components of infrared spectrometer. The instrument produces a beam of
infrared radiations from a hot wire and by means mirrors divides into two parallel beams of
equal intensity radiations. The sample is placed in one beam and other beam is used for
reference. The beams then pass into monochromator which disperses each into continuous
spectrum of infrared light. The monochromator consists of rapidly rotating sector (beam
chopper) that passes two beams alternatively to a diffraction grating. The slowly rotating
diffraction grating varies the frequency or wavelength of radiations reaching the detector. The
detector senses the ratio between the intensities of the reference and sample beams. In this
way detector determines the frequencies which have been absorbed by the sample and
frequencies which are unaffected by the light passing through the sample. After the signal
from detector is amplified, the recorder draws the resulting spectrum of the sample on a chart.
The main part of IR spectrometer are as follows a) radiation source, b) Monochromator, c)
sample d) detectors
a) IR radiation sources: IR instruments requires a source of radiant energy which emits IR
radiations, which must be steady, intense enough for detection and extends over the desired
wavelengths. Various sources of IR radiation are as follows. i) Nernst glower ii) Incandescent
lamp, iii) Mercury arc, iv) Tungsten lamp, v) glober source vi) Nichrome wire etc.
Monochromators: In the infra-red region glass and fused silica transmits very little. therefore,
prism made from alkali and alkaline earth halides are preferred to use such as potassium
bromide, sodium chloride or cesium iodide.
Sample and Sample cells: To determine the infrared spectrum of a compound, one must place
the compound in sample holder or cells, cells must be constructed of ionic substances such as
glass or plastics which strongly absorbs in the infrared region. IR spectroscopy has been used
to characterize solid, liquid or gas samples.
a) Solids: To obtain IR spectra of solids, a finely ground sample has been mixed with
powdered KBr and mixture is pressed under high pressure. Under high pressure Kbr melts
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and seals the sample into matrix. The resulting KBr pellate is inserted into instruments to
record IR. In another method nujol mull involves grinding the compound with mineral oil
(Nujol) to create finely divided sample dispersed in mineral oil. The thick suspension is placed
between two plates and held in instruments.
b) Liquids: A drop of liquid organic compound is placed between a pair of polished sodium
chloride or potassium bromide plates referred to as salt plates. When the places are squeezed
gently a thin liquid film form between them. Then it is inserted in to instrument to record
spectrum.
c) Gas: Sampling of gas is more or less similar to liquid sampling.
5.9.5.4: Uses of Infra-red spectrum:
Since every type of bond has different natural frequency of vibration and because two of the
same type of bond in two different compounds are in two slightly different environments
therefore no two molecules of different structure have exactly the same infrared absorption
spectrum. Although some of the frequencies absorbed in two cases might be same but in no
case two different molecules will have identical infrared spectra. By comparing the infrared
spectra of two substances thought to be identical we can tell whether, they are in fact, identical.
If their infrared spectra coincide peak by peak, in most cases two substances will be identical.
A second more use of infrared spectrum is to determine structural information about
molecule. The absorption of each type of bond (N-H, C-H, O-H, C-X, C=O, C-O, C-C C=C, C
C, CN and so on) are regularly found only in certain small portions of the vibrational infrared
region. A small range of absorption can be defined for each type of bond. Outside this range
absorption are normally due to some other type of bond. For instance, any absorption in the
range 2500-3700 cm-1 is due to O-H, N-H, C-H functional groups. Similarly, the appearance of
absorption peak at 2100 – 2500 cm-1 shows the presence of C C or CN bond is known as
triple bond region. The IR absorption peak at 1500 -1600 cm-1 shows the presence of aromatic
C=C stretching frequency as shown in following figure 5.18.
Figure: 5.18: Region in the infrared spectrum
The important IR absorption band at 1700 cm -1 clearly indicates the presence of a carbonyl
group (C = O) of aldehydes or ketones. Following table gives the most common occurring
frequencies for the different functional groups in the molecule.
Table 5.3: Organic functional group and their IR absorption frequencies.
Functional Group
Water O-H Stretch
Frequency (cm-1)
3700- 3100
Intensity
strong
Module 5: Spectroscopic methods and applications
Alcohol O-H Stretch
Carboxylic acid O-H Stretch
N-H stretch
 C-H stretch
= C-H stretch
-C- H aldehyde
CN stretch
CC stretch
C=O aldehyde
C=O anhydride
C=O ester
C=O ketone
C=O amide
C=C alkene
C=C Aromatic
3600-3200
3600-2500
3500-3350
3300
3100 -3000
2900-2800
2250
2260-2100
1740-1720
1840-1800, 1780-1740
1750-1720
1745-1715
1700-1500
1680-1600
1600-1400
159
strong
strong
strong
strong
weak
variable
strong
variable
strong
Weak, strong
strong
strong
strong
weak
weak
How IR spectra looks? - Following figure 5.19 represents the typical IR spectrum. For. E.g
IR spectrum of butanal is shown below. Here, we can clearly see that C-H (aldehyde
stretching) around 2700 cm-1 and C=O (aldehyde) stretching at around 1720 cm -1. In this way
data can be interpreted from IR spectrum.
Figure 5.19: Representation of IR spectrum.
5.9.5.5: Applications:
a)
b)
c)
d)
e)
Identification of functional group and structure elucidation.
Identification of substances
Studying the progress of the reaction
Detection of impurities
Quantitative analysis
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Let’s check the take away from this lecture
1) Which of the following source of light is not used in infrared region
a) Nernst Glower
c) Xenon arc lamp
b) Nichrome Wire
d) Mercury arc
2) Infrared spectroscopy can be used for
a) Identification of functional group and structure elucidation.
b) Detection of impurities
c) Studying the progress of the reaction
d) All of the above
3) CN stretching vibrations takes place at __________
a) 2250 cm-1
b) 2750 cm-1
c) 3200 cm-1
d) 1220 cm-1
4)  C-H stretching vibrations takes place at______
a) 2250 cm-1
b) 3300 cm-1
c) 2800 cm-1
d) 1520 cm-1
Exercise
Q.1
Give the principle, construction and working of IR spectroscopy.
Q.2
How infra-red spectrum can be used for interpretation of structure of organic molecule?
explain in details.
Questions/Problems for practice:
Q.3
State the applications of Infra-red spectroscopy.
Learning from this lecture: Learners will be able to know the components of IR spectrometer.
Conclusion:
Analysis of any chemical product/material is very important. Spectroscopic analysis is one
of the tools in chemical/drug/petrochemicals/paper/pulp etc. industries. By using
spectroscopic techniques like UV-Visible spectroscopy and Vibrational spectroscopy limits
of Impurities in chemical products can be kept in specified limit.
Add to Knowledge:
Summary of Some spectroscopic techniques:
i)
Electron Spin Resonance Spectrophotometry
Principle
Detection of magnetic moment associated with unpaired electrons.
Main Uses
Research on metalloproteins, particularly enzymes and changes in the
environment of free radicals introduced into biological assemblies, e.g.
membranes.
ii)
Flame (Emission and Absorption) Spectrophotometry
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161
Principle
Energy transitions of outer electrons of atoms after volatilization in a flame.
Main Uses
Qualitative and quantitative analysis of metals, particularly in clinical biochemistry.
Emission techniques are for routine determination of alkali metals. The absorption
technique extends the range of metals that may be determined and the sensitivity
iii)
Raman Spectrophotometry
Principle
Atomic vibrations involving a change in polarizability.
Main Uses
Qualitative analysis and fingerprinting of purified molecules of intermediate size.
Mainly used in research.
iv)
Mass Spectrophotometry
Principle
Determination of the abundance of positively ionized molecules and fragments.
Main Uses
Qualitative analysis of small quantities of material (10-6 to 10-9 g), particularly in
conjunction with gas liquid chromatography. Mailnly used in research, but has
high potential for the rapid determination of the primary structure of peptides.
v)
Nuclear Magnetic Resonance Spectrophotometry
Principle
Detection of magnetic moment associated with an odd number of protons in an
atomic nucleus.
Main Uses
Research into the structure of organic molecules of molecular weight less than 20,000
daltons.
vi)
Spectrofluorimetry
Principle
Absorbed radiation emitted at longer wavelengths.
Main Uses
Routine qualitative and quantitative analysis, enzyme analysis and kinetics, and
detection of changes in protein conformation. More sensitive at lower concentrations
than visible and u.v. absorption spectrophotometry .
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5.10: Set of Questions for FA/CE/IA/ESE
Short Answer Questions (2 Marks)
1) What is electromagnetic radiations? State the important properties of it.
2) What are the characteristics of sample that can be viewed by scanning electron
microscopy?
3) Give distinguishing features between colorimetry and spectrophotometry.
4) What is the basis of spectroscopy?
5) What do you mean by wavelength, frequency and wavenumber? Give the correlation
equation between them and explain the term involved.
6) Enlist the stretching and bending vibration observed in vibrational spectroscopy.
7) Which part of electromagnetic radiation is used for colorimetry and vibrational
spectroscopy study? Mention the wavelength of that regions.
8) What are the key differences between stretching and bending vibration?
9) What is the advantages spectrophotometer holds over colorimeter?
10) For structure elucidation physical methods are preferred over chemical methods,
why?
11) Analyst wants to do functional group analysis, Which technique he should prefer to
use? Give its principle.
12) What will happen, if radiation will interact with matter? Obtain an expression for an
energy of the unit radiation called photon.
13) Which technique is used to find the concentration of unknown solution in liquid state?
Give its principle.
14) What are the energy level changes will take place? If molecule will absorb radiation in
the UV-Vis region? Explain.
15) What are the energy level changes will take place? If molecule will absorb radiation in
the IR region? Explain.
Short Answer Questions (3 Marks)
1) Highlight the important differences between colorimeter and spectrophotometer?
2)
State the important applications of vibrational spectroscopy.
3) Name the laws and along with its statement give its mathematical expression which
governs the absorbance of solution.
4) A lab analyst wants to do structure determination of chemical compound. In his lab
both physical and chemical methods of analysis are available. Which method of
analysis he should prefer? And why?
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5) A ferrous ammonium sulphate solution of unknown concentration is placed into a
spectrophotometer. A student finds that the solution’s absorbance is 0.920. The molar
absorptivity of ferrous ammonium sulphate is 7.35 Lit mol-1 cm -1 and the path length
of the solution container is 5.00 cm. What is the concentration of solution
Long Answer Questions (5 Marks)
1) Explain construction and working UV-Vis spectrophotometer with clean labelled
diagram.
2) State and derive Beers-Lamberts law.
3) If molecule will be irradiated with UV-VIS light, what are the transitions will takes
place at molecular energy levels? Explain each of them in details with the help of
suitable energy level diagram.
4) What will happen if CO2 molecule will be bombarded with IR radiation? Explain all
modes of vibration along with pictorial representation which CO2 will undergo.
5) A solution of KMnO4 of concentration of 2.8 x 10-5 shows an absorbance of 1.1 at its
wavelength of maximum absorbance. Calculate the concentration of unknown KMnO4
solution which shows an absorbance of 0.81 in the same cuvette.
Very Long Answer Questions (10 Marks)
1) Give the principle of infrared spectroscopy. Explain its construction and working of
infrared spectrophotometer with clean labelled diagram and mention its applications.
2) i) When 0.005 M solution is placed in 4 cm length cell showing an absorbance of 0.25.
What will be the absorbance of solution if it is placed in 1 cm path length cell?
ii) A solution of K2Cr2O7 of concentration of 3.8 x 10-5 shows an absorbance of 2.1 at
its wavelength of maximum absorbance. Calculate the concentration of unknown
K2Cr2O7 solution which shows an absorbance of 0.92 in the same cuvette.
3) One of the analyst wants to do functional group estimation. Which technique he
should prefer to use? Give construction and working of that technique with clean
labelled diagram and mention its applications.
4) One of the analyst wants to do the estimation of unknown KMnO4 solution. In this
regard answer the following questions
i) Which technique he should prefer to use? And why?
ii) Give construction and working of that technique with clean labelled diagram.
iii) What are the other areas where this technique is find applications?
5) Use of colorimetry and spectrophotometry techniques are based on which laws? Give
its statements along with mathematical expression. Discuss advantages of
spectrophotometer over colorimeter for chemical analysis. Discuss construction and
working of single beam spectrophotometer with neat diagram.
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Self-Assessment:
1) What makes it possible to use spectroscopy technique as a tool to elucidate the
structure of molecule?
2) Enlist the basic components used in spectrophotometer and explain each of them in
detail?
3) What makes double beam spectrophotometer best choice over single beam
spectrophotometer?
4) State the important applications of IR spectroscopy.
5) What will happen if NO2 molecule will be irradiated with IR radiation? Explain all
modes of vibration along with pictorial representation which NO2 will undergo.
References:
1)
Analytical Chemistry, 7th Edition, by Gary Christian
2)
Textbook on Analytical Chemistry by Manali Publication, T. Y. BSc, University of Pune
3)
Textbook on Analytical Chemistry by Nirali Publication, T. Y. BSc, University of Pune
4)
Textbook on Organic Chemistry by Manali Publication, T. Y. BSc, University of Pune
Module 5: Spectroscopic methods and applications
165
Self-evaluation
Name of
Student
Class
Roll No.
Subject
Module No.
Tick
S.No
Your choice
1.
2.
3.
4.
5.
Do you understand the various properties of
electromagnetic radiations?
o
Yes
o
o
No
Yes
o
No
Do you know the different components of
colorimeter and spectrophotometer?
o
Yes
o
No
Do you understand how structure can be
elucidated for organic molecule?
o
Yes
o
No
Do you understand module ?
o
Yes, Completely.
o
Partialy.
o
No, Not at all.
Do you understand the Beer Lamberts law
and numerical on it?
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Module 6
Stereo Chemistry
Lecture: 39
6.1. Motivation:
This module will help the student to know how molecules structure can be arranged in the various
possible ways. This module focuses on stereoisomers and relationship between enantiomers,
diastereomers, meso structures and racemic mixtures. This module also tells us about various
nomenclature system available for isomers.
6.2. Syllabus:
Lecture
no
Content
Duration
SelfStudy
39
Introduction of Stereochemistry, Optical Isomerism,
Optical Activity, Specific Rotation
1 Lecture
2 hours
40
(Numerical Based on Specific rotation) Elements of 1 Lecture
symmetry Chirality/Asymmetry, Molecules with two
similar and dissimilar chiral-centers,
2 hours
41
Optical isomerism in tartaric acid and 2,3 dihydroxy 1 Lecture
butanoic acid, Enantiomers, Diastereoisomers, meso
structures, racemic mixture.
2 hours
42
Representation of Sterioisomers by flying wedge, Fischer 1 Lecture
projection.
2 hours
43
Geometrical Isomerism, Geometrical Isomerism in 1 Lecture
oximes, Nomenclature of Stereoisomers: D&L system,
2 hours
44
R-S Configuration E-Z nomenclature
1 Lecture
2 hours
45
Newman projection, Conformation analysis of alkanes 1 Lecture
(ethane and n-butane); Relative stability with energy
diagram
2 hours
6.3. Weightage: 16-18 Marks
6.4. Learning Objectives: Students should be able 1. To understand the optical activity and symmetry elements responsible for molecule to be
optically active.
2. To state whether molecule is Chiral/optically active or not.
3. To identify Enantiomers, Diastereoisomers and meso structures from the given set of
compounds.
4. To represent given molecule in Fischer/Flying Wedge representation.
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5. To assign D/L and R/S nomenclatures to given compounds.
6. To state the stable/unstable conformation of ethane and butane.
6.5 Theoretical Background:
Chemistry which is dealing with properties of molecules in relation to spacial arrangement of
atoms or groups is called stereochemistry.
For simple molecules such as methane, ethane & ethene (C2H4) the molecular formula provides
enough information to work out the structure of compound however for larger more complex
molecules this is not the case because the atoms can be arranged in many different ways.
6.6. Abbreviations:
SOR = Specific Optical Rotation
6.7. Formulae:
∝
1] The specific optical rotation: [∝]𝑡𝜆 = 𝐶.𝑙
In above formula,
[∝]𝑡𝜆 = The specific optical rotation, t = temperature, λ = Wavelength of plane polarized light, α =
Optical rotation, c = concentration (gm/ml), l = length of polarimeter tube containing solution in
dm.
Unit for specific optical rotation = (degree) gm-1 ml dm-1
6.8. Key Definitions:
i) Stereoisomers: - This are the compounds in which atoms are joined in the same order but the
position of the atoms in space are different.
ii) Enantiomers: Enantiomers are the stereoisomers which are non-superimposable mirror images
of each other.
iii) Diastereomers: Diastereomers are the stereoisomers that are not mirror images of each other.
iv) Optical isomerism: The stereoisomers of a substance which are similar in physical & chemical
properties but differ in their action towards the plane polarized light are called optical isomers &
the phenomenon is called optical isomerism.
6.9. Course Content:
6.9.1: Classification of Isomers:
Molecules that have the same molecular formula but differ in way in which their atom are
arranged are called isomers. Isomers may have different functional groups, different carbon
skeletons or just different shapes. It is important to understand the shape of molecules because it
influences the physical & chemical properties & the way the molecules acts in biological system.
The presence of different isomers is called isomerism. Isomerism is the term used to describe the
existence of different isomers. These are two main classes of isomers depending on the different
way in which the atoms are arranged in the isomers. Classification of isomer is shown in figure
6.1a
1. Structural (Constitutional) Isomers: - This are the compounds in which atoms are joined
together in different ways (order).
Module 6: Stereochemistry
168
2. Stereoisomers: - This are the compounds in which atoms are joined in the same order but the
position of the atoms in space are different.
Structural isomers are further subdivided into three types.
i)
Chain
ii) Position
iii) Functional Group
Similarly, Stereoisomers are also further subdivided into two types i.e.
i) Conformational
ii) Configurational
Figure 6.1: Classification of isomers
1.
Chain isomers are molecules in which the carbon chain is connected in different ways. E.g. npentane & 2-methyl butane is e.g. of chain isomers they both have M.F. C 5H12 but whereas
pentane is straight chain alkane whereas 2-methyl butane has branched chain (Figure 6.2 I &
II). The different structure of the alkanes leads to different properties e.g. 2- methyl butane has
lower B.P. than pentane.
2. Position isomers: - This are the molecules in which one or more functional groups is at different
position on the carbon chain for e.g. there are two possible straight chain chloral alkanes with
the molecular formula C4H9Cl (Figure 6.2 III &IV). These isomers arise because the Cl functional
group is situated on one of the two carbon atoms. Another important example relates to
benzene derivatives containing two or more substituents for e.g. dichloro benzene → There are
three ways of arranging positions of Cl Functional groups on benzene ring (Fig. 6.2 V, VI & VII).
3. Functional group Isomers: - This are the isomers with same molecular formula but different
functional groups e.g. propan-1-ol & methoxy ethane (Figure 6.2 VIII & IX) which have
molecular formula (C3H8O). In addition, having different physical properties (such as B.P.) have
quite different chemical properties also because of their dissimilar functional groups.
Figure 6.2: Examples of structural isomers
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6.9.1.1: Stereoisomers:
The atoms in stereoisomers are connected in the same way but are arranged different in space.
These are two kind of stereoisomers called conformational & constitutional isomers.
Conformational isomers are usually interconverted rapidly at room temperature by rotation about
carbon- carbon bonds. In constant, configurational isomers do not usually interconvert at room
temperature since this normally involves breaking & reforming bonds. There are two types of
configurational isomers E & Z isomers & isomass with chiral center.
Conformational isomers: - Stereoisomers separated by relatively low energy barrier (< 60 KJ mole1) are easily interconvertible at ambient temperature are called conformational isomers or
conformers. Unless a conformation is held rigid by a small ring or double bonds, a molecule can
have an infinite no. of conformations, but only one configuration Or New geometrics or shape
produced due to C-C bond rotation or flipping of ring systems or interconversion of non-planar
arrangements called conformations. Following figure 6.3 shows some examples of conformational
isomers. For example, conversion of axial chlorocyclohexanone into equatorial
chlorocyclohexanone (I), conversion of gauche form of ethane into equatorial form (II), and
inversion of amines (III). So, in all this example shown here, there is only interconversion is there
no breaking of bond is there.
Configurational isomers: - Stereoisomers separated by high energy barrier (> 100 K.J mole-1) are
stable & isolable at room temperature. These are called configurational isomers.
They are classified as enantiomers & diastereomers
Enantiomers → Enantiomers are the stereoisomers which are non-superimposable mirror images
of each other. For. e.g 2-butanol (Fig 6.3 (V))
Diastereomers → Diastereomers are the stereoisomers that are not mirror images of each other.
Following examples of Cis 2 butene-Trans 2 butene (Fig 6.3 (V)) and meso tartaric acid – with one
of the forms of active tartaric acid shows the diastereomeric relationship. (Fig 6.3 (VI))
Module 6: Stereochemistry
170
6.9.1.2: Optical Isomerism:
The stereoisomers of a substance which are similar in physical & chemical properties but differ in
their action towards the plane polarized light are called optical isomers & the phenomenon is called
optical isomerism.
6.9.1.2.3: Plane polarized light
Ordinary light consists of waves vibrating in all planes perpendicular to the direction of
propagation whereas plane polarized light is light in which all wave vibrations have been filtered
out except for those in one plane. Ordinary (non-polarized) light consists of many beams vibrating
in different planes as shown in figure 6.4 (i).
Figure 6.4: i) Ordinary light and ii) plane polarized light
Whereas plane polarized light consists of only those beams that vibrate in the same plane as shown
in figure 6.4(ii) Plane polarized light is created by passing ordinary light through a polarizing
device. Such devices transmit selectively only that component of light beams having electrical &
magnetic field vectors oscillating in a single plane. The plane of polarization can be determined by
an instrument called a polarimeter. Following figure 6.5 shows the components of polarimeter.
Figure 6.5: Components of polarimeter
Monochromatic (single wavelength) light is polarized by a fixed polarized next to the light sources
which convert ordinary light to plane polarized light. A sample cell holder is located in line with
light beam followed by a movable polarized (analyses) and an eyepiece through which light
intensity can be observed. In modern Instruments an electronic light detector takes the place of
human eye.
An optically active compound is the one which rotates the plane of polarized light. An optically
active substance which rotates the plane of polarized light to the right (clockwise) is said to be
dextrorotatory (d) & it has been denoted by optical rotation sign (+) sign. An optically active
substance which rotates the plane of polarized light to the left (anticlockwise) is said to be
levorotatory (l) & it has been denoted by optical rotation sign (-) sign. For example, an enantiomer
Chemistry Sem – I/II
171
of alanine (amino acid) which rotates the plane of polarized light in clockwise and anti-clockwise
direction can be written as (+) alanine and. (-) alanine respectively.
The magnitude of rotation is an intensive property of the molecule. The observed rotation depends
on
1. Concentration of the sample compound.
2. The length of sample tube.
3. The temperature.
4. Wavelength
Specific rotation is the degree of rotation by 1.00 gm of compound in 1.00 ml of solution measured in a tube
with 1 decimeter path length.
The amount of rotation is also dependent on the wavelength (typically the D line of sodium, 589.3
nm) & solvent.
The specific optical rotation of any compound can find by following formula,
∝
[∝]𝑡𝜆 = 𝐶.𝑙
In above formula,
t = temperature, λ = Wavelength of plane polarized light, α = Optical rotation, c = concentration
(gm/ml), l = length of polarimeter tube containing solution in dm.
Unit for specific optical rotation = 0(degree) gm-1 ml dm-1
Let’s check the take away from this lecture
1) Compounds which have different arrangements of atoms in space while having same atoms
bonded to each other are said to have
a) position isomerism
c) chain isomerism
b) functional group isomerism
d) stereoisomerism
2) Which of the following can make difference in optical isomers?
a) heat
c) polarized light
b) temperature
d) pressure
3) Which of the following is not classified as structural isomers?
a) Chain
c) Position
b) Functional
d) Conformational
Exercise
Q.1
What is isomers? Give the classification of isomers with examples.
Questions/Problems for practice:
Q.2
What do you understand by the term dextrorotatory and laevorotatory? explain it.
Learning from this lecture: Learners will be able to classify the isomers and understand
dextrorotatory and levorotatory terms.
Lecture: 40
Learning objective: In this lecture learners will be able to solve the numerical on specific optical
rotation.
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172
6.9.1.3: Numericals:
1. Numerical: - 0.3 gm of cholesterol was dissolved in 15 ml of CHCl3 taken in 10 cm polarimeter
tube observed optical rotation was (α) = -0.380. what is specific optical rotation?
∝
−0.380
[∝]𝑡𝜆 = 𝐶.𝑙 = 0.3/15 𝑥 1 (10 cm = 1 dm)
1
= -0.380 x 0.3 x 0.1
= -3.8 x 5
= -19 0 gm-1 ml dm-1
2. A pure enantiomer has an observed optical rotation of -0.820 when measured in 1 dm at a
concentration 0.3 gm / 10.0 ml. calculate specific rotation for this molecule?
∝
−0.82
Sol: - [∝]𝑡𝜆 = 𝐶.𝑙 = 0.3/10 𝑥 1
=
−8.2
0.3
= -27.30 gm-1 ml dm-1
3. What will be the concentration of pure enantiomer which has an observed optical rotation of 0.920 and specific optical rotation of – 25.4 gm-1 ml dm-1 when measured in 1 dm polarimeter
tube?
Sol: 𝛼 = -0.920, [∝]𝑡𝜆 = - 25.4 gm-1 ml dm-1, l = 1 dm
∝
∝]𝑡𝜆 = 𝐶.𝑙
- 25.4 =
−0.92
𝐶𝑥1
−0.92
(10 cm = 1 dm)
c = −25.4 𝑥 1 (10 cm = 1 dm)
= 0.036 gm ml-1
6.9.1.4: Origin of Optical Isomers:
As we have already discussed optical isomers are the compounds with the same molecular formula
but differ in the way they rotate plane polarized light. The substances which rotates plane polarized
light are called optically active substances. The phenomenon of rotating plane of polarized light is
called optical activity. The factors responsible for optical activity is that the molecule should have
asymmetry or dissymmetry. The difference between chiral and achiral molecules can be explained
on the basis of the plane of symmetry. If all the attached group to the central carbon atom are
different then there is no plane of symmetry. Such a molecule is known as a chiral or asymmetric
molecule & a compound is dissymmetric in nature if it does not have the following elements of
symmetry
i) Plane or symmetry
ii) Centre of symmetry
iii) An alternating axis of symmetry
i) Plane of symmetry: - A plane of symmetry is a plane which divides molecule (or object) in to
two halves which are mirror images of each other. When a molecule has plane of symmetry it
will not be optically active even though it has chiral carbon. In following figure 6.6, 2,3dibromobutane, 1,2-dimethylcyclobutane 1,3-dimethylcyclobutane and
acetylene are not
optically active because of plane of symmetry.
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Figure 6.6: Plane of symmetry
ii) Centre of symmetry: - Centre of symmetry is an imaginary point in a molecule from which the
similar groups stay opposite at equal distance is called Centre of symmetry. Following examples of
cyclobutene-1, 2, 3, 4-tetracarboxylic acid and 1, 3-dibromo-2, 4-dimethylcyclobutane are optically
inactive because they are having center of symmetry.
Figure 6.7: Centre of symmetry
iii)
Improper or Alternating Axis of Symmetry: - A molecule which rotated around the axis by
an angle 360/n (n is an integer) then reflected across the plane perpendicular to the axis of rotation
an identical structure result. This is called improper or alternate axis of symmetry.
The value of n defines the number of fold symmetry thus n=2, 3 & 4 respectively represents rotation
of 1800, 1200, 900 which is then respectively called two-fold, three-fold or four-fold symmetry.
For. e.g. Rotate 1,2,3,4 tetra methyl cyclobutene (I) (Figure 6.8) about axis through 90 0. A reflection
taken perpendicular to the axis of rotation gives a mirror image (III) which is same as structure (I),
so the compound is said to have four-fold (S4 =360/90 = 4) axis of symmetry.
Figure 6.8: Alternating axis of Symmetry 1,2,3,4 tetra methyl cyclobutene (I)
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Similarly, ∝ trauxilic Acid (1,3 trans carboxylic 2,4 trans diphenyl cyclobutene) has two-fold (S2
=360/180 =2) alternating axis of symmetry. In following figure 6.9, ∝ trauxilic acid has rotated
through 1800 which gives image (II) when its reflection will be observed on mirror perpendicular
to the axis of rotation it gives mirror image (III) which is same as structure (I) so the compound is
said to have two-fold (S2 =360/180 = 2) axis of symmetry.
Figure 6.9: Alternating axis of Symmetry in ∝ trauxilic Acid
6.9.2.1: Chirality:
Chirality refers to object which are not super imposable on (cannot be made to coincide with)
their mirror image. A Chiral molecule is one that is not-super impossible on its mirror image.
Achiral compounds (objects) can be super imposed on their mirror. sp3 hybridized carbon
atoms possessing four different substituents display this property due to their tetrahedral
geometry. A chiral carbon is one that has four different group attached to it.
Thus, for carbon to be chiral it must follow that i) A carbon atom is sp3 hybridized and ii) there
are four different groups attached to the carbon atom. for.e.g 2- bromobutane (Fig. 6.10 (I)) and
Lactic acid (II) which is optically active as it satisfies above two conditions.
Figure 6.10: Chirality
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All molecule containing one Chiral Centre’s are Chiral, but molecules are achiral though they
contain multiple Chiral Centre’s. For e.g Mesotartaric acid (Fig. 6.10 (III)) is having two chiral
centers still it is optically inactive because of presence of plane of symmetry. Further molecules can
be Chiral even in the absence of Chiral Centre’s e.g. some allenes & substituted biphenyls (Fig. 6.10
(IV & V)). Thus, presence or absence of chiral center does not guaranteed chirality to a molecule.
Let’s check the take away from this lecture
1) Which of the following is the definition of chirality?
a) The superimposability of an object on its mirror image
b) A molecule with a mirror image
c) The non-superimposability an object on its mirror image
d) A molecule that has a carbon atom with four different substituents
2) Which symmetry element makes the given compound achiral?
a) Plane of symmetry (POS)
b) Center of symmetry (COS)
c) Axis of symmetry (AOS)
d) Alternating axis of symmetry (AAOS)
3) State whether the given compound is a chiral molecule.
a) True
b) False
Exercise
Q.1
Explain the world Chirality with suitable examples.
Questions/Problems for practice:
Q.2
Explain with examples how improper or alternating axis of symmetry affects the optical
activity.
Learning from this lecture: Learners will be able to understand what makes molecule optically
active and optically inactive.
Lecture: 41
Learning objective: In this lecture learners will be able to explain different terms like
enantiomers, diastereomers, meso compounds and racemic mixture.
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6.9.2.2: Optical isomerism in Tartaric Acid: Tartaric acid contains two Chiral Centre’s (C) which are identical. All the four groups attached to
2nd & 3rd carbon atoms are same as shown in figure 6.11.
Structure I & II are non-superimposable mirror images, so they are enantiomers of each other & are
optically active. Similarly, structure III & IV are also mirror images but by turning (IV) upside down
by 180° in plane of the paper. It can be exactly super imposed on (III). Therefore structure III & IV
are identical molecules & hence they are not enantiomers.
Figure 6.11: All possible forms of tartaric acid
This means tartaric acid exists only these isomeric forms. They are represents & designated as
follows as shown in figure 6.12.
Figure 6.12: Actual forms of tartaric acid
On the basis of specific rotation four types of tartaric acid are recognized they are as follows:
i) (+) or d tartaric Acid (I): - Both the chiral centre are acting in the same direction. So, it rotates
the plane of polarized together to the right or clockwise direction. It is called dextrorotatory.
ii) (-) or l tartaric acid (II): - Both the chiral center is acting in same direction. It rotates plane of
polarized light to the left or anticlockwise direction it is called levorotatory.
iii) Meso tartaric acid (III): - It possesses a plane of symmetry passing between 2nd & 3rd carbon.
Hence it is achiral molecule so optically inactive.
The optical inactivity is due to internal compensation.
iv) dl or (±) tartaric Acid: - When equimolar proportion of d & l forms of tartaric acid are mixed
together the resulting solution becomes optically inactive & is called (±) or dl tartaric acid.
Optically inactivity is due to external compensation.
6.9.2.3: Enantiomers:
Enantiomers have identical structure but differ in spatial arrangement of atoms or groups around
chiral center. Examples of enantiomers have shown in figure 6.3 (I) i.e 2- butanol and pair of active
tartaric acid as given in figure 6.3 (III)
i) They cannot be interconverted in to each other without breaking bonds.
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ii) They are related as objects & mirror images i.e. the reflection of one molecule in mirror
compounds to the structure of another molecule.
iii) They cannot be super imposed hence they are different molecules.
iv) They bear identical chemical properties except their action against optically active reagents.
v) They differ in the rate of reaction with other optically active regents.
vi) Equimolar mixture of pair of enantiomers forms optically inactive mixture called racemic
mixture.
6.9.2.4: Diastereomers: Non superimposable stereoisomers of an optically active compounds
which are not mirror images of each other are called diastereomers. E.g. 3-Chlorobutane-2-ol.
CH3
CH3
CH3
CH3
H
OH
HO
H
H
H
Cl
Cl
H
Cl
OH
H
HO
H
H
Cl
CH3
CH3
CH3
CH3
(I)
(II)
(III)
(IV)
Figure 6.13: Isomers of 3-chloro butan 2-ol
All these structures represent optically active stereoisomers of 3 Choro-butane-2-ol. Structure I, II
and III, IV are mirror images of each other. Therefore, they form enantiomeric pairs. But structure
I is not mirror image of III or IV and vice versa. So (I - III & I - IV) form two pairs of diastereomers.
Similarly, structure II is not mirror image of III & IV. So (II - III) & (II- IV) form two more pair of
diastereomers. Molecules containing only one chiral center do not have diastereomers.
Diastereomers contains more than one chiral center. That is for compound to be diastereomeric it
should have more than one chiral center. A molecular containing n number of dissimilar chiral
center have 2n number of pairs of diastereomers.
Properties of Diastereomers:
1. Diastereomers may or may not be optically active.
2. Diastereomers possesses different physical properties like M.P. B.P. densities & specific
rotation etc.
3. They show similar but not identical chemical properties.
6.9.2.5: MESO Compounds:
Meso compounds are compounds that contain chiral carbons but can be superimposed on their
mirror image. Meso compounds are compounds that have chiral carbon (2) within the molecule,
but the compound is optically inactive. Actually, meso compounds have two chiral carbons and
there is a plane of symmetry that will have half the molecule on one side of plane of symmetry
while other half which is mirror image of first half on the other side of plane.’
COOH
COOH
H
OH
H
OH
COOH
≡
HO
H
HO
H
COOH
Figure 6.14: Mesotartaric acid
The meso compound is optically inactive because as the plane polarized light passes through the
molecule, half the molecule will rotate the light to the right & other half of the molecular will rotate
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the light back in other direction by same number of degrees. This means that the light exists the
sample with no apparent rotation.
6.9.2.6: Racemic mixture:
A mixture consisting of equal parts of any pair of enantiomers is called a racemic mixture & it is
designated (+/-). A racemic mixture does not rotate plane polarized light because (+) rotation
caused by one enantiomer is cancelled by rotation in the opposite direction by the (-) enantiomer.
A solution of racemic mixture of enantiomer is optically inactive. Clearly, unequal mixture of two
enantiomer will have lower optical rotation than a pure enantiomer & strength of this rotation
depends on enantiomeric excess. (i.e.) of the mixture.
Let’s check the take away from this lecture
1) Which of the following is capable of existing as a pair of enantiomers?
a) 2-methylpropane
c) 3-methylpentane
b) 2-methylpentane
d) 3-methylhexane
2) Which of the following can exist as diastereomers?
a) Lactic acid
b) 1-Butene
c) 2-Butene
d) Ethane
3) Diastereomers are
I) Non-Superimposable II) Non mirror images
III) super imposable
IV) mirror images
a) I & II
b) III & II
c) I & IV
d) III & IV
Exercise
Q.1
Write a short note on i) Enantiomers, ii) Meso Compounds
Questions/Problems for practice:
Q.2
Write a short note on i) Diastereomers, ii) Racemic Mixture
Q.3
Explain the optical isomerism in tartaric acid
Learning from this lecture: Learners will be able to understand the difference between various
terms such as enantiomers, Diastereomers, Meso compounds and racemic mixture.
Lecture: 42
Learning objective: In this lecture learners will be able to represent stereoisomers by various
means of representation like Flying Wedge, Fischer representation etc.
6.9.3: Representation of Stereoisomers:
Majorities of molecules are three dimensional. Properties of these are described by taking three
perpendicular planes on axis. Two dimensional molecules can be presented on plane of paper. But
for three dimensional it is not possible to present in 2D plane, because plane of paper is 2D.
Therefore, one need to develop a method to modify 3D molecule in to 2D in order to present on
plane of paper or board. Presentation of 3d molecules, on 2d plane by following certain rules are
called as molecular representation.
6.9.3.1: Flying Wedge Representation or Wedge Representation/Projection
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i) Normal line
H3 C
CH3
ii) Heavy wedge
H 3C
CH3
iii) Broken wedge
H3 C
CH3
iv) Wavy lines
H3C
CH3
Normal lines will show groups or atoms arrangement in to the plane of paper or board. Heavy
wedge will show orientation or arrangement of groups of atoms, near to observer, in front of plane
or above the plane. Broken Wedge will indicate groups or atoms orientation away from the
observer or behind the plane or below the plane. Wavy lines- will indicate stereochemistry, not
specified, i.e atom or group above or below the plane.
In following figure 6.15 of flying wedge representation, in structure (I) a and b are nearer to
observer or above the plane and X, Y are away from the observer or below the plane. In structure
(II) a is away from the observer, b is nearer to observer and X, Y are in the plane. Similarly, in
structure (III) Y is nearer to observer or above the plane, b and X are below the plane or away from
the observer and a is in the plane.
Figure 6.15: Flying Wedge Representation
6.9.3:2: Fischer Representation/Formulae
Fischer developed a two-dimensional plane projection formula for three dimensional molecules,
these projections are particularly useful for carbohydrates & amino acids and can also be used to
represent many chiral molecules. Fischer projections are particularly useful for compounds with
two or more stereocenters.
Rules:
i) Identify the no of carbons in the molecule
ii) Keep maximum no carbons on vertical line
iii) Try to place highly oxidized, carbon at the top of vertical line.
iv) Each cross represents carbon or stereocenters (meeting point of vertical and horizontal line is
the cross in Fischer projections)
v) If asymmetric center is present, use crosses only for asymmetric center.
vi) All the Fischer projections are eclipsed, zero dihedral angle.
vii) Always remember 180° rotation in plane produces identical structure.
viii) Always remember 90° rotation in plane produces identical structure.
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ix) Any two groups mutually exchange at chiral center in Fischer produces enantiomers, if
molecule is having only one chiral center, more than one chiral center produces enantiomers.
x) Any three groups mutually exchange at chiral center produces identical (No change of
configuration)
xi) At chiral center group of vertical line away from the observer, produces a horizontal line near
to observer.
xii) In mirror images of Fischer, horizontal group has to be opposed.
Following figure 6.16 shows the examples of Fischer representation. In first, two examples Fischer
projection of ethane (I) and tartaric acid (II) has converted in to wedge representation. In next four
examples lactic acid (III), alanine (IV), glyceraldehyde (V) and mandelic acid’s (VI) Wedge
representation has converted in to Fischer representation.
Figure 6.16: Examples of Fischer and Wedge Representation
In following figure 6.17 some more examples of Fischer projections were given where different
isomers were produced after rotation at 90° (structure I & II) and 180° (structure III to VI). Here we
can observe that when molecule rotates to 90° on plane of paper it produces enantiomers and when
molecule rotates to 180° on plane of paper it produces identical image. It has been illustrated with
following examples of lactic acid, Alanine and Tartaric acid.
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Figure 6.17: Examples of Fischer Representation
Let’s check the take away from this lecture
1) If we rotate a molecule at 90°, what it will produce?
a) identical
c) Diastereomers
b) enantiomers
d) none of the above
2) If we rotate a molecule at 180°, what it will produce?
a) identical
c) Disteriomers
b) enantiomers
d) none of the above
3) What is the significance of heavy lines in Flying Wedge formula
a) Groups of atoms are away from observer
b) Groups of atoms are nearer to observer
c) Groups of atoms may nearer or away from observer
d) None of the above
Exercise
Q.1
Give the flying Wedge structure for the following compounds
a)
b)
c)
Cl
H
H3C
I
Br
I
COOH
H
Br
H
NH2
CH3
Questions/Problems for practice:
Q.2
What will happen if meso and active tartaric will rotate at 180°, explain with structure.
Learning from this lecture: Learners will be able to understand how to convert Flying wedge
structure in to Fischer form and vice versa.
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Lecture: 43
Learning objective: In this lecture learners will be able distinguish maleic and fumaric acid.
6.9.4: Geometrical Isomerism:
The stereoisomers in which atoms or groups are held in different spatial positions due to restricted
rotation about some bond are called geometrical or cis-trans isomers.
This phenomenon which gives rise to geometrical isomers or cis-trans isomerism. It is special kind
of stereoisomerism in which molecule acquires different stable configuration because of hindered
rotation in molecule. It is shown by substituted define & sing compounds, oximes etc.
Geometric Isomerism in Maleic & Fumaric Acid: - Maleic acid & Fumaric acid are geometric
isomers of butene dionic acid (HOO C-CH = CH-COOH). Butene dionic acid exists in two isomeric
forms due to restricted rotation across C=C bond axis. They cannot be easily interconverted. The
two forms differ in most of their physical properties but resemble in chemical properties. So, they
exist as individual compounds.
In maleic acid the two -COOH groups (hence two H atoms) are on the same size of double bond.
This isomer is termed as cis isomer (Figure 6.18 (I)).
Figure 6.18: Maleic and Fumaric acid
In fumaric Acid two -COOH groups (hence two -H atoms) are on opposite side of double bond
(Figure 6.18 (II)). The isomer is termed trans isomer. The physical & chemical properties are useful
to distinguish the cis & trans conformation.
i)
ii)
The M.P, B.P. and stability of Cis isomer is lower than that of trans isomer.
The density, solubility, dipole moment. R.I. & heat of combustion of cis form is higher than that
of trans form.
iii) The two isomers can be distinguished & separated by anhydride formation.
Maleic acid (cis form) readily forms anhydride on heating only at 413k. Fumaric Acid (trans form)
remains unaffected at this temperature, because two COOH groups are for opposite from each other
due to trans configuration. Both reactions are shown figure 6.19.
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Figure 6.19: Reactions of Maleic and Fumaric acid on heating
In Following figure 6.20 some more examples of olefinic geometric isomers have been given.
Figure 6.20: Additional examples of geometric isomers.
6.9.4.1: Geometrical isomerism in oximes (>C= N-): Oximes (aldoximes a=H & ketoxime a= R) are crystalline compounds formed by the condensation
of aldehydes & Ketone’s with hydroxyl anime. In oximes both C & N are sp2 hybridized nitrogen
uses two halves filled sp2 hybrid orbitals to bind with C & OH (Figure 6.20). While the third sp2
orbital containing lone pair of electrons (nonbonding orbital) stays in the plane of C, N & OH.
Figure.6.20: Molecular Orbital structure of >C=N bond of Oxime.
Molecular orbital structure of C= N bond of oximes. The unhybridized, half-filled P- orbitals of
C=N overlap laterally forming п – bonds. Thus, the C & N atom are joined by a double bond. The
OH group & nonbonding orbital of N stay opposite to the line joining of C & N. The double bond
(C=N) restricts the movement of atoms or group attached so they remain fixed in space giving rise
to geometrical isomerism. This type of isomerism is shown by oximes of aldehydes &
unsymmetrical ketones.
Isomers of oximes are named using the prefixes syn & anti instead of cis & trans respectively. In
aldoximes these prefixes indicate the relationship between the hydrogen atom & hydroxyl group,
when H & OH lie on the same side of double bond then the configuration is known as syn & if they
lie on opposite side of double bond it is named as anti. Thus, in figure 6.21 shows the various
examples of acetaldoxime, benzaldoximes and ketoximes.
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Figure.6.21: Various examples of oximes
Thus, structure (V) is called syn methyl phenyl ketoximes, implies the CH3 & OH groups on same
side of double bond. It may also be named as anti-phenyl methyl ketoxime because Ph & OH group
are on opposite sides. Similarly, structure (VI) can be named as syn-phenyl methyl ketoxime or
anti-methyl phenyl ketoxime because of same reason. The syn & anti forms have different physical
& chemical properties.
6.9.5: Nomenclature of Stereoisomers: 6.9.5.1: D & L system: i) Draw the molecule in the form of Fischer configuration.
ii) Select a bottom most asymmetric Centre.
iii) At bottom most asymmetric Centre is more electronegative group or atom at right side then
configuration is D.
iv) At bottom most asymmetric Centre, if more electronegative group or atom at the left side
then configuration is L.
v) Relation between D &L is enantiomeric or diastereomeric.
e.g. Following Figure.6.22 shows the various examples of D-L relationship at single chiral centers.
Figure.6.22: D-L relationship at single chiral center
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e.g. Following Figure.6.23 shows the various examples of D-L relationship at multiple chiral
centers.
Figure.6.23: D-L relationship at multiple chiral center
Let’s check the take away from this lecture
1) Maleic acid & Fumaric acid are geometric isomers of _______?
a) butene dionic acid
c) butane dionic acid
b) butene trionic acid
d) butane trionic acid
2) _________isomer is able to form a anhydride on heating at 413K.
a) Fumaric acid
b) Maleic acid
c) malic acid
d) none of the above
3) Relation between D and L lactic acid is
a) Enantiomeric b) Diastereomeric
d) none of the above
c) Identical
Exercise
Q.1
What is geometrical isomerism? Discuss geometrical isomerism in maleic and fumaric acid.
Questions/Problems for practice:
Q.2
Give syn -anti nomaniclature for following compounds
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Learning from this lecture: Learners will be able to state the properties and differences of maleic
and fumaric acid. Learner will also able to give syn and anti-nomenclature for oximes.
Lecture: 44
Learning objective: In this lecture learners will be able assign R, S, D, L and E, Z nomenclature
to various molecules.
6.9.5.2: R-S Configuration (Absolute Configuration)
A most advanced & superior system was developed by Cahn. Ingold & Prelog to represent the
absoulate configuration. It is based on actual three dimensional formula of a compound. It imvolves
following steps.
R → RECTUS → Clockwise
S → SINISTER → Anticlockwise
Rules:1.
2.
3.
4.
Identity the asymmetric centre in the molecule.
Give the priority for the attached atoms or groups according to their atomic numbers.
Higher atomic no-atom/groups gets top priority.
After assigning priority for groups at asymmetric centre, see the priority dissection (higher to
lower).
5. Its priority direction, clockwise configuration → R priority direction, anticlockwise,
configuration → S.
Sub Rules: 1. Keep least priority group away from the observer.
2. After keeping least priority group away from the observer. See the priority direction.
3. It Fischer projection given, least priority, on vertical line, select priority direction according to
sequence of atomic numbers, then assign configuration R & S.
4. It least priority group or atom on horizontal line, try to keep this least priority atom or groups
on vertical line, by doing even no. of mutual exchange of pair of groups or reverse the
configuration.
5. At asymmetric center, its isotope of elements present give priority according to their matters.
6. At asymmetric centre, if directly attached atoms of groups same, select next atom in sequence
then decide priority order.
7. At asymmetric centre if unsaturated groups present do the duplication or triplication.
8. In the priority R → (Precedes) S
i.e. R configuration group gets more priority over S configuration group.
Or Cis precedes trans group
Or Z precedes E group.
9. Priority Order: - I > Br > Cl > SO3H > SH > F > OCOR > OR > OH > NO3 > NO2 > NR2 > NHR
> NH2 > COOR > COOH > COR > CHO > CH2OH > CN > PH > CR3 > CHR2 > CH2R > CH3 >
D>H
Examples
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Consider the first example in following figure, where chiral center will be following priority like I
> Br > CH3 > H. So here least priority group is on vertical line therefore assigned configuration is
R.
Figure.6.25: Examples of R-S nomenclature
Here, in the following figure 6.26, example (I) shows the least priority atom (H) is on horizontal
line & which is nearer to observer. But according to R/S rules it must be away from the observer
must be away from the observer therefore should be kept on vertical line. Therefore, for this
purpose even no of mutual exchange has to be done as shown in following figure 6.26 (II). There
are two possible way of mutual exchange (a & b) as shown in figure 6.26 (II). This will be one
of the ways to predict R-S nomenclature for such molecules and other way is just simply reverse
the configuration.
Another important point we can note from R- S nomenclature is that, Enantiomer’s will have
always opposite configuration at all asymmetric Centre. i.e. R is enantiomer of S or vice versa.
In following examples (Figure 6.26 III- IV & V-VI) configuration changes from R→ S or S → R
as least priority group is on horizontal line.
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Figure.6.26: Additional examples of R-S nomenclature.
R-S nomenclature for Meso tartaric Acid: - In meso compound configuration of chiral center is having
opposite configuration as we observe it in figure 6.27 (I). If two isomers of compounds are mirror
images of each other, then relationship between them is enantiomers. For e.g. structure II and III in
following figure 6.27 shows the enantiomeric relationship. So here we can say that relationship
between RR and SS tartaric acid is enantiomeric. i.e. RR Tartaric acid and SS tartaric acid are
enantiomers of each other’s.
Figure.6.27: Relationship between isomers of tartaric acid
If in two isomers of compound at same chiral center same configuration is there and at other chiral
center if opposite configuration is there then the relationship of isomer is diastereomers. For e.g.
structure I and II and similarly structure I and III are not mirror images of each other therefore they
are diastereomers of each other. So here we can say that relationship between RS and RR is
diastereomeric and similarly relationship between RS and SS is also diastereomeric.
R-S nomenclature for Stilbene Dibromide: From, following example (Figure 6.28) of stilbene dibromide
structure I and II and similarly structure III and IV are mirror images of each there therefore they
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are enantiomers of each other. Similarly structure I and III and structure II and IV are not mirror
images of each other therefore they are diastereomers of each other.
Figure.6.28: Relationship between isomers of stilbene dibromide
This, in short, we can summaries for stilbene dibromide that SR  RS and RR  SS relationship is
enantiomeric whereas SR (I)  RR (III) and RS (II)  SS (IV) relationship is diastereomeric.
Thus, relationship of various isomers of stilbene dibromide can be shown as.
6.9.5.3: E & Z systems: The cis - trans terminology is inadequate to describe the geometrical isomers of olefins compounds
containing all four non- equivalent substituents. Cis- trans Nomenclature not possible in following
figure 6.29 (V) and (VI) example because similar substituent is not at double bond. In order to
differentiate these types of geometrical isomers E & Z nomenclature is introduced.
For e.g.
a
a
a
b
a
a
a
c
b
b
b
a
b
c
b
a
Cis (I)
Trans (II)
Cis (III)
Trans (IV)
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190
c
a
d
Not possible to give Cis -Trans
Nomaniclature
and
b
d
(v)
b
c
(vI)
Figure.6.29: Possibility and Non possibility of Cis-Trans nomenclature.
Rules: 1. Give the priority for the groups at each sp2 carbon A/C to R-S nomenclature rules.
2. On double bond or unsaturation, its same priority groups on same side nomenclature or
configuration is Z (zussamen meaning together)
3. If same priority groups on opposite side nomenclature or configuration is E (Entgegen s)
meaning opposite.
Following figure 6.30 shows the various examples of E and Z isomers.
Figure 6.30: Various examples of E and Z isomers
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E-Z nomenclature is not only applicable for C=C bond system but also applicable for other system
as well. For e.g. butanone oxime and acetophenone Oxime has shown in following figure 6.31. (ε. z
configuration)
Figure 6.31: E and Z isomers of butanone and acetophenone Oxime
Let’s check the take away from this lecture
1) Correct sequence of following group according to priority order is
a) CHO > COOR > CN > PH
c) PH >COOR > CHO > CN
b) CN > PH > COOR > CHO
d) COOR > CHO > CN > PH
2) The relationship between E & Z isomers of 2 butene is?
a) Enantiomeric
b) diastereomeric
c) meso
d) racemic
3) Among the following which of the following will not show cis-trans isomerism?
Exercise
Q.1
What is E & Z isomerism? Explain, why its need arises?
Questions/Problems for practice:
Q.2
Give R-S nomenclatures for the following compounds
Q.3
Give E-Z nomenclatures for the following compounds
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Learning from this lecture: Learners will be able to assign R-S and E-Z nomenclatures to given
compounds.
Lecture: 45
Learning objective: In this lecture learners will be able draw various possible forms of ethane.
6.9.6: Newman Projection:
In this projection, the molecule is viewed along the bond joining of two carbon atoms as shown
in following figure (6.32, (I)). The front carbon atom is shown by three lines (three atoms) which
come together in Y shape. The back carbon is shown by circle with three bonds pointing out
from it as shown in following figure (6.32, (II)). By this projection, it is easy to observe
relationships of atoms or groups at front or back carbons. This is one of the best representations
for conformational analysis.
Figure. 6.32: Front and back view of carbon for C- C bond (I) and Newman projection (II).
Rules:
i) See the molecules along the C-C bond axis.
ii) Carbon nearer to observer (front carbon) shown as dot.
iii) Carbon away from the observer (back carbon) shown as circle.
iv) Observer cannot see back carbon atom.
v) Place three lines with 120° of separation at front and back carbons
vi) Molecule must be with minimum two carbon atoms.
6.9.6.1: Representation of Newman projections:
Eclipsed Conformation (I): In this conformation, the C-H bonds of front and back carbon are
aligned with each other with dihedral angle of 0° as shown in following figure (6.33, I).
Gauche Conformation (II): This is the confirmation in which the relationship between two
atoms or groups whose dihedral angle is more than 0 o (i.e., eclipsed) but less than 120o (i.e.,
the next eclipsed conformation) as shown in following figure (6.33, II).
Staggered conformation (III): In the staggered conformation the C-H bond on back carbon lie
between those on front carbon with dihedral angle of 60° following figure (6.33, III)
Chemistry Sem – I/II
193
Figure 6.33: Representation of various forms of Newman projections.
Rotation about C-C bond is possible by both clockwise and anticlockwise directions (front or back)
in one step with minimum 60° or multiple of 60°. Following figures (6.34) shows the examples of
conversion of Fischer projection in to Newman projections and conversion of one form into another.
4
A
C
3
C
2
1
B
C
B
C
C
H
B
B
60°
back
C
A
A
B
AA
H
H
60°
back
H
A
C
AB
(IV) Eclipsed
(III) Gauche
(II) Eclipsed
(I) Fischer
B
C
A
B
60°
back
A
B
C
B
C
H
H
B
C
60°
back
B
B
C
H
60°
back
C
A
C
60°
back
A
A A
A
(VIII) Starting Configuration
B
H
B
C
AC
A
(V) Staggered (B, B)
(VI) Eclipsed
(VII) Gauche
B
Figure 6.34: Fischer projection in to Newman projection & various conformers by Newman.
Following figure (6.35) gives the example of (2S, 3S) butane 2, 3 diols. Where Fischer projection has
been converted into Newman projection and further possible conformers have shown for it by
rotating back carbon at 60°.
OH
4
3
HO
2
H
1
OH
H
CH3
OH
H
H
H
OH
60°
back
H
CH3
H
CH3 CH3
CH3
H
(I) Fischer
H
60°
back
OH
CH3H
H
CH3
(IV) Eclipsed (OH)
(III) Gauche (OH)
(II) Eclipsed
OH
CH3 H
OH
60°
back
CH3
H
OH
H
OH
H
H
H
60°
back
CH CH3
H 3
(VIII) Starting Configuration
H
OH
H
CH
H 3
60°
back
OH
CH3
CH3
(VII) Staggered
OH
CH3 OH
H
(VI) Eclipsed
H
60°
back
OH
H
OH
CH3
(V) Gauche (OH)
Figure 6.35: Possible conformers of 2S, 3S butane 2, 3 diols by Newman projection.
6.9.6.2: Conformations: As we have already discussed conformations are different forms of molecule, related by simple
rotation about single bond. For example, alkanes of two or more carbons can be twisted into a
number of different three-dimensional arrangement by rotation about carbon single bond. The
different spatial arrangements that molecule can adopt due to rotation about 6 bonds are called
conformations & hence conformational isomers or conformers.
Module 6: Stereochemistry
194
Such inter conversations are very rapid, so that sample of given molecule may exists in many
different conformations. Different conformers can interconvert by rotation around single bonds,
without breaking chemical bonds.
How conformational isomers arise? - The existence of more than one conformation is due to
hindered rotation of sp3 hybridized carbon bonds. Let us explain this by taking the simple example
of ethane which has carbon – carbon sigma bond is cylindrical or symmetrical in nature & hence
permits free rotation about its axis. Hence the energetic barrier to rotation about sigma bonds is
generally very low.
6.9.6.3: Conformation of Ethane:
Two of the remarkable conformation of ethane are eclipsed conformation & staggered
conformations as per shown in figure 6.36. While all other intermediate conformation is called
skewed. Eclipsed bond is characterized by dihedral angle of 00. With the dihedral angle around 600
the spatial relationship is gauche & when it is 1800 the relationships are anti. Gauche or anti
relationships between bonds on adjacent atoms can only exists in staggered conformations.
The staggered conformation of ethane are the low energy forms while the eclipsed conform actions
represents transition states. The energy difference between two staggered conformers is very close
to about 3 kcal / mol (12kJ / mol). The destabilization associated with the eclipsing of bonds on
adjacent atoms is called torsional strain. Torsional strain is the force that oppose the rotation of one
part of the molecule about a bond when the other part is held right. The torsional strain in eclipsed
ethane is due to slight repulsion of electron pairs of adjacent C-H bonds as they rotate past one
another in converting from one staggered to another.
Three pairs of eclipsed bonds produce about 3Kcal / mol (12 KJ / mol) of torsional strain in ethane.
Thus, in an alkane in general each pair of eclipsed bonds that it contains leads to an energy cost of
about 1Kcal/mol (4.2 KJ/mol). This torsional strain is primarily due to electron pair repulsions
when the bonds are eclipsed.
The activation energy for rotation about the C-C bond in ethane is small, the thermal energy from
the surrounding is sufficient to cause staggered conformation of ethane to interconvert millions of
times each second at room temperature.
Figure 6.36: Conformational energy diagram for ethane.
Chemistry Sem – I/II
195
6.9.6.4: Conformation of Butane:
In butane it is the rotation about the C2-C3 bond that is of most interest since the relative position of
the two methyl groups is important. Butane has not only eclipsed & staggered conformation but
also have other forms that vary in the relative orientation of the methyl group.
a) Fully Eclipsed (Syn): - The structure shown at 00 is fully eclipsed, that is both methyl groups
are alighted & are interacting as shown in figure 6.37 (I).
b) Gauche: - The two staggered forms with the methyl groups in closer proximity (60 0) are
gauche conformation. As the front methyl group is rotated 600 in structure I, a gauche
conformation is produced as shown in figure 6.37 (II). The two staggered forms with the
methyl groups in closer proximity (600) (II & VI) are gauche conformation.
c) Partially eclipsed: - Another 600 rotation of conformation II produces an eclipsed version of
the gauche conformation. Further rotations generous an equivalent eclipsed gauche
conformer (at 2400), another gauche form (3000) & finally the eclipsed form at 3600.
Figure 6.37: - The anti and gauche conformations of butane.
Energy diagram for rotation about the C-C bond in butane (conformational analysis) as with ethane
the eclipsed conformations are higher energy than the staggered groups as eclipsed conformers are
in close proximity with the molecule, therefore the potential energy is at its highest value.
Figure 6.38: - The anti and gauche conformations of butane.
Module 6: Stereochemistry
196
When the dihedral angle = 60° or 300° (Gauche conformation) in butane, the methyl groups are
further apart and therefore the potential energy drops by 4.1Kcal/mol due to stearic hindrance
between the two methyl groups. Such an interaction referred to as gauche-butane interactions
because butane is first alkane to discovered to exhibit such an effect. When the dihedral angle =
120° or 240° (eclipsed conformation) in butane the methyl groups align themselves with hydrogen
on the same plane, resulting in the butane having a higher potential energy than the gauche
conformations though still having a lower potential energy than the syn conformation (dihedral
angle = 0 or 360°) by 1.5 kcal/mol. The staggered conformation (when the dihedral angle = 180°)
where the two methyl groups are far away from one another as possible. i.e. anti-conformation, is
the lowest energy arrangement.
Figure: Conformational energy diagram for butane.
Let’s check the take away from this lecture
1) In butane, for syn conformation dihedral angle is?
a) 0°
b) 360°
c) both of the above
d) none of the above
2) In butane, for anti-conformation dihedral angle is?
a) 0°
b) 90°
c) 180°
d) 300°
3) In butane which of the following conformation is having lowest energy?
Chemistry Sem – I/II
a) Eclipsed
197
b) antiperiplanar
c) Gauche
d) None of the above
Exercise
Q.1
Explain all possible conformations of butane.
Questions/Problems for practice:
Q.2
Explain conformational analysis of butane with conformational energy diagram.
Learning from this lecture: Learners will be able to understand the difference between various
conformations of butane.
Conclusion:
Stereochemistry may seem like a minor subject because differences between stereoisomers are
usually indistinct. In nature, however, and most significantly, in biological systems such as the
human body, these indistinct differences have extensive implications. Most drugs for example, are
often composed of a single stereoisomer of a compound, and while one stereoisomer may have
positive effects on the body, another stereoisomer may be toxic. Because of this, a great deal of work
done by synthetic organic chemists today is in devising methods to synthesize compounds that are
purely one stereoisomer. Therefore, understanding of stereochemistry is very important for
synthetic organic chemist.
Add to Knowledge: An often-cited example of the importance of stereochemistry relates to the
thalidomide disaster. Thalidomide is a pharmaceutical drug, first prepared in 1957 in Germany,
prescribed for treating morning sickness in pregnant women. The drug was discovered to be
teratogenic, causing serious genetic damage to early embryonic growth and development, leading
to limb deformation in babies. Some of the several proposed mechanisms of teratogenicity involve
a different biological function for the (R)- and the (S)-thalidomide enantiomers. In the human body
however, thalidomide undergoes racemization: even if only one of the two enantiomers is
administered as a drug, the other enantiomer is produced as a result of metabolism. Accordingly, it
is incorrect to state that one stereoisomer is safe while the other is teratogenic. Thalidomide is
currently used for the treatment of other diseases, notably cancer and leprosy. Strict regulations and
controls have been enabled to avoid its use by pregnant women and prevent developmental
deformations. This disaster was a driving force behind requiring strict testing of drugs before
making them available to the public.
6.10: Set of Questions for FA/CE/IA/ESE
Short Answer Questions (1-2 marks)
1) What is enantiomers? Explain with suitable example.
2) What is diastereomers? Explain with suitable example.
3) Explain the term Plane of symmetry with suitable example.
4) Explain the term Geometrical isomerism with suitable examples?
5) Illustrate the following terms with suitable examples chiral centre and chiral molecules.
Module 6: Stereochemistry
198
6) How isomers are classified? Give their classification.
7) Explain the term Chirality with the help of Suitable examples.
8) What are the conformers? Give any three examples of it
9) One has to check, molecule is optically active or not, which conditions he should check?
10) Give the conditions of molecular asymmetry/ chirality
11) Predict following molecules are optically active or not?
i) Acetylene ii) Cis 1,3 dimethyl cyclobutane
iii) Trans 1,3 dimethyl cyclobutane
12) A pure enantiomer has an observed optical rotation of -0.92 when measured in 1 dm at a
concentration of 0.4 gm/10 ml calculate specific rotation for this molecule.
13) One of the analyst wants to identify, whether the molecule is dextrorotary or laevorotatory.
Which technique he should use and explain the term dextrorotary or laevorotatory.
14) Draw the possible stereo isomers of the following compounds and assign absolute configuration
to each of them.
i) CH3CH(OH)CH3
ii) C6H5CHNOH
iii) CH3CH(CN)CH2CH3
15) Predict following compounds will show geometrical isomerism or not?
i) CH3CH=CH2
ii) (CH3)2C=CHOH iii) Cl2C= CHCH3
Short Answer Questions (3 Marks)
1) Write a notes on a) Enantiomers, b) Disteriomers, c) Chiral centres
2) Define optical isomerism and explain Optical Isomerism in tartaric Acid.
3) Explain the optical isomerism in tartaric acid.
4) Discuss sterioisosomerism in 2 butene dioic acid.
5) How improper or alternating axis of symmetry is used to find the compound is optically active
or not? Explain with suitable example.
Long Answer Questions (5 Marks )
1) Write a short notes on i) Enantiomers ii) Chiral centres iii) E-Z nomaniclature iv) Meso tartaric
acid and Disteriomers.
2) Explain geometrical isomerism of oximes. Draw the geometrical isomers of the following oximes.
Give their names.
i) butaraldoxime
ii) methyl phenyl ketoxime
iii) benzyl ethyl ketoxime
iv) formaldoxime
3) What is Geometrical isomers? How physical and chemical properties are useful for
distinguishing maleic and fumaric acid explain with reaction wherever applicable.
4) What is optical isomerism? Which technique is useful for finding compound is optically active
or not. How plane polarised light is produced, explain it with the help of clean labelled diagram.
5) Name the symmetry element to determine the chirality of molecule. Explain why 1,2,3,4
tetramethyl cyclobutane is optically inactive? While trans 1, 2 dimethyl cyclobutane is optically
active?
Very Long Answer Questions (10 Marks)
1) What is stereochemistry? Give classification of isomers and explain it with suitable examples.
2) Write a short note on
i) Centre of symmetry
ii) Plane of symmetry
iii) Improper or alternating axis of symmetry
iv) Chirality
v) Meso Compounds
Chemistry Sem – I/II
199
3) State with reasons following molecule will show geometrical isomerism or not? and If yes give
structure of their isomers wherever applicable.
i) CH3C (OH) =C (CH3)2
ii) (CH3)2C = C (C2H5)2
iii) H3C C (Br) = C(CH3)2
iv) CH3C2H5 C = C (C2H5)2
(v) H3C-CH = CH-CH3
4) Why there was need arises for introducing Z and E nomenclature system?
Give the Z and E nomenclature for following compounds.
5) Give the R and S nomenclature of following compounds.
References:
1)
Basic Stereochemistry of Organic Molecule by Subrata Sen Gupta
2)
Stereochemistry of Organic Compounds: Principles and Applications by D. Nasipuri
3)
Stereochemistry: Conformation and Mechanism by Prof. P. S. Kalsi
4)
Basic Organic Stereochemistry, by Ernest L. Eliel and Michael P. Doyle
Module 6: Stereochemistry
200
Self-Assessment:
1) What is geometrical isomerism? Discuss geometrical isomerism in maleic and fumaric acid.
2) How conformational isomers are different than configurational isomers explain it with
suitable examples.
3) 0.9 gm of cholesterol was dissolved in 25 ml of CHCl3 taken in 100 mm polarimeter tube
observed optical rotation was (α) = -0.580. What is specific optical rotation?
4) Predict following molecules are optically active or not?
i) Acetylene
ii) Trans 1,3 dimethyl cyclobutane iii) Cis 1,3 dimethyl cyclobutene
5) Predict following compounds will show geometrical isomerism or not?
i) CH3CH=CHCH3
ii) (CH3)2C=CHCl
iii) F2C= CHCH3
Chemistry Sem – I/II
201
Self-evaluation
Name of
Student
Class
Roll No.
Subject
Module No.
Tick
S.No
Your choice
1.
2.
3.
4.
5.
Do you understand the various properties of
electromagnetic radiations?
o
Yes
o
o
No
Yes
o
No
Do you know the different components of
colorimeter and spectrophotometer?
o
Yes
o
No
Do you understand how structure can be
elucidated for organic molecule?
o
Yes
o
No
Do you understand module ?
o
Yes, Completely.
o
Partialy.
o
No, Not at all.
Do you understand the Beer Lamberts law and
numerical on it?
A. CO Mapping with Revised Bloom Taxonomy Level
Sr. No.
Course Outcome
Revised Bloom
Taxonomy
Level
CO1
Analyse microscopic chemistry in terms of atomic and molecular
orbitals and intermolecular forces.
L1, L2, L3
CO2
Apply the knowledge of instrumental method of analysis for
analysis of various samples.
L1, L2, L3
CO3
Understand and apply principles of catalysis and its application in
maintaining green matrix of reactions.
Understand electromagnetic spectrum used for exciting different
molecular energy levels in various spectroscopic techniques
L1, L2, L3
Understand and apply bulk properties and processes using
thermodynamic considerations .
L1, L2, L3
CO4
CO5
L1, L2, L3
CO6
L1, L2, L3.
Understand the stereochemistry and determination of structures of
organic compounds
B. CO and PO Mapping
PO
CO
CO1
CO2
CO3
CO4
CO5
CO6
PO1
PO2
H
H
H
M
H
H
M
H
H
H
M
H
PO3 PO4
M
M
H
L
M
M
PO5
PO6
M
H
M
H
H
M
H
H
M
H
L
M
PO7
PO8
PO9
PO10
L
M
M
L
L
H
H
H
H
H
L
PO11
PO12
H
M
M
H
L
L
M
C. CO, PO and PSO Mapping
PO
CO
CO1
CO2
CO3
CO4
CO5
CO6
PSO1
PSO2
PSO3
H
H
H
H
H
H
H
H
H
H
H
H
M
H
M
L
L
L