Elements of Molecular Spectroscopy [1 ed.] 9781906574413, 9781906574055

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Elements of Molecular Spectroscopy P S SINDHU

NEW ACADEMIC SCIENCE

New Academic Science

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Elements of Molecular Spectroscopy P S SINDHU Reader Dept of Chemistry Delhi University, India

NEW ACADEMIC SCIENCE

New Academic Science Limited The Control Centre, 11 A Little Mount Sion Tunbridge Wells, Kent TN1 1YS, UK www.newacademicscience.co.uk • e-mail: [email protected]

Copyright © 2012 by New Academic Science Limited The Control Centre, 11 A Little Mount Sion, Tunbridge Wells, Kent TN1 1YS, UK www.newacademicscience.co.uk • e-mail: [email protected]

ISBN : 978 1 906574 41 3 All rights reserved. No part of this book may be reproduced in any form, by photostat, microfilm, xerography, or any other means, or incorporated into any information retrieval system, electronic or mechanical, without the written permission of the copyright owner. British Library Cataloguing in Publication Data A Catalogue record for this book is available from the British Library Every effort has been made to make the book error free. However, the author and publisher have no warranty of any kind, expressed or implied, with regard to the documentation contained in this book.

Preface A first course on molecular spectroscopy requires change of perception that molecular energy levels are continuous but quantized energy levels; transition between which gives rise to a spectrum. Spectroscopy can be introduced through quantum mechanics but it is a lengthy and difficult approach. The other alternative is to make students conversant with spectra pictorially making spectroscopy easier and attractive. This approach involving visual images at the molecular level will excite every reader. Further, to invoke the interest of the reader and keep him engrossed in the subject, every chapter has a large number of solved problems. The approach followed in this book is quite innovative. As molecular spectroscopy is a mathematical-based subject, a total pictorial presentation is included. I have tried to make the subject understandable. Molecular spectroscopy finds applications in determining the structure of molecules and study of intermolecular interactions. Vibrational and electronic spectroscopy as applicable to the elucidation of structure of organic molecules have been introduced in Chapter 6 and 8 respectively. Application of crystal field theory and the ligand field theory to the prediction of electronic spectra and molecular structure of coordination complexes have been discussed in quite details in Chapter 9. PMR and ESR spectra have been introduced along with spin-spin interactions in Chapter 10 and 11. Chapter 12 has been prepared from latest research work where molecular spectroscopy has been used as a tool to the research problems. This gives the readers an idea as to how wide and divergent spectroscopy has its applications. Whenever spectroscopy has to be used as a tool, the user is invariably not confident with its basics and thus lacks confidence in its desired applications. This book is an effort in this direction. Every chapter has a number of solved problems besides additional good problems to solve. SI units have been used throughout the text. For every chapter some useful Internet sites have been given in Appendix IV. I am also thankful to my publisher for undertaking the publication of this book. P.S. Sindhu

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Contents 1 Quantization of Energy........................................................................... 1–21 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11

Introduction ...................................................................................................................... 1 Bohr model of hydrogen atom .......................................................................................... 1 Idea of wave-particle duality ............................................................................................ 7 Heisenberg uncertainty principle ...................................................................................... 9 Probability ....................................................................................................................... 10 Schrödinger wave equation ............................................................................................ 11 Significance of wave function, ψ .................................................................................. 14 Wave equation for hydrogen atom ................................................................................. 14 Spectra of hydrogen atoms ............................................................................................ 17 Use of atomic term symbols to describe atomic spectra .............................................. 17 Wave equations for some systems ................................................................................. 20

2 Introduction to Molecular Spectroscopy ............................................ 23–41 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10

Spectroscopy .................................................................................................................. 24 Nature of radiation .......................................................................................................... 26 Born-Oppenheimer approximation .................................................................................. 27 Absorption, emission and scattering spectra .................................................................. 30 Spectrophotometers ........................................................................................................ 30 Regions of electromagnetic radiations ............................................................................ 32 Intensity of spectral lines ................................................................................................ 33 Width of spectral lines .................................................................................................... 35 Scattering spectra—Raman spectroscopy ..................................................................... 38 Importance of molecular spectroscopy ......................................................................... 40 Problems ......................................................................................................................... 41

3 Particle in a Box .................................................................................... 43–49 3.1 Particle in one-dimensional box ...................................................................................... 43 3.2 Quantization of energy .................................................................................................... 44

❖ viii / Contents ❖

3.3 Zero point energy............................................................................................................ 45 3.4 Particle in a three-dimensional cubical box .................................................................... 47 3.5 Degeneracy ..................................................................................................................... 48

4 Rotational Spectra ................................................................................. 51–68 4.1 4.2 4.3 4.4 4.5 4.6

Introduction .................................................................................................................... 51 Diatomic molecules as a rigid rotator and energy levels ................................................ 52 Selection rules ................................................................................................................. 53 Effect of isotopic substitution ........................................................................................ 56 Intensity of spectral lines ................................................................................................ 58 Rotational Raman spectrum ............................................................................................ 62 Problems ......................................................................................................................... 67

5 Vibrational Spectra ............................................................................... 69–86 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9

A vibrating diatomic molecule ........................................................................................ 69 Selection rules ................................................................................................................. 72 Effect of isotopic substitution ........................................................................................ 75 Interaction between IR-radiation and vibrating molecules ............................................. 75 Anharmonicity in vibrations ............................................................................................ 76 Vibrational Raman spectra .............................................................................................. 80 Polyatomic molecules ..................................................................................................... 81 Carbon dioxide ................................................................................................................ 83 Water ............................................................................................................................... 84 Problems ......................................................................................................................... 86

6 Group Frequency Concept.................................................................. 87–106 6.1 6.2 6.3 6.4 6.5

Introduction .................................................................................................................... 87 Regions ........................................................................................................................... 91 Effect of hydrogen bonding ........................................................................................... 94 Application of IR spectra in structure determination ..................................................... 97 Sample preparation for spectra recording .................................................................... 103 Problems ....................................................................................................................... 105

7 Electronic Spectra ............................................................................. 107–123 7.1 7.2 7.3 7.4 7.5

Born-Oppenheimer approximation................................................................................ 107 Electronic transitions in heteronuclear diatomic moleclues ......................................... 107 Homonuclear diatomic molecules .................................................................................. 111 Franck-Condon principle .............................................................................................. 113 Dissociation energy ...................................................................................................... 115

❖ Contents / ix ❖

7.6 Dissociation energy and temperature ........................................................................... 117 7.7 Dissipation of energy by excited molecules ................................................................. 118 7.8 Phosphorescence .......................................................................................................... 121 Problems ....................................................................................................................... 122

8 Electronic Transitions in Organic Molecules ................................ 125–144 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8

Various bonds and transitions ....................................................................................... 125 Some important terms used in electronic transitions ................................................... 129 Absorption due to ethylene chromophore .................................................................... 131 Acetylenic and benzenoid chromophores ..................................................................... 135 Carbonyl chromophore ................................................................................................. 135 Solvent effects on electronic spectra ........................................................................... 137 Colour and constitution ................................................................................................ 139 Sample preparation ....................................................................................................... 141 Problems ....................................................................................................................... 142

9 Transition Metal Complexes and their Electronic Structures ...... 145–167 9.1 Crystal field theory ....................................................................................................... 145 9.2 Crystal field stabilization energy (CFSE) for weak and strong octahedral fields and pairing energies ............................................................................................. 148 9.3 Factors affecting magnitude of 10Dq or ∆ο ................................................................. 150 9.4 CFSE for tetrahedral symmetry ................................................................................... 152 9.5 Ligand field theory ........................................................................................................ 153 9.6 Electronic absorption spectra of complexes ................................................................ 154 9.7 Energy level diagrams ................................................................................................... 159 9.8 Jahn-Teller effect .......................................................................................................... 161 9.9 Jahn-Teller tetragonal distortion in octahedral symmetry ............................................ 162 9.10 Square planar coordination ........................................................................................... 165 Problems ....................................................................................................................... 166

10 Proton Magnetic Resonance .......................................................... 169–186 10.1 10.2 10.3 10.4 10.5 10.6

Magnetic moments ....................................................................................................... 169 Larmor precession ........................................................................................................ 170 Techniques and instrumentation ................................................................................... 172 Chemical shift ............................................................................................................... 174 Spin-spin interaction—Qualitative treatment ................................................................ 179 Solved examples ........................................................................................................... 182 Problems ....................................................................................................................... 185

❖ x / Contents ❖

11 Electron Paramagnetic Spin Resonance ....................................... 187–192 11.1 11.2 11.3 11.4

Introduction .................................................................................................................. 187 Hyperfine structure ....................................................................................................... 188 ESR spectra of radicals/ions with equivalent protons.................................................. 190 ESR spectra of aromatic radicals/ions ......................................................................... 191 Problems ....................................................................................................................... 192

12 Some Applications of Molecular Spectroscopy ............................ 193–197 • Nuclear magnetic resonance ......................................................................................... 193 • Microwave spectroscopy ............................................................................................. 194 • Infrared and Raman spectroscopy ............................................................................... 194 • Ultraviolet spectroscopy ............................................................................................... 196

Appendix I

: Photochemistry ........................................................................................ 199

Appendix II

: Time management ................................................................................... 213

Appendix III : References ............................................................................................... 217 Appendix IV : Some very good internet sites on spectroscopy ................................. 219 Appendix V

: Some relationships between commonly used units ............................ 220

Index .................................................................................................................................... 223

CHAPTER

1

Quantization of Energy 1.1

INTRODUCTION

In ancient times men had guessed that there were people, animals, land etc. beyond which there existed a world of the ultra small before it was actually discovered. Thinkers had meditated on the way nature had produced the world out of something quite formless. How was it, they queried, that it came to be inhabited by its great diversity of things. Nature might have worked like a builder that makes a large house out of small stones. Then what are these stones? Is there no limit to this division and subdivision of matter? Are these particles so small that even nature is no longer able to break them up? The answer was YES, so said the ancient philosophers. These particles were given the name ‘atom’. Their chief property was that no further division is possible. The word ‘atom’ is Greek, means ‘nondivisible’. What did an atom look like? In those times this question remained unanswered but as more experimental results accumulated, the understanding of atomic structure became more and more clear. An atom is made up of three basic particles, electrons, protons and neutrons. Their names and charges are as follows: Particle

Mass

Charge

Proton (mp)

1.672 × 10–27 kg

1.602 × 10–19 C

Neutron (mn)

1.674 × 10–27 kg

No charge

Electron (me)

9.109 × 10–31 kg

–1.602 × 10–19 C

= 1/1836 mp

1.2

BOHR MODEL OF HYDROGEN ATOM

On heating a body it emits electromagnetic radiation and when whole of this radiant energy is absorbed by another body, the system will be in thermal equilibrium. A body that absorbs all the radiations is called black body. Radiation emitted by a body in equilibrium with matter at a particular temperature is called black body radiation. At low temperatures, radiations of long wavelength are emitted. At higher temperature the amount of energy emitted is much greater and its principle component lies in the infrared region. At still higher temperature, black body glows dull red, then white and afterwards blue

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Elements of Molecular Spectroscopy

and the total amount of energy radiated increase dramatically. Classical mechanics could not account for this energy distribution in black body radiation at different temperatures. Max Planck derived the correct expression on the basis of Quantum Theory of Radiation. According to this theory: 1. Radiant energy is emitted or absorbed discontinuously in the form of tiny bundles of energy known as quanta. 2. Each quanta is associated with a definite amount of energy E depending upon the frequency of radiation, the two being related by the equation E = hν where E is energy in joules, ν is frequency of radiation in reciprocal seconds (s–1) and h is a fundamental constant known as Planck’s constant with a numerical value of 6.626 × 10–34Js. 3. A body can emit or absorb energy only in whole number multiples of quantum i.e., 1 hν, 2 hν, 3 hν… . Energy in fractions of quantum cannot be lost or absorbed. This is known as quantization of energy. Based on these postulates, Planck obtained the following expression for energy density of black body radiation. E(ν)dν =

8πhν 3 c

3

×

dν exp(hν/kT ) − 1

This equation adequately accounts for black body radiation intensity at all wavelengths obtained at different temperatures. An atom has a minute but massive positively charged body called nucleus at its centre. All the protons and neutrons are contained in the nucleus. Since the mass of an atom is entirely due to the presence of protons and neutrons, it is concluded that almost the entire mass resides in the nucleus. The positive charge of the nucleus is also due to the presence of protons. The electrons of atoms are distributed around the nucleus in a way about which we shall talk later on. Since the mass of the electron is 1/1836 the mass of proton, it does not contribute anything to the mass of the atom. The electrons carry charge equal but opposite to that of protons. As the atom is neutral, the number of electrons must be equal to number of protons in the nucleus. Most of the space around the nucleus is empty except for the presence of extremely minute electrons. However, as electron movement around nucleus is very fast, they cover almost all the space around the nucleus, and thus atom appears as a sphere. Bohr (1913) adopted these ideas for the hydrogen atom, a quite different system, and tried to give it a structure. He postulated Atomic Theory as: (i) An atom consists of a nucleus (containing protons and neutrons) surrounded by revolving electrons. The electrons move around the nucleus in a circular path called orbits. These orbits were numbered 1, 2 …. which turned out to be the principal quantum number of an orbit. (ii) The coulombic force of attraction between the electrons and nucleus hold the atom together. (iii) Even though an electron in a particular orbit is constantly accelerated, yet each orbit in an atom has a discrete energy. Each orbit is called a stationary state. As long as an electron is in a stationary state, it does not radiate any energy. (iv) Mathematical condition determines the size of an orbit. It states that the angular momentum of an electron is an integral multiple of h/2π, where h is a Planck’s constant. i.e.,

mvr = nh/2π

...(1.1)

where n = 1, 2, 3…. and is called the Principal Quantum Number. These numbers happen to be the same numbers as stated in the first postulate.

Quantization of Energy

3

(v) Emission or absorption of radiation energy takes place only if an electron jumps from one stationary state of energy E1 to another stationary state of energy E2. The frequency of this radiation is given by ν = (E2 – E1)/h Using these postulates let us try to give a structure to an atom.

(i) Radius of an Atomic Orbit Consider a nucleus with a positive charge + Ze and an electron with a negative charge – e revolving around nucleus with a velocity v, in a circular orbit of radius r. There are two forces: electrostatic force of attraction and centrifugal force, acting on an electron. Centrifugal force =

Electrostatic force of attraction =

Ze 2 4πε0 r 2

mv 2 r

. In order that an electron continues to move in a

particular stationary orbit, these two forces should balance one another. i.e.,

Ze 2 mv 2 = r 4πε0 r 2

1 2 1 Ze 2 mv = ...(1.2) 2 2 4πε 0 r According to the fourth postulate, the angular momentum of an electron is an integral multiple of h/2π. i.e., or

mvr =

nh 2π

nh 2πmr Substituting the value of v from equation 1.3 into 1.2 gives v=

r=

n 2 h2 ε0 Ze 2 mπ

...(1.1) ...(1.3)

...(1.4)

For a hydrogen atom in its ground state Z = 1, n =1 so that its radius is

r=

(6.626 × 10 −34 Js) 2 (8.85 × 10 −12 CNm −2 ) (1.672 × 10 −10 C) 2 (9.101 × 10 −11 kg )(3.141)

= 5.29 × 10–11 m = 52.9 pm This value is in very good agreement with the experimental value and was the first achievement of the Bohr model. From equation 1.4 the radius of the nth orbit can also be calculated as ...(1.5) rn = n2 r1 = 52.9 n2 pm

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Elements of Molecular Spectroscopy

(ii) Velocity of Electron Substituting the value of r from equation 1.4 in equation 1.2, we get v2 =

or

Z 2e4 4n 2 h 2 ε 20

Ze 2 2nh ε 0 Velocity of electron in hydrogen atom in ground state comes to v=

v=

e2 2h ε 0

...(1.6)

...(1.7)

On substituting the values of e, h, ε0, this relation gives a value of v = 2.188 × 106 ms–1. This high speed of the electron makes the atom appear spherical.

(iii) Energy of Stationary States The total energy of the electron in a stationary state is equal to the sum of its kinetic and potential energies.

1 Ze 2 2 m v − E= 2 4 πε 0 r Substituting the value of kinetic energy from equation 1.2 into 1.8, so that 1 Ze 2 Ze 2 1 Ze 2 − =− 2 4πε 0 r 4πε 0 r 2 4 πε 0 r Substituting the value of r from equation 1.4 in the above equation, we get E=

E= −

me 4 Z 2 8ε 02 n 2 h 2

...(1.8)

...(1.9)

...(1.10)

In this equation except n, all other parameters are constants; hence in a particular orbit electron has a discrete energy that can be calculated. The negative sign in this relation implies that the electron is bound to the nucleus by attractive force so that energy has to be supplied to the electron so as to get separated from nucleus. As ‘n’ increases, the numerical value of the energy decreases, but on account of the negative sign the actual energy will increase. This implies that outer orbits have greater energy than inner ones. Further, the electron and nucleus are infinitely far apart when n = ∞ , and E = 0 so that the atom get ionized. As they move close together, they are attracted and the energy of the system becomes less than zero, i.e., negative. Thus, the energy of an electron in a stationary state is negative as compared to the energy of a free electron.

(iv) Spectral Lines of Hydrogen Atom The energy of the hydrogen atom when the electron is in the n1th orbit E1 = −

me 4 Z 2 1 8ε 02 h 2 n12

and the energy of the atom when the electron is in the n2th orbit.

Quantization of Energy

5

E2 = −

me 4 Z 2

1

8ε 02 h 2

n 22

Applying the Bohr’s fifth postulate radiation energy (hν) with frequency ν is emitted when an electron jumps from the n2th to the n1th orbit and thus is given by hν = E2 – E1 =

me 4 Z 2  1 1 − 2 2 2  2 8ε 0 h  n1 n 2 

For the hydrogen atom, Z = 1 and ν=

where

me 4 8ε 02h 3

1 me 4  1 1 1 − 2  = RH  2 − 2  2 3  2 8ε 0 h  n1 n2   n1 n2 

...(1.11)

is called the Rydberg constant, RH with a value of 3.289 × 1015 Hz.

RH /c = 109724 cm–1. The electron in an atom keeps on moving in an orbit but when it is heated it absorbs energy and jumps from one energy state to another energy state. The excited electron returns from a higher energy level to one of the lower energy levels when an emission spectrum is obtained. Relation 1.11 gives the frequency of light so emitted. The five series of the hydrogen atom spectra are known by giving different values to n1 and n2 in equation 1.11. These spectral lines can be generated as shown in Table 1.1.

or

Table 1.1: Generation of spectral lines from equation 1.11

Name of spectra series

Spectral region of series

Frequency and value of n1 and n2

Lyman Series

UV Region

 1  ν = RH 1 − 2   n2 

n2 = 2, 3, ...

Balmer Series

Visible Region

1 1  ν = RH  − 2   4 n2 

n2 = 3, 4, 5, ...

Paschen Series

Near IR Region

1 1  ν = RH  − 2   9 n2 

n2 = 4, 5, ...

Brackett Series

IR Region

1 1 ν = RH  − 2  n2 = 5, 6, ... 16 n2 

Pfund Series

Far IR Region

1 1 ν = RH  − 2  n2 = 6, 7, ...  25 n2 

These transitions are shown diagrammatically in Fig. 1.1.

6

Elements of Molecular Spectroscopy Volts 14

cm–1

n

13.53 13

12

3

11

2

Balmer Series

10

6562.79 4861.33 4340.47 4101.74 3970.07 3889.05 3835.39 3797.90 18751.1 12818.1 Paschen Series 10938 10049.8 9546.2 4.05µ Brackett Series 2.63µ Pfund Series 7.40µ

0 6 5 4

9

10,000

20,000

30,000

40,000

8 50,000 7

60,000

5

Lyman Series 1215.66 1025.83 972.54 949.76 937.82

6

70,000

4 80,000 3 90,000 2

1,00,000 1

0

1

1,10,000

Fig. 1.1: Energy level diagram of the H atom and various transitions as depicted from Bohr structure of hydrogen atom

Spectral lines so obtained are in very good agreement with experimental results.

Quantization of Energy

7

Limitations of Bohr’s Theory 1. It fails altogether to give a quantitative explanation for spectra of atoms having more than one electron. 2. Bohr model picked up one hydrogen atom and tried to give it a structure. It does not deal with a collection of hydrogen atom . Hence this treatment does not lay a base to calculate the intensity of spectral lines. 3. It cannot explain the fine spectrum of hydrogen atom. In most cases, what were considered to be single lines, proved to be clusters of closely spaced lines detected by more sophisticated instruments. (i) On placing hydrogen in a magnetic field and then recording spectra, the single line in the earlier spectra split up into a number of closely spaced lines. This splitting is known as the Zeeman effect. (ii) It failed to explain the Stark effect. The effect of an electrostatic field on spectral splitting is called the Stark effect. 4. It could not lay a basis for the periodic table of elements. With further developments about the nature of subatomic particle, the idea of orbits of an electron was jeopardized and hence contradiction regarding the problems of instantaneous jump of the electron from one orbit to another was automatically eliminated.

1.3

IDEA OF WAVE-PARTICLE DUALITY

Classical physics acquaints us with two types of motions—corpuscular motion and wave motion. The first type is characterized by localization of the object in space as shown by a trajectory motion. The second type is characterized by delocalization in space. Localized objects do not corresponds to the wave motion. In the world of macro-phenomena, the corpuscular and wave motions are clearly distinguished. These usual concepts, however, cannot be transferred to quantum mechanics. The strict demarcation between the two types of motions is considerably obliterated in micro-particles. The motion of micro-particles is characterized simultaneously by wave as well as corpuscular properties. If macro-particles and waves are considered as two extreme cases of motion of matter, micro-particles must occupy in this scheme a place somewhere in between. They are not purely particles, they are not purely wave like, but they are something qualitatively different. Moreover, how much it is particle in nature and how much is wave like, depends on the conditions under which the micro-particle is considered. While in classical physics a corpuscle and a wave are two mutually exclusive extremities, these extremities at the level of micro-phenomena, combine dialectically within the framework of a single micro-particle. This is known as wave-particle duality. As early as 1917, Einstein suggested that quanta of radiation should be considered as particles called photon possessing not only a definite energy but also a definite momentum i.e., ∆E = hν and p = hν/c. His suggestion was based on his interpretation of photoelectric effect. When high frequency light falls on a metal surface electrons are emitted. This emission occurs only if the frequency of incident light is greater than a threshold value characteristic of the metal. Intensity of incident radiation plays no part in the emission. As the frequency is increased beyond the threshold value, kinetic energy of the emitted electrons increase linearly to the frequency. These facts could be explained as follows.

8

Elements of Molecular Spectroscopy

Light of frequency ν can impart energy only in discrete amounts of magnitude hν. A light beam of frequency possess a energy nhν, which could be regarded as containing n light corpuscles, called photons. These photons of threshold value collide against the metal electrons and knock them off leading to emission of electrons. In case photons have energy less than threshold value, no emission is expected. However, if photons have energy higher than threshold value, the energy excess to threshold value will be transferred to emitting electrons in the form of kinetic energy. In this interpretation intensity of radiation plays no part. In 1924, de Broglie suggested that duality should not only be extended to radiation but also to the micro-particles and the idea was confirmed by de Broglie himself in 1927 with his discovery of electron diffraction as shown by KCl crystal. For electrons the crystal lattice served as a diffraction grating, while studying the passage of electrons through a thin foil, Davison and Germer observed characteristic diffraction rings on the detector screen. Measurement of distances between diffraction rings for electrons of a given energy confirmed the de Broglie theory that every electron of momentum p is associated with a wave of wavelength λ given by λ =

h h = mv p

...(1.12)

Such waves do not move with the velocity of light but with a velocity v and are called metallic waves. Like electromagnetic waves, metallic waves are propagated in an absolute void; hence they are not mechanical waves. Since these waves move with the velocity of particle in motion, they are not electromagnetic waves. Let us take the examples of metallic waves.

PROBLEM 1.1 With the help of the Bohr model of the hydrogen atom one can calculate the velocity of an electron in the ground state from equation 1.7, which comes to be 2.188 × 106 ms–1. Assuming the mass of electron = 10–31 kg one can calculate the de Broglie wave associated with it. SOLUTION

λ=

6.6 × 10 −34 Js (10

−31

−1

kg ) (2.188 × 10 ms ) 6

= 3.01 × 10 −9 m

This is quite small and 10–9 corresponds approximately to the wavelength of X-ray, which can be easily detected. Thus particle nature is characterized by the momentum and wave nature by the wavelength. The two terms are inversely related to each other. For a macroscopic particle, which has a large value of momentum, the wavelength as calculated from the de Broglie relation is too large to be determined by experiment. For such a case the wave nature may be completely ignored and thus particle has corpuscular nature governed by classical mechanics. On the other hand, all atomic particles have dual character. X-rays pass through the crystal almost unimpeded while electrons are totally absorbed even in a millimetre thick crystal. Metal foils do show the electron diffraction pattern. Even when an electron beam was directed at a small angle to the crystal, a diffraction pattern was observed. If it were produced by a small number of diffracted electrons, at first glance it would appear that the electrons impinged randomly on the plate. But there is one thing that attracts our attention. We measure the aperture of the diaphragm from which the electrons emerged and project the outline onto the target. It would seem that all the electrons should fit inside this outline, no matter how randomly they had fallen on the photographic plate. However, some of the hits are far outside the boundary line.

Quantization of Energy

9

Every electron that hits the photographic plate, decompose Ag2S2O3 to Ag2S thereby leaving a black spot. If the number of hits on the target is small there are closely bunched black places. If a line is drawn through these places small rings appear. However, these rings are not well defined, but improve as the number of electrons striking the plate increases. Thus graph of electron hits on photographic plate is not a figment of imagination. It reflects the existence of a real wave of wavelength (h/mv) associated with an electron moving with velocity v.

1.4

HEISENBERG UNCERTAINTY PRINCIPLE

Consider a measurement of position of an atomic particle of mass m. If it has to be located within a distance ∆x then light with a wavelength of the size of a particle should be used to illuminate it. For the particle to be seen, a photon must collide in some way with the particle, for otherwise the photon will pass right through and the particle will appear transparent. The photon has a momentum, p = h/λ and during the collision some of the momentum will be transferred to the particle. The very act of locating the particle leads to a change in its momentum. If a particle has to be located more accurately, light of even greater momentum or smaller wavelength should be used. As some of the photon’s momentum is transferred to the particle in the process of locating it, the momentum change of the particle becomes greater. A careful analysis of this process was carried out by Warner Heisenberg who showed that it is not possible to determine exactly how much momentum is transferred to the electron. This means that if a particle has to be located within a region ∆x, then this causes an uncertainty in momentum of particle. Heisenberg was able to show that if ∆p is the uncertainty in the momentum, then ∆x . ∆p ≥

h 4π

...(1.13)

The smaller the ∆x, the greater the ∆p and vice versa, but the product of the two is always given by equation 1.13. This amounts to that it is impossible to measure simultaneously the position and momentum of a particle with arbitrary precision. Such an interpretation may mean that the uncertainty relation is responsible for limitations associated with the process of measurement. One might be led to assume that a micro-particle possesses a definite coordinate as well as definite momentum, but the uncertainty relation does not permit us to measure them simultaneously. This may be an erroneous conclusion. Thus uncertainty relation is not that which creates certain obstacles, to the understanding of micro phenomena but it reflects certain peculiarities of the objective properties of micro particles. The following two examples demonstrate the numerical consequence of the uncertainty principle.

PROBLEM 1.2 Calculate the uncertainty of position of an automobile of mass 500 kg moving with a speed of 50 + 0.001 km hr–1.

SOLUTION The uncertainty of position is ∆x =

1.05 × 10−34 Js
= = 3.779 × 10−2 m 4πm∆v 2(5 × 102 kg) (2.77 × 10−4 ms −1 )

This is a very small distance and wholly negligible as compared to the mass and velocity of the automobile.

10

Elements of Molecular Spectroscopy

PROBLEM 1.3 What is the uncertainty of momentum of an electron in an atom so that ∆x is 52.9 pm? SOLUTION ∆p =

1.05 × 10 −34 Js
= = 9.9 × 10 −25 kg ms −1 2∆x 2(52.9 × 10 −12 m)

Because of p = mv and mass of electron is 9.11 × 10–31 kg, this value of ∆v corresponds to: ∆v =

∆p 9.9 × 10 −25 kg ms −1 = = 1.087 × 10 6 ms −1 m 9.11 × 10 −31 kg

Compare this uncertainty in velocity with velocity of electron in hydrogen atom, which is 2.188 × 106 ms–1. This is a very large uncertainty in speed and cannot be neglected. These two examples show that although the Heisenberg Uncertainty principle is of no consequence for macroscopic bodies it has very important consequence in dealing with the atoms and subatomic particles. This is similar to the conclusion drawn for the application of de Broglie relation between wavelength and momentum.

1.5

PROBABILITY

In classical mechanics the position and velocity of any particle for any instant of time can be predicted with absolute certainty provided the forces acting upon it at that instant are known. However, it was evident almost immediately that one could not apply classical mechanics directly to the motion of gas molecules. The reason is quite simple. Even small volumes of gas, say 1 ml, contain 1019 molecules. Now to give an accurate picture of their motion would require writing and solving 1019 equations of motion. The gas molecules are never at rest, they are constantly colliding with other molecules bouncing off some, running into others and these events occur millions of times every second. It is preposterous even to imagine writing Newton’s equation for all molecules. Million of years would be spent in just writing down the equations. More millions and millions of years in solving them. Meanwhile others would have replaced all these motions. In search for a reasonable way out, scientists saw that they should not be interested in the motion of each individual molecule of gas colliding with other molecules with unbelievable rapidity. Rather, should their interest lie in the state of the entire mass of gas, its temperature, density, pressure and other characteristic properties? In other words, there is no need to determine the velocities of the separate molecules. All characteristics of the state of the gas should refer to the whole system of molecule as an assembly. Now mainly the mean velocity of the gas molecules determines these characteristics. For example, the higher the velocity, higher the temperature. If in the process the gas does not change its volume then there will be an increase in the pressure with rise in temperature. But to learn these relationships accurately, one had to find some way to determine the mean velocity of the molecules. Here was where the probability came in. The behaviour of a large assemble of molecules can be described statistically where random molecular motion has a definite form and hence every collision, every individual motion of a molecule could be described by Newtonian law and that if one desired to solve millions of millions of equations, he could express these motions with absolute precision and without any kind of mean values. We do

Quantization of Energy

11

not do that; of course in principle it could be done! We determine the motion of a gas by means of probability laws, but underlying them are the exact laws of Newtonian Mechanics. Due to the wave particle dual behaviour and uncertainty principle electrons, atoms and molecules cannot follow the laws of Newtonian mechanics. This amounts to saying exact motion of electrons cannot be traced even in principle. Hence quantum mechanics, also called wave mechanics, has been developed in which basic probability laws are incorporated but modified by wave particle duality and uncertainty principle. This will be explained by the following two examples: 1. Electron diffraction: The electron diffracted by a crystal hit upon the photographic plate to give dark concentric rings but not all the electrons are unique here. There are certain greyish places between the darkest and lightest sections. A mean number of electrons impinge on these portions. We see this very clearly on the distribution curve of hits in our shooting game. An electron leaves its source, passes through the diaphragm, is reflected from the crystal and is moving towards the photographic plate. Where will it hit the plate? Classical mechanics calculates the angle, distance and velocity with great accuracy and says, “HERE, which is usually not where it hits at all”. Wave mechanics says, “I do not know exactly but the greatest probability is that it would have hit the darkest rings, there is less probability that it would have hit the grey sections and it is hardly at all that it may have impinged on the light rings”. 2. Structure of atom: Bohr’s treatment say that electrons are moving in circular rings called orbits in which their exact position and velocity at any time can be located. But according to wave mechanics say we are not interested in exact position and velocity and thus need not tell how an electron is moving. Rather we say that there is certain probability of finding electrons in a particular orbital. This amounts to saying that every electron has some most probable distribution in an orbital and thus behaviour of these orbitals give characteristic properties to constituting atom.

1.6

SCHRÖDINGER WAVE EQUATION

Classical mechanics deals with observable parameters such as position and momentum as function of time or sometime of each other and Newton’s laws of motion enable these functions to be determined. Quantum mechanics recognizes that all the information about the system is contained in its wave function and that, in order to extract the information about the value of an observable parameter, some mathematical operation must be done on the function. This is analogous to the necessity of doing an act, an experiment on the system in order to make a measurement of its state. Quantum mechanics really boils down in making correct selection of the operation appropriate to the observable parameter. In the simple quantum mechanics that concerns us is it turns out that the right way to determine the momentum from a wave function is simply to differentiate it and then multiply the result by h/2πi, where i = − 1 . Thus, gradient of wave function at a particular point determines the momentum. The operator that extracts the position turns out to be simply multiplication by x, but this, as you can imagine, is deceptively simple. Once we know what the operators are for the dynamical variables of position and momentum, we can set up the operators for all observable parameters, because these can be expressed as a function of two basic variables. Thus, the kinetic energy in classical mechanics is a function of the momentum namely p2/2m, therefore, the corresponding operators can be obtained by ∂ 2 replacing p2 by (h/2πi ) . This shows that curvature of the wave function determines the kinetic energy. ∂x

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Elements of Molecular Spectroscopy

With the discovery of uncertainty principle and wave particle dualism, orbit concept became redundant and it became necessary to formulate a new mechanics to explain structure of atom. For this Schrödinger formulated wave mechanics, which is used like any other law of nature. The solutions to the wave equation are called wave functions, which give a complete description of the system and quantized energy values. Our concern in this book will be to evaluate quantized energy levels only and not to stationary state wave functions. This mechanics showed that fundamental laws of nature are not dynamic but are statistical and thus probabilistic form of casualty, while the classical determinism is just its limiting case. The first wave equation for a particle of mass m moving in space was given in the form of a statement as ∂ 2ψ ∂x

or

2

+

∂ 2ψ ∂y

2

+

∂ 2ψ ∂z

2

+

∇ 2ψ +

2m




( E − V )ψ = 0

2m ( E − V )ψ = 0


The Schrödinger equation for any system can be set up in the following ways: 1. A system of particles can be represented in terms of generalized coordinates qi and generalized ∂q j ( with j = 1, 2, . . . . , N) . This representation is described by a function velocities q j = dt ψ(q1, q2, …, qh), which determines all measurable quantities of the system. Any general law so formulated will be independent of the coordinate system. 2. In wave mechanics we deal with linear operators. An operator is a symbol that tells you to do something to whatever that follows. For example, consider dy/dx to be the d/dx operator, operating on the function y (x). An operator A is said to be linear if A [c1 f1 (x) + c2 f2 (x)] = c1 A f1 (x) + c2 A f2(x)

...(1.13a)

Where c1 and c2 are possibly complex constants. Clearly the differential and integrate operators are linear because these operators satisfy the 1.13a condition, i.e., d/dx [c1 f1 (x) + c2 f2 (x)] = c1 d/dx f1 (x) + c2 d/dx f2(x)

∫ [c1 f1 (x) + c2 f2 (x)]dx = c1 ∫ f1 (x)dx + c2 ∫ f2(x)dx The square operator on the other hand is a nonlinear because it does not satisfy the 1.13a condition SQR [c1 f1 (x) + c2 f2 (x)] = c12 f 12 (x) + c22 f 22 (x) + 2c1 c2 f1 (x) f2 (x) ≠ c12 f 21 (x) + c22 f 22 (x)

3. Hamiltonian operator Hop, appropriate to the problem, is written in terms of generalized coordinates and momenta. The Hamiltonian may be defined as sum of kinetic energy and potential energy as Hop =

p 2j

∑ 2 m + V (q j )

...(1.14)

j

Various Hamiltonians for the same problem differ only in the form of potential energy term.

Quantization of Energy

13

4. In quantum mechanics, momentum pj is defined as pj =

∂ h ∂ = − i
∂q j 2πi ∂q j

...(1.15)

where h is Planck’s constant,
= h/2π and i = − 1 . Replace each momentum pj in equation 1.14, wherever it occurs by such operators (Eq. 1.15) and we get quantum mechanical Hamiltonian. Involvement of momenta and Planck’s constant represent the involvement of de Broglie wave character of particle and Uncertainty principle. The quantum mechanical Hamiltonian is as follows:



Hop =


2 ∂ 2 

∑  − 2m ∂q 2j  + V(q j )

... (1.16)   5. Select an appropriate wave function, ψ and when Hamiltonian operates upon it, Schrödinger wave equation is obtained i.e., Hop ψ = E ψ The differential equation so obtained is j




2 ∂2 ψ  + V(q j )ψ = E ψ 2 j 

∑  − 2m ∂q j

... (1.17)

This differential equation is called Schrödinger wave equation, for a so-called stationary state of the system i.e., one whose energy does not vary with time. Equation 1.17 is the characterization equation of the operator, Hop and E is the eigen value of Hop associated with the eigen function ψ. 6. Eigen functions: Some operations or combination of operations and functions are such that when operation is done the same function is regenerated, but perhaps multiplied by a number. Thus differential of the function exp2x give 2exp2x which is the same function multiplied by the number 2. When this occurs the function is said to be an eigen function of the operator (in this case differential operator) and the numerical factor (2 in the example) is called the eigen value of the operator. Wave equation has the form Hψ = Eψ where Hamiltonian H is differential operator and ψ is the wave function. This has the form of an eigen value equation with the energy E playing the part of eigen value and wave function as eigen function. The wave function represents a state of the system, and so ψ is often termed the eigen state. 7. The wave function is generally a complex quantity. The integral of probability density over N particles and thus 3N coordinates is equated to one i.e.,

∫ . . . . ∫ ψ ψdq1 dq2 . . . dq3N *

or

∫ . . . . ∫ ψ ψdτ *

=1 = 1 where dτ = dq1, dq2, ... , dq3N

The condition is called the normalization condition.

...(1.18)

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Elements of Molecular Spectroscopy

1.7 SIGNIFICANCE OF WAVE FUNCTION, ψ Wave equation so written is solved subject to the condition that wave function ψ be a continuous function, which is not allowed to have singularities so that the normalization integral may diverge. The wave function contains all the information about the dynamical properties of system. If the wave function is known, all the observable properties of the system in that state may be deduced by performing appropriate mathematical operation. The interpretation of ψ is based on a suggestion made by Born. The Born interpretation draws an analogy with wave theory of optics in which square of the amplitude of an electromagnetic wave is interpreted as intensity of radiation which amounts to number of photons present. The analogy for particles is that wave function is an amplitude whose square indicates the probability of finding the particle at each point of space. The Born interpretation of ψ is, therefore, that ψ*(x)ψ*(x)dx is proportional to probability of finding the particle in an infinitesimal region between x and x + dx.

1.8 WAVE EQUATION FOR HYDROGEN ATOM A system which consists of a positively charged nuclei and electron moving about it is found in the hydrogen atom as well as in ions of He+, Li+2, Be+3 etc. According to Coulomb’s law, the force between a pair of charged particles is operative with magnitude F = –eZe/4πεr2, where –e is the charge of electron, Ze the nucleus charge and r the distance between the particles. The potential energy resulting from this force is r

∫

V = − Fdr = − ∞

Ze 2 4πε 0 r

...(1.19)

Now we are in a state to write down Hamiltonian for hydrogen like atoms. The Hamiltonian for hydrogen like atoms is described as H = KE of proton + KE of electron + PE of hydrogen atom =

1 1 ( p x2 + p 2y + p z2 ) electron + V ( p x2 + p 2y + p z2 ) proton + 2me 2m p

Since proton is 1836 times as heavy as electron, it may not be moving from its position, i.e., it may be assumed to be stationary and hence kinetic energy of proton may be assumed to be zero. So Hamiltonian reduces to: H=

1 ( p x2 + p 2y + p z2 ) + V 2me

...(1.20)

Replacement of momentum by quantum mechanical operator (given by equation 1.15) gives quantum mechanical Hamiltonian. H= −

∂2 ∂ 2  Ze 2
2  ∂ 2 + + − 2µ  ∂x 2 ∂y 2 ∂z 2  4πε0 r

...(1.21)

When it operates upon a wave function ψ, we get Schrödinger wave equation for hydrogen atom.

Quantization of Energy

15

 ∂2 ∂2 ∂ 2  2µ  Ze 2   ψ=0 + + ψ + + E  2 ∂y 2 ∂z 2  4πε 0 r 
2   ∂x 

...(1.22)

This equation in Cartesian coordinates cannot be solved. However, it can be solved if it is in polar coordinates. In spherical coordinates equation 1.22 becomes

1 ∂  r 2 ∂ψ  1 ∂2ψ 1 ∂  ∂ψ  2µ  Ze2  sin E + + θ + + ψ=0   r ∂r  ∂r  r 2 sin 2 θ ∂φ2 r 2 sin θ ∂θ  ∂θ 
2  4πε0 r 

...(1.23)

With centre of mass of the electron and nucleus as the origin of coordinates, µ is the reduced mass of the atom, E is its total energy. Since wave equation (1.23) involves the use of momentum operator, hence it includes de Broglie dual character of atomic particles. Further, since it has Planck’s constant, it incorporates Heisenberg uncertainty principle. 1. In the equation 1.23, ψ function can be written as a product of functions Rnl (r) Θ (θ) Φ (φ) written more briefly as Rnl Θ Φ which gives three differential equations. Solution of these differential equations, give three solutions corresponding to three coordinates r, θ and φ. Relation 1.24 between principal quantum number, n, and energy, E is given by: E=–

µZ 2e 4 8ε 20 n 2 h 2

n = 1, 2, 3,......

...(1.24)

Since principal quantum number n appears only in radial wave function Rnl (r), the energy of atoms is related to the distance of the electron from the nucleus and not to the angular momentum. However, the angular momentum has its importance in selection rules, which states that n may change by any number but l must change by ±1 and m may change by 0, ±1 in a transition. These two factors were not there in Bohr theory. The value of energy given by equation 1.24 is same as was given by Bohr theory in equation 1.10 but the operating mechanics is different. 2. The orbital angular momentum comes from function Θ part and solution of corresponding differential equation gives the solution as: L=

l (l + 1)


...(1.25)

where l = 0, 1, 2, …… (n – 1). The values calculated from equation 1.25 are the only exact values of angular momentum L. 3. Though the magnitude of orbital angular momentum are known, yet the orientations of orbital momentum with respect to an external reference comes from function Φ part and solution of corresponding differential equation is L z = m
. ...(1.26) where m = – l, – (l – 1), …..0, …….. (l – 1), l As long as there is no external magnetic field, all orientations of the angular momentum L possess the same energy. For one value of l there can be 2l+1 values of m, as for example, when l = 2, the parameter m can have five values 2, 1, 0, –1, –2. This means that for a

16

Elements of Molecular Spectroscopy

magnitude of angular momentum L = 6
, in a magnetic field only five orientations are allowed. Similar solutions are obtainable for any one-electron atom problem. The principal quantum number n and azimuthal quantum number l decide radial distributions of the electron and thus the values of r. The permitted values of these numbers are: Principal quantum number, n Azimuthal quantum number, l

1 0

2 0, 1

3 0, 1, 2

4 0, 1, 2, 3

The function Θ depends only on angle θ, therefore, describes the electron distribution as a function of angle θ. These functions again depend upon two quantum numbers l and m. Though the permitted values of m are 0, ±1, ±2… ±l, the Θ functions depend only on the magnitude of l. The orientation of orbitals is decided by the value of quantum number m. Thus, the total wave function ψ which constitutes what is known as orbital dependence on the quantum numbers n, l and m, i.e., different ψ functions for different orbital have different values of n, l and m and hence different behavior of the different electrons in an atom. It is customary to designate the values of l by letters as given below: Values of l

0

1

2

3

4

5

Designation of atomic orbital

s

p

d

f

g

h

The magnetic quantum number describes the Z-component of the angular momentum of the electron through equation 1.26. The energy of the electron depends only on the value of n and not at all on l and m. Thus, all ψ functions with same value of n but different values of l and m are degenerate as follows:

n=1

l=0

m=0

n=2

l=0

m=0

l=1

m = +1

E1 =

µe 4 8ε 02 h 2

E2 =

1 E1 4

E3 =

1 E1 9

m=0 m = –1 n =3

l=0 l=1

m=0 m = +1 m=0 m = –1

l=2

m = +2 m = +1 m=0 m = –1 m = –2

Quantization of Energy

1.9

17

SPECTRA OF HYDROGEN ATOMS

Atomic spectra are obtained when transition of electrons from one wave function (or orbital) to another wave function takes place. A more rigorous quantum mechanical study of transition between quantum states indicates that certain restrictions in the change in the values of l and m must be satisfied. The transitions, which do not follow these restrictions, are forbidden. These restrictions are referred to as selection rules.

Selection Rules (i) n may change by any integer i.e., ∆n = any value (ii) l must change by ±l value i.e., ∆l = ±1 (iii) m may change by ±1 or not at all i.e., ∆m = 0, ±1 For example, if an electron changes its principal quantum number from n = 2 to n = 1, it must go from a state of l = 1 to l = 0, i.e., the transition 1s ← 2p is allowed. The transition 1s ← 2s where ∆l = 0 is forbidden. Similarly, 2s ← 2p, 2p ← 3s, 2p ← 3d are allowed transition but 2s ← 3s, 2p ← 3p, 2s ← 3d are not allowed. Some of the hydrogen atom transitions are given in Table 1.2. Table 1.2 Series name

Allowed transition

Lyman series Balmer series

n1 = 1 n1 = 2

n2 = 2, 3, 4,… n2 = 3, 4, 5,…

Paschen series

n1 = 3

n2 = 4, 5, 6,…

1s ← n2 p 2s ← n2 p 2p ← n2s 2p ← n2d 3s ← n2 p 3p ← n2s 3d ← n2p 3p ← n2d 3d ← n2 f

1.10 USE OF ATOMIC TERM SYMBOLS TO DESCRIBE ATOMIC SPECTRA Atomic term symbols are sometimes called spectroscopic term symbols because atomic spectral lines can be assigned to transitions between states that are described by atomic term symbols. For example, consider atomic hydrogen. Exact solution of Schrödinger wave equation for hydrogen can be obtained from equation 1.24 as En =

µe 4 8ε 20 n 2 h 2

It is peculiar to the simple 1/r Coulombic potential of a hydrogen atom that the energy depends only on the principal quantum number. An electron in a 3s, 3p or 3d orbital, for example, has the same energy E3 in equation 1.24. As n increases coupling between spin angular momentum and orbital angular momentum takes place. This introduces a new quantum number J. States with different values of J will have different energies and thus have different term symbols.

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Elements of Molecular Spectroscopy

Term symbol that takes into account the spin orbit coupling is written as follows: 1. First take the l’s of each electron outside the closed shell and from these calculate orbital angular momentum vector L. The magnitude of L can equal is (l 1+l 2), (l1 +l2 –1)… (l1 – l2). Each L can have 2L + 1 various values. A notation grew up in which different L values are described according to symbol. L Symbols

0 S

1 P

2 D

3 F

4 G

5 H

6 I

2. The spin of each electron outside the closed shell are coupled to give resultant spin, Sˆ. 3. Each L and Sˆ vector is coupled to get resultant J vector. The particular state is said to be a multiple or can have a multiplicity equal to the number of J values. The energy state with quantum number L, Sˆ and J is indicated by a code symbol, which is called term symbol as: 2 Sˆ +lL ˆ J

Term Symbol

For example, 4D7/2 means L = 2 and thus d state; Sˆ = 3/2 or 2 Sˆ + 1 = 4 and J = 2 + 3/2 = 7/2. The electronic configurations and corresponding term symbols for various states of atomic hydrogen are given in Table 1.3. Table 1.3: The first few electronic states of atomic hydrogen Electronic configuration

Term symbols

Energy / cm–1

1s

1s 2 S1/2

2p

2p 2 P1/2

82258.917

2s

2s 2S1/2

82258.942

2p

2p 2 P3/2

82259.272

3p

3p 2 P1/2

97492.198

3s

3s 2S1/2

97492.208

3p, 3d

3p 2P 3/2, 3d 2 D3/2

97492.306

3d

3d 2D 5/2

97492.342

4p

4p 2 P1/2

102823.835

4s

4s 4p 2S1/2

102823.839

4p, 4d

4p 2P 5/2, 4d 2 D3/2

102823.881

4d, 4f

4d 2 D5/2, 4f 2 F3/2

102823.896

4f

4f 2F 7/2

102823.904

000

Quantization of Energy

19

Let us use Table 1.3 to take a closer look at the hydrogen atom spectrum. In particular, let us look at Lyman series which is the series of transitions from n = 1 state to states of higher n. Rydberg formulae Table 1.1 can be used to calculate the frequencies of the lines in the Lyman series. The frequencies of lines in the Lyman series are given by 1   ν = 109677.8 1 − 2  cm–1  n 

n = 2, 3,......

which gives the following results: Table 1.4: Calculated Lyman series of lines of hydrogen atom Frequencies / cm–3

Transition 1→2

82258.20

1→3

97491.18

1→4

102822.73

1→5

105290.48

Table 1.3 shows three states for n = 2 and so we do know which state to use to calculate the transition frequency into the ground state 1s 2S1/2. The changed selection rules for allowed transitions incorporating spin orbit coupling are: ∆L = ±1

and

∆J = 0, ± 1

In the ∆L = 0 case, the transition from a state with J = 0 to another state with J = 0 is forbidden. Thus Lyman series of atomic hydrogen for the allowed transitions are: np

2P

1/2

→

1s 2S1/2

np

2P

2/2

→

1s 2S1/2

No other transitions into the 1s 2S1/2 ground state are allowed. The frequencies associated with the 2 → 1 transitions can be computed from Table 1.3 and their values are: ν = (82258.917 – 0.000) cm–1 = 82258.917 cm–1 and

ν = (82259.272 – 0.000) cm–1 = 82259.272 cm–1 respectively.

Thus, n = 2 to n =1 transition occurs at a frequency ν = 82258.20 cm–1 if we ignore spin orbit coupling consist of two closely spaced lines. These closely spaced pair of lines are called doublet, and so do we see that under high resolution, the first line of Lyman series is a doublet. Table 1.3 shows that all the lines of Lyman series are doublets and that the separation of the doublet lines decreases with increasing n. The increased spectral complexity caused by spin orbit coupling is called fine structure. Similarly, lines in the 3d 2D to 2p 2P transition can be calculated. There are two 2p states in atomic hydrogen 2p 2P3/2. The transition from 3d 2D states into 2p 2P1/2 are: 3d 2D3/2 → 2p 2P1/2

ν = (97492.306 – 82258.917) = 15233.389 cm–1

3d 2D3/2 → 2p 2P3/2

ν = (97492.306 – 82259.272) = 15233.034 cm–1

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Elements of Molecular Spectroscopy

3d 2D5/2 → 2p 2P3/2

ν = (97492.342 – 82259.272) = 15230.070 cm–1

3d 2D5/2 → 2p 2P1/2

ν = (97492.342 – 82258.917) = 15233.425 cm–1

Note that the 3d 2D5/2 → 2p 2P1/2 transition is not allowed because ∆J = 2.

1.11 WAVE EQUATIONS FOR SOME SYSTEMS 1. Particle in a One-Dimensional Box A particle of mass m is constrained to remain strictly in a one-dimensional box of length, l with no seeping into or through the walls of the container. It is so called because the confinement can be achieved by arranging a zero potential energy within the box but to rise perpendicularly to infinity outside it. Thus, particle will have only kinetic energy with in the box. So, Hamiltonian for such a system will have only kinetic energy term i.e.,

p2 1  h ∂  =   H = kinetic energy = 2m 2m  2πi ∂x  =

Wave equation

−h

2

∂2

...(1.27)

8π 2 m ∂x 2

Hψ = Eψ

or

–

h 2 ∂ 2ψ 8π 2 m ∂x 2

= Eψ

...(1.28)

Solution to this wave equation E=

h 2n2 8ml 2

where n = 1, 2, ......

...(1.29)

2. For a Rigid Rotator For a rigid rotator, again, the molecule is continuously rotating and hence potential energy may be taken equal to zero and moment of inertia, I = µr2. This relation reduces two-body problem to one-body problem. Thus entire energy is the kinetic energy of body. The Hamiltonian may be defined as Hop = −

h2  ∂ 2 ∂2 ∂2   2 + 2 + 2 2µ  ∂x ∂y ∂z 

...(1.30)

The Schrödinger wave equation for rigid rotator is given

−


2  ∂ 2 ψ ∂ 2ψ ∂ 2 ψ  + +   = Eψ 2µ  ∂x 2 ∂y 2 ∂z 2 

...(1.31)

It is convenient to transform the above equation into spherical coordinates r, θ and φ. This transformation is lengthy, hence only transformed expression is given in equation (1.32).

−

1 1 ∂  ∂ψ  ∂ 2ψ 
2  1 ∂  2 ∂ψ  θ + + r sin       = Eψ 2µ  r 2 ∂r  ∂r  r 2 sin θ ∂θ  ∂θ  r 2 sin 2 θ ∂φ 2 

...(1.32)

Quantization of Energy

21

Taking 1/r2 as a common term so that µr2 = I = moment of inertia of molecule. Since in this ∂ψ =0 treatment, it is assumed that molecule behaves as rigid rotator hence the first term involving ∂r and equation 1.32 reduces to

−

1 ∂ 2ψ  ∂ψ 
2  1 ∂   sin θ +   = Eψ 2I  sin θ ∂θ  ∂θ  sin 2 θ ∂φ 2 

...(1.33)

Solution of equation 1.33 gives

h E = J( J + 1) = BJ (J + 1) cm–1 hc 8π 2 Ic where B =

h 8π Ic 2

...(1.34)

cm −1 and is called rotational constant. J is called rotational quantum number which

can have integral values i.e., J = 1, 2, 3…...

3. Linear Harmonic Oscillator Harmonic oscillators occur in classical mechanics when restoring force on a body is proportional to displacement. A force = – kq implies the existence of a potential energy U = 1/2 kq2 where k is called force constant and is characteristic of a bond, and q is the displacement. Since it is linear oscillator, its kinetic energy is confined to one dimension. Hence Hamiltonian may be defined as H=


2  ∂ 2  1 2 − kq 2µ  ∂q 2  2

...(1.35)

When this Hamiltonian operates upon wave function ψ, it gives wave equation for harmonic oscillator.

∂ 2ψ ∂q 2

+

1 2 8π 2 µ . E − 2 kq  ψ = 0 2   h

...(1.36)

The solution to this differential equation gives quantized oscillator energy levels, is given by E=

h k 1 V +  2π µ  2

where V is vibrational quantum number confined to the integral values i.e., V = 0, 1, 2…. This implies h k , when the oscillator is in its lowest energy 4π µ state, with V = 0 all the energy cannot be removed from an oscillator.

the existence of a zero point energy for V = 0 of

!❖"

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CHAPTER

2

Introduction to Molecular Spectroscopy Matter is made up of atoms, which have neutrons, protons and electrons with diameter as small as 10–15 m. Atomic nuclei carry positive charge and are the heavy part of an atom. Electrons carry negative charge, which are continuously moving around the nucleus. Since diameter of these particles is very small as compared to size of atom, these particles may be regarded as point charges. Despite all these, speed of electron around the atomic nucleus is so high that it shields the nucleus in such a way that atom behave as if it is electrically neutral. The fast motion of electron gives its volume, a spherical structure and electrical neutral nature. However, if right kind of force is applied on electron motion, spherical atom may become distorted. As one millilitre of a solid has 1023 atoms consequently force may originate from bulk matter or from outside in numerous ways, that may distort the spherical structure. This will create positive and negative charge centres with in an atom. We may say that atom has polarized. Similar statement hold good for molecules. As the numbers of electrons are different in different molecules, their way to balance the positive charge will be different; consequently the forces that these atoms and molecules exert on each other will be different. This gives rise to different properties like dipole moment, polarization, polarizability, dielectric constant and nature of inter and intramolecular forces. Consequences of these properties is that different molecules: • Show different chemical behaviour. • Non-ideal behaviour of gases, solutions and electrolytes. • Change in chemical reaction with the change of solvent or catalyst. • Common ion effect on solubility, dissociation and various equilibriums. All these properties have a relationship to the movement of electrons in a molecule though their precise manner of motion is never our concern. Hydrogen chloride, HCl (g) has a covalent bond between hydrogen and chlorine atom. On passing HCl gas through water, it ionizes to give proton and chloride ions. Water is a polar molecule, which exerts an electric field on HCl molecules, thereby polarizes the electronic motion in such a way that leads to its ionization. Chloroform, another polar compound shows a dielectric constant of 3.1 below –64°C, the melting point of chloroform, while above this temperature dielectric constant lies in the range 7.0–5.0. In solidstate polar chloroform molecules orient themselves in such a way as to cancel their mutual molecular fields. In other words, in solid state there is strong intermolecular interaction between the chloroform molecules. The moment it goes from solid to liquid state, the intermolecular distance increases. Molecules

24

Elements of Molecular Spectroscopy

move out of the force field of adjoining molecules and thus are free to orient themselves. This gives rise to a sharp increase in dielectric constant. In either case, the force field is different in solid and liquid and thus movement of electron changes and thus such a behavior. Alkenes react with strong aqueous solution of halogen acids to give alkyl halides. The order of reactivity is Hydrogen iodide > Hydrogen bromide > Hydrogen chloride The relative reactivity depends on the ability of HX to donate an H+ to form the carbonation, the ratedetermining step. HI being the strongest halogen acid is most reactive. Further, Markownikoff studied many reactions of this kind and formulated the rule called Markownikoff’s rule. It states that the negative part of the addendum (molecule to be added) adds to the carbon containing less number of hydrogen atoms. For example, when hydrogen iodide is added to propylene it gives isopropyl iodide. Due to hyperconjugation, methyl group is electron repelling; propylene will polarize the molecule as under

To this polarized molecule, hydrogen iodide addition takes place in two steps, as below:

Clearly this is a case of polarized state of molecule that follows the above rule. The microscopic point of view is taken care of in molecular spectroscopy and quantum mechanics where individual atoms and electrons in a molecule are dealt with. In macroscopic point of view, average of these interactions over a large number of molecules is considered. In the process microstructures disappear and, thus, it may be quite fair to neglect the peculiar individual intermolecular effects revealed in microscopic survey. This makes the corresponding macro equations look natural. Further, to reduce intermolecular effects to minimum a good spectrum is recorded in gaseous state or in dilute solutions. The entire science deals with physical processes or chemical processes. Situation arise when the detailed structure of matter may not be understood even then attempt is made to give microscopic interpretation of the results. Finer the microscopic interpretation, better are the conclusion. It is this microscopic viewpoint that throws light on fundamental physical and chemical processes and relate the subject as a whole. The microscopic point of view only prepares us for the more intimate detailed processes to be fully understood in the nature of our world. Further science of today looks at every process-taking place in a system at molecular level and molecular spectroscopy immunize every reader to look at that level.

2.1

SPECTROSCOPY

Energy changes in molecules can only be by discrete amount and hence molecules can exist only in certain states. These states are separated by a finite difference of energy called quantum of energy ∆E. Another way of saying it is that a fixed amount of energy ∆E can make the molecules to go from one energy state E1 to another state of energy E2. Only a transition of the molecule from energy state E1 to energy state E 2 can lead to absorption or emission of radiation of fixed quantum ∆E = | E2 – E1 |. This energy is directly related to the frequency of radiation and is called the condition of quantization, i.e., Excited molecule = Molecule + Quantum of energy E2 E1 ∆E

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25

| E2 – E1 | = ∆E = hν The quantum of energy, ∆E is emitted or absorbed as the equilibrium, is moving to right or left and results are an emission or absorption spectrum. In both cases, it is a result of interaction of radiation and molecules that gives rise to spectra. This also modified the theory of light to include that the energy of radiation can change only by an integral number of quanta. Thus atoms, molecules and light are so intimately related that a section concerning radiation without referring extensively to their structure and vice versa cannot be thought of. Earlier Einstein had hypothesized that a quantum of light energy hν as photon particles. A light beam of frequency ν and energy in hν could be regarded as containing n photons. Each photon of wavelength λ carries an energy hν and a momentum hν/c. When a photon of wavelength λ is absorbed by molecule it goes from state E1 to state E2, which is reflected in its spectra. Certain metal ions show characteristic flame test. On placing metal ions in flame, it gets reduced to metal atom, which is excited by flame. When the excited atom falls back to ground state, emits a radiation that lies in visible region and thus can be seen. We say that these atoms have spectra in visible region. As coloured flame is a consequence of transition from excited state to ground state, it is a case of emission spectra. The characteristic flame test shown by certain metal is a result of certain quantized energy levels, transition between which give a characteristic colour. Had the energy levels been continuous all metals should have shown coloured flame but not a characteristics flame ? Even in metal ions like Zn+2, Al+3 the quantized energy levels are there but the energy difference between first excited state and ground state is so large that radiation emitted does not lie in visible region hence does not impart colour to the flame. Take the case of copper which imparts green colour (ν = 5.7 × 1014 Hz) to flame. Energy associated with a flame = kT. Had the energy levels been continuous for ions/atoms to show a flame test, this energy, kT should be equal to energy of green radiation, or temperature required for this excitation should have been

hν 6.626 × 10 −34 × 5.7 × 1014 = = 27368 K k 1.38 × 10 − 23 But the maximum temperature that a burner can give is 1100°C = 1473 K. Since with 1473 K flame, energy of 27368 K cannot be attained, hence something is missing. The reasoning say that all copper atoms get excited and when they fall back they emit green colour, which is not the case. Copper atoms have quantized energy levels. On placing copper in flame, copper atoms distribute to various energy levels. This distribution is given by the Boltzmann distribution. =

− hν

ni = e kT no Fraction of molecules in excited state to emit green colour can be calculated i.e., − ni =e no

6.626 × 10−34 × 5.7 × 1014 1.38 × 10−23 × 1473

= 8.53 × 10–9

Thus, nearly 10–7 per cent of the copper atoms will go to excited state, which when fall back on ground state emit green colour. Atoms redistribute themselves so that 10–7 per cent again go to excited state and making the flame green. Since emission is a continuous process, redistribution is also a continuous process and thus a good green flame is obtained.

26

Elements of Molecular Spectroscopy

Two things should be apparent: (i) The energy levels are not continuous but quantized in nature. (ii) Spectra do not come from isolated atom but is shown by a collection of atoms or molecules. A quantum mechanical calculation gives absolute values to energy levels, which cannot be measured. However, if these energy levels are fixed, the energy difference between any two energy levels will also be fixed. This energy difference can be measured, as spectra and if this agrees with quantum mechanical energy difference we say that our calculated results are correct. In other words, spectroscopy is an eye of quantum mechanics and two have to go together. As a spectrum reflects a statistical distribution of atoms or molecules over a range of energy levels, probability has to be an integral part of spectroscopy. One of the main reasons for the failure of Bohr theory of hydrogen atom was that he took one hydrogen atom and tried to interpret its spectra.

2.2

NATURE OF RADIATION

A light ray consists of oscillating electric, ∈ and magnetic, B fields. The directions of both fields are in the perpendicular planes to the direction in which light is propagated. The wave field has no wave component in its direction of propagation of radiation. A monochromatic beam of light is made up of a large number of rays and thus can be regarded as the resultant wave of a number of rays of light in which every ray has the electrical (or magnetic) vector in different planes. A Nicole prism filters the light beam with its electrical (or magnetic) vectors confined to a particular plane only, which do not change direction as the wave move on. This light is said to be plane-polarized in which vector orientation remains constant with varying amplitudes. Plane-polarized light has an important property, i.e., when it falls on a second Nicole prism; it will pass with undiminished intensity only if the polarizing axes of the two prisms are parallel to each other. Any other orientation of these axes will further decrease the intensity until the axes are perpendicular when intensity approaches zero. Light rays are electromagnetic in nature and as the name implies both an electric field (∈) and a magnetic field (B) is associated with it. These fields oscillate in a periodic manner, sinusoidal, at mutually perpendicular directions and to the direction of propagation of the radiation. In Fig. 2.1, a radiation with electric and magnetic fields oscillate in fixed planes in space is shown thereby giving it a wave character. The periodic behaviour of the electric and magnetic field at a time, t in a beam is given by

Fig. 2.1: Electric field, ∈-component in the X-Z plane and magnetic field, B-component in the Y-Z plane of plane polarized light

Introduction to Molecular Spectroscopy

27

z  ∈ = A cos 2πν  t −  c  

B=

ε z  cos 2πν t −  µ  c

...(2.1)

where ∈ and B are the X and Y components of electric and magnetic fields respectively, A is the amplitude of the wave, c velocity of light, ν frequency of light related to velocity of light as given by equation 2.2. Subsequently, ε and µ are dielectric constant and permeability of the medium and z is the phase difference. In a monochromatic plane polarized beam, different rays have different z values but all rays have electrical field in same plane. However, in a laser plane polarized beam, different rays have not only electrical field in the same plane but also have same z value. These radiation waves travel with the velocity of light c having a wavelength λ and frequency ν. The frequency represents the number of wave crests or wavefronts passing a given point in a unit time, and is expressed in cycles per second or simply seconds–1. The reciprocal seconds (s–1) is called hertz (Hz). In spectroscopy wavelength are expressed in a variety of units, chosen so that in any particular range the wavelength does not involve large power of tens. Thus, in microwave region wavelength is measured in cm or mm, while in infrared it is usually given in micrometer (µm)—formerly called micron—where 1 µm = 10–6 m. In visible and ultraviolet region λ is still often expressed in Angstrom although the proper SI unit for this region is nanometer (nm). 1 nm = 10–9 m = 10 A The relation between the wavelength and frequency of light wave is λν = c ...(2.2) 8 –1 where c is velocity of light and is equal to 3 × 10 ms . To illustrate the values that occur for λ and ν, consider the blue radiation with wavelength λ = 4.7 × 10–7 m. The corresponding frequency is ν = c/λ = 3 × 108/4.7 × 10–7 = 6.38 × 1014 Hz Such inconvenient large numerical values forced the spectroscopist to introduce a third quantity called wave numbers, ( ν ), which is simply given by the relation ...(2.3) ν = l/λ Units of wave numbers are unit’s inverts of wavelength and are smaller than the frequency by a factor of c, expressed in cm–1. Wave number is a permitted non-SI unit.

2.3

BORN-OPPENHEIMER APPROXIMATION

An atom has nuclei around which electrons are distributed giving it spherical structure. Energy distribution makes it go to various electronic states or make atom to go for translational motion. There is no other way by which energy could be retained by atom. Hence, atomic spectra are a consequence of electronic transition. The moment when two atoms combine spherical symmetry is lost and thus can make some more motions in addition to electronic transition. Each motion has its own quantized energy levels and thus a spectra.

28

Elements of Molecular Spectroscopy

An isolated molecule possesses: (i) Translational energy by virtue of the motion of the molecule as a whole. (ii) Rotation of the molecule about an axis passing through centre of gravity of molecule is associated quantized rotational energy. Transitions between these energy levels give rise to rotational spectra. (iii) Periodic displacement of atoms in a molecule from their equilibrium positions is associated with it quantized vibrational energy levels. Transition between which give rise to vibrational spectra. (iv) Electronic energy associated with transition of electrons between various electronic energy levels in a molecule (v) Nuclear energy and energy due to nuclear and electron spins. These various types of energies associated with different motions of the molecule are different and thus independent of one another. As a first approximation, the total energy of a molecule can be expressed as the sum of the constituent energies, that is ...(2.4) Etotal = Etrans + Erot + Evib + Eelec In other words, molecules have different quantized energy levels for different motion. Consider two energy states of a system E1 and E2, subscripts 1 and 2 are referred invariability with some quantum numbers. A transition from E1 to E2 can occur provided an appropriate amount of energy ∆E = E1 to E2 is absorbed by the system to give corresponding spectrum. The time required for various transitions are as follows: Nature of transition

Electronic

Vibrational

Rotational

Nature of spectrum

Electronic spectra

Vibrational spectra

Rotational spectra

Time required

10–8 s

10–13 s

10–10 s

These times are so different that as an approximation they do not interact with one another. A change in the total energy as a result of electronic transition in a molecule is or

∆E = ∆Ee + ∆Ev + ∆Er ∆ε = ∆εe + ∆εv + ∆εr cm–1

...(2.5)

Translational energy is so small that it is taken to be continuous. As a consequence when we talk of change in energy that term disappears. The approximate orders of these changes are ∆εe = ∆εv × 103 = ∆εr × 106 Another way of writing equation 2.5 is as follows: ∆E = [ E ′e′ − E ′e ] + [E ′v′ − E ′v ] + [E ′r′ − E ′r ]

...(2.6)

These differences in energies give rise to spectra in different region as ∆E = νelec (ultraviolet) + νvib (infrared) + νrot (microwave) h

...(2.7)

Introduction to Molecular Spectroscopy

29

Since these regions are so far apart, electronic transition can be recorded in a molecule as if in a static nuclear framework, vibrational spectra independent of rotational motions and rotation giving a rotational spectrum. Due to this large difference in energy, the three spectra do not interact with one another. This is called Born-Oppenheimer approximation. This amounts to that every electronic energy level comprises a number of vibrational levels and each vibrational level consists of several rotational levels as shown in Fig. 2.2.

Fig. 2.2: Energy levels of a diatomic molecule (the actual spacings of electronic levels are much larger and those of rotational levels are much smaller than those shown in the figure).

At this point it is essential to elaborate a little more to the quantum nature of energy levels. In kinetic theory of gases, thermodynamics energy was taken to be continuous in nature while in quantum theory and spectroscopy we say that energy levels are quantized in nature. These two contradictory bases may be explained as follows:

30

Elements of Molecular Spectroscopy

In kinetic theory of gases and thermodynamics we deal with macro systems involving large number of energy levels say 1023 resulting from translational motion. This makes the energy levels to come so close to one another so that they behave as if energy states are continuous. The energy levels of even translational motion are quantized but the energy levels are so near to one another that they behave as if energy levels are continuous. This is the basis of presumption that energy levels are continuous and that makes kinetic theory of gases and thermodynamics simpler, easier to understand, easier to carry out various calculations and thus make all these topics beautiful.

2.4

ABSORPTION, EMISSION AND SCATTERING SPECTRA

For the same system three different spectras can be obtained: emission spectra, absorption spectra and Raman spectra.

Emission Spectra Atoms or molecules are subjected to intense heat or electric discharge so as to absorb the same to become ‘excited’ so as to be in state E2. On returning to their lower energy state atoms or molecules may emit radiation. Such emission is the result of a transition of an atom or molecule from an excited state to one of lower energy, usually the ground state. This excess energy is emitted as a photon and the corresponding frequency is recorded for the emission spectrum. If the transition is from energy state E2 to E1, the emission spectrum shows a line at a frequency ν given by equation (2.8). E2 – E1 = ∆E = hν

...(2.8)

Absorption Spectra The absorbing sample is placed between the source of light in the frequency range being studied and the spectrophotometer, which records the percentage of light absorbed against the range of frequencies to give the absorption spectrum. This is possible if the molecule initially in energy state E1 can be excited to energy state E2 following the relation given by equation 2.8. Raman Spectroscopy It is a technique, which explores energy levels of molecules by the scattering of light. Photons of frequency νin are scattered by collision with the molecules of the sample, when new frequencies are added, because photons can acquire or lose energy during inelastic collisions (Fig. 2.3). If the light excites the molecules, during the collision molecules withdraw some energy from the photons and so scattered light emerges with a lower frequency (νin – ν). The lines so obtained are called Stokes lines. If the light photons collide against some excited molecules, these molecules may give up this energy and the emerging photons will have higher frequency, viz., (νin + ν). The lines so obtained are called Antistokes lines. Raman spectroscopy has recently undergone considerable development because of the availability of lasers, which are so intense that small samples, and shorter exposures give a good spectrum. 2.5

SPECTROPHOTOMETERS

The schematic diagram of the apparatus for absorption spectroscopy as well as Raman spectroscopes is given in Fig. 2.3. The radiation source in an electronic absorption spectrometer is a tungsten filament, which gives out intense white light, and for ultraviolet region the source is a hydrogen discharge lamp. The radiation

Introduction to Molecular Spectroscopy

31

source for infrared region is a heated ceramic filament coated with rare earth oxides. A klystron or more commonly, a semiconductor device called Gunn diode is used to generate microwave radiation. An oscillating electric current in a wire coil generates the radio-frequency radiation.

Fig. 2.3: Schematic representation of absorption spectrometer and Raman spectrometer. In the former source, sample and detector are collinear while in Raman spectrometer, source is perpendicular to sample and detector.

The variation of absorption with frequency is determined by analyzing the spectral radiation by means of dispersing element, which separates different frequencies into rays that travel in different directions. The simplest dispersing element is a glass or quartz prism but a diffraction grating is more widely used. A diffraction grating consists of a glass or ceramic plate into which fine grooves have been cut about 1000 nm apart (a spacing comparable to the wavelength of visible light) and covered with a reflective aluminum coating. The grating causes interference between waves reflected from its surface and constructive interference at specific angles that depend on the frequency of the radiation being used. In this way wavelength of light is directed into specific direction. The third component of spectrometer is the detector, a device that converts the spectral radiation into an electrical signal that is passed on to a recording device operating synchronously with the analyzer, thus producing either a trace on a chart recorder or a computer recording the spectrum. Common radiation sensitive detectors are semiconductors. The radiation is chopped by a shutter that rotates in the beam so that alternating signal is obtained from the detector. A modulator is introduced to convert the signal to an alternating current. This procedure enables more AC electronics to be employed at the recording stages. In the microwave region, the source frequency is varied and analyzer is not necessary. The highest resolution is obtained when sample is gaseous at very low pressure so that collision broadening is reduced. In order to increase the accuracy path length must be very long, say of the order of metres. These long paths are achieved by multiple passage of the beam between two parallel mirrors at each end of the sample cavity. For infrared spectra, the sample is typically a liquid held between windows of sodium chloride (which is transparent to 700 cm–1) or potassium bromide (down to 400 cm–1). Other ways of preparing the sample include grinding it into a paste with Nujol, hydrocarbon oil, and pressing it into a KBr solid disk. In Raman spectrophotometer intense monochromatic radiation consists of a large spiral discharge tube with mercury electrodes is allowed to fall perpendicularly to the cell containing a gaseous or liquid

32

Elements of Molecular Spectroscopy

samples. The most intense line emitted from mercury discharge tube is of wavelength 435.8 nm serves as exciting line. The scattered light is observed at right angles to the direction of incident radiation. Most of the Raman spectra in literature were obtained using photographic recording. The Raman spectra of gases are generally weaker than those of liquids. It is necessary to use very long discharge lamps and cells containing mirrors at both ends arranged so as to increase the effective path length of cell. However, now laser lamps have replaced mercury lamps. It is a radiation of very high intensity in a narrow beam with a well-defined frequency. Further, for recording a Raman spectra photographic plates have been replaced with very sensitive photomultiplyer tubes, and direct photoelectric recording has become possible. Linear response, excellent reproductibility and convenience are the chief advantages of this method with improved intensity. Using a laser Raman spectrometer, the spectrum of a sample can be recorded quite quickly and has become an extremely important analytical tool.

2.6

REGIONS OF ELECTROMAGNETIC RADIATIONS

Figure 2.4 illustrates the region into which electromagnetic radiation has been divided. The division is made for our convenience and nature of different instrumentation used in different regions. The boundaries between the regions are by no means precise, although the molecular processes associated with each region are different:

Fig. 2.4: The electromagnetic spectrum

1. Radio frequency region –105–1010 Hz or 3 km–3 cm, wavelength: The energy change with the change of spin of nucleus or electron is of the order 0.001 to 10 J mol–1 and falls in this region. 2. Microwave region –1010–1012 Hz or 3 cm–0.3 mm, wavelength: Energy changes due to molecular rotations are observed in this region as rotational spectra with ∆E of the order of 100 J mol–1. 3. Infrared region –1012–1014.5 Hz or 0.3 mm–950 nm wavelength: Molecular vibrations give one of the most valuable spectroscopy and are observed in this region. Energy separation associated with molecular vibrations is in the order of 104 J mol–1. 4. Visible and Ultraviolet regions –1014.5–1016 Hz or 950 nm–30 nm wavelength: Energy separation for the transition between the valence electrons is observed in this region and is of the order of some kJ mol–1. 5. X-ray region –1016–1018 Hz or 30 nm–300 pm wavelength: Energy changes involving the inner electron excitation of a molecule, could be of the order of 105 kJ mol–1. These energy changes associated with a spectra are observed in X-ray region.

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33

6. γ-ray region –1018–1010 Hz or 300 pm wavelength: Energy changes involve the rearrangement of nuclear particles, having energies of 109–1011 Jg–1. Typical wavelength, frequencies, wave number and energies of photons of visible light are given in Table 2.1. Table 2.1: Typical wavelength, frequencies, wave number and energies of visible light

Wavelength, λ in nm

Frequency, ν in Hz

Wave number ν in cm–1

Energies E in kJ einstein–1

Red

660

4.55 × 1014

1.51 × 104

1.81 × 102

Orange

620

4.8 × 1014

1.6 × 104

1.9 × 102

Yellow

580

5.2 × 1014

1.7 × 104

2.1 × 102

Green

530

5.7 × 1014

1.9 × 104

2.3 × 102

Blue

470

6.4 × 1014

2.1 × 104

2.5 × 102

Violet

420

7.1 × 1014

2.4 × 104

2.8 × 102

400–200

(7.5–15.0) 1014

(2.5–5) 104

(3.0–6.0) 102

Name of light

Near Ultraviolet

2.7

INTENSITY OF SPECTRAL LINES

A recorded spectra is a study of variation of intensity of radiation absorbed/emitted with frequency. Such a study is always on a collection of molecules. Transition of molecules from energy Ε1 → E2 involves absorption/emission of radiation. A typical absorption spectra is shown in Fig. 2.5.

Fig. 2.5: A typical absorption spectra, showing maximum absorption at wavelength λmax. AB corresponds to width of spectra.

34

Elements of Molecular Spectroscopy

The intensity of absorption can be expressed as transmittance (T), which is defined as the ratio of the intensity of the radiation transmitted from the sample (I) to that of the radiation incident on the sample (I0) i.e., ...(2.9)

T = I/I0 Intensity of absorption is more conveniently expressed in terms of absorbance (A) i.e., A = log (1/T ) = log10 (I0/I)

...(2.10)

Absorbance of a band is related to the sample thickness (l) and concentration (c) of the absorbing species. The relation is expressed in the form of Beer-Lambert law as below: A = log10 I0/I = εcl

...(2.11)

Absorbance is a dimensionless quantity. Concentration c is usually expressed in mol dm–3 and path length (l) in cm, hence extinction coefficient ε, has the units of dm3 mol–1 cm–1. If we use SI units of mol dm–3 for concentration and m for path length, the units of ε will be m2 –1 mol . We can obtain the values of ε in m2 mol–1 units from those in dm3 mol–1 cm–1 units in the following manner: ε = dm3 mol–1 cm–1 = 10–3 m3 mol–1 (10–2 m–1) = 10–1 m2 mol–1. Values of ε in SI units can therefore, easily be obtained from published values in dm3 mol–1 cm–1 by dividing the numerical quantity in latter units by 10. λmax value: The value of wavelength at which absorption maximum occurs is called the λmax and is expressed in nanometer (nm). 1 nm = 10–9 metres. εmax value: It is known as molar absorptive or molar extinction coefficient and is a measure of intensity of absorption. The ε value is characteristic of a particular compound at a given wavelength. Usually for the wavelength of maximum absorption λmax molar extinction coefficient εmax is expressed. Absorption bands with εmax value > 103 m2 mol are considered to be high intensity or strong bands whereas those with εmax values < 102 m2 mol–1 are known as low intensity or weak bands. There are two main factors with which the intensity of spectral lines are affected: (i) Transition Probability, (ii) Population of state. Transition probability: The intensity of spectral lines arising from a molecular transition between a pair of stakes i and j with wave function ψi and ψj depends on transition probability, Rij. It is defined as Rij ∝ ∫ ψi Μ ψj dτ ...(2.12) Larger the value of Rij more intense will be the spectral lines. The conditions under which the integral fail to vanish for various possible pairs of states of a given system are known as selection rule values. A transition between a pair of energy levels for which Rij vanishes is said to be a forbidden transition and may appear as a weak spectral line. It is this integral which gives rise to selection rule for various spectra. At a much lower level of sophistication selection rule allows us to decide between which levels of transition will give rise to spectral lines. Population in state: If transition from 2nd energy level to a third energy level are equally probable, then the most intense spectral line will arise from the level, which has the greater population. Boltzmann’s distribution gives this population distribution. If no is number of molecules with energy Ei then the number of molecules nj with energy Ej is nj = no e − ( E j − E i ) /kT

...(2.13)

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35

k is the Boltzmann constant. Obviously, before molecules are excited from state ψi this state must be populated. Larger the population of the state ψi more intense will be transitions. However certain transitions are not allowed by selection rule and invariably of low intensity though have high population in ground state.

2.8

WIDTH OF SPECTRAL LINES

Spectral transitions are functions of the measured position in terms of frequency, wavelength or wave numbers. The spread of spectra over a range of frequency is called width of spectra. The width (∆ν) of a spectral band is usually defined in terms of the half bandwidth, that is width of band at half height. By definition, frequency ν = n/t i.e., n waves that pass in t seconds to observer, the spectrophotometer, but it is not possible to measure n and t with more precision than the resolving power of the instrument. This width inherent in the instrument in any transition is minimum, beyond that the sharpening is not possible. The following factors contribute to it. Heisenberg uncertainty principle: Natural line broadening is closely related to the uncertainty principle δεlower . δtlower =

h 4π

where δεlower represents uncertainty in the lower energy level from which transition occur and δtlower the uncertainty in the time for which the energy state is occupied. An analogous relation holds good for upper energy state. The uncertainty in frequency associated with lack of precision in defining upper and lower quantum states are taken to be additive and determined by the expression. h δν = (δεupper + δεlower) = δε

h or δt = (4π δν)–1. 4π Take the case of electronic spectra where lifetime of a quantum state is of the order of 10–8 second, which gives an uncertainty in frequency of the order of 108 Hz. As the electronic spectra is observed in the region 1014–1016 Hz, comparatively this uncertainty is negligible. Radio-frequency (108 –109 Hz) with other extreme of the spectra where spin resonance spectroscopy lies. The energy difference in energy levels is very small within region. Excited electron spin has a lifetime of 10–7 second and thus uncertainty in frequency is 107 Hz. This value is comparable to the radio frequency range and thus uncertainty effect is always there in spin resonance spectroscopy. h δν . δt =

Doppler Broadening Doppler effect shifts the wavelength of light emitted by a moving atom. If the gas molecules are moving and radiation is viewed in x-direction, the wavelength of light emitted by molecule with a component of velocity vx in the x direction is vx   λ = λ0  1 ±  c  

...(2.14)

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Elements of Molecular Spectroscopy

The plus sign corresponds to an atom receding from the observer and minus sign to an atom approaching the observer. Molecules move towards as well as away from the observer in a continuous range of velocities, thus even a perfectly monochromatic radiation moving in this way would appear to emit a range of frequencies. This is called Doppler broadening of spectral line and is given by equation (2.14a) ∆νDoppler =

2ν o  2kT log 2   c  M 

...(2.14a)

where k is Boltzmann constant, c velocity of light, M molar mass of gas and T temperature of gas.

Collision Broadening The molecules emitting or absorbing radiation in a gas undergo collision. In each collision, there is certain possibility that a molecule initially in an excited state to make a radiationless transition to a lower state, so that the life time of excited state will be decreased. This will add certain frequencies to a monochromatic beam, thereby broadening the spectra and is given by ∆νcollision =

6 R

P σ MT

...(2.15)

where σ is the effective cross-sectional area, P is pressure of gas and R gas constant. Like Doppler broadening, collision broadening is inversely proportional to the square root of molar mass. However, it is inversely proportional to the square root of temperature. With this inverse relation to temperature measurement made at lower temperature decreases Doppler broadening but increases collision broadening and vice versa with change of temperature. So a spectra is recorded over a range of temperature and where minimum width is obtained, at that temperature Doppler broadening and collision broadening balance one another so that best spectra is recorded at that temperature. The spectral data are usually presented as graphs in which extinction coefficients are in the ordinate and wavelength or frequency as the abscissa. The spectra of pure compounds are also plotted with the absorption intensity in terms of the absorbance or log10 Io/I as ordinate and the wavelength increasing from left to right as abscissa. The units most commonly employed for abscissa are wavelength in nm, frequency in Hertz (Hz) or wave numbers cm–1 as the abscissa. Absorption bands of substance in solution on a wave number scale tend to be more symmetrical and more uniformly spaced along the abscissa than plots on a wavelength scale. The plots on a wavelength scale tend to compress the short wavelength bands and expand the long wavelength bands as compared to the wave number plots. The spectrum of phenanthrene is shown in Fig. 2.6 plotted on different sets of coordinates. The low intensity structure at long wavelength end is apparent in Fig. 2.6 (d) and (f ) is practically obscured in the plot using molar absorptivity, [Fig. 2.6(b)]. The greatest advantage of the log plots (a, c, e) is the expansion of these weak bands, although they result in some distortion and obliteration of fine structure. The different molecules have different electronic structure and thus have a different spectrum. Coordinate axis are so selected that spectral peaks are most vividly represented. As in the case of phenanthrene, Fig. 2.6(d) is probably the best spectral representation of this molecule.

Introduction to Molecular Spectroscopy

Frequency, Hz

Fig. 2.6: The spectrum of phenanthrene in various coordinate systems

37

38

Elements of Molecular Spectroscopy

2.9 SCATTERING SPECTRA—RAMAN SPECTROSCOPY When a photon of any frequency νo impinges on a molecule, electrons are forced to oscillate at the frequency of the incident radiation and a radiation of frequency νo is emitted. Such elastic collisions between photons and molecules result in a change of direction of photons but no change of energy. The change of direction of photons is called scattering of the radiation. Scattering without a change in frequency is known as Rayleigh scattering. Both the blue colour of the sky and the red colour of the setting sun are due to Rayleigh scattering of sunlight by air molecules. These sky colours in the early morning and in the evening when colour is visible but sun is not visible show that photons have changed the direction. In the daytime when we are sitting in shade, we feel as if we are sitting in lighted place. Since light rays travel in straight line, and in shade source of light is not visible, light photons have changed the direction and reach us through scattering. Sir C.V. Raman on February 28, 1928, discovered that all scattered radiation’s were not of the same frequency as the incident radiation νo, but changed scattered frequencies were also present. This change in frequency composition of the scattered radiation is termed as Raman scattering after the name of the discoverer. This involves inelastic collisions between photons and molecules which change direction as well as frequency of the incident radiation to νj = νo ± νM, where νM is the rotational or vibrational frequency of the molecule and νj is the observed scattered radiation frequency. The plot of intensity of the scattered radiation as a function of frequency νj falls into two sections: 1. Stokes lines 2. Anti-Stokes lines Stokes lines: If a molecule M accepts a quantum of energy which is just sufficient to raise it to an excited level M* so that frequency of scattered radiation decreases νj = νo– νM

...(2.16)

Such Raman lines are called Stokes lines. They can be summed up as hνo + M → M* + hνj with h (νo– νj ) > 0 Anti-Stokes lines: An excited molecule M** on collision with the incident photon can give a quantum of energy to fall to its lower energy state so that the frequency of scattered photon νj is higher than the frequency of the incident radiation νj = νo + νM

...(2.17)

Such lines fall on the higher frequency side of Raleigh lines and are called anti-stokes lines. They can be summed up as hνo + M**→ hνj + M* with h(νo – νj) < 0 Since the number of molecules in excited state are far less than in ground state, hence intensity of anti-Stokes lines is far less than Stokes lines. Figure 2.3 also shows schematically a broad arrangement of recording a Raman spectrum. One of the essential requirements is a highly monochromatic source capable of giving high irradiance

Introduction to Molecular Spectroscopy

39

at the sample. The gas laser meets these requirements perfectly, and in addition, provides radiation, which is self-collimated and plane-polarized. Such a monochromatic beam of radiation having a narrow line width is made to fall on the sample and light scattered in the perpendicular direction is made to pass through dispersing system. The radiation then passes through a detection device to give the required spectrum. In a spectrograph the detector is a photographic plate in the spectrometer. The detector is invariably a spectra with photomultiplier tubes and direct photoelectric recording has become very suitable for detection of low-level signals. A molecule is made up of an assemblage of positively charged nuclei embedded in a cloud of negative electricity is electrically polarizable. Electric vector ∈ of light waves interacts with molecules and electrons are forced into oscillations thereby inducing a dipole moment P in the molecule, as depicted in Fig. 2.7. Thus molecular polarizability is a quantity representing molecular distortion under an electric field associated with radiation source. Here permanent dipole moment of the molecule is not important at all. The magnitude of the induced dipole moment P depends on the strength of the electric field ∈ of the incident radiation and the polarizability of the molecule. This induced dipole moment, P is linearly related to electric field (∈) i.e.,

P ∝∈

P=α∈

or

...(2.18)

where α is a constant for a molecule termed as the polarizability. Since an oscillating dipole radiates energy, it is recorded as a Raman spectrum. The more easily deformed the electron shell in a molecule, the more is its polarizability and thus greater the induced dipole moment by a given field.

Fig. 2.7: Polarization induced by the electric field of the electromagnetic radiation

PROBLEM 2.1 The IR spectra of water vapors show three fundamental vibration peaks as sharp absorption wave numbers as follows H — O — H symmetrical stretching 3650 cm–1, H — O — H bending mode 1595 cm–1 and H — O — H asymmetrical stretching 3755 cm–1. Predict the position of first two Raman lines if the water vapor is irradiated with (a) sharply well defined mercury light of wavelength 435.8 nm and (b) Ruby laser of wavelength 694.3 nm (14493 cm–1).

SOLUTION The stokes and anti-stokes lines are expressed in wave numbers or hertz. So converting wavelength to wave number ν (Hg-line) =

ν (Ruby laser) =

1 435.8 × 10 −9 1 694 × 10 −9

= 22946 cm–1

= 14403 cm–1

40

Elements of Molecular Spectroscopy

(a) When irradiated with mercury light, Stokes lines H — O — H stretching line will appear at 22946 – 3650 = 19296 cm–1

Rayleigh line = 22946 cm–1. Anti-Stokes lines H — O — H stretching line will appear at 22946 + 3650 = 26596 cm–1

H—O—H bending line will appear at 22946 – 1595 = 21351 cm–1

H—O—H bending line will appear at 22946 + 1595 = 24541 cm–1

(b) When irradiated with a Ruby laser, Stokes lines H — O — H stretching line will appear at 14403 – 3650 = 10753 cm–1

Rayleigh line = 14403 cm–1 Anti-Stokes lines H — O — H stretching line will appear at 14403 + 3650 = 18053 cm–1

H — O — H bending line will appear at 14403 – 1595 = 12808 cm–1

H — O — H bending line will appear at 14403 + 1595 = 15998 cm–1

As in any other Raman spectrum, Stokes lines have wavelengths greater than the Raleigh line wavelength and anti-Stokes lines have a wavelength less than this value on the other side. Since the former is accompanied by an increase in molecular energy (which can always occur subject to certain selection rules) while the latter involves a decrease (which can only occur if the molecule is originally in an excited vibrational or rotational state), Stokes lines are generally more intense than anti-Stokes. The spread of the entire spectrum usually is very small and so the incident light should be monochromatic with very accurately defined wavelength. Since the intensity of scattered radiation is low, the radiant source should be of very high intensity. These purposes are best served with any continuous wave laser only.

2.10 IMPORTANCE OF MOLECULAR SPECTROSCOPY Quantum mechanical calculations give the energy value of the various molecules energy levels. There is no experimental way by which to verify these absolute energy values. On the other hand, molecular spectra give transition energies between various energy levels. If the difference between energy levels as calculated by quantum mechanics agree with the corresponding spectral values implies that calculations are correct. So molecular spectroscopy is the tool to the verification of quantum mechanical calculations. In other words, the two are so interrelated that without one other cannot be appreciated. Thus two together give a deep insight to various molecular processes that can take place in a molecule. Molecular spectroscopy helps us to give electronic structure of the molecule and thus explain the various possible reactions that can take place. From the energy difference values and with the help of statistical mechanics complete partition function for the molecules involved in a chemical reaction can be written. From this partition function one can calculate reaction rate constant for such reactions. In other words, reaction rate constants and thus rates of various chemical reactions are related to the molecular energy levels. This also implies equilibrium constants and thus various thermodynamic quantities can also be calculated from these energy levels. If a molecule shows absorption spectra at frequency ν Hz, it implies that this radiation is interacting with this molecule. This implies that a molecule can be photochemically active for a light of frequency ν, only if it shows an absorption peak at this frequency. In case this light does not interact with molecule, it may be photochemically inactive. In other words, molecular spectroscopy, an experimental branch, is the heart of entire physical chemistry.

Introduction to Molecular Spectroscopy

41

Without indepth knowledge of various molecular processes, molecular interactions cannot be understood. Manuals are available that may help you in identification of the molecular structure from molecular spectra. However, without deep understanding of the subject the excitement associated with the subject is lost and thus one may never be confident of implied results. So a complete grasp in molecular spectroscopy is a must for understanding of any chemical system.

PROBLEMS 1. Discuss the factors that determine intensity and width of spectral lines. 2. State Born-Oppenheimer approximation. Explain. 3. Why are anti-Stokes lines less intense than Stokes lines? Explain. 4. Calculate the energy in joules per quanta, joules per mole, and in electron volts of photons wavelength 300 nm. 5. Calculate the energy per photon, energy per mole of photons, moment of photons when their wavelength is 400 nm. 6. An atom of oxygen is expelled from a molecule of haemoglobin in a way that fixes its position parallel to the principal plane of haemoglobin with an uncertainty ∆y of 0.15 nm. What is the uncertainty of momentum ∆p, along the y-coordinate axis? 7. The colour shown by some simple transition metal states is a case of emission or absorption spectrum. Justify your answer. 8. The colour of flowers is a result of thermal excitation of flower molecules or some other. Discuss and explain. 9. What is the uncertainty in momentum if we wish to locate an electron within an atomic nuclei, say that ∆x is approximately 50 pm? What is the consequence of high momentum value? 10. The pure rotational energy levels of CN have a value 2.87 and 8.4 cm –1. Calculate absorption frequency for this transition. 11. The two lowest rotational energy levels of CO have value of 3.86 and 11.58 cm–1 and lowest vibrational energy levels as 1071 and 3213 cm–1. Calculate the absorption frequency for rotational and vibrational spectra. Compare these energies to thermal energy kT at room temperature and comment on results.

D❖E

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CHAPTER

3

Particle in a Box In this chapter reasoning is through numerical values as obtained in different situations.

3.1

PARTICLE IN ONE-DIMENSIONAL BOX

This is the simplest quantum mechanical problem. Here a particle of mass m is confined to move in a one-dimensional box of length a, having infinitely high walls. It is assumed, that potential energy of the particle is zero everywhere inside the box, that is U = 0.

U=0

Fig. 3.1: Particle in a one-dimensional box

The total energy possessed by the particle is kinetic energy and is confined to move in one dimension so that Hamiltonian = H = KE + PE =

p x2 +0 2m 2


2  ∂  ...(3.1)   2m  ∂x  Wave equation is obtained when this Hamiltonian operator is made to operate on a wave function ψ i.e., Ηψ = Eψ

=–

44

Elements of Molecular Spectroscopy

−

or

D 2 ∂ 2ψ = Eψ 2m ∂x 2

...(3.2)

Solution to this wave equation is E=

n2h2 8ma 2

where n = 1, 2, 3, ......

...(3.3)

Expression 3.3 gives energy of the particle in a box and it should be committed to memory. For n =1 E1 =

h2

...(3.3a)

8ma 2

So that for different values of n, energy values are as follows: n E

1 E1(eq. 3.3a)

2 4E1

3 9E1

4 16E1

5 25E1

6 36E1

Since energy depends upon the quantum number n, which can have any integer value, so the energy levels of particle in a box are quantized.

3.2

QUANTIZATION OF ENERGY

This will be reflected by a numerical example. An electron is constrained to move in an infinitely small one-dimensional box of width 0.1 nm. Calculate the energies for first four energy levels and exhibit it in a suitable energy level diagram.

n 2h 2

mass of electron, m = 9.1 × 10–31 kg 8ma 2 a = 0.1 nm = 10–10 m. Substituting the values in this equation, From equation 3.3, En =

En =

n 2 (6.626 × 10 −34 Js −1 ) 8(9.1 × 10 −31 kg ) (10−10 m)

= n2 60.23 × 10–19 Joules Energies of electron corresponding to different values of n are as follows: n=1

E1 = 60.23 × 10–19 J

n=2

E2 = 22 E1 = 240.94 × 10–19 J

n=3

E3 = 32 E1 = 542.11 × 10–19 J

n=4

E4 = 42 E1 = 963.76 × 10–19 J

The above energies are represented in the form of an energy level diagram in Fig. 3.2. Note that the energy levels are sufficiently apart and thus their quantization is distinctly evident. Another example can be for a gas molecule. Carbon monoxide is constrained to move in an infinitely small one dimensional box of width 0.7 nm (average mean free path of gases). Calculate the quantized energy values. Compare this with the kinetic energy of any gas at 300 K.

Particle in a Box

45

Mass of the molecule of CO =

28 6.023 × 10 23 × 1000

= 0.215 × 10–26 kg

Energy for CO molecule as particle in a box problem. En =

n2h2

=

n 2 (6.626 × 10 −34 ) 2

− 26 −10 2 8ma 2 8(0.215 × 10 )(7 × 10 m) = n2 × 0.521 × 10–22 J. For different values of n, value of En n = 1 E1 = 52.1 × 10–24 J n = 2 E2 = 240.8 × 10–24 J ...(3.4) –24 n = 3 E3 = 460.8 × 10 J n = 4 E4 = 819.92 × 10–24 J These distinct values show that energy levels are quantized. Also kinetic energy of CO molecule at room temperature 1 1 = kT = × 1.38 × 10–23 × 300 = 207 × 10–23 J ...(3.5) 2 2 Since the values as given in relation (3.4) and (3.5) are comparable, one can conclude that kinetic energy of CO and for that matter any gas is quantized but can be taken as continuous. As this energy is used up for the translational motion, we say that energy levels in translational motion is continuous as taken in kinetic theory of gases and thermodynamics.

  

Fig. 3.2: Energies of an electron moving in a one-dimensional box

3.3

ZERO POINT ENERGY

We will talk more about zero point energy in Chapter 5. Energy corresponding to n = 1 viz., called zero point energy.

h 8ma 2

is

46

Elements of Molecular Spectroscopy

The uncertainty principle requires that if a particle is confined within a region, then it must possess kinetic energy. For example, since the position of the particle is uncertain to no more than D implies that there must be an uncertainty in width of box. i.e., ∆x = a, the condition ∆x .∆p = 2 D at least. This suggests the presence of kinetic energy of the order of momentum of the order ∆p ≈ 2a 2 D ∆p ≈ which is about the magnitude of the zero point energy. Since zero point energy is at least 2m 8ma 2 finite and not equal to zero it means that particle inside the box is not at rest even at 0 K. Another way of looking at it is that if the wave function is to be zero at the walls, both smooth, continuous, yet not zero everywhere then it must have curvature; but if it is curved the particle possess kinetic energy. Therefore, if the particle is in cavity it must possess kinetic energy. If the box is made very large as if walls of the one-dimensional box are removed, particle becomes free to move without any restriction on the value of potential energy. The energy value given by equation 3.3 is not quantized. Hence we conclude that a freely moving particle without any restriction has continuous energy spectrum as is taken in translational motion. This amounts to saying that restrictions put over movement of particle quantizes its energy level. Since in quantized state, by Pauli exclusion principle no two particles can have all the quantum numbers same, hence they are not expected to collide with one another. Further, as the length of the box increases, the energy value goes on decreasing so much that it behaves as if continuous. In other words, as the length of box increases when particles go from quantized energy state to continuous energy state, or particle behaves as free particle. This is what happens in gaseous state, molecules have no restriction and hence collide against one another, thereby showing many other properties. The separation between neighbouring quantum levels is ∆E = En + 1 – En = (2n + 1) h2/8ma2

...(3.6)

and this decreases as length of container increases. The separation ∆E decreases to extremely small values as the length of container is increased. It follows that a free particle moving in an unbound region of space has unquantized translational energy levels. For this reason atoms and molecules free to move in laboratory size vessels may be treated as though their translational energy were unquantized. Consider a situation in which a ball of 100 g is confined to 1 m box moving with velocity 0.001 ms–1. En =

n 2 (6.626 × 10 −34 ) 2 8(0.1 kg )(1 m) 2

KE = E1 =

= n2 5.49 × 10–67 J

1 1 mu2 = × (0.1) (0.001)2 = 5 × 108 J 2 2

...(3.7) ...(3.8)

Equating equations 3.7 and 3.8 n2 × 5.49 × 10–67 = 5 × 10–8 or n2 = 9.1 × 1058 or n = 3.01 × 1029. Notice the fantastical high value of the quantum number n. To see if the energy levels of ball are quantized, let us calculate the difference in successive energy levels those characterized by the quantum number n and n + 1.

Particle in a Box

47

∆E = E n + 1 – En = (2n + 1)h2/8ma2 = (2n + 1) × 5.49 × 10–67 n = 3.01 × 1029 ∆E = (2 × 3.01 × 1029 + 1) (5.49 × 10–67)

with

...(3.6)

neglecting 1 as compared to 1029 so that ∆E = 2 × 3.01 × 1029 × 5.49 × 10–67 = 33.04 ×10–38 J This implies E1 = 5 × 10–8 J E2 = 5 × 10–8 + 33.04 × 10–38 = 5 × 10–8 J Since E1 and E2 are practically same. It is not possible to observe quantization of energy levels of macroscopic system like ball. Calculate first few allowed energy levels for an electron (m = 9.1 × 10–31 kg) confined in a onedimensional box about the size of atom a = 3 × 10–10 m. The energy state for any value of n will be En = Ground state energy Second energy level Third energy level

n=1 n=2 n=3

n 2 (6.626 × 10 −34 Js) 8(9.1 × 10

− 31

)(3 × 10

−10

m)

= n 2 6.7 × 10 −19 J.

10–19

E1 = 6.7 × J –19 E2 = 26.87 × 10 J E3 = 60.3 × 10–19 J

The energy evolved when the electron undergoes a transition from E2 to E1 as well as from E3 to E2. ∆E1 = E2 – E1 = 20.1 × 10–19 J = 1210.41 kJ mol–1 ∆E2 = E3 – E2 = 35.5 × 10–19 J = 2047.5 kJ mol–1 If these amounts of energies are emitted as single photon the wavelength of the emitted radiation λ1=

(6.626 × 10−34 )(3 × 108 ms −1 ) hc = = 99 × 10–9 m = 99 nm. ∆E1 20.1 × 10 −19

λ2=

(6.626 × 10 −34 )(3 × 108 ms −1 ) hc = = 69.4 × 10–9 m = 69.4 nm. ∆E 2 (35.5 × 10 −19 )

This radiation corresponds to electronic transition that falls in ultraviolet region. In the transition shown above only those transitions for which ∆n = +1 is taken. Such restriction on the change of quantum number is called selection rule.

3.4

PARTICLE IN A THREE-DIMENSIONAL CUBICAL BOX

Consider the motion of a particle of mass m confined to a three-dimensional cubical box with edge of length a and volume equal to a3. The potential energy is zero inside the box but infinite outside the box.

48

Elements of Molecular Spectroscopy

The wave equation for the particle is
2  ∂ 2 ψ ∂ 2 ψ ∂ 2 ψ  + + = Eψ 2m  ∂x 2 ∂y 2 ∂z 2 

...(3.9) .

On solving, we get 2 2 2 E = ( nx + n y + n z )

h2

...(3.10)

8ma 2

where nx, ny, nz = 1, 2, 3,…...

3.5

DEGENERACY

It is seen from equation 3.9 that the total energy depends upon the sum of the squares of three quantum numbers. It is evident that groups of different states, each specified by a unique set of quantum numbers, can have the same energy. In such a case, the energy levels and the corresponding independent states are said to be degenerate. Consider for instance the energy levels having energy = 14h2/8ma2. There are size combination of nx, ny and nz, which can give this value of energy. nx

1

1

2

3

2

3

ny

2

3

1

1

3

2

nz

3

2

3

2

1

1

This energy level is therefore 6 fold degenerate i.e., its degeneracy is equal to 6. Proceeding in this manner we can calculate degeneracy of the energy levels of a particle in a three-dimensional cubical box. The results are shown in Fig. 3.3.

Fig. 3.3: Energy in units of h2/8ma2 and degeneracy of a few states of a particle in a three-dimensional cubic box

Thus for a certain integral set of values of nx, ny and nz, E can have different values as given in Table 3.1.

Particle in a Box

49

Thus from Table 3.1 we observe that there can be different eigen states with same energy. Such eigen states are said to be degenerate. If there are two eigen states with same energy, the states are said to be doubly degenerate or its degeneracy is two, if there are three such states then it is threefold degenerate, and if six such states then it is sixfold degenerate and so on. Table 3.1: Allowed energy states for a particle in three-dimensional box in terms of nx, ny and nz

n2

nx

ny

nz

Wave function

3

1

1

1

111

1

6

2 1 1

1 2 1

1 1 2

211 121 112

3

9

2 2 1

2 1 2

1 2 2

221 212 122

3

11

3 1 1

1 3 1

1 1 3

311 131 113

3

12

2

2

2

222

1

14

3 3 2 2 1 1

2 1 1 3 2 3

1 2 3 1 3 2

321 312 213 231 123 132

6

= >

Degeneracy

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CHAPTER

4

Rotational Spectra 4.1

INTRODUCTION

In a monoatomic gas, atoms are associated with translational motions and electronic transitions that give rise to atomic spectra. The moment atoms combine to form molecule, spherical symmetry is lost and molecule can rotate and vibrate also. Molecular rotation and vibrations are associated with its own energy levels, transitions between which give rise to rotational spectra, vibrational spectra respectively which are not exhibited by atoms. Heteronuclear diatomic molecules, which are always associated with dipole moment, interact with electromagnetic radiation and thus show a rotational spectrum. As is evident from Fig. 4.1, the rotation of HCl plus and minus charges change places periodically and component of dipole moment fluctuates regularly. This fluctuation is plotted in the lower half of Fig. 4.1 and it is seen to be exactly similar in form to the fluctuating electric field of radiation. Thus interaction between radiation and molecules can occur if radiation frequency matches rotational frequency of molecule. This can happen only if rotational energy levels are quantized, energy can be absorbed or emitted, to give rotational spectrum. A rotational spectrum is observed in the microwave (0.3 to 30 cm–1) and far infrared (30 to 100 cm–1) region. In homonuclear diatomic molecules like N2, O2 charge separation as in HCl is absent hence electrical vector of radiation cannot interact with such molecules and thus pure rotational spectra is not observed. In other words, pure rotational spectra will be observed only if molecule has a permanent dipole moment. Direction of rotation Direction of dipole

Vertical component of dipole

time

Fig. 4.1: The rotation of a diatomic molecule, HCl, showing the fluctuation in the dipole moment measured in a particular direction

52

Elements of Molecular Spectroscopy

Molecules can rotate in all orientations. However, these rotations can be conveniently resolved into three mutually perpendicular rotational components, passing through the centre of gravity of the molecule, which are taken as principal axes of rotation. This will give every molecule three principal moments of inertia, one about each axis, usually designated as Ia, Ib, Ic. For linear molecules like HCl, CO2, CSO, acetylene Ic = Ib and Ia = 0. The moment of inertia Ii of a rigid rotator system about its centre of gravity is given by I i = Σmi ri2

...(4.1)

where ri is the perpendicular distance from the mass mi to the axis. In a diatomic molecule AB, in which atoms A and B have atomic masses mA and mB are separated by an internuclear distance r. This bond length does not change as the molecule rotates and such molecular rotations are termed as rigid rotators. Let mA be at a distance rA from centre of gravity (Fig. 4.2) and mB at a distance rB, then rA + rB = r and moments mA rA = mB rB which gives

rA =

mB r m A + mB

and

rB =

mAr m A + mB

...(4.2)

Fig. 4.2 2 2 But its moment of inertia, I = m A rA + m B rB

=

where

4.2

m A mB2 r 2 (m A + mB ) 2

+

...(4.3) m B m A2 r 2 (m A + m B ) 2

=

m A mB r 2 = µr 2 m A + mB

...(4.4)

m A mB = µ is called the reduced mass. m A + mB

DIATOMIC MOLECULES AS A RIGID ROTATOR AND ENERGY LEVELS

The relation I = µr2 has reduced the two body problem to one body problem. Rotating diatomic molecule will have zero potential energy so that Hamiltonian will be Hopp = −

∂2 ∂2 
2  ∂ 2  2 + 2 + 2 2µ  ∂x ∂y ∂z 

...(4.5)

Rotational Spectra

53

The Schrödinger equation for a rotator is given −


2  ∂ 2 ψ ∂ 2ψ ∂ 2 ψ  + +   = Eψ 2µ  ∂x 2 ∂y 2 ∂z 2 

...(4.6)

It is convenient to transform the above equation into spherical coordinates r, θ and φ. This transformation is lengthy, only transformed expression is given below.

−

∂  ∂ψ  ∂ 2ψ  1 1
2  1 ∂  r 2 ∂ψ  + 2 + 2 sin θ  2  = Eψ   ∂θ  r sin 2 θ ∂φ 2  2µ  r ∂r  ∂r  r sin θ ∂θ 

...(4.7)

Taking 1/r2 as a common term so that µr2 = I, moment of inertia of molecule. Since in this treatment, it is assumed that molecule behaves as a rigid rotator i.e., during the course of rotation internuclear ∂ψ = 0. Thus equation 4.7 reduces to distance does not change hence the first term involving ∂r
2  1 ∂  1 ∂ 2ψ  ∂ψ   sin θ +   = Eψ 2I  sin θ ∂θ  ∂θ  sin 2 θ ∂φ  Solution of this equation 4.8 gives −

...(4.8)

E h = 2 J(J + 1) = BJ(J+1) cm–1 hc 8π Ic

...(4.9)

h

with cm–1 as its unit. J is called rotational 8π 2 Ic quantum number which can have integer values i.e., J = 0, 1, 2….. Since h, 8π2 and c are all constant, B reduces to 27.993 × 10–47/ I. Here I = µr2 and µ is reduced mass as defined in equation 4.4 and r is distance between atom A and atom B in the molecule AB. The energy levels as given by equation 4.9 are called rotational states of the molecule and will be in cm–1. For different values of J, energy states will have different values. where B is called rotational constant and is equal to

Rotational quantum number, J Energy values, E/hc in

cm–1

0

1

2

3

4

0

2B

6B

12B

20B

As distinct energy values are obtained for different values of J, we say that rotational energy levels are quantized.

4.3

SELECTION RULES (a) The necessary condition for observing a rotational spectrum is that electromagnetic radiation should interact with molecule which is only possible if molecule has a dipole moment. (b) Transition between these energy levels give rise to the rotational spectra. Since different molecules will have different internuclear distance r and different reduced mass, consequently different rotational constant thus different rotational energy states. Selection rule for rotational transitions is ∆J = ±1 ...(4.10)

54

Elements of Molecular Spectroscopy

So that for rotational absorption spectra, ∆J = +1 i.e., only J → J + 1 transitions are allowed. i.e., E1/hc = BJ(J + 1), E2/hc = B(J + 1)(J + 2) so that

(E2 – E1) / hc = ν = 2B(J + 1) cm–1

...(4.11)

And for emission spectra, ∆J = –1 i.e., only J + 1 → J transitions are allowed

ν = 2B J cm–1

...(4.12)

A stepwise raising of rotational energy results in the absorption spectrum consisting of lines at 2B, 4B, 6B cm–1 etc. The rotational energy levels and corresponding transitions are shown in Fig. 4.3 and spectrum is drawn schematically at the bottom of this figure where equally spaced lines, separated by 2B cm–1 are observed.

Fig. 4.3: Energy levels scheme and allowed transitions for a rigid rotator diatomic molecule and the spectrum that result

Rotational Spectra

55

The values of B can be determined very accurately from the separation of lines from which interatomic distances can be calculated. The things will become clear with the Microwave spectra of the molecule CH that showed the following lines in cm–1: 28.914 57.829 86.743 115.658 144.572 Since the spectrum is observed in microwave region, it corresponds to the rotational spectra of this highly unstable molecule. Consecutive line spacing as shown by the spectra should be 2B, which is as follows: 28.914 57.829 86.743 115.658 144.572 Line position in cm–1 2B 28.915 28.914 28.915 28.914 Average value of 2B = 28.914 cm–1 Rotational constant, B = 14.475 cm–1 Substituting values of all constants give B = h/8π2Ic = 27.993 × 10–47/I I of CH molecule = 27.993 × 1047/14.457

= 1.9362 × 1047 kg m2

Reduced mass of CH = mCmH/(mC + mH)N = 12 × 1 / (12 + 1) 6.06 × 1023 = 0.1523 × 10–23 gm = 1.523 × 10–27 kg r2 = I/µ = 1.9362 × 10–47 / 1.523 × 10–27 = 1.271 × 10–20 m2 r = 1.127 × 10–10 m = 0.1127 nm

PROBLEM 4.1 In a very bad spectrum of carbon monoxide following lines were observed in cm–1: (a) (b) (c) (d)

7.7252 11.587 23.1756 27.0382 Calculate the rotational energy levels for the CO molecules. Assign the observed lines to various transitions. Calculate the line position of the missing lines. Calculate the internuclear distance in CO molecule.

SOLUTION The consecutive line spacing is shown in the second row of table: Line spacing ν in cm–1

7.7252

Consecutive line spacing 2B

11.587 3.8626

23.1756 11.587

27.0382 3.8626

J = ν /2B – 1

0

1

2

3

4

5

6

7

Energy levels BJ(J+1)

0 0

2B 3.8626

6B 11.3878

12B 23.1756

20B 38.626

30B 57.939

42B 81.115

56B 108.15

Transitions Line positions E2–E1/hc Consecutive line spacing 2B

0→1

1→2

2→3

3→4

4→5

5→6

6→7

2B 3.8626

4B 7.752

6B 11.388

8B 15.4504

10B 19.313

12B 23.175

14B 27.037

3.8626

3.8626

3.8626

3.8626

3.8626

3.8626

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Elements of Molecular Spectroscopy

The lowest value of consecutive line spacing is 3.8626 cm which may be taken as 2 B. So rotational constant of CO should be 2B = 3.8626 cm–1, B = 1.9313 cm–1 Energy levels of a diatomic molecule can be calculated by equation E/hc = BJ (J + I) with J = 0, 1, 2 = 1.9313 J (J + 1) cm–1 Varying the value of J in this equation gives energy level values for CO as given in row four of the table. Spectral lines will correspond to transition J → J + 1 as shown in row five. Calculate the difference in consecutive energy values (E2 – E1) as given in row six. The values correspond to line position of exact spectra of CO. A diagram with this data similar to Fig. 4.3 can be drawn for CO molecule, showing energy levels line positions and spectra. This may be taken as an exercise. By comparison of numerical values in row one and row six one can see that the observed lines correspond to the 1 → 2, 2 → 3, 5 → 6 and 6 → 7 transitions. The missing transitions and positions are as follows: 0→1 3→4 4→5 3.8626 15.4504 19.313 cm–1 To calculate internuclear distance, B for CO = 1.9313 cm–1 Moment of inertia of CO, I = 27.993 × 10–47/1.9313 = 14.35 × 10–47 Reduced mass of CO = mC mO /(mC + mO)N = 12 × 16/(16 + 12) 6.06 × 1023 = 1.13 × 10–23 g = 11.32 × 10–27 kg r2 = I/µ = 14.357 × 10–47/11.32 × 10–27 = 1.268 × 10–20 m2 r = 1.126 × 10–10 m = 0.1126 nm

4.4 EFFECT OF ISOTOPIC SUBSTITUTION On replacing one of the atoms in a molecule by its isotope, the molecule will be chemically identical with the original one without any change in nature of chemical bond. As the internuclear distance in molecules are practically determined by the electronic structure of the chemical bond the isotopic molecules should have the same internuclear distances. This is also confirmed by various experimental results. However, since the reduced masses are different the rotational constant B will be different. This should reflect a change in the energy levels. The heavier species will show a smaller separation 2B between the lines than the lighter ones 2B. Such a study is particularly of interest if one of the atoms is hydrogen. Study is carried out when hydrogen is replaced by its isotope deuterium. This will be explained with HCl as example: BHCl/BDCl = µDCl/µHCl = {mDmCl/(mD+ mCl)/{mH+ mC1/(mH + mCl)} Assuming

mH + mCl ≈ mD + mCl BHCl/BDCl = mD/mH = 2 BDCl =

1 B 2 HCl

Rotational Spectra

57

Thus rotational constant of CD will be half the constant of CH molecule. The corresponding rotational lines position will be as follows: Various transitions

0→1

1→2

2→3

3→4

4→5

C–H line position in cm–1

28.914

57.829

86.743

115.658

144.572

C–D line position in cm–1

14.457

28.914

43.371

57.829

72.742

This approximation holds good with hydrides. In other molecules full ratio of the reduced masses have to be taken. As in the case of NO (BN14O = 1.6707), if nitrogen is replaced by its isotope N15, then BN15O/BN14O = µN14O/µN15O = 0.96444 so that BN15O = 0.96444 × BN14O = 0.96444 × 1.6707 = 1.6113 cm–1 BN15O (Expt.) = 1.6125 cm–1 Similarly, in hydrogen molecule if one of the hydrogen is replaced by deuterium atom then BHD/BH2 = µH2/ mHD = 3/4 or

BHD = 3/4 × 60.809 = 45.606 cm–1 BHD (Expt.) = 45.606 cm–1 Even in these two cases the results are in very good agreement with the calculated values.

P ROBLEM 4.2 The pure rotational lines of HF molecule are represented by the equation 40.88(J + 1)cm–1, where J has integral values. Calculate (a) moment of inertia, (b) internuclear distance of HF, (c) the rotational constants when hydrogen is replaced by D and T in HF molecule, (d) line of maximum intensity in HF spectra.

SOLUTION (a) & (b) This equation 40.88 (J + 1) is similar to equation 4.11 and varying the value of J can generate line position of spectra. 2B 4B 40.88 81.76

6B 122.64

8B 183.52

10B 224.40

The difference between any two consecutive lines will be 2B so that, we have 2B = 40.88 cm–1 or B = 20.44 cm–1 Now B is given by h/8π2 Ic so that I = 2.642 × 10–47/20.44 = 1.369 × 10–47 Reduced mass of HF = 1.562 × 10–27 kg r2 = I/µ = 1.369 × 10–47/1.562 × 10–27 = 0.872 × 10–20 m2 r = 0.936 × 10–10 m = 0.093 nm. (c) The substitution of H by D or T does not affect internuclear distance but only change the reduced mass. Since B is inversely related to I, we write BHF/BDF = mD/mD = 2, so that BDF = 1/2 × 20.44 = 10.22 cm–1,

BHF/BDF = mT/mH = 3 BTF = 1/3 × 20.44 = 6.81 cm–1

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Elements of Molecular Spectroscopy

However, if exact reduced masses are taken then the rotational constant is obtained. B′DF = 10.756 cm–1

and

B′TF = 7.53 cm–1.

These values agree better with the experimental values, which are BDF (Expt.) = 11.01 cm–1 and BTF (Expt.) = 7.693 cm–1 (d) Line of max. intensity (as will be done in Section 4.5) 1 = 1.75 [Jmax ]HF = 0.5896 300/ 20.44 + 2 Rounded to nearest integer = 2 Transition which will show maximum intensity = 2 → 3 Corresponding line with maximum intensity = 122.63 cm–1.

4.5

INTENSITY OF SPECTRAL LINES

The prime requirement for finding the relative intensities of spectral lines is knowledge of the probabilities of transition between the various energy levels. As mentioned earlier the selection rule (Equation 4.10) for rotational spectra, ∆J = ±2, ±3 was forbidden. In other words, transition probability of all change with ∆J= ±1 is almost same — all spectral lines should have equal intensity in Fig. 4.4. However, one notices two features, these are: (i) the variation in the amount of radiation absorbed i.e., in the intensity components of the band, (ii) spacing between the components i.e., 2B is not in fact exactly constant throughout the spectra.

Fig. 4.4: Rotational spectrum of HCl molecules at 25°C

The relative probabilities of transition between various energy levels gives the relative line intensities with selection rule ∆J = ±1. The larger the number of molecules in a state the more intense should be

Rotational Spectra

59

the spectral line. Consider an imaginary case when all molecules are in the ground state for J = 0 then the intrinsic transition probability for single molecule to move from J = 0 state to J = 1 is the same as a single molecule moving from 1→2 or 2→3 or J = n→n + 1 or J = n + 1→n. This process of transition will continue to take place till an equilibrium states is reached. This will give rise to a distribution of molecules over the entire range of energy levels. As the spectrum is recorded over an assemblage of molecules which are distributed over the different energy states and since the intrinsic transition probabilities are identical consequently the line intensity will be directly proportional to the initial number. This distribution of molecules is given by Boltzmann distribution. NJ / N0 = exp −( E J − E 0 ) / kT = exp–hcBJ (J + 1)/kT

...(4.13)

where B is the rotational constant in cm–1, h = Planck’s constant and c = velocity of light. A simple calculation shows how NJ varies with J. For example, taking typical value of B = 10.59 cm–1 for HCl at room temperature T = 300 K, the relative population in the J = 1 and J = 2 are as follows: For J=1

for

N1/N0= exp(–10.59 × 6.63 × 10 = exp(–0.1006) = 0.9 J=2

–34 × 3 × 1010 × 1 × 2/1.38 × 10–23 × 300)

–34

10

–23

N2/N0= exp(–10.59 × 6.63 × 10 × 3 × 10 × 2 × 3/1.38 × 10 × 300) = exp(–0.3018) = 0.74 we see there are a little less number of molecules in J = 1 but for J = 2 it has further decreased to 74%. Similar calculation for HCl and HBr were carried out for entire range of J and plotted NJ vs J as shown in Fig. 4.5. It shows more rapid decrease of NJ/N0 with increasing J and with larger B. This implies that J = 0 → J = 1 line should be the most intense but spectra of HCl (Fig. 4.4) does not correspond to this conclusion. Experimental results show that line intensity does not decrease exponentially. Rather starting from J = 0 →1 the intensity of the spectral line increases with the increase of J and reaches a maxima after which there is a fall. This may be due to some additional features in the Boltzmann distribution that must be introduced to obtain the necessary population. Let us look at the rotating molecule again. A rotating molecule in an electric field of electromagnetic radiation is the one which tries to line up the molecule in the direction of the field. The presence of the field reveals an additional quantum restriction of total angular momentum. The rotational energy states of molecule are related to its rotational angular momentum by 2 EI = J (J + 1) h / 2π which shows that angular momentum is quantized too. Now, not only the total angular momentum is quantized but so also its Z-component is quantized with MJ = ±J, ± (J – 1), ……0 Pz = MJ h/2π, However, unlike energy, angular momentum is a vector quantity, consequently has an orientation φ taken along the axis about which rotation occurs. The component of angular momentum along the reference direction is given by quantum number MJ. This puts a restriction on the orientation of angular vector. The number of orientation is 2J + 1. The allowed orientation of the field direction are given by

P =

φ = cos–1 Pz/P = M J / J(J + 1)

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Elements of Molecular Spectroscopy

Fig. 4.5: The Boltzmann distribution of molecules in nondegenerate rotational energy level for HCl and HBr at 300 K

For J = 1 the vector length 1(1 + 1) = 1.42 and three values of Mj along a reference direction are +1, 0, –1 and φ = 45, 90, 135: making energy level three-fold degenerate. Similarly for J = 2 and J = 3 vector lengths are 2.45 and 3.365 with MJ = ± 2, ±1, 0 and MJ = ±3, ±2, ±1, 0 this makes these energy levels five-fold and seven-fold degenerate respectively. Thus, every rotational energy levels in the absence of an electric and magnetic fields is (2J + 1) degenerate. The behaviour in the presence of the field reveals that for each allowed rotational energy there are 2J + 1 different states. Thus, for energy level corresponding to the rotational quantum number J, the quantum restriction on orientation reveals it to have 2J + 1 state. We say that the Jth energy level has a multiplicity of 2J + 1or that it is (2J + 1)-fold degenerate. As Boltzmann distribution gives us the number of molecules per state: for a (2J + 1) degenerate state distribution has to be multiplied by this factor so that − ( E − E ) / kT Intensity ∝ NJ /No = (2J + 1) exp J 0

= (2J +1) exp [ − hcBJ ( J +1) /kT ]

...(4.14)

For increasing values of J this expression shows an initial increase in population because the 2J + 1 term increases and the exponential term remains near unity. However, at high J values the energy term EJ – E0 becomes large and the exponential decreases faster than the 2J + 1 coefficient increases. These two factors so operate that equation 4.14 lead to the population of the rotational energy levels. The intensities of the rotational spectra are reflection of population in that level, therefore, should be proportional to the populations of the rotational energy levels. Thus, the population distribution

Rotational Spectra

61

carries over into the intensity variation. This has been calculated for HCl and HBr and plotted as shown in Fig. 4.6. This variation is very much similar to spectral variation of intensity as shown in Fig. 4.4. At the maximum intensity in Fig. 4.6 if tangent is drawn it will be parallel to X-axis and thus dI/dJ should be zero. Differentiate equation 4.14 with respect to J. dI/dJ = 2 exp–hcJ(J+1)B/kT – (2J + 1) [hcB (2J + 1)/kT] exp.–hcBJ(J+1)/kT Equating

dI/dJ = 0 in this equation so that 2 – (2J + 1)2 hcB/kT = 0 (2J + 1)2 = 2kT/Bhc

or

Jmax =

kT / 2Bhc +

1 2

In this equation k, h, c are all constants so that Jmax = 0.5896 T/B +

...(4.15)

k / 2hc = 0.5896 and equation 4.15 reduces to 1 2

...(4.16)

The variation of the population and thus intensity for HBr with B = 8.4648 cm–1 and HCl with B = 10.5934 cm–1 with change of J is shown in Fig. 4.6.

Fig. 4.6: Intensity distribution of diatomic molecules of rotational spectral lines. The diagram has been drawn taking values of B = 10.5934 cm–1 for HCl and B = 8.4648 cm–1 for HBr at 300 K

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Elements of Molecular Spectroscopy

[Jmax ]HCl = 0.5896

1 300 / 10.5934 + 2

[Jmax ]HBr = 0.5896

= 2.367 Rounding to nearest integer = 3 Corresponding transition = 3 → 4 Line of Max. Intensity (Expt.) = 4 → 5 (Fig. 4.4)

1 300 / 8.4648 + 2

= 3.0 Rounding to nearest integer = 3 Corresponding transition = 3 → 4 (Expt.) 3→4 (not shown here)

The calculated value and experimental result are in good agreement with one another.

Fig. 4.7: Working arrangement in an absorption spectrophotometer

The arrangement of observing a rotational absorption spectra is something as shown in Fig. 4.7. Line joining the source and observer forms the z-axis of the system and a sample gas is placed between the two. Gas molecules are free to rotate and can have all possible orientations, but each orientation will not interact with radiation falling over molecules and thus will not show a rotational spectra. Only those molecules in the rotational energy level J and has one of the 2J +1 orientation along the path of beam that acts as the z-axis flick from energy state J to J +1 and thus contribute to rotational absorption spectra.

4.6

ROTATIONAL RAMAN SPECTRUM

For a diatomic molecule (homonuclear or heteronuclear) or any linear molecule rotational energy levels are given by E/hc = B (J +1) J,

J = 0, 1, 2, ......

….(4.9)

As the perturbation operator for Raman effect involves polarizability and not the dipole moment as in rotational spectra, significantly different selection rules arise. These are summarized in Table 4.1 for diatomic molecules. In all cases ∆J = 0 the Rayleigh line which defines the centre of the rotational spectrum will be there.

Rotational Spectra

63

Table 4.1: Selection rules for pure rotational Raman spectra

Molecule

1. Linear

Microwave spectra, Selection rule C∞v, only ∆J = ± 1

Raman spectra, Selection rule

Equation in cm–1

C∞v, and D∞h,

Stokes lines

∆J = ± 2

J+2←J ν = ν 0 – B(4J + 6) Anti-Stokes lines J←J+2 ν = ν 0 + B(4J + 6)

It may be noted that both Stokes and anti-Stokes lines correspond to S branch lines. The two quantum rotational jump behaviour stems form the fact that transitions depend on polarizability involving two dipole transitions–one for the photon coming in and one for the photon going out, but transitions originated because of the existence of rotational states. This suggests the selection rule as ∆J = 0, ± 2. Raman rotational spectra of homonuclear diatomic molecules is of particular interest here, which was not observed in pure rotational spectra showing thereby that homonuclear diatomic molecules do rotate and have rotational energy states. Further ∆J = J' – J'' is difference between the J. Value corresponding to symbols of branches S With the selection rules, ∆J = +2

R +1

Q 0

P –1

O –2

respectively. However, ∆J = –2, –1 will not be there in the rotational Raman spectrum. Only ∆J = +2 need be considered here which gives two series of lines. Stokes lines J → J + 2 and anti-Stokes lines J + 2 → J. It may be pointed out that many authors wrongly use S branch lines for anti-Stokes lines and O-branch lines for Stokes lines inspite of defining both series by ∆J = +2. Thus S-branch and R-branch lines are given by S-Branch lines R-Branch lines

ν Stokes = ν 0 – B (4J + 6) ν anti-Stokes = ν 0 + B (4J + 6) ν = ν 0 ± 2B (J + 1)

}

...(4.17) ...(4.18)

where the plus sign refers to anti-Stokes lines and the minus to Stokes and ν 0 is the wave number of the exciting radiation. Figure 4.8 shows a series of nearly equidistant lines on either side of the Rayleigh line with a line spacing of 4B. Homonuclear molecules like H2, D2, N2 show Raman rotational lines corresponding to the S-branch. Rayleigh line is a very sharp line and is used as band head which can very easily be identified. The first rotational Stokes and first rotational anti-Stokes lines are separated by 12B or so to say 6B on either side of Rayleigh lines. Subsequent lines will be separated by 4B, which is twice the value as observed in pure rotational spectra. This will be illustrated with an example.

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Elements of Molecular Spectroscopy

Fig. 4.8: Allowed Raman transitions between the energy levels of a rigid rotator diatomic molecule and the spectrum which arises from them

PROBLEM 4.3 The pure rotational Raman spectrum of N2 was obtained when gas contained at a pressure of the atmosphere in a specially constructed cell and excited by mercury light of 22946 cm–1 when following lines were observed: 22910, 22918, 22926, 22934, 22946, 22958, 22966, 22974 and 22982 cm–1 Draw the spectrograph to show the line position and thus assign each line to the transition.

SOLUTION Figure 4.9 shows N2 spectrograph like when it is excited by mercury lamp of 22946 cm–1. The broad line at 22946 cm–1 corresponds to the Rayleigh line. On the left hand side lines with frequency less than 22946 cm–1, are the Stokes lines and lines on the right hand side are anti-Stokes lines as

Rotational Spectra

65

observed. As N2 is a homonuclear diatomic molecule, all these lines correspond to S-branch with ∆J = +2, in accordance to Equation 4.17 the first Stokes and first anti-Stokes lines are separated by 6B = 12 cm–1 from Rayleigh lines. Subsequent lines on either side of Rayleigh line are separated by 4B = 8 cm–1. Since no line is missing these lines can be identified to transition indicated in this spectrograph blindly. In this discussion intensity of lines have not been touched, only line position was only considered. The rotational constants as evaluated give the following values: B = 2.00 cm–1

I = 13.9965 × 10–4 kg m2

r = 0.11008 nm

Fig. 4.9: Rotational Raman spectrogram of N2 in which line positions, in cm–1 are indicated

Another very good example will be on hydrogen molecule. Rotational Raman spectrum H2 molecule excited with mercury lamp of 22946 cm–1 has been observed. Some of the lines are as follows: Line positioning, cm–1

Transitions

21850.7 243.4

4B

5←3

Stokes line

243.4

4B

4←2

Stokes line

243.4

4B

3←1

Stokes line

365.4

2←0

Stokes line

365.1

6B Rayleigh line 6B

0←2

Anti-Stokes line

243.4

4B

1←3

Anti-Stokes line

243.4

4B

2←4

Anti-Stokes line

243.4

4B

3←5

Anti-Stokes line

22094.1 22337.5 22580.9 22946 23311.1 23554.5 23797.9 24041.3

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Elements of Molecular Spectroscopy

All these lines correspond to S branch lines where ∆J = 2. The Rayleigh line lies at 22946 cm–1. The lines less than this value correspond to Stokes lines and lines more than 22946 correspond to antiStokes lines. The first Stokes line and first anti-Stoke line are separated by 12B = 730.2 cm–1 so that B = 60.85 cm–1. The subsequent lines are separated by 243.3 cm–1 = 4B on either side. This again gives a value of B = 60.85 cm–1. These lines correspond to the various transition as indicated in the above table. As H2 is a homonuclear diatomic molecule with selection rule ∆J = +2, only S branch lines are observed. From the value of B internuclear distances can be evaluated which are as follows: I = 27.993 × 10–47/B = 27.993 × 10–47/60.85 = 46.00 × 10–49 Reduced mass = µ = r2

mH mH = 82.50 × 10–29 kg (mH + mH )

46.00 × 10−49

= 0.5575 × 10–20 m. 82.50 × 10 −29 r = 0.746 × 10–10 m = 0.0746 nm. = I/µ =

The Stokes lines will be more intense than anti-Stokes lines. However, due to nuclear spin, Raman spectra show a peculiar behaviour in the intensity variation. This variation is dependent on nuclear spin also which has not been discussed here. We have been only concerned with the line positions only in this Raman rotational spectra, discussion.

Transition 5←3 4←2 19430 ± 190.62 148.26 line position or line position =10.59 × 18 14

3←1 105.9

2←0 63.54

2→0 63.54

3→1 105.9

4→2 148.26

10

6

6

10

14

5→3 190.62 18

Fig. 4.10: Expected spectrogram of HCl

PROBLEM 4.4 Rotational energy levels of HCl can be generated from equation E/hc = 10.59 J (J + 1) cm–1. Generate the rotational spectrum as well as rotational Raman spectrum with a Rayleigh line at 19430 cm–1. SOLUTION From the equation E/hc = 10.58 J (J + 1). By varying the value of J, rotational lines can be generated from equation ν = 2B (J + 1) = 2 × 10.59 (J + 1) = 21.18 (J + 1) cm–1. Some of the transitions and corresponding line positions of pure rotational spectra of HCl are as follows:

Rotational Spectra

67

J

0

1

2

3

4

5

Line position in cm–1

21.18

42.36

63.54

84.74

105.9

127.06

Transition

0→1

1→2

2→3

3→4

4→5

5→6

These lines are equally spaced with 2B = 21.18 cm–1.

Raman Spectrum The Rayleigh line will be at 19430 cm–1 in Raman spectrum. On its left hand side will be Stokes lines which will be less than 19430 and on the right hand side are anti-Stokes lines which will be more than 19430. By varying the value of J in equation 4.17 with B = 10.59 cm–1 rotational Raman spectrum of HCl is generated (Fig. 4.10).

PROBLEMS 1. Which of the following molecules are expected to be microwave active and why? H2, Cl2, HCl, NO, CO, CO2, H2O, H2S, O2, N2, CCl4 2. The solution to the Schrödinger wave equation to a rigid rotator gives the energy as E = [J(J + 1)h2/8π 2I] joules. Express this in terms of energy in wave number units. Give the expression for rotational constant, B in wave numbers. 3. Explain why only molecules having permanent dipole moment exhibit rotational spectra. What is the selection rules for the rotational transition of diatomic molecules. 4. The reduced mass was a quantity that appeared at several points in this chapter and it is important to know how to calculate it. Calculate the moment of inertia of (a) 1H35 Cl, (b) 2H35 Cl, (c) 1H37 Cl with r = 127.45 pm. 5. Calculate the angles at which the angular momentum of a rotating molecule be oriented w.r.t. a reference direction for J = 3. Draw the diagram to show various orientation that molecule can have for J = 3. 6. Calculate the degeneracies of the rotational levels of a diatomic molecule with energies h/8πIc and 6h/8πIc; where I = moment of inertia. 7. Which spectra can help you to calculate the bond length in hydrogen molecule. Outline the method to calculate the bond length in H2 by use of this spectra? 8. H2 does not show a pure rotational spectrum whereas HF does. Comment. 9. Rotational spectrum of a molecule shows a series of equidistant lines spaced 0.84 cm–1 apart. Calculate: (i) ν of J = 2 → J = 3 transition, (ii) most intense transition at 300 K. 10. Some of the rotational spectral lines of AB (g) at 300 K at 86.7, 115.6, 144.5, 173.4 cm–1. Assuming it as a rigid rotator, draw the rotational energy levels for AB and label them indicating the transitions to which these lines belong. Find out the rotational constant and moment of inertia. 11. Raman rotational spectra of H 2 was observed with Hg light with Rayleigh line at 2294 cm–1. Stokes lines were observed at 2258.1, 2233.7, 2209.4, 2185.1 cm–1. Draw the rotational energy levels diagram for hydrogen molecule and show the Stokes and anti-Stokes transitions.

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Elements of Molecular Spectroscopy

12. CN+ has a bond length of 0.129 nm. Calculate in wave numbers the frequency of first four lines in its microwave spectrum. 13. In the rotational spectrum of NO molecule, the consecutive lines were separated by 3.344 cm–1. Calculate (i) I, (ii) µ, (iii) ro for NO molecule. 14. In a rotational spectrum of HCl an average consecutive line separation of 20.6 cm–1 was observed. Calculate the bond length in HCl. Summarize the factors which are responsible for intensity of lines. 15. The frequency separation between two consecutive lines for 12C16O and 13C16O rotational spectra are 3.842 and 3.673 cm–1 respectively. Assuming the bond lengths remains unchanged, calculate the exact atomic mass of 13C. Given 12C = 12 and 16O = 15.9994. 16. 2B for 79Br 19F was found to be 0.7143 cm–1. Calculate the (i) I, (ii) ro and (iii) ν of the transition J = 9 to J = 10. 17. The first two lines in the rotational spectrum of respectively. Calculate ro.

12C16O

appears at 1.15 × 155 and 2.30 × 105 MHz

18. Raman rotational spectrum of HBr showed the following lines: 19345.2, 19379.12, 19480.88 and 19514.80 cm –1 with Rayleigh line at 19430 cm –1 . Calculate (i) rotational constant B, (ii) Assign these values to various transitions. 19. A molecule has a moment of inertia of 6.65 × 10–46 kg m2. Determine the rotational transition for this molecule which has transition energy equal to kT/2 at 300 K, where k is the Boltzmann constant. 20. Rotational lines of H79Br can be generated from equation ν = 16.62 (J + 1) cm–1 by varying J. Draw the rotational energy diagram and spectra showing first four transitions (i) Which rotational line in H79Br spectra will be most intense at 27°C?, (ii)What will be the rotational constants for D79Br? 21. Copper monohalides CuX exist mainly as polymer even in the gas phase, so it was difficult to get spectra of the monomer. This was overcome by passing the halogen gas over chips of copper heated to 1000–1100 K. For CuBr the J = 13–14, 14–15, 15–16 transitions occurred at 84421.34, 90449.25, 96476.72 MHz. Calculate the rotational constant and bond length of CuBr.

a b

CHAPTER

5

Vibrational Spectra 5.1

A VIBRATING DIATOMIC MOLECULE

The interaction of infrared radiation with molecular vibrations gives the infrared spectrum. If the average position and orientation of a molecule remains constant but the distances between the atoms in the molecule change, molecular vibrations are said to take place. Vibrational spectrum is observed experimentally, as Infrared as well as Raman spectra but the physical origin of the two types of spectra are different and thus will be discussed separately. Vibrational transitions occur in the infrared region (100 to 4000 cm–1). A convenient frequency unit universally employed is number of waves per centimeter or wave numbers in reciprocal centimeters (cm–1) or SI unit of wavelength in micrometer (1 µm = 10–6 m).

Fig. 5.1: The vibrating system of two particles m1, m2 and a massless spring showing the displacement x1 and x2 and the restoring force, f

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Consider diatomic molecule as a conservative system of two mass points m1 and m2, joined by a massless perfectly elastic spring. The only forces are those due to the interaction potential (U). The particles move only along the bond axis of the system with displacement x1 and x2 from equilibrium position as shown in Fig. 5.1. These point masses start from equilibrium position gets stretched, stops for sometime, gets compressed and stops there for sometime, comes back to the equilibrium position. This makes one complete vibration. The frequency ν with which molecule vibrates is given by ν =1/2π k / µ ...(5.1) where k is called the force constant characteristics of a bond, µ is the reduced mass of the molecule. The time for which vibration stops may be very small but it stops which gives rise to a potential energy (U) to the molecule. The magnitude of the forces that restores each particle to the equilibrium position is proportional to the extent of compression or extension of the bond (spring) i.e., Restoring force, f ∝ – (x2 – x1) or f = – k (x2 – x1 ) = –kq ...(5.2) where k is characteristic of a bond called the force constant. The negative sign indicates that the restoring force acts in a direction opposite to the displacement as shown in Fig. 5.1. Also, when q is positive, it corresponds to extension, and compression gives a negative value of q. The work that must be done to displace the atoms by a distance dq is –fdq. This work is forced in the system as potential energy dU so that dU = – fdq ...(5.3)

Fig. 5.2: Potential energy curve, energy levels and infrared transitions of the harmonic oscillators diatomic molecule. The short vertical lines representing the transitions spread apart from one another in a horizontal direction for the sake of clarity only. The abscissa for the broken curve is the displacement from the equilibrium position (minimum). This curve is for diatomic molecule HCl, where ν = 2991 cm–1.

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71

Total potential energy can be calculated by integration of equation 5.3 with potential energy equal to zero at equilibrium position. U

q

q

1

∫ dU = U = −∫ f dq = ∫ kq dq = 2 kq 0

0

2

0

1 2 kq ...(5.4) 2 On substituting the value of force constant, k = 4π2µνosc in equation 5.4, gives U = 4π2µ νoscq2 ...(5.5) This shows that in a simple harmonic motion, the potential energy is proportional to square of the displacement. The potential energy curve is parabolic in form as shown in Fig. 5.2 is obtained.

or

U=

Linear Harmonic Oscillator Harmonic oscillations occur in classical mechanics when restoring force on a body is proportional to 1 2 kq , where k is called force 2 constant and is characteristic of a bond, and q is the displacement. Since it is linear oscillator its kinetic energy is confined to one dimension.

displacement. A force –kq implies the existence of a potential energy U =

Kinetic energy for a one-dimensional vibration = Potential energy for harmonic oscillator = U =

1 −h . ∂    2µ  2πi ∂q 

2

1 2 kq 2

Hamiltonian for simple harmonic oscillator H=

− h2 ∂ 2

8π µ ∂q Schrödinger wave equation Hψ = Eψ

or

2

2

+

1 2 kq 2

1 ∂ 2 ψ 8π 2µ   + 2  E − kq 2 ψ = 0 2 2 ∂q h  

...(5.6)

...(5.7)

The solution of this equation vanishes at infinity and is single-valued and finite but the energy is discontinuous that changes in by integral value of the vibrational quantum number V given by

h k 1 1  V +  E = hν  V +  = 2π µ  2 2 

...(5.8)

The quantum energy level, with simple harmonic oscillator as a model, is equidistant and has been represented by horizontal lines in Fig. 5.2. It is convenient to express vibrational energy level in wave numbers, cm–1 as follows: 1  −1  E = G ( V ) = ν  V + cm 2 hc 

where G (V) is called term values, ν the vibrational frequency in wave numbers.

...(5.9)

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For various integer values of V following G (V) are obtained: V G (V)

0

1

2

3

4

5

1 ν 2

3 ν 2

5 ν 2

7 ν 2

9 ν 2

11 ν 2

These values show that energy levels are equally spaced. Each energy level is separated by other consecutive level by ν cm–1. This is also shown in Fig. 5.2 which also shows that vibrational energy levels are quantized. 1 h ν (in Joules) or 2 ν /2 (in cm–1). It is called zero point energy and has no counterpart in the classical mechanics. This implies that when oscillator is in its lowest energy states, i.e., V = 0, all the energy cannot be removed from an oscillator. This is also in accord with uncertainty principle i.e., at 0° K even when translational, rotational motion have been frozen, molecule is still vibrating with zero point energy. Hence uncertainty of the position of molecules still exists due to zero point energy. Eliminating all energy, and, therefore, momentum would imply an infinite uncertainty in position but the molecule is confined by potential 1 energy. Zero point energy for polyatomic molecules will be hν per vibrational mode. 2

Zero point energy: Energy value (equation 5.8) for V = 0 is not zero but

5.2

SELECTION RULES

The dipole moment of a vibrating molecule will obviously be a function of the equilibrium internuclear distance. A vibrational transition is IR-active only if vibration leads to a change in the dipole moment. This condition is fulfilled only for diatomic molecules with unequal nuclei where the displacements x2 and x1 are not equal, as is the case shown in Fig. 5.1. Thus, in all cases of vibrations, change in dipole moment is expected and vibration should be IR-active. Such a behaviour is seen even in case of polyatomic molecules where vibrations are IR-active which involve a change of dipole moment. On the other hand, in homonuclear molecules such as N2, O2 & H2, displacements x2 and x1 are equal, hence, there is no change in dipole moment consequently no emission or absorption IR-spectrum is expected. However, an isotopic substituted diatomic molecule, say HD, displacements x2 and x1 will not be equal because of the different point masses and thus will involve a change in dipole moment and thus will show an IR spectrum. The selection rule for simple harmonic motion is ∆V = ± 1 This selection rule as per equation 5.9 gives the absorption frequency as 3 1    E ′′ − E ′  −1  hc  =  V ′ + 2  ν −  V′ + 2  ν = ν cm      

Transition between any two energy levels (equation 5.9) should give rise to an absorption peak. Since these energy levels are equally spaced, thus transition energies will be same and thus only one line is expected. The whole thing can be illustrated with NO molecule as an example. The energy level of NO molecule can be generated from the relation

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73

1  −1 G (V) = 1904  V +  cm ...(5.9a) 2  By substituting various integral values of V in this equation, energy level values can be calculated.

Zero Point Energy Zero point energy of NO molecule (V = 0) = 1904/2 = 952 cm–1 . This may be converted to Joules by multiplying with hc i.e., 952 × (6.626 × 10–34 Js × 3 × 1010 cm s–1) = 18923 × 10–24 J

Absorption Frequency of NO 1 = 1904 cm–1 2 As a harmonic oscillator NO molecule show absorption at ν = 1904 cm–1 = 1904 × 3 × 1010 = 5712 × 1010 Hz The energy ∆E associated with photon of 1904 cm–1 frequency is ∆E = hν = 6.606 ×10–34 × 5712 ×1010 = 37847 × 10–24 J Also force constant can be calculated from equation 5.1

G(3/2) – G(1/2) = 1904 × 3/2 – 1904 ×

k = 4π2µν2,

i.e.,

where µ of NO =

14 × 16 30 × 6.06 × 10 23

= 12.49 ×10–27 kg

ν = 5712×1010 Hz Thus k = 4π2 × 12.49 × 10–27 × (5712 × 1010)2 = 1609.4 N cm–1 Force constant k is a qualitative measure of how strong are forces of attraction between two atoms in the molecule. All the molecules must be distributed to various vibrational energy levels in accordance to Boltzmann distribution law. With this law, number of molecules in V = 1 at room temperature (27°C) can be calculated.

N v =1 N v =0

= exp − ∆E / kT = exp

 37847 × 10 −24  −  − 23 × 300   1.38 × 10

= e–9.1428 = 0.000107. It is seen that less than 1% of NO molecules are in V = 1 state and still smaller number in higher energy levels. Similar will be the situation in other molecules. Thus population in V = 1 and higher energy levels is negligibly small. With the absence of molecules in V = 1 and higher energy levels, transition from V = 1 → 2 and higher levels do not contribute to the peak V = 0 → V =1 transition. In other words, one absorption peak observed in vibrational spectra is a result of V = 0 → 1 transition only. This transition is called fundamental frequency. Let us go back to Fig. 5.1. The atom moves along the bond axis. These point masses start from equilibrium position, gets stretched, stops there for a moment, get compressed stops there for sometime and comes to the equilibrium position. This makes, one complete vibration. When the atoms stop for a moment, kinetic energy becomes zero. The entire energy is converted to potential energy i.e., Eosc = U.

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1  1  V +  hν = kq 2 2 2 

So that

...(5.10)

1  hν  2 V +  ...(5.10a) 2 k  This relation is used to calculate q as well as potential energy. This will be done for NO as an example.

or

q=±

1  2  V +  37847 × 10−24 2  q= ± 1609.4 = ± (2V + 1) × 0.00489 nm Also

U=

...(5.10b)

1 .4 π 2 µ ν 2 q 2 2

1  = 2 (3.141)2 (12.49 × 10–27)  V +  (0.00489 × 10–9)2 2   = 3.78 ×10–20  V + 

1  2

...(5.10c)

From equation 5.10b and 5.10c one can calculate q and U for various values of V. Some of these calculated values are as follows: V = q (m) U=

0 ± 4.85 ×

1 10–12

1.89 × 10–20

± 8.4 ×

2 10–12

5.67 × 10–20

± 10.84 ×

3 10–12

± 12.83 ×

9.45 × 10–20

4 10–12

13.23 × 10–20

± 14.55 × 10–12 17.01 × 10–20

On plotting U versus q, parabola similar to Fig. 5.2 will be obtained. Energy levels as calculated from equation (5.9a) can be drawn on this parabola.

Comparison of Thermal Energy with Vibrational Energy Energy associated with vibrational transition lie in infrared region. This energy for NO molecule as calculated earlier is ∆E = 3784 × 10–24 J= 3.78 × 10–20 J Now imagine we are sitting before a heater, the temperature of which is 800 K. Translational energy associated with this temperature =

3 kT 2

3 × 1.38 × 10–23 × 800 2 = 1.656 × 10–20 J =

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75

As these two values are comparable, we can conclude: (i) heater or hot air blower emits infrared radiation, (ii) the feeling of warm air falling on us is through molecular vibrations. Molecules coming in contact with heater get excited and go to various vibrational energy levels. Collision of these excited molecules against our body makes these molecules to come to ground state but in the process heat is transferred to our body. That is how we feel the warmth of heater. At this point, it may be noted that if air were made up of N2 and O2 only infrared radiation would not have interacted with these molecules. The reason being that in their molecular vibrations no change in dipole moment would have taken place, hence these molecules will not be excited by the infrared radiation. Consequently, we would not have the feeling of heat from a heater. This amounts to saying that warm air that we feel is due to the CO2, H2O and various pollutant molecules. In the same way warming up of earth’s from sun radiation is through H2O, CO2 and pollutant molecules only.

5.3

EFFECT OF ISOTOPIC SUBSTITUTION

As force constant is dependent on electronic distribution, isotopic molecules will have same force constant and hence frequency shift should depend on relative change of masses. In a molecule AB, if A is substituted by its isotope A' then νAB =

1 k , 2π µ AB

ν AB ν AB = = ν A′B ν A′B

ν A′B =

1 k 2π µ A ′B

...(5.11)

µ A′B µ AB

...(5.12)

This relation is very different from the one obtained in rotational spectra. Substitution of hydrogen atoms by deuterium produces the largest frequency shifts because of large change in the relative masses of the molecule. A study of the spectra of isotopic molecules is useful in the spectral assignments. For a diatomic molecules X–H,

νHX νDX =

µ DX = µ HX

mD = 2 = 0.7245 mH

...(5.13)

For some of the molecules with their isotopic substitution are as follows:

νH–F

νHCl

νHBr

ν HI

cm–1

4138

2991

2649

2309

ν D–X in cm–1

2998

2145

1884

1639

Ratio ν H–X/ ν D–X

0.7245

0.717

0.7112

0.7098

ν H–X in

5.4

INTERACTION BETWEEN IR-RADIATION AND VIBRATING MOLECULES

A vibrating dipolar molecule is like a fluctuating electric dipole component, a stationary alternating electric field is produced whose magnitude changes periodically with a frequency equal to that of the

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vibrational frequency. This stationary alternating electric field (shown in Fig. 5.3) interacts with moving electric field of electromagnetic radiation. In homonuclear diatomic molecule, the stretching and compression of the bond does not change the dipole moment and hence no fluctuating dipolar electric field is produced thereby implying non-interaction between radiation and molecules. On the other hand, in heteroatomic molecule AB the electric dipole oscillates with the vibrational frequency and hence radiation of this frequency can be absorbed. The energy absorbed is enough to excite the molecule to a higher allowed vibrational energy level. If the frequency of the radiation is not equal to the vibrational frequency of the molecule under consideration there will be no interaction between the two.

Fig. 5.3: The vibration of the AB molecule showing the fluctuation in the dipole moment

5.5

ANHARMONICITY IN VIBRATIONS

It should be apparent from Fig. 5.2 that potential energy and, therefore, the restoring force increases to infinity with increasing distance from equilibrium position. Therefore, it places no limit on how far a bond can be stretched, while in a molecule when atoms are at a great distance from one another, the attractive force approaches zero and at this stage the bond will break. The corresponding potential energy should attain a constant value. Further, as can be seen in any spring, force of compression and that of stretching are not equal. This factor should make the variation of potential energy (U) with internuclear distance asymmetrical in nature as shown in Fig 5.4. This figure shows the variation of potential energy with internuclear distance for HCl molecule and this kind of potential energy function is called Morse potential. The dotted lines in the figure gives the variation of potential energy for simple harmonic motion.

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77

Using the Morse potential, Schrödinger wave equation for molecular vibrations can be written, solution of which gives us the energy value of the vibrational energy level as G (V) = E/hc = ν (V + 1/2) – ν x (V +1/2)2 ...(5.14) –1 where ν is vibrational frequency in cm as given by equation 5.1. V is vibrational quantum number and can have all integer values, i.e., 0, 1, 2, 3 …., x is quite small and is called anharmonicity constant. Addition of a square term in equation 5.14 makes the energy levels not equally spaced rather makes the energy levels to converge. Such a comparison is clearly shown in Fig. 5.4, where both the variation of potential energy Harmonic oscillator is illustrated along with Morse potential. It is apparent from the figure that for smaller values of V, the two curves coincide with one another. In this chapter since we will be dealing with lower values of V, for the present we can assume that anharmonicity constant is equal to zero.

Fig. 5.4: Morse potential function for HCl. The broken line is a parabola for simple Harmonic oscillator that form best approximation to Morse potential near the minima.

The selection rule also changes to ∆V = ±1, ±2, ±3, ...(5.15) where plus sign corresponds to absorption spectrum and minus to emission spectrum. The fundamental absorption band νvib also called the normal band, corresponds to ∆V = 1, V = 0 → 1, i.e., G (1) – G (0) = ν With

1 k ν = 2πc µ

...(5.16) ...(5.17)

Bands corresponding to ∆V = ± 2, ±3 and transition V = 0 → 2 and V = 0 → 3 are called first and second overtones, given by

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V=0→2

G (2) – G (0) = 2 ν

...(5.18)

V = 0 →3

G (3) – G (0) = 3 ν

...(5.19)

Fundamental band will be of maximum intensity, while first and second overtones will have considerable less and still lesser intensities. Also, if the spectrum is recorded at a higher temperature, there may be quite a few molecules in the first, second or higher vibrational energy levels. The transition from higher energy levels with the selection rule ∆V = ±1 are called hot bands. The first hot bands correspond to transition V = 1 →2 so that G(2) – G (1) = ν And the second hot band is V = 2 → 3 as G (3) – G (2) = ν

...(5.20a) ...(5.20b)

However, hot bands are not observed at room temperature. Therefore, only one line of high intensity will be observed and this frequency is same as given by equation 5.17. Such a simple result is obvious from Fig. 5.2, where some of the expected energy levels are shown. These energy levels are equally spaced, transition energy between any two neighbouring states will require some energy and thus only one line will be observed. The absorption frequency of HCl molecule is observed at 2991 cm–1 = 3 × 1010 × 2991 = 8973 × 1010 Hz The energy of this quantum of radiation is hν = 6.62 × 10–34 × 8973 × 1010 = 5.9455 × 10–20 J. It is this energy that must correspond to energy difference. Even for V = 0 → 1 transition at 300 K population of HCl molecule in V = 1 can be calculated by Boltzmann distribution.

N v =1 N v =0

=e

−

hν kΤ

= exp

 − 5.9455 × 10−20    − 23 × 300   1.38 × 10

= exp[–14.288] = 6.23×10–7 µ = 1.6148 × 10–27 kg for HCl molecule ν = 8973 × 1010 Hz k = 4π2 1.6148 × 10 –27 × (8973 × 1010)2 = 515 N n–1 (SI units) for HCl molecule Results for some more molecules are compiled in Table 5.1.

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Table 5.1 : The position of the fundamental vibrational absorption of some diatomic molecules, their force constants, reduced masses, values of amplitude of vibrations in the ground state V = 0

Molecule

ν osc /cm–1

k/Nm–1

Reduced masses, µ /kg

qmax /cm

HF

4138

9742

1.60156 × 10–27

9.18 × 10–10

HCl

2991

515

1.6148 × 10–27

10.76 × 10–10

HBr

2649

398

1.5968 × 10–27

11.5 × 10–10

HI

2309

310.0

1.647 × 10–27

12.16 × 10–10

CO

2169.81

1904.6

11.39 × 10–27

4.757 × 10–10

CO+

2214

1983

–

4.708 × 10–10

NO+

2379.42

2512

–

4.337 × 10–10

NO

1904

1609.4

12.495 × 10–27

4.847 × 10–10

NO–

1363

824.45

–

5.73 × 10–10

CN

2068.56

16.31

10.733 × 10–27

5.017 × 10–10

CP

1239.67

784.4

104.37 × 10–27

5.602 × 10–10

It shows that less than 1% of the molecules are in the V = 1 state and still smaller number in higher levels. It shows that in experiments at room temperature only transition from the ground vibrational state V = 0 are of major importance. However, if spectra is recorded at higher temperature say at 1000 K, population in V = 1 state increases to 0.014 making transition V = 1 → 2 viable. Since this transition is observed at higher temperature it is called hot band. According to equation 5.20a & 5.20b, hot band should appear at the same frequency as fundamental, because anharmonicity constant was taken to be equal to zero. If that is taken in hot bands will be observed at a lower frequency. As an example, HCl hot band values are given below: Fundamental band

1st hot band

2nd hot band

3rd hot band

Transition

0→1

1→2

2→3

3→4

Vib. freq. in cm–1

2885.4

2779

2675

2568.6

This increasing difference is enough to emphasize that hot bands will appear as separate peaks from fundamental bands. Similar results are obtained in other molecules. Table 5.1 shows that rigidity of a chemical bond, measured by the force constant which is reflected in the amplitude of vibration. Thus, it is seen that greater the rigidity, lower is the amplitude, e.g., compare HF value with those of the HI molecule. Any change, which enhances bond order in a molecule, increases its force constant and thus increases its frequency and vice versa. As an example, compare CO and CO+, NO+, NO and NO– and other values. The treatment that is given here should not lead to misconception that rotational or vibrational spectra is shown by covalent bonded compounds only. All ionic compounds (solids) have to have its vapor state. For example, ionic solids like NaCl in its vapor state is in the diatomic form i.e., NaCl, shows a rotational spectra. However, for ionic solids, its state does not decide if it will show a vibrational

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spectra or not. Ionic solids do show vibrational spectra and all that has been said above is applicable on ionic compounds as well. A large number of investigations have been made on the vibrational spectra of alkali halides. Ionic crystals of alkali halides have intense absorption in the far infrared region. When considering vibrations for alkali halides it is found that positive and negative ions vibrate in the same way as HCl in gaseous state. The absorption coefficient in the alkali halides vary over the frequency range 1–500 cm–1 by a factor of 104, but samples of thickness 1 cm to 1 µ are necessary. The absorption spectra of thin films produced by subliming alkali halides on crystal quartz are shown in Fig. 5.5. It may be noticed that on cooling, the peaks shift to higher frequency due to contraction of the lattice. MgO is a typical case having the same structure as NaCl but with double charged ions. Its main absorption occurs at 410 cm–1 at 100 K while partially ionic crystal ZnS has a absorption peak almost as high as that of the alkali halides at 277 cm–1.

5.6

VIBRATIONAL RAMAN SPECTRA

Vibrational energy levels are given by equation 5.8; transition between these energy levels will give Raman spectra. The selection rule for Raman spectra, if vibrations are simply harmonic, is that vibrational

Fig. 5.5: The far infrared absorption spectra of thin films of alkali halides (Sublimed films on quartz) recorded at 90° K (–––––) and 300° K (-------)

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81

quantum number V changes by ±1. A classical picture will elucidate why the rotational quantum number changes by two but the vibrational quantum number by one. In rotational spectra of a molecule, the dipole moment is restored to an indistinguishable orientation after a complete rotation of the molecule, but polarizability is restored to an indistinguishable orientation twice on a rotation. In a vibration the dipole moment and polarizability return at the same time to an indistinguishable position and thus vibrational spectra and Raman vibrational spectra both have ∆V = ±1. Second, in order that a molecule show Raman vibrational spectra, it is essential that vibration should produce a change in polarizability. This shifts the spectral lines by νM = (EV+1 – EV)/hc on either side of the Rayleigh’s line. The line position of Rayleigh line is denoted by ν 0 and other Raman lines are given by ν 0 ± ν M . Unlike IR spectra where line position has a fixed value for a molecule, in Raman spectrum line position are dependent on ν0. Minus sign corresponds to Stokes line while plus sign gives the anti-Stokes line. All diatomic molecules show a change in polarizability and exhibit a vibrational Raman spectrum. This includes the homonuclear diatomic molecules. The relation will give the spectral line ν = ν0 ±

1 k cm −1 2 πc µ

where µ is the reduced mass and k, the force constant of molecule. At room temperature more than 99% of the molecules are in the ground vibrational state (V = 0) and the remaining molecules are distributed in the other states. Therefore, the intensity of the Stokes line (0 → 1) is much greater than that of the anti-Stokes line (1 → 0). Similar reason can be given for very low intensity of Raman hot bands and Raman overtones. As these two types of spectra generate no new information, much interest has not developed for diatomic molecules.

5.7

POLYATOMIC MOLECULES

Though the change in going from diatomic to polyatomic molecules can be regarded as ones of degree only. The situation for general molecule containing N atoms is sufficiently different to require a new approach and even new concepts. A molecule with N atoms will have 3N degrees of freedom for making various types of motions. The molecule can make translational motion in any direction, but all these motions can be resolved into three motions which are directed along the three coordinate axis, thereby giving three independent movements (x, y, z), and thus taking away three degrees of freedom. Similarly, rotation about three perpendicular axes accounts for another three degrees of freedom. However, for the linear molecule rotation about its axes is no rotation, so rotation about other two perpendicular axes accounts for another two degrees of freedom. The residual (3N–5) movements for linear molecules or (3N–6) movements for nonlinear molecules or degrees of freedom must involve only internal vibrations. Of the (3N–6) normal modes of vibration, (N–1) may be described as bond stretching vibrations. The remaining (2N–5) modes (2N–4) (in linear) are angle-bending vibrations. More explicitly, the total vibrational energy can at any time be expressed as the sum of not more than 3N–6 terms of the form 1/2kq2 (Equation 5.4), where k is a force constant. These individual terms add up to represent the vibrational modes of the frequency into a very complicated pattern. They are the simplest and smallest number of independent vibrations by vector addition of which any possible vibrational oscillation within the structure can be compounded. For N = 6, e.g., CH3OH the number of such vibrations are 12. The molecule vibrates with all the 12 vibrations simultaneously making real movement of the molecule quite complicated. Since this motion is a result of vector addition of each

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molecular vibration, a reverse vector process can resolve it. The complex molecular motion of the molecule is resolved into individual vibrations by a mathematical process called Normal Coordinate Analysis. The complication of the spectrum when each such vibration may give an absorption peak at its own frequency is obvious. Two features help in resolving this situation. Firstly, only those vibrations, which involve a change in dipole moment, will interact with infrared radiation and thus show an absorption line. Secondly, symmetry in molecular structure may lead to pairs or even triplets of vibrations coinciding in frequency. Examples of the operation of these factors are methane CH4, (3N – 6 = 9) which shows only two fundamental absorption bands in infrared region. Both the vibrations (see Fig. 5.6) ν 3 and ν4 are triply degenerate vibrations. This means that three vibrations of ν 3 mode are νO

νP 

νP¡

νP = OSPT ¢ªO

νO = PWOT ¢ªO

νQ 

νQ¡

νQ¢

_O νQ = QNPN ¢ª O

νR 

νR¡

νR = OQNT ¢ªO

Fig. 5.6: Normal vibrations of a tetrahedral CH4 molecule

νR¢

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83

identical and thus show only one peak. Even then this make up to six vibrations. The other three vibrations involve no change in dipole moment and hence are IR inactive. It may be mentioned here that methane has a zero dipole moment and in ν 3 and ν4 vibrational modes dipole moment appears and thus IR active. Similar condition applies to Raman spectrum but there the dipole factors are replaced by molecular polarizability. The necessary condition for a vibration to be Raman active is that there should be a change in polarizability. Raman spectra of methane shows in addition to two peaks observed in IR spectra, it shows two more peaks, ν1 (nondegenerate) and ν2 (double-degenerate). In other words, all the nine vibrations involve change in polarizability and hence are Raman active. These nine vibrations are represented in Fig. 5.6. Now if two of the hydrogen atoms are replaced by chlorine to get methylene chloride, the symmetry of the molecule decreases and all the nine vibrations become nondegenerate. Raman spectra show nine lines for these nine vibrations while the infrared spectra shows even absorption peaks. Another example is that of benzene, C6H6( 3N – 6 = 30). These thirty vibrations are made up of ten nondegenerate and ten double degenerate vibrations. Infrared spectra of benzene show four absorption peaks, three of which are double degenerate. Since benzene has centre of symmetry, none of the infrared peaks are observed in Raman spectra. It shows seven lines, five of which are double degenerate. The two together accounts for nineteen vibrations. Eleven vibrations of benzene neither involve any change of dipole moment nor any change of polarizability and thus infrared as well as Raman inactive. High symmetry of benzene molecule makes it behave in this way. By fundamental absorption, it corresponds to an actual molecular vibration frequency, say ν1 = ν, and not an overtone (2 ν1 , etc.) or a combination tone, ( ν1 + ν2 , ν1 – ν2 etc.). The intensity of these subsidiary features is usually far less than (about one-tenth) that of the fundamentals.

5.8

CARBON DIOXIDE

As a general rule of Thumb “Fundamental vibration with highest value is asymmetric stretching and with minimum value correspond to bending mode”. This rule applies to all types of molecules. Carbon dioxide is another molecule of interest, which has a linear symmetrical molecule belonging to the point group D∞h with zero dipole moment. It should show four fundamental modes (3N – 5 = 4) two of which are bond stretching and one double degenerate angle-bending mode as shown below.

Symmetric stretching ν1 =1340 cm–1 IR-inactive Raman-active

Bending mode ν2 = 667 cm–1 IR-active Raman-inactive

Asymmetric stretching ν3 = 2349 cm–1 IR-active Raman-inactive

These three vibrational modes 1340, 667 and 2349 cm–1 were assigned to ν1 , ν2 , ν3 . As a rule of thumb the highest value is assigned to asymmetrical stretching and lowest to bending mode. The ν1

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Elements of Molecular Spectroscopy

symmetric stretching vibration is IR inactive because both the carbon–oxygen bonds are extended equally but in opposite direction and so dipole moment remains zero. The ν3 is IR active in which one bond is extended, the other is compressed and vice versa leading to a change in dipole moment. The ν2 vibration involves angle-bending vibration, one in the plane of paper and other in a plane perpendicular to this plane. In this vibration molecule goes from linear to bent structure and thus develop a dipole moment and thus IR active. This vibrational mode is double degenerate. Polarizability may be looked upon as proportional to the volume of molecule. Thus the change of polarizability means change in volume. In ν1 there is increase in bond length and thus change in volume, hence Raman active. On the other hand in ν3 one bond is stretched and the other is compressed, hence involve no change in volume thus Raman inactive. ν2 involves bending but no change in volume, no change in polarizability and thus Raman inactive. Fundamental vibrational frequencies of some other molecules with similar geometry are given in Table 5.2. Table 5.2

Molecular

ν1

ν2

ν3

360

70

413

N3–

1345

652

2030

CS2

658

397

1533

NO2+

1400

667

2360

HgCl2

5.9

Fundamental vibrational frequencies in wave numbers

WATER

Another molecule of interest can be water, H2O. It has a triatomic, symmetrical bent structure, which gives it dipole moment. It has a bent structure and should have 3N – 6 = 3 × 3 – 6 = 3 fundamental vibrational modes, two (N – 1 = 2) of which are bond stretching and one (2N – 5 = 1) angle-bending mode. The water vapors show three sharp absorption peaks at 3755, 3650 and 1595 cm–1. All three are Raman as well as IR-active. As all the three fundamentals are both Raman as well as IR-active molecule, has bent structure with no centre of symmetry. The one with highest value, 3755 cm–1 corresponds to asymmetrical stretching and one with lowest value 1595 cm–1 as bending. The third corresponds to symmetrical stretching vibration. These vibrations may be represented as follows:

Vibrational Spectra

85

The vibrations ν1 , ν2 are labelled as symmetric because the symmetry operation takes the vibrating molecule to an equivalent configuration. Similar reasoning makes ν3 asymmetric stretching. As all the three vibrations are associated with change of dipole moment hence IR-active. During the symmetric stretching, the molecules as a whole increases and decreases in size. Now when a bond is stretched, bond electrons are less firmly held by nuclei and so the bond becomes more polarizable. Thus, the polarizability of water may be expected to decrease while the bonds stretch and increase while they compress. Thus symmetrical stretching will involve a change of polarizability and thus Raman active. On the other hand, while undergoing the bending motion there is continuous change of bond structure hence polarizability also change and thus Raman active. Finally, we have asymmetric stretching motion ν3 , whether both size and shape remain approximately constant but the direction of the major axis changes markedly. Thus all three vibrations involve changes in polarizability and all are Raman-active. Table 5.3

ν1

cm–1

ν2 cm–1

ν3 cm–1

Gas

3651.7

1595

3755.8

H2O liquid

3219

1627

3445

Solid

3400

1620

3220

H2O in dioxane

3518

1638

3584

In the liquid state, the water molecule will be hydrogen bonded when each hydrogen will be shared by two oxygen, as a result ν1 and ν3 will decrease but ν2 increase. Similarly, dissolving the water in dioxane when there is formation of hydrogen bonding between water and dioxane and is responsible for the shift of the stretching modes to lower frequencies and for bending mode shift in frequencies too higher values. The experimental results for water in different states are summarized in Table 5.3. Some other molecules with structure similar to water are given in Table 5.4. Table 5.4

Fundamental vibrational frequencies in wave numbers Molecule

ν1

ν2

ν3

H2S

2615

1183

2684

Cl2O

688

320

969

ClO2

943

445

1111

NO2

1320

648

1621

NO2–

1328

828

1661

SO2

1151

518

1360

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Elements of Molecular Spectroscopy

PROBLEMS 1. The vibrational energy levels of 1H81 Br are given by the equation G(V) = 2650 (V + 1/2) – 50 (V+1/2)2 where V is the vibrational quantum number. Calculate (i) fundamental vibrational frequency, first hot band and first overtone, (ii) vibrational frequency for the transition V = 32 to V = 33, (iii) the fundamental vibrational frequency for 2H81 Br. 2. For NO molecule fundamental vibrational frequency = 1904 cm –1 and reduced mass = 12.495 × 10–27 kg. Calculate (i) zero point energy, (ii) force required to increase the bond length by 0.01 nm assuming the bond obeys Hooke’s law. 3. Raman vibrational lines (Stokes) for H2 is observed at 18545 cm–1 with Hg lamp but no line is observed in IR spectra. However, on replacing one of the hydrogen atom by Deuterium, HD molecule also show IR spectra. Explain. 4. The four vibrational energy levels of a diatomic molecule were found to lie at 2990, 5672, 8344.9 and 10924.8 cm–1. Calculate the zero point energy, fundamental frequency and first hot band, for this molecule. 5. Zero point energy is not zero but some finite positive value. Further how does this corroborate with the Heisenberg Uncertainty principle. 6. Which of the following are expected to exhibit IR spectra or be active in IR region? H2, O2, HCl, H2O, CO2. Explain. 7. Write short notes on: (i) Zero point energy, (ii) Hot bands, (iii) Stokes and anti-Stokes lines. 8. Account for the fact that at room temperature, the line resulting from V = 0 is the most intense than those resulting from higher values of V. However, in microwave spectrum the line resulting from the transition from J = 0 is not necessarily the most intense. 9. Assuming that the force constants of OH and OD are the same, calculate the fundamental vibration frequency of OD if that of OH is 3737 cm–1. 10. In order to increase the intensity of anti-Stokes lines of any substances, should the sample be heated or cooled? Explain. 11. What is Raman effect? Explain the origin of Stokes and anti-Stokes lines. 12. Fundamental vibrational frequency of CO and SbBr are 2169 and 243 cm–1 respectively. At 300 K, CO shows only one absorption peak corresponding to V = 0 to V = 1 transition, whereas SbBr shows two absorption peaks corresponding to V = 0 to V = 1 as well as V = 1 to V = 2 transitions. Explain. 13. At what temperature would the population of AB molecule V = 1 level be half the population of the V = 0 level, given the fundamental vibrational frequency of AB as 4395 cm–1. 14. The vibrational energy levels of 88SrH are given by the relation G(V) = 1206.2 (V + 1/2) – 17.0 (V + 1/2)2 Calculate the fundamental vibrational frequency, zero point energy, second hot band and second overtone for SrH molecule and thus calculate force constant of the bond. At 300 K and 3000 K, what are the intensity of the first hot band relative to fundamental bands? 15. How will you distinguish between the overtones and the hot bands of a spectrum?

a b

CHAPTER

6

Group Frequency Concept 6.1

INTRODUCTION

A molecule with N atoms exhibits 3N – 6 fundamental vibrations, each imposed over other making molecular motion quite complex. However, by applying a mathematical process called Normal Coordinate analysis, this complex motion is resolved into 3N – 6 vibrations each corresponding to one spectral peak. In principle this mathematical process can be applied to molecule of any complexity, but then it becomes difficult to visualize how such a complex molecule will vibrate, and thus difficult to use IRspectroscopy as a tool to determine molecular structure. It is for such complex molecules that a nonmathematical concept of group frequency has been developed. The normal modes of molecular vibrations can be divided into two types: (i) Skeletal vibrations (ii) Characteristic group vibrations. The skeletal vibrations involve displacement of all atoms to some extent, absorption peaks of which fall in the range 1400–700 cm–1. A complex pattern of bonds may occur in the infrared spectrum of a compound, which could be identified as due to a particular skeletal structure. This will be a particular skeletal structure. The second type of normal modes is known as characteristic group vibrations in which displacement of only a small portion of the molecule independent of the rest of the molecule take place. For example, OH group of alcohols absorb between 3200–3600 cm–1, ketones C O group absorbs around 1700 cm–1, C—H alkanes 2850–2960 cm–1, Alkenes 3020–3080 cm–1, Alkynes 3300 cm–1. Thus, a terminal alkyne, RC C—H is characterized by its C—H stretching band at 3300 cm–1. Such groups may be assumed to be a diatomic like molecule made up of RC2 as one body and hydrogen as another. The movement of the heavy part is so small to be assumed as if it remains almost stationary while the light atoms like hydrogen atom’s movement stretches and compresses the bond as RC2—H →

← →

Stretched

RC2←H Compressed

Vibrational frequency for such a diatomic like molecule can be given

ν=

mH mRC 2 1 k and µ = m + m 2π µ H RC 2

...(6.1)

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Elements of Molecular Spectroscopy

Further assuming mRC 2 ≈ mH + mRC 2 so that

k 1 m 2π H If the force constant of the bond is independent of the nature of the molecular moiety, the group frequency to a very good approximation is independent of the rest of the molecule. Thus group stretching frequency such as OH, NH, NH2, CH2, CH3, C O etc. when compared with many molecules is fairly constant. This is the basis of the group frequency concept. At this stage a word of caution may be added that group frequency concept may not be applicable to small and lighter molecules like HCN, H2O, SO2, SiF2, C2H2, C2H4, C2H6 where the body of the molecule is not so heavy that it be stationary. As a matter of fact such molecules do not vibrate in violation and independent of one another. Rather each vibrational mode is imposed on other and molecule vibrates in a complex way. However, by applying a mathematical process called Normal Coordinate Analysis, this complex motion of molecules is resolved into individual vibrations and gives the right results. Further, some books, apply the group frequency concept over such molecules and explain the difference due to coupling of vibration. This approach is not acceptable because it amounts to saying that differences in results is due to coupling and where results agree there is no coupling which is wrong. ν=

Many commonly occurring functional groups such as —CH3 ,

C

O, —NH 2 etc. have

characteristic group frequencies when they are present in a molecule. For example, all compounds having a —CH3 group show absorption bands in the region of 2950 cm–1 (stretching) and 1400 cm–1 (bending). Similarly, all compounds having the C

O group have a strong band at about 1700 cm–1.

Some such group frequencies are listed in Table 6.1, which are also known as Group Frequency Charts. This division into charts will help to use this concept more effectively. Table 6.1: Group frequencies charts CHART 1: HYDROCARBON CHROMOPHORE

Group I. C—H Stretching (a) Alkanes and Alkyl groups (b) Alkenes (c) Alkynes (d) Aromatic compounds II. C—H Bending (a) Alkane, C—H Alkane, CH2 Alkane, (CH3)3 Isopropyl split—two peaks of equal intensity Alkane, tert-butyl, gem dimethyl (doublet)

Intensity

Range/cm–1

(m–s) (m) (s) (v)

2962–2853 3100–3000 3300 3030

(w) (m) (m) (s) (m) (s)

1340 1485–1445 1470–1430 and 1385–1370 1395–1385 and 1365

Group Frequency Concept

89

(b) Alkene, monosubstituted (vinyl) Alkene, disubstituted, cis Alkene, disubstituted, trans Alkene, disubstituted, gem Alkene, trisubstituted (c) Alkyne (d) Aromatic substitution type: Five adjacent hydrogen atoms Four adjacent hydrogen atoms Three adjacent hydrogen atoms Two adjacent hydrogen atoms One hydrogen atom III. C—C Multiple Bond Stretching (a) Alkene, nonconjugated Alkene, monosubstituted (vinyl) Alkene, disubstituted, cis Alkene, disubstituted, trans Alkene, disubstituted, gem Alkene trisubstituted Alkene, tetrasubstituted Conjugation (b) Alkyne, monosubstituted Alkyne disubstituted (c) Aromatic

(s) (s) (s) (s) (s) (m) (s) (s) (s) (s)

995–985 915–905 1420–1410 690 970–960 1310–1295 895–885 and 1420–1410 840–790 630

(v, s) (v, s) (v, s) (v, m) (v, m) (v, w)

750 and 700 750 780 830 very characteristic 880

(v) (m) (m) (m) (m) (m) (w) Intensity increase (m) (v, w) (v) (m)

1680–1620 1645 1656 1675 1653 1669 1669 Lower than ≈1610 cm–1 2140–2100 2260–2190 1580–1600 1450–1500

CHART 2: CARBONYL CHROMOPHORE

Group Carbonyl Stretching C I. Ketones (a) Saturated (b) α, β-Unsaturated (c) Aryl (d) α-diketones β-diketones Enolic OH Stret

Intensity

Range/cm–1

O stretching bands are 10–20 cm–1 higher in solution (s) (s) (s) (s) (s)

1725–1705 1685–1665 1700–1680 1750–1700 1640–1520 3450–2870

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Elements of Molecular Spectroscopy

II. Aldehydes (a) Saturated aliphatic and side chain aromatic (s) α, β-unsaturated (s) Aryl (s) (b) C—H stretching (m) (one or two bands) III. Esters and Lactones Saturated, acrylic (s) Saturated cyclic (a) δ-lactones (and larger rings) (s) (b) γ-lactones (s) (c) β-lactones (s) Vinyl ester type (s) (a) α, β-unsaturated and aryl (s) (b) α-ketoesters (s) (c) β-ketoesters (enolic) (s) For all types of esters stated above, one or two bands appear in the region. Many other bands appear in this region but difficult to assign with certainty IV. Carboxylic Acids (a) Saturated aliphatic acids (s) α, β-unsaturated aliphatic acids (s) Aryl acids (s) (b) Hydroxyl stretching (bonded), several bands (w) (c) Carboxylate anion stretching anti-symm (s)

1740–1720 1705–1680 1715–1695 2900–2820 and 2775–2700 1750–1735 1750–1735 1780–1760 1820 1800–1770 1730–1717 1755–1740 1650 1300–1050

1725–1700 1715–1690 1700–1680 3000–2500 1610–1550 and 1400–1300

CHART 3: HYDROXY COMPOUNDS AND ETHERS

Group I. Alcohols and Phenols (a) O—H stretching Free O—H, Primary alcohol Intermolecularly hydrogen bonded Single bridge compounds Polymeric association Intramolecularly hydrogen bonded Single bridge compounds Carbohydrates Extremely intense around 3300

Intensity

Range/cm–1

(v, sh) 3700–3600 (changes on dilution) (v, sh) 3550–3450 (s, b) 3400–3300 (no change on dilution) (v, sh) 3570–3450 (s) 3320–2750 (v) broadband 3200–3400

Group Frequency Concept

91

(b) O—H bending and C—O stretching (coupled) Frequency of these bands decreases on dilution Primary alcohols Secondary alcohols Tertiary alcohols Carbohydrates Phenols II. Ethers (C—O stretching) Dialkyl ethers Alkyl vinyl ethers or Alkyl phenyl ethers

(s) (s) (s) (s) (s) (s) (s) (s) (s)

1050 and 1350–1260 1100 and 1350–1260 1150 and 1410–1310 1030–980 1200 and 1410–1310

(s) (s) (s)

1150–1070 1270–1200 and 1075–1020

Abbreviations: s = strong, m = medium, w = weak, v = variable, b = broad, sh = sharp, anti-symm = antisymmetrical vibration and symm = symmetrical vibration.

6.2

REGIONS

Now we will talk of the characteristic regions in an infrared (IR) spectrum, which will help us in identifying the functional groups. We know that IR region lies between 600–4000 wave numbers, cm–1. The whole region is divided into four regions that will facilitate identification of functional groups in a polyatomic organic molecule. The four distinct regions are given below: Region 1 4000–2500

Region 2 cm–1

2500–2000

Region 3 cm–1

2000–1500

cm–1

Region 4 1500–600 cm–1 Skeletal vibrations

Region 1: 4000–2500 cm–1 The absorption occurring in this region is due to stretching of OH and CH bonds and bands invariably point out the presence of O—H, N—H or C—H bonds. The O—H stretching band occurs in the range 3700–3600 cm–1 when not involved in hydrogen bonding. It is relatively broadband in the 3200–3600 cm–1. The O—H group of alcohols and phenols also show absorption in this region. The O—H stretching frequency of carboxylic acids, shall be discussed under hydrogen bonding effect in Sec. 6.3. The C—H stretching from aliphatic compounds occur in the range of 3300–2850 cm–1. This vibration is characteristic of the hybridization of the carbon holding the hydrogen: 2800–3000 cm–1 for tetrahedral carbons, 3000–3100 cm –1 for trigonal carbons (alkenes and aromatic ring) and at 3300 cm–1 for diagonal carbon (alkynes) as moderately broadbands. A little care has to be exercised

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Elements of Molecular Spectroscopy

while identifying C—H stretching, because in aromatic compounds it occurs as shoulder around 3030 cm–1. The antisymmetric and symmetric stretching of methylenes ( CH2) and methyl (—CH3) groups can be seen between 2965 and 2880 cm–1. C—H stretching absorption usually appears as doublet at 2900–2820 and 2775–2700 cm–1. This doublet is a result of interaction between the C—H stretching fundamental and overtone of C—H deformation.

Region 2: 2500–2000 cm–1 The compounds containing triple bond absorbs between 2300 and 2100 cm–1 with weak intensity. The triple bond is a high strength bond and hence high force constant. Since the vibrational frequency is directly proportional to the square root of force constant, the triple bond absorbs at higher frequency as compared to double and single bonds.

Region 3: 2000–1500 cm–1 All compounds having double bonds show absorption in this region. The groups which exhibit significant absorption in this region are

C

C

and

C

O. The C

O stretching is recognized as an intense

band observed between 1800 and 1900 cm–1 (Chart 2–Table 6.1). This intense absorption helps in identifying carbonyl group in organic compounds. The C C stretch is much weaker in character and can be noticed around 1650 cm–1.

Region 4: 1500–600 cm–1 (Fingerprint region) This region is significant for identifying skeletal vibrations. Assignment of bands below 1500 cm–1, especially in the region 1450–900 cm–1, is a little tedious. Even structurally similar molecules exhibit different absorption patterns. This is the reason that this region is referred as fingerprint region. Many of the single bonds such as C—C, C—O also show absorption bands in these regions. C—O stretching shows a broad but strong band in the range 1000–1200 cm–1. Exact frequency depends on the nature of alcohol as: 1° R—OH 1050 cm–1 3° R—OH about 1150 cm–1 –1 2° R—OH about 1100 cm Ar—OH about 1230 cm–1 Phenols also show these bands but C—O stretching appears at somewhat higher frequency. Ethers show C—O stretching frequency but the OH band is absent. Carboxylic acids and esters show C—O stretching but give characteristic absorption of carbonyl group.

C—H deformation (i) The C—H deformation band is observed between 1400–1000 cm–1 as an intense band. (ii) Both alkenes and aromatic rings show Carbon-Hydrogen bending vibration in which hydrogen atoms moves in plane and out of plane also. For alkenes out of plane bending gives strong bands in the region 800–1000 cm–1, exact location depends upon the nature and number of substituents. For aromatic rings C—H bending give strong absorption in the region 625–870 cm–1. (iii) In substituted benzene, the spectral pattern in this region give information whether it is monosubstituted or 1, 2; 1, 3 or 1, 4 disubstituted. This should be amply clear from the spectrum shown in Fig. 6.1.

Group Frequency Concept

(a)

(b)

(c)

Fig. 6.1: Infrared spectrum of (a) Toluene, (b) p-xylene, (c) m-xylene. In addition to the above bands disubstituted benzene show some characteristic bands, ortho (735–770 cm–1), meta (680–710 and 750–810 cm–1) and para (790–810 cm–1)

93

94

Elements of Molecular Spectroscopy

This is due to C—H bonds lying adjacent to these substituted positions appear distinctively in 850–690 cm–1 region. The position of absorption band position varies with substitution pattern as follows: o-disubstituted 735–770

m-disubstituted 680–710, 750–810

p-disubstituted 790–810

monosubstituted aromatic compounds show C—H deformation in the range 730–770 and 690–710 cm–1. (iv) Compounds containing gem-dimethyl group —CH(CH3)2 (isopropyl group) show a doublet in the range 1370–1395 cm–1. Tert. Butyl group shows unsymmetrical doublet at 1370 (s) and 1390 cm–1 (m). (v) The compounds containing a chain of at least four methylene groups show a band at 720 cm–1. Fig. 6.1 shows IR spectrum of some aromatic hydrocarbons. Now, what do we look for in IR spectrum of hydrocarbon. To begin with we can readily tell whether the compound is aromatic or purely aliphatic. Aliphatic absorption is strongest at higher frequency and is essentially missing below 900 cm–1; aromatic absorption is strong at lower frequencies (C—H out of plane) between 650–900 cm–1. In addition, an aromatic ring will show C—H stretching at 3000–3100 cm–1 often these C—C stretching absorbs at 1500–1600 cm–1 while C—H in plane in the 1000–1100 cm–1 region.

PROBLEM 6.1 The infrared spectrum of 2–methyl propanol has the following absorption bands: 3400 cm–1 (broad) 2960 cm–1 (medium) –1 1450 cm (medium) 1380 & 1370 (sharp and doublet) –1 1050 cm (sharp) Indicate the group responsible for these bands.

SOLUTION 3400 cm–1

O—H stretching

2960 cm–1

C—H stretching of CH3 and

1380 & 1370 cm–1

C—H bending of

1050

cm–1

CH2 group

CH2 group

O—H bending and C—O stretching.

Note: All alkanes show C—H (str.) between 3000–3100 cm –1 and out of plane (strong) bending between 800–1000 cm–1

6.3 EFFECT OF HYDROGEN BONDING Hydrogen bonding can occur in systems containing a proton donor group (X—H) and a proton acceptor group (Y), provided, the s orbital of the donor can overlap effectively with the p orbital of the acceptor group. Atoms X and Y are electronegative atoms but Y atom must possesses lone pair of electrons. The common proton donor groups in organic molecules are carboxyl, hydroxyl, amino or amido

Group Frequency Concept

95

groups. Common proton acceptor atoms are oxygen, nitrogen and the halogens. Unsaturated groups with ethylenic linkage can also function as proton acceptors. Hydrogen bonding alters the force constants of both the groups and, stretching and bending frequencies are altered. Because of hydrogen bonding, the X—H stretching bands move to lower frequencies with increased intensity and band widening. The stretching frequency of the acceptor group, for example, of the C—O group is also reduced but to a lesser degree than the proton donor group. The X—H bending vibration usually shifts to higher frequency, when hydrogen bonding occurs. This shift is less pronounced compared to the stretching frequency shift.

There can be intermolecular or intramolecular hydrogen bonding. Intermolecular hydrogen bonding (or association) may result in dimer molecules or in polymers. For example, carboxylic acids form dimers (I) in organic solvents like benzene. The proportion of dimer is more in benzene but there is no dimer formation in dioxane. Intramolecular hydrogen bonding is possible in systems where 5 or 6 membered ring formation is possible as in II, III with a non-associating solvent (CCl4, CS2, CHCl3).

Intermolecular hydrogen bonding (or association) is affected by concentration change with a non-associating solvent and the temperature change, whereas these factors do not influence intramolecular hydrogen bonding. As a result of dilution or temperature change IR spectral pattern of such associated molecules is altered. For instance, dilution reduces the possibility of dimer or polymer formation. So at low concentrations, intermolecular hydrogen bonding possibility becomes less and the intensity of the corresponding absorption band also becomes less. This has been illustrated in Fig. 6.2 for ethanol. The IR spectra of both 10% and 1% ethanol in CCl 4 show the effect of concentration of intermolecular H-bonding. In Fig. 6.2(a) the 10% ethanol solution has one sharp absorption at 3640 cm–1 (A) and a strong broad absorption at 3340 cm–1 (B). Two such peaks are also noticed in Fig. 6.2(b), for 1% ethanol. However, in 1% ethanol, the sharp band at 3640 cm–1 (A) has increased intensity and, to balance this, the intensity of the broadband at 3340 cm–1 is less. From these considerations, the sharp band is due to monomeric form of ethanol (O—H) bond unassociated and broadband due to the polymeric form (O—H bond associated). The amount of monomer or unassociated alcohol has increased on dilution resulting in the gain of intensity of sharp band (at 3640 cm–1) relative to the broadband (at 3300 cm–1). More polar solvents such as acetone or benzene will influence OH absorption. The π-ring behaves as Lewis base and form hydrogen bond to acidic hydrogen. As a result OH absorption frequency is lowered by 40–100 cm–1 when the spectrum is recorded in benzene.

Elements of Molecular Spectroscopy 100

100

80

80

Transmittance (%)

Transmittance (%)

96

60 A

40

20

60

40

20 A

B

B 0 4000

3500

3000 –1

Wave number/cm (a)

0 4000

3500

3000

Wave number/cm

2500 –1

(b)

Fig. 6.2: Infrared spectrum of ethanol (a) 10% V/V in CCl4, (b) 1% V/V in CCl4

The effect of temperature on intermolecular H-bonding is similar to the one noticed in the case of concentration. The association of molecules is prevented at higher temperatures and hence an increase in the intensity of sharp band at 3640 cm–1 is noticed. It is desirable to record the spectra of some simple molecules under different physical states or concentrations. In very dilute solution or in the vapor phase, molecular association effects are minimized, whereas in solid state or in concentrated solution, molecular association effects are considerable. For example, in the IR spectrum of pure liquid acetic acid show C O stretching at 1718 cm–1; while in the vapor phase, two carbonyl bands are observed at 1733 cm–1 and 1786 cm–1. The C O absorption in the liquid state is attributed to the dimer only, whereas the two bands at 1733 cm–1 and 1786 cm–1 in the vapor state are due to the dimer and monomer, respectively. The absorption caused by free O—H stretching of a carboxylic acid is observed near 3550 cm–1 whereas the bonded O—H in the dimeric form absorbs in the region 2700– 2500 cm–1. Also when a carbonyl group is involved in intramolecular hydrogen bonding, it has lower stretching frequency. For instance, acetophenone has carbonyl absorption at 1700 cm–1 whereas 2-hydroxyacetophenone (III), which has intramolecular hydrogen bonding, has carbonyl absorption band at 1635 cm–1, showing a decrease of 65 cm–1.

Group Frequency Concept

97

The existence of keto and enol forms can be understood in terms of the effect of hydrogen bonding on IR spectral frequencies. For instance, ethyl acetoacetate which has both keto and enol forms shows absorption characteristic of both the forms 1724 cm–1 keto carbonyl 1748 cm–1 ester carbonyl

}

keto form IV

1650 cm–1 ester carbonyl

enol form V

H O C CH3

O

O C CH2

O

C2H5

C CH3

O C

C O

C2H5

H IV

V

Hydrogen bonding in enol and chelates is particularly strong and the observed O—H str. Frequencies may be very low ≈ 2800 cm–1. Since these bonds are not easily broken, dilution by an inert solvent, may not show O—H str., at low concentration.

P ROBLEM 6.2 The infrared spectrum of butanol (liquid film) has single broadband between 3500–3200 cm–1. When it is diluted with CCl4, it shows additional band near 3650 cm–1. Explain the reason.

SOLUTION The liquid film means pure liquid was taken for spectral study without adding any solvent. The single broadband of the pure alcohol is due to associated —O—H group. On dilution, the free O—H group stretching also appears and thus two bands are observed.

6.4

APPLICATION OF IR SPECTRA IN STRUCTURE DETERMINATION

The infrared spectra find a wide range of applications. For instance, structure determination, quantitative analysis, hydrogen bonding studies and conformational studies are some of the fields where infrared spectra play a vital role. In this section, the role of IR spectra in arriving at the structure of simple organic compounds will be discussed. The matching of the IR spectrum of an unknown compound with those of known compounds provides the best means of structure identification of organic compounds. The finger print region could be particularly useful for this purpose. Infrared correlation charts such as Table 6.1 could be used to identify the hydrocarbon skeleton and the functional groups present in a molecule. To be sure of the structure of a compound, data from IR spectra should be supported by other evidences like its chemical, physical or other spectroscopic data (NMR, UV etc.) Certain observations in the interpretation of a spectrum are: 1. Always begin at the high frequency end of the spectrum and use the finger print region for confirmation. 2. Total interpretation of the spectrum is seldom required 10–20% bands can usually be assigned with certainty.

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Elements of Molecular Spectroscopy

3. Place more reliance on negative evidence, which is the absence of absorption in a particular region than on positive evidence. 4. The limits of published range of group frequencies are often defined by extreme cases of electron influences. A compound with no such influence is expected to absorb towards the centre of the range quoted.

I. Hydrocarbon Skeleton Look for IR data indicating the presence of alkane, alkene, alkyne and aromatic residues. For this compare the IR spectral data of the compound with the entries under Hydrocarbon chromophore in Chart 1 of Table 6.1 and find the answer for the following questions: (i) What is the type of hydrocarbon skeleton? (ii) Are there any special features in the IR spectrum, which throw light on substitution pattern? You can arrive at the hydrocarbon skeleton using C—H stretching, C—H bending and C—C multiple bond stretching absorption. Look for specific indications regarding the type of aromatic substitution, cis-trans isomers, presence of gem-dimethyl groups etc.

II. Functional Groups The nature of functional groups can be decided from the answers to the following questions: (i) Does the compound have carbonyl group(s)? Look for sharp IR band(s) in the region 1800–1600 cm–1. If you find one such, answer the question (a) to (d) given below: (a) Does the spectrum have C—H stretching characteristic of an aldehyde? (b) Does the spectrum have C—O stretching characteristic of an ester (or lactone)? (c) Does the spectrum have O—H stretching characteristics of —COOH group? (d) Could it be a ketone only? Remember that the compound could have two carbonyl groups even. Note down the inferences and find the answers for the following questions also: (ii) Does it show O—H stretching, O—H bending and C—O stretching bands characteristics of alcohols or phenols? (iii) Does it show only C—O absorption without O—H stretching band (which is characteristic of ethers)? (iv) Does the spectrum show any special features which bring out the presence of tautomers, intramolecular hydrogen bonding, intermolecular hydrogen bonding etc.? The answers to the question (iv) provide us information regarding hydrocarbon skeleton and the functional groups present in the compound. Let us work out three problems using the above approach.

PROBLEM 6.3 A compound of molecular formula C7H8 has IR bands at the following frequencies in wave numbers, cm–1– 3050, 2900, 1600, 1500, 1430, 1380, 850, 700. Suggest a possible structure for the compound.

SOLUTION We have to look for the answers for only the questions relating to the hydrocarbon skeleton, since the molecular formula suggests that it is a hydrocarbon and it cannot have any specific functional group.

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99

Hydrocarbon skeleton (Chart 1)

The presence of aromatic ring is indicated by the presence of bands at 3050 cm–1 (aromatic C—H stretching) and at 1600 cm–1 and 1500 cm–1 (C—C multiple bond stretching of the aromatic ring). The presence of —CH3 and CH2 groups is hinted by the absorptions at 2900 cm–1 (aliphatic C—H stretching) and at 1430 cm–1 and 1380 cm–1 (aliphatic C—H bending). Substitution type: The absorption at 750 cm–1 and 700 cm–1 show the presence of monosubstituted benzene ring (C6H5). Since the molecular formula is C7H8, we can infer that it has a —C6H5 group and a —CH3 group (C7H8, C6H5 — CH3). Hence, the possible structure of the compound is

CH 3 This needs confirmation through NMR data.

PROBLEM 6.4 A compound has molecular formula C4H8O. The NMR spectrum shows the presence of two methyl groups and one methylene group. Further NMR spectrum excludes the possibility of aldehyde group. The IR spectrum shows three prominent bands: 2941–2850 cm –1 (medium), 1716 cm–1 (sharp) and 1459 cm–1 (medium). Deduce the structure of the compound.

SOLUTION Hydrocarbon skeleton (Chart 1): The hydrocarbon skeleton is of alkane type since C—H stretching (2941–2850 cm–1) and bending vibrations (1459 cm–1) indicate the presence of one or more —CH3 and

CH2 groups.

Functional group (Chart 2 and 3): The absorption of 1716 cm–1 shows the presence of

C

O

groups. The molecular formula indicates the presence of only one oxygen atom and this, in conjunction with IR band at 1716 cm–1 indicates that the compound should be an aldehyde or a ketone. Since NMR spectrum excludes the presence of aldehyde group, the structure of the compound is

This structure is in line with NMR spectral data which shows the presence of two —CH3 groups and one

CH2 group.

PROBLEM 6.5 A compound of molecular formula C9H10O2 yields the following IR data: 3022 cm–1 (weak) 2940 cm–1 (weak)

2940 cm–1 (weak) 1605 & 1504 cm–1 (weak)

1730 cm–1 (sharp) 1060 (sharp) 835 (sharp)

NMR data provide information regarding the presence of two —CH3 groups and one group. Suggest possible structure.

C6H4

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Elements of Molecular Spectroscopy

SOLUTION The structure can be arrived at as follows: Hydrocarbon skeleton (Chart 1): The absorption at 3022 cm–1, 1605 cm–1 and 1503 cm–1 bring out the presence of aromatic ring. Further the absorption at 2940 cm–1 is suggestive of the presence of one or more —CH3 and/or CH2 groups. Substitution type: The band at 835 cm–1 is indicative of p-disubstitution. Functional group (Chart 2 and 3): The absence at 1730 cm–1 and 1060 cm–1 indicate the presence of ester group. On the basis of available IR and NMR data, we can conclude that the compounds has (i) p-C6H4 group, (ii) two methyl and (iii) one ester group. Possibly one of the methyl group is attached directly to the benzene ring and another methyl group forms part of the ester group, i.e., either as —COOCH3 or CH3COO—. As it is, the two possible structures are:

More evidences are required to take a clear decision regarding structure.

PROBLEM 6.6 A compound has the molecular formulae C4H10O. It has no prominent absorption bands above 3000 cm–1, in the region 2900–1500 cm–1 or below 100 cm–1. The main absorption bands with no doublet are given below: 2970 cm–1 1450 cm–1 1370 cm–1 1100 cm–1 Suggest the possible structure for the compound. SOLUTION The C—H stretching absorption at 2970 and C—H bending absorption at 1450 cm–1 and 1370 cm–1 indicate the presence of —CH3; and CH2 groups. Absence of doublets around 1370 cm–1 rule out the possibility of isopropyl or t-butyl branching. Functional group: The absorption at 1100 cm–1 indicate ether linkage. The two possible structures are given below: CH3—CH2—O—CH2—CH3

CH3—O—CH2—CH2—CH3

Exact structure can be confirmed by recording the high-resolution NMR spectra of the compound.

PROBLEM 6.7 Identify a given hydrocarbon C7H16 containing seven carbon atoms from its spectrum reproduced in Fig. 6.3. SOLUTION Since the compound is a hydrocarbon, it is useful to begin with C—H stretching frequency near 3000 cm–1. Since there is no band above this frequency, three types of hydrocarbons may be ruled out viz., aromatic compounds (C—H stretching near 3030 cm–1), alkynes (a strong band at 3300 cm–1) and alkenes (C—H stretching lies between 3040 and 3010 cm–1). The main bands at 2900 cm–1 clearly arise from C—H bond in saturated hydrocarbon with a small peak at 2872 cm–1 on the side of main band is due to symmetrical C—H stretching in a methyl group. As this band is not particularly strong, it suggests that only a small number of CH3 groups are present.

Group Frequency Concept

101

Fig. 6.3: Spectrum I

Presence of a weak absorption from 1500–2500 cm–1 confirms the absence of acetylene or olefin C—H bond. A strong band at 1450 cm–1 arises from the methylene vibration which is very strong in all aliphatic hydrocarbons. At 1460 cm–1 the anti-symmetrical deformation of methyl group is also found due to overlapping of —CH2 vibration. So no conclusion about —CH3 group could be drawn. Absence of sharp doublet peaks at 1175–1165 and 1255–1245 cm–1 rules out the presence of isopropyl and t-butyl group in the alkane. All this lead us to conclude that hydrocarbon C7H16 has a linear structure and may be heptane. However, exact structure of heptane cannot be predicted from IR spectrum only.

% Transmittance

PROBLEM 6.8 Identify the organic compound from its IR spectrum (Fig. 6.4).

B

D

E F

A C 4000

3000

2000

1500

1000

Wave number (cm –1)

Fig. 6.4: Spectrum II

SOLUTION The infrared spectrum (Fig. 6.4) shows a strong band at 3350 cm–1 (A) with shoulder (B) at 3600 cm–1, these strongly suggest the presence of an —OH group. This is confirmed by a strong band at 1050 cm–1 (C) (see chart 3). This band is a result of OH bending coupled to C—O stretching. The position of this band suggests that the compound is a primary alcohol. The presence of bands between 2000–3100 cm–1 (D) and between 1400–1450 cm–1 (E) and absence of absorption ascribed to an aromatic nucleus between 3000–3100 cm–1 and 1500–1600 cm–1 indicate that the compound is an aliphatic alcohol.

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PROBLEM 6.9 How would you differentiate between the following pairs of compounds using infrared spectra as a tool?

COOH (i)

COOH and

(ii)

OH

CH3CH 2CHO and CH2 =CHCH2OH

OH

SOLUTION (i) The absorption peaks for OH and COOH in ortho-hydroxy benzoic acid will be 3500– 3450 cm–1 (OH intramolecular hydrogen bonded) and 3300–2500 cm–1 for COOH group, 755 for 1 : 2 disubstituted and C—H stretching will be obtained. Further hydrogen bonding will be altered as a result of dilution or temperature change and thus alter the IR-spectrum. In 3-hydroxy benzoic acid the absorption peak will be observed at 780, 700, 800 cm–1 for C—H stretching ; 3610 cm–1 for phenolic OH and 3300–2500 cm–1 for COOH. Dilution will not alter IR-spectrum showing there is no hydrogen bonding. The spectra of the given compound be recorded and compare with this data to identify the compound. (ii) CH 3CH 2CHO will show absorption bands at 1740–1720 cm –1 for C O stretching; 400–1000 cm–1 for C—CHO skeletal. CH2 CH—CH2OH will have an absorption band at 1050 cm–1 for primary alcohol (Chart 3) and 905 for vinyl group, CH2—C—H (C—H stretching 3100–3000 cm–1 and for C C double bond. Record the IR spectra of the given compound and compare with this data to identify the structure of the compound.

PROBLEM 6.10 An organic compound with molecular formula C8H6 decolorizes bromine in carbon

Absorbance

tetrachloride and gives a white precipitate with ammoniacal silver nitrate solution. Its IR-spectrum is shown in Fig. 6.5. Identify the compound.

4000 3500 3000 2500 2000 1800 1600 1400 1200 1000 800 600 400 200

Wave number (cm–1)

Fig. 6.5: Spectrum III

SOLUTION (i) Reaction with Br2 shows that it is a unsaturated compound. (ii) Reaction with ammoniacal silver nitrate shows it to be alkyne.

Group Frequency Concept

103

(iii) Peak at ~3300 cm–1 shows the presence of alkyne group. (iv) Peak at 3030 suggests the presence of aryl group. This is further confirmed by peak (~1500 cm–1) due to C—C stretch. (v) Peak at 770 cm–1 suggests it to be monosubstituted benzene. Structure of the compound C8H6 deduced from ‘i to v’ is phenylacetyl C6H5—C CH.

6.5

SAMPLE PREPARATION FOR SPECTRA RECORDING

Wide ranges of techniques are available for mounting the sample in the beam of spectrometer. These sampling techniques depend on whether the sample is a gas, a liquid or a solid. Intermolecular forces vary considerably in passing from solid to liquid to gas, and the spectrum will normally display the effect of these differences in the form of frequency shifts or additional bands, etc. It is, therefore, most important to record on a spectrum the sampling technique used.

Infrared Spectrum Gases

The gas sample is introduced into a gas cell, typically as shown in Fig. 6.6. It has infrared-transparent windows (for example, NaCl) and allow the cell to be mounted directly in the sample beam. In a modified form, the use of internal mirrors permits the beam to be reflected several times through the sample (multi-pass gas cell) to increase sensitivity. For a regular class, in place of NaCl windows, a cellophane paper can be used for window, which also gives reasonable good spectrum.

Fig. 6.6: Gas cell used in infrared sampling technique

Solutions in organic solvents

Examining a substance in solution (usually 1–5 per cent) is a very important method in infrared work and, in principle; it is almost of universal application since most compounds are soluble in some solvent. Drawbacks arise since all solvents have some absorption bands in the infrared region and frequently non-absorbing solvents will not dissolve the material to be examined. CCl4 and CS2 between them cover the whole range 5000 to 666 cm–1 but may turnout to be poor solvents for certain compounds.

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Chloroform and bromoform are useful since they are transparent from 2500 to 1250 cm–1 where it is otherwise difficult to find a satisfactory solvent. Solvents can be rapidly dried by bubbling dry air (air through P2O5 or from liquid air) through the liquid. Since many organic compounds absorb intensely in the infrared, a very small path length is needed when dealing with pure substances. This reduces the accuracy of quantitative measurements since not only must the path length be carefully measured but also any variation over the cell area must be avoided. Solvent may shift band frequency and change intensity as well on changing solvents of different polarity. This is particularly true of C O bands and hydrogen bonded systems. Aqueous solutions

The intense absorption by water throughout much of the infrared region seriously restricts the use of aqueous solutions for measurements. In addition, many of the usual window materials are ruled out because of their water solubility. Calcium fluoride is usable to 1111 cm–1 while the rather fragile barium fluoride extends the range to 833 cm–1. However, silicon and germanium windows, bloomed to reduce the reflection loss are now becoming available and the window situation has become much easier. To obtain spectra from aqueous solution, use of double-beam spectrometer is essential, so that water absorption can be carefully cancelled out. From the visible to 4000 cm–1 relatively long path length glass cell can be used but elsewhere it is impracticable. Usable regions are 2860 to 1720 cm–1 and 1540 to 950 cm–1 for water and 2130 to 1250 cm–1 and below 1140 cm–1 for the rather more transparent heavy water. At 1111 cm–1 0.075 mm of water transmits only 2 per cent of the incident radiation and it is, therefore, necessary to employ a spectrometer with negligible stray light. A lowresolution grating instrument is best for this type of work since opening the slits wide can offset the low transmittance. The solute concentration must exceed one per cent in water: pH variation may assist. Alkali halide disc

In most of the cases KBr disc is prepared which must be dried in an oven at 110°C. In this method about 1 mg of finely ground sample is mixed thoroughly into a 200 mg of high purity KBr. The mixture is further grounded and a transparent disc is made by placing a suitable quantity in a 15 mm diameter die (Fig. 6.7) provided with a highly polished flat anvil, B, inserting a similar plate C, and finally applying a force of about ten tons on the plate. Either a hand operated screw or a hydraulic press may be used, and after about a minute the die can be taken apart and the KBr disc is removed. Points to be noted are the evacuation of the die through A to remove water vapor, which if trapped in the disc causes it to break up. The technique is extremely useful, but sometimes marked peculiarities with abnormal spectra are encountered when KBr is given a heavy grinding. Liquid mulls—internal standard

A spectrum of a solid material can be obtained by adding a drop of medicinal paraffin, also called nujol to a finely ground sample, and placing some of the slurry between transparent plates or without a thin spacing washer. Paraffin has absorption bands in the range 2800–3050 and 1280–1500 cm–1 due to (—CH2) and (—CH3) groups but much of the range 5000 to 666 cm–1 is relatively free from any other absorption. The paraffin greatly reduces the scatter of radiation, which the powder would give. The closer the refractive indices of paraffin and solid the smaller be the scatter. As bands of interest matter may be obliterated by those of paraffin, a fully fluorinated hydrocarbon, ‘Fluorolube’, or other hydrogen free liquid can replace nujol. To avoid anomalous scattering effects the sample material should be ground to fineness such that the wavelength should exceed the average particle size.

Group Frequency Concept

105

O' mng

C

Sample

B A

Fig. 6.7: Die for alkali halide discs

Preparation from melt

Thin layers of low melting point solids may be prepared by melting a small quantity between transparent windows on a hot plate. Synthetic polymers will, in many cases, flow into thin layers if pressed at a sufficiently high temperature. Casting from solution

A solution of the sample may be poured over a horizontal plate of glass or infrared transparent material in an oven. After solvent evaporation a thin film remains and, if it is a polymeric substance, may usually be stripped of the glass plate. Alternatively, the film may be examined on the NaCl supporting plate itself. Polymers and various waxy materials give excellent spectra this way.

PROBLEMS 1. Rationalize the observed C O stretching frequency in the following compounds: (ii) Acetone 1725 cm–1 (i) Potassium acetate 1575 cm–1 (iii) Acetic acid 1721, 1776 cm–1 (iv) Acetic anhydride 1825,1754 cm–1 2. In the infrared spectrum of penta-3-en-2-one, CH3CHCH (C O)CH3, two bands were found for carbonyl, absorption at 1670 and 1700 cm–1. Explain with reason. 3. Account for the fact that acetyl acetone exhibits a band 1613 and 1725 cm–1 due to carbonyl stretching mode and also a broad weak band at 3000–2700 cm–1 for OH stretching. 4. Discuss briefly the underlying principle for identifying the structure of an organic compound by IRspectroscopy.

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5. How will infrared spectral data be useful to distinguish the compounds in the following pairs?

(i)

HO

O

OH and

C O

OH (ii)

C

CH 3

and HO

O

C

CH3

O

6. Account for the fact that infrared carbonyl frequency in salicylic aldehyde is lower than that of m-hydroxy benzaldehyde. 7. Give the type of IR absorption bands you will be looking for to distinguish between the following pairs of compounds: (i) CH3—C C—CH3 and CH3—CH2—C CH (ii) CH3—CH2OH (iii) CH3—CH2CH2CH3 (iv) CH3CH2CHO (v) C6H5—C C—CH ( vi) CH3COOH

and and

CH3COOH CH3CH2CH

and

CH3COCH3

and and

C6H5—CH CH2 CH3COOC2H5

CH2

(vii) CH3CH2CH2OH and (CH3)2CHOH 8. An organic compound having molecular formula C8H8O has a strong IR absorption band near 1690 cm–1. Assign a structure to it. Different isomers of the formulae C8H8O are: (i) C6H5CH2CHO (ii) C6H5COCH3 (iii) HO—C6H4—CH—CH2 (iv) C6H5—CH—CH2 O (v) C6H5O—CH

CH2

9. Two isomers A and B of molecular formula C6H6O give IR absorption band near 1650 cm–1 and 1710 cm–1 respectively. Assign structural formulae consistent with their IR absorption bands to A and B. 10. IR spectrum of acetone gives two maxima due to C—H vibrations at 1360 cm–1 and 3000 cm–1. Identify the stretching and bending bands.


❖

CHAPTER

7

Electronic Spectra Electronic spectra involve transition of electron from one energy level to another. Electrons are also involved in bond formation in a molecule and thus distributed in these energy levels. Apparently electronic spectra should be very intimately related to the electronic configuration and thus MO energy levels. The two have to be discussed in conjunction with one another. As an approximation consider each bond in isolation, so the complete electronic spectrum of a molecule will be sum of spectra of each bond. Further, properties of each bond like bond length, bond order, dissociation energy, dipole moment and many others, will be affected by every electronic transition. This makes whole thing very complex but a great deal of information about the molecule is contained within it. As the electronic configuration of every molecule is different, so energy levels and electronic spectra have to be different thereby giving the message that other associated properties will be different. The message is– Each molecule requires a separate discussion.

7.1

BORN-OPPENHEIMER APPROXIMATION

Electronic spectra involve change of at least three quantum numbers simultaneously viz., electronic, vibrational and rotational quantum numbers. Applying the Born-Oppenheimer approximation that energy difference between rotational (ER), vibrational (EV) and electronic (Ee) is so large that for all practical purposes it can be assumed that one type of transition does not interact with other. This is called BornOppenheimer approximation and we will be strictly following this. This also amounts to that electronic energy level comprises a number of vibrational level and each vibrational level consists of several rotational levels, though in the present context we will not be so concerned with them.

7.2

ELECTRONIC TRANSITIONS IN HETERONUCLEAR DIATOMIC MOLECULES

An electronic spectrum is transition of electron from one orbital to another. Such transitions may be allowed or forbidden depending on the selection rules. To correlate the pure electronic spectra with the MOs is quite tedious. So a broad pictorial standpoint to understand it will be given. Further, the simultaneous change of the electronic, vibrational and rotational energies complicates this molecular spectrum. There is no general selection rule, which may be restricted to vibrational-electronic transitions. Electronic spectra will be introduced with the help of hydrogen atom. Around the nucleus of hydrogen atoms a series of orbitals are assumed with only two forms of spins of an electron. So one

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Elements of Molecular Spectroscopy

electron can have one of the two spins in any orbitals. The electron occupying different orbital represents the different energy levels of the atoms. These orbitals are the probability distribution of electrons about a point in space, the s-orbital is spherical in shape without any nodal plane, the p-orbital is dumbbell type with one nodal plane, and the d-orbital has two nodal planes. In the atomic electronic transitions, electron goes from one orbital to another. This will be associated with an increase or decrease in number of nodal planes (∆n). The selection rule states that the change in number of nodal planes (∆n) should be equal to ±1. This is the selection rule for transition taking place in hydrogen atom. Absorption of light by hydrogen atoms leads to the transition of electrons from 1s-orbital to 2p-orbital thereby changing the number of nodal planes by one. However, 1s →2s transitions are forbidden because no change in ‘transition moment’ is involved. Similarly, 1s →3d transition is not allowed because it involves a change by two nodal planes i.e., ∆n = 2. In general, transition allowed by selection rule are of high intensity while forbidden transition are of low intensity. The energy levels of the atom are accurately calculable from the Schrödinger equation.

Fig. 7.1: (a) MO energy diagram for diatomic AB, (b) Contour diagram of the MO so formed

Electronic Spectra

109

In general, for any diatomic molecule AB, molecular orbitals energy levels are shown in Fig. 7.1 where B is more electronegative than A the energy difference will be different for different molecules but order will be same. Further, the more electronegative atom has lower energy and is always shown on the right hand side. nA and nB are nonbonding electrons coming from A and B atoms. σ and π are bonding molecular orbitals while σ* and π* are anti-bonding molecular orbitals, pie orbitals are doubly degenerate MOs. On the right side of these energy levels are shown the contour diagrams for the corresponding molecular orbital. It may not be possible in this type of book to give details as to how these are obtained. For any molecule the total number of electrons involved in the formation of bond are calculated. These electrons are filled in these MOs following Aufbau principle and Pauli exclusion principle. Electronic transition takes place between highest filled orbital to next vacant orbital. This will be explained with the help of some examples. CO molecule: As oxygen is more electronegative than carbon, A atom will be oxygen while B atom will be carbon. Electronic configuration of O and C are as follows: Oxygen, atom A Carbon, atom B 1s2, 2s2, 2p2 1s2, 2s2, 2p4 As 1s2 will have very low energy in both the atoms and thus will not participate in the bond formation. This leaves number of electrons participating in bond formation = 6 + 4 = 10. Filling these ten electrons in these MOs, of Fig. 7.1 gives an electronic configuration as:

nO2 , σ 2 , π 4 , nC2 So, the electronic transition will take place within nC → π*. Since this involves a change in nodal planes by one hence the transition is allowed. This corresponds to transition energy of 48686.7 cm–1. The highest filled MO lies in nC2 , which is centered more on carbon and leads to formation of metal carbonyls. The dative bond is formed through carbon to metals. Inhalation of CO leads to C—Fe bond in hemoglobin as in carbonyl. This bond is so strong that oxygen cannot replace CO in hemoglobin and it becomes ineffective as oxygen carrier. Excess inhalation of CO means larger numbers of hemoglobin molecules are lost which means congestion and that may lead to death. At this point another term may be defined, called bond order. 1 Bond order = (No. of bonding electrons – No. of anti-bonding electrons) 2 Larger the bond order more stable will be the molecule, smaller the internuclear distance. It may be noted that nonbonding electrons make no contribution to bond order. Thus bond order in CO will be (6 – 0)/2 = 3. In the CO+ and CO– molecules total number of electrons are 9 and 11, thereby giving bond order 3 and 2.5. The properties are as follows: CO CO– CO+ Transition nC → π* nC → π* nC → σ* ∆n +1 +1 +2 Transition allowed Allowed Allowed Forbidden transition Transition energy 20733.3 cm–1 48686.7 cm–1 – Change in Bond order 3 → 2.5 3 → 2.5 2.5 → 2.5 Bond length Stability

1.115 A 1.128 A – CO– less stable than CO which is as stable as CO+

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Elements of Molecular Spectroscopy

Some energy is used up in unpairing the electrons; consequently the transition energy in CO is higher than CO+ though transitions take place between the same states. In the molecule CO– both the transition states have two nodal hence it is a forbidden transition and is not allowed hence no value was given. Further, it may be noted here that forbidden transition show no change in bond order. NO molecule: In this molecule also oxygen will be atom A and nitrogen as atom B. The total number of electrons participating in the bond formation is 11. So that electronic configuration of NO will be

nO2 , σ 2 , π 4 , n N2 π*1 Since it has an unpaired electron, NO is paramagnetic in nature. The electronic transition corresponds to π*→σ* and is observed at 43965 cm–1. In this transition there is no change in nodal planes and thus should not be allowed. This transition is observed because of spin orbit coupling. Consider two other molecules, results of those are compiled as follows: NO NO– NO+ Transition nN → π* π* → σ* π* → σ* Diamagnetic Paramagnetic Paramagnetic ∆n +1 Zero Zero Allowed Forbidden Forbidden Transition energy 52190 cm–1 44248 cm–1 6050 cm–1 Change in Bond order (6 – 0)/2 = 3 → 2.5 (6 – 1)/2 = 2.5 → 2.5 (6 – 2)/2 = 2 → 2 Bond length 1.063 A 1.151 A 1.5957 A These results show that as bond order decreases bond length increases and transition energy also decreases. As NO+ has a bond order 3 is most stable followed by NO while NO– is least stable. As CO and NO+ has same electronic structure, both form dative bonds with metals while NO which has unpaired electron in antibonding pie orbital does not form compounds like carbonyls. As NO molecule has an unpaired electron, it pairs up by forming a N—N bond and thus giving a diamagnetic dimer, N2O2. Due to similar reason, when NO comes in contact with oxygen it gets converted to NO2 molecule. CN molecule: In this molecule nitrogen will be atom A and carbon be atom B. The total number of electrons participating in bond formation is nine so that electronic configuration will be nN2 , σ 2 , π 4 , n1C Electronic transition will take place within nC →σ*. In this transition there is change in number of nodal planes by one, hence transition is allowed. This corresponds to transition energy of 25797 cm–1. The highest filled MO lies on carbon atom with n1C but is unpaired and thus very reactive. This leads to formation of dimer structure (CN)2 in which formation of sigma like bond takes place between the two carbon atoms. The bond order in CN is equal to 3. The other properties are as follows: CN+ CN CN– Transition π→nC nC → π* nC → π* Not allowed Allowed Allowed 25797 cm–1 – Transition energy 31381 cm–1 Change in Bond order 3 → 2.5 3 → 2.5 3 → 2.5 Bond length 1.1729 A 1.1718 A 1.15 A Stability All equally stable Since some energy is used up in unpairing the electrons, consequently energy in CN+ is higher than CN.

Electronic Spectra

111

On passing HCN gas through water, it dissociates to give CN– ion and is an irreversible process. CN– has electronic configuration similar to CO, show similar behaviour. The paired electron is centered on carbon which form strong dative bonds with metal ions to give cyanide complexes. Since CO and CN– have similar electronic structures, similar d-d splitting in metal atom/ion is observed. HF molecule: There are no nonbonding and pie electrons of the type are observed in other molecules. The molecule has only a sigma bond and three lone pairs of electrons centered on fluorine. The contour diagrams for such molecules will also be different. So, electronic configuration of HF is σ 2 nF6 and electronic transition corresponds to nF → σ*. Quite similar behavior is observed in HCl, HBr, HI, the essential difference lies in the additional completed shells and the changes in principle quantum numbers. This only increases the bond length but decreases transition energies though bond order in each case is 1. Bond length and transition energies of some molecules are as follows: Bond length Transition energy, cm–1 HF HCl HBr HI

7.3

0.9168 A 1.2745 A 1.425 A 1.609 A

84776 76322 67663 55874

HOMONUCLEAR DIATOMIC MOLECULES

These molecules will not have any nonbonding orbital (Fig. 7.1) as found in CO and NO molecules, A and B will have same energies but without any hybrid structures. This has been redrawn for convenience in Fig. 7.2, rest will be the same. However, there is a strong interaction between s and p orbitals in the bond formation. Atom

Molecule

Atom

3σ* 1π*

2p

+ –

– +

–

–

+

+

2p

3σ

–

+ –

2σ* 2s

–

+

1π

–

+ 2s

2σ (a)

+ (b)

Fig. 7.2: (a) MO energy diagram for the second row homonuclear diatomic, (b) Contour diagram of the MO so formed

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Elements of Molecular Spectroscopy

Nitrogen molecule: In nitrogen molecule 1s and 2s electrons do not participate in the molecule formation and thus have six bonding electrons. This give rise to a definite electronic configuration to the molecule.

π 4 , 3σ 2 , π*0 with transition corresponding to 3σ → π*. This corresponds to a weak absorption peak observed at 49754 cm–1. This involves no change in number of nodal planes; hence the transition is not allowed, thus a weak absorption peak is observed. Nitrogen molecule has bond order of three with bond length 1.0977 A. N2 N2– N2 + Transition 3σ → π* 3σ → π* π* → 3σ* ∆n 0 0 1 Transition 9015 cm–1 49754 cm–1 – Bond order 2.5 3 2.5 Bond length 1.1164 A 1.0977 A 1.19 A Oxygen molecule: In this molecule 1s and 2s orbital do not participate in molecule formation thus have 8 bonding electrons. The electronic configuration will be

π 4 , 3σ 2 , π*2 Since there are two unpaired electrons oxygen molecule is paramagnetic in nature. With this electronic configuration electronic transition should correspond to π* → σ*. As there is change in number of nodal planes this transition should be observed. However, there is strong spin orbital coupling consequently the lowest absorption observed at 34320 cm–1 is not pure π*→ σ* transition. The bond length in oxygen molecule is found to be 1.207 A with bond order 2. As is expected this bond length is larger than nitrogen molecule. O2 O2– O2 + Transition ∆n Absorption peak Bond order Bond length

π* → σ* 1 40668 cm–1 2.5 1.1164 A

π* → σ* 1 34320 cm–1 2 1.207 A

π* → σ* 1 25000 cm–1 1.5 1.35 A

Halogen molecules: In this case only uppermost ten p electrons are involved in bond formation. This gives electronic configuration of halogen molecules MO as (np – π)4 (np – σ)2 (np – π*)4 (np – σ*) Since there are four pie bonding electrons and four pie anti-bonding electrons, which cancels the bonding effect in the pie orbital. As a result these pie electrons behave as nonbonding electrons. The deepening of colour in halogen from F2 to I2 can be understood from their MO configurations. The colour of these gases correspond to pπ*– pσ* transition. The lone pair electron on each halogen atom occupying pπ* orbital absorbs a photon, when one of the pπ* electron is raised to the next higher pσ* orbital. The larger the energy gap between π*–σ* shorter the wavelength required for affecting the transition. Conversely, the smaller the energy gap between π*–σ* level the longer the wavelength, in the visible region of spectrum required to bring about these transitions. Since in these cases nonbonding electron is in π* orbital, so n → σ* or π* → σ* are one and same thing. This energy difference goes on decreasing as we go from F2 and I2. Thus iodine is purple, bromine scarlet red, chlorine yellowish-

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113

green and fluorine pale yellow. As π* → σ* is not an allowed transition, these gases have light colours. However, N2 and O2 are colourless while halogens are coloured, bond order is one and thus because energy difference between n → σ* is small, electrons are easily excited by visible light to σ* which when fall back to ground state gives it a colour. In N2 and O2 bond orders are 3 and 2 respectively thus electrons are held with a strong force. As a consequence radiation emitted in σ* → π* does not lie in visible region but in ultraviolet region thus N2 and O2 are colourless. F2 Absorption ≈ 17240 cm–1 –1 peak ν cm λ, nm 580 Colour Yellow Bond order Bond length

7.4

1 1.412 A

Cl2

Br2

I2

O2

17170 cm–1

13818 cm–1

10100 cm–1

34320 cm–1

582.75 Yellowishgreen 1 1.987 A

723.7 Scarletred 1 2.281 A

990 Purple

291.4 Colourless

1 2.666 A

2 1.207 A

FRANCK-CONDON PRINCIPLE

The Franck-Condon principle states, “The time required for molecular vibrations (≈10–12 s) is much larger than that required for electronic transition (≈10–15 s). The electronic transition in a molecule takes place so rapidly in comparison to vibrational motion that immediately afterwards the nuclei still have very much the same relative position and velocity as before the transition”. It is because nuclei are much more massive than electrons so that in the course of an electronic transition the electronic redistribution occur while the nuclei stay at their initial separation. The electronic transition rapidly builds up charge density in new regions, and this new charge density exerts a force on the nuclei. This new force drives them into oscillation, and thus swinging backwards and forwards from their original separation. The stationary equilibrium separation of the nuclei in the initial state of the molecule, therefore, becomes the turning point of the vibration of the excited electronic state. Consider the molecular potential energy curves drawn in Fig. 7.3 they relate to different electronic states of the different molecules. Before the absorption occurs the molecule is in the ground vibrational state of its lower electronic state. The form of the lowest vibrational wave function shows that the most probable location of the nuclei is at their equilibrium separation re. It follows that most electronic transition will occur while the nuclei are at this separation. When an electronic transition occurs, the molecule is excited to the state represented by the upper curves. According to the Franck-Condon principle the nuclei framework remains constant during the actual transition and so we may imagine the energy of the molecule as rising up the vertical lines marked in Fig. 7.3. This is the origin of the expression vertical transitions, which is used to denote an electronic transition that occurs without change of nuclear geometry. The vertical transition cuts through several vibrational levels of the upper electronic state. The place where the arrow head touches the upper potential energy curve is the one in which the nuclei are most probably at the initial separation re and so this is the most probable level for the termination of the excitation. It is not, however, the only vibrational level at which the excitation may end, because several of its neighbours also have a high probability of having nuclei at the separation re. Therefore, transition takes place to all the vibrational levels in this region, but most intensely to the level having a wave function that overlaps the initial vibrational state wave function most favourably. The vibrational structure

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of the spectrum depends on the relative displacement of the two molecular potential energy curves, and a long progress of vibration (a lot of vibrational structure in the spectrum) is stimulated if the two states are significantly displaced. The upper curve is usually, but not always displaced to longer bond length, because excited states usually have more anti-bonding character than ground states.

Fig. 7.3: Potential energy curves explaining the intensity distribution in the absorption spectra as stated by Franck-Condon principle

The width of electronic absorption bands in liquid samples is due to the unresolved vibrational structure. This structure arises because electronic transitions are accompanied by changes of the vibrational state of the molecules, the vibrational structure can be resolved in gaseous state or solid state and be used to obtain information about the vibrational characteristics of excited state molecules. This will be explained with three examples.

1. CN molecule In this case ground state as well as in excited state internuclear distance is similar as is reflected by the position of potential energy minima of two curves [Fig. 7.3(a)]. As most of the vertical transitions originate from V" = 0 (ground state) to V' = 0 (excited state) line corresponding to V" = 0 →V' = 0 be the most intense. Lesser number of vertical transitions may touch the upper curve at V' = 1, 2, 3, but will be less intense. Hence the shape of spectra will show maximum intensity for the transition V" = 0 → V' = 0 followed by lines of less intensity on its right hand side.

Electronic Spectra

115

2. CO molecule Excited state internuclear distance of CO is a little larger than ground state, as a consequence this potential energy curve has shifted to the right hand side [Fig. 7.3(b)]. Vertical transition from ground state touches the excited state at V' = 2 and not V' = 0 as in CN molecule. Since maximum transitions will be from V" = 0 →V' = 2, this line will be most intense. The rest of lines will be spread on the left and right of this line with lesser intensity.

3. I2 molecule The internuclear distance in the excited state of iodine is quite large as compared to its ground state, consequently excited state potential energy curve has shift quite a bit to right hand side from ground state [Fig. 7.3(c)]. This shift is so much that the point where vertical transition from ground state touches the excited state, the molecule dissociates, which is reflected by a continuum in the spectra. On the left hand side of this continuum some lines with lesser intensity do appear.

7.5

DISSOCIATION ENERGY

Dissociation energy D0 corresponds to the heat of reaction → A + B AB  It corresponds to the work done to dissociate the molecule from the lowest level (V = 0) of the ground electronic state into atoms in the ground state. As dissociation of molecule takes place at higher values of vibrational quantum numbers, anharmonicity constant has a significant role. Rather it plays a major role in the dissociation of molecules, actual dissociation energy of a molecule can be calculated. Using equation 5.15 in which anharmonicity constant is included and thus vibrational energy levels are not equally spaced. 2

1 1   ...(5.15) G (V) = ν  V +  − xe ν  V +  2 2    As the value of vibrational quantum number V increases, the second term in this equation becomes more dominant and energy difference between the consecutive energy level decreases. As in the case of CO molecule ν = 2169.8 cm–1 and ν xe = 13.288 so that

1 1   G(V) = 2169.8  V +  − 13.288  V +  2 2  

2

...(7.1)

G(2) – G(1) = 2117.85 cm–1 G(3) – G(2) = 2090.1 cm–1 G(1) – G(0) = 2142.02 cm–1 This shows that as V increases energy difference goes on decreasing. So much so that for V = 30 to V = 31 transition energy difference comes to 1345 cm–1. This is almost half the fundamental frequency. This may be looked at in conjunction to potential energy curve (Fig. 5.4). As V increases, vibrational stretching go on increasing and thus internuclear distance goes on increasing , hence energy difference goes on decreasing. In other words, energy levels go on converging as molecule gets more and more stretched. A stage comes when the energy difference between two consecutive energy level becomes zero. At this point molecule is so much stretched that bond breaks. At this point the molecule is said to dissociate. Vibrational energy for this energy level can be written as 1 1   G(V) = ν  V +  − xe ν  V +  2 2  

2

3 3   G (V + 1) = ν  V +  − xe ν  V +  2 2  

2

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Elements of Molecular Spectroscopy

So that G (V + 1) – G (V) = ∆G = ν [1 – 2xe (V + 1)] The separation ∆G goes on decreasing till a continuum appears when ∆G→0 so that ν [1 – 2xe (V + 1)] = 0 or Vmax = 1/2xe –1 For this value of vibrational quantum number, V the molecule will dissociate. On substituting this value of Vmax after rounding to the next lowest number, in equation 5.15 the energy corresponding to dissociation of molecule can be calculated. Thus for CO molecule Vmax = 2169.8/2 × 13.288 – 1 = 80.64 ≈ 81 i.e., CO molecule will dissociate for a value of V = 81. Substituting this value in equation 7.1 dissociation energy can be calculated which will be 88576.5 cm–1 = 10.982 eV. This method has an extra advantage that this simple method helps one to calculate dissociation energy of molecules in different excited states even.

P ROBLEM 7.1 Anharmonicity constant and vibrational frequency of LiH is 16.5 × 10–3 and 1405.65 cm–1 respectively in an excited electronic state. Calculate the dissociation energy of LiH molecule in this excited state.

SOLUTION Vmax =

1

2 × 16.5 × 10 −3 Writing equation 5.15 specific to LiH gives

– 1 = 29.3 ≈ 29.

1 1   G (V) = 1405.65  V +  − 1405.5 × 16.5 × 10 −3  V +  2 2   On substituting V = 29 in this equation gives dissociation energy of LiH G (V) = 41466.67 – 20189.8 = 21276.87 cm–1 = 2.638 eV = Dissociation energy (1 eV = 8065.47 cm–1).

2

...(7.2)

PROBLEM 7.2 CuO in its excited electronic state has anharmonicity constant = 6.92 × 10–3 and vibrational frequency of 640 cm–1. Calculate the spectroscopic and thermochemical dissociation energy of CuO.

SOLUTION Vibrational energy levels can be given by equation 2

1 1   ...(7.3) G (V) = 640 V +  − 640 × 6.92 × 10 −3  V +  2 2    1 – 1 = 71.25 Vmax = 2 × 6.92 × 10−3 Dissociation energy can be calculated from equation (7.3) with V = 71 G (V) = 640 × 71.5 – 4.43 × 71.5 × 71.5 = 23445.1 cm–1 = 2.867 eV = Dissociation energy (spectroscopic) Thermochemical dissociation energy = Spectroscopic dissociation energy + Zero point energy Zero point energy of CuO = 640/2 = 320 cm–1 Thermochemical dissociation energy = 23445.1 + 320 = 23765.1 cm–1 = 2.94 eV.

Electronic Spectra

7.6

117

DISSOCIATION ENERGY AND TEMPERATURE

The treatment given until now was as if we have been talking of individual molecule and dealing with individual molecule. However, we always deal with an assemblage of molecules where intermolecular interactions, distribution of molecules in various energy levels play a major role in their behaviour making the things still interesting. This will be explained with some example. The molecular model suggests that as more and more energy is provided to a system, the molecule vibrates more vigorously and thus go to a higher vibrational energy state. When the molecule reaches the highest vibrational state it dissociates. This energy can be provided by a flame even. The energy associated with a flame may be taken as kT and when this energy is equivalent to dissociation energy, D the molecule should dissociate i.e., kT = D or T = D/k This relation may be used to calculate the temperature at which the molecule will dissociate. Let us carry out this with example. The dissociation energy of Cl2, I2 are 3.974 × 10–19 J and 2.472 × 10–19 J respectively, calculate the temperature at which it will dissociate. Molecule Cl2 I2 Dissociation Temp.

3.974 × 10−19 = 28797 K 1.38 × 10−23

2.472 × 10 −19 = 17916 K 1.38 × 10−23

Similar results will be obtained in other molecules. This result says that all molecules will go to excited state and will dissociate. However, these figures have created more confusion because such temperatures cannot be attained in a laboratory hence it may not be possible to dissociate. In actual practice we have an assemblage of gas in which molecules are distributed to various energy states by Boltzmann distribution law. On heating gas molecules go from one energy state to another by this law. If temperature of flame were 1000 K the number of molecules in excited state that may lead to dissociation can be calculated. I2 Molecule Cl 2 Nv/No = e

−

3.974 × 10−19 1.38 × 10−23 × 1000

= 3.11 × 10−13

=

−

e

2.472 ×10−19 1.38 × 10−23 × 1000

= 1.661 × 10−8

Percentage of molecules in dissociation state = 1.661 × 10–6 = 3.11 × 10–11 If one mole of a gas is taken, then number of molecules in dissociation state will be 18.7 × 1010 and 9.999 × 1015 molecules of Cl2 and I2 respectively. This means so many molecules go to that excited state which lead to dissociation even at 1000 K. As heating is an ongoing process and after so many molecules dissociate, redistribution of molecules in energy levels takes place and more molecules go to the highest vibrational state and dissociate. In this way the process goes on. Since percentage of molecules at any time in dissociation state in I2 and Cl2 are 10–6 and 10–11 respectively which mean that I2 is 105 times more populated in the dissociation state thus less stable than Cl2. Further, it may be noted that neither two dissociation energy values differ in that proportion nor can we draw the above conclusion from those values. Similar reasons can be given for some reactions not taking place at room temperature but do take place at higher temperature. Similar mechanisms lead to the breaking of the bond there and thus subsequent reaction.

118

7.7

Elements of Molecular Spectroscopy

DISSIPATION OF ENERGY BY EXCITED MOLECULES

Light absorption leads to the excitation of molecules from the ground to excited states. But what is the fate of this energy? This section deals with the different modes of deactivation of excited states. The half-lives of the excited states vary down from 10–9 to 10–4 s. The reverse process corresponding to the emission of radiation of the same frequency is one of the simplest mode of losing energy, as is observed in emission spectra. Another process is known as pre-dissociation. A third fate of the absorbed energy may lead to direct photodecomposition. It occurs when the vibrational states in the excited state are continuous and is signalled with a continuous absorption in the spectrum. Absorbed light may be degraded into thermal motion or it may be remitted. Such emission from an excited molecule is called fluorescence or phosphorescence. These two types of emission may be distinguished by the mechanism that generates them.

Fig. 7.4: The process leading to fluorescence and shift of latter to longer wavelength

The mechanism of fluorescence is illustrated in Fig. 7.4. The incident light excites the molecule from the ground state, O. By the Franck-Condon principle this excitation should be vertically upwards along OD, producing a compressed excited molecules, which vibrates with a period of 10–13 s as

Electronic Spectra

119

shown by the upper curve. Vibrational excitation is thus coupled to electronic excitation. As molecules in the liquid state have a collision frequency of ~ 1012 times/s, solvent molecules collide with the excited molecules and rob them of their excess vibrational energy so that they reach at point E. Here two things may occur. One, the solvent may carry away the electronic energy and deactivate the molecule. This is possible if the energy levels of the solvent and excited molecules match each other, for then there may be a resonant transfer of radiation to the solvent, which then fritters away its excitation into thermal motion. An alternative mode of decay is fluorescent decay along line EF, remitting light of whose mean frequency is proportional to the length EF and falls back into a lower electronic state. This emitted light is what is called fluorescence. Such a picture of the process enables one to visualize the following points: 1. Fluorescent light is always of smaller frequency than the incident light because EF is always less than OD. 2. The fluorescence spectrum shows a vibrational structure giving the vibrational frequency of the molecule in the ground state. This is in contrast to the electronic state. This follows that the fluorescent spectrum and electronic spectrum should appear as mirror images of each other, but this description should be taken literally. 3. The fluorescence spectrum is independent of the wavelength or absorbed light (within limits), since the fluorescence transition is from E downward, whatever higher level such as D has reached on light absorption. 4. The time for an electron to stay in excited state prior to the ground state is about 10–6 and 10–10 s. 5. Spin states of the involved state should not change i.e., all involved states are either singlets or triplets. The conditions for fluorescence are much more favourable in aromatic and particularly in condensed hydrocarbons. These compounds absorb at relatively long wavelengths and have very rigid structure, making internal conversion quite difficult. Some of the experimental materials have been assembled in Table 7.1. Table 7.1: Fluorescence shown by some substituted hydrocarbon in hexane solution

Compound

Wavelength region of fluorescence in nm

Compound

Wavelength region of fluorescence in nm

Benzene

270–310

Naphthalene

300–360

Toluene

270–310

Anthracene

370–460

o, m, p-Xylene

270–320

Phenanthrene

280–470

Styrene

290–365

Fluorine

300–370

Chlorobenzene

275–345

Biphenyl

290–360

Hexamethylbenzene

280–330

Triphenylmethane

280–340

Benzoic acid

310–390

o-Hydroxy benzoic acid

375–480

Aniline

310–405

Dimethylaniline

325–405

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Elements of Molecular Spectroscopy

Substitution in aromatic hydrocarbon naturally shifts the wavelength of fluorescence in agreement with the effect of the same substitution on the absorption spectrum. Alkyl substitution has little effect. Chlorine and bromine weaken the fluorescence and iodine completely inhibits it. The simplest aromatic heterocyclics, pyridine, pyrrole, furan and thiophene do not show fluorescence. A similar situation results if emission occurs from one of the excited electronic state reached by internal conversion. The process then would be as that shown in Fig. 7.5. Even in this case the emission process that connects the two states has same number of unpaired electrons—two singlet states or two states with half-lives in the range 10–9 –10–4 s.

Internuclear distance

Fig. 7.5: Fluorescence following the internal conversion

Another type of fluorescence invariably accompanying irradiation of some compounds but not due to the excited molecules occurs when the compound undergoes a photochemical reaction or decomposition. Many carbonyl compounds, photochemically dissociate to form acryl radicals, which in turn, dimerize. These dimeric biacryls show fluorescence in the visible region and give the appearance that the original carbonyl compound show fluorescence. The fluorescence intensity increases with time suggesting that a fluorescing species is formed during irradiation. In general compounds absorbing only in the far UV or middle UV, decay fluorescence because predissociation deactivates excited states. Consequently, it is no surprising that aliphatic and simple nonconjugated olefins, hydrocarbons and their substitute derivatives do not show fluorescence.

Electronic Spectra

7.8

121

PHOSPHORESCENCE

The electronic excitation is accompanied by a vibrational excitation and this vibrational energy is transferred to the surrounding molecules by the type of process described in fluorescence. If the vibrational deactivation is not too fast, another process may intervene. Let there be a triplet state of the excited molecules. This is illustrated as curve T1 (Fig. 7.6). There is probability of a molecule switching from the singlet state to the triplet as it loses energy and is named to result from intersystem crossing. At the bottom of energy curve, T1, it finds itself trapped; it cannot radiate its electronic energy and drop to the ground state, which involves singlet-triplet-singlet transitions. It cannot go back to the crossing point, because the collisions with the lattice cannot supply that much energy. It cannot give up its electronic energy to the surrounding molecules by a radiationless transition because the vibrational deactivation involves smaller energy and is thus weak. If the foregoing description were true, the molecule would be stuck in this triplet state. In actual practice, the molecule does come down to the ground state.

Intersystem crossing

Fig. 7.6: The mechanism of phosphorescence

The fallacy is that the singlet-triplet transitions are not strictly forbidden. If it were so, the molecule would have been unable to cross from the singlet state S1 to the triplet state T1. The fact that it did cross implies that there is enough spin-orbit coupling present to break down the singlet-triplet

122

Elements of Molecular Spectroscopy

selection rule and so this becomes weakly allowed. However, as it is only weakly allowed, the transition T1 →S0 is slow and may persist even after the illumination has ceased. Thus, phosphorescence occurs under the following conditions: 1. There is a suitable triplet state in the vicinity of the excited singlet state. 2. There is sufficient spin-orbit coupling to induce intersystem crossing. 3. There is sufficient time available for the molecule to cross from one curve to the other. This means that the vibrational deactivation must be slow so that the molecule is quenched and taken below the point where the curves intersect before intersystem crossing interaction has time to operate. 4. The wavelength of the emitted light is longer than the fluorescent emission. 5. Since singlet-triplet transition is not allowed, consequently, the transition is slow. This makes phosphorescence emission long after the irradiating light is switched off. This does not happen in fluorescence. The involvement of the triplet state an unequivocally be studied by applying electron-spin resonance to the phosphorescent state. From this, the time of deactivation can also be obtain. Phosphorescence is most easily attained in a solid state rather than in the liquid state. In solid state dissipation of energy by virtue of collision is minimized and time to the excited state gets increased. The result is increase in probability of intersystem crossing and consequently to phosphorescence. Solids like rigidity and thus phosphorescence can also be attained by absorption of molecule on a surface or by using a Micelle to stabilize the molecules.

PROBLEMS 1. Justify the following statements: (i) The vibration structure of the electronic absorption spectrum is characteristic of the higher electronic levels whereas that of fluorescence spectrum is characteristic of the lower electronic state. (ii) Fluorescence or phosphorescence radiation has a lower frequency than the incident radiation. (iii) Phosphorescence is more intense in solid samples than in solutions. (iv) The amount of phosphorescence depends on the presence of a heavy atom. (v) Phosphorescence unlike fluorescence cannot be studied in liquid phase. 2. The values of ν and ν xe for upper and lower electronic states of AB molecule are 1515.61, 17.25; 2170.21, 13.46 cm–1 respectively. Calculate the dissociation energy of AB in these two states. 3. The anharmonicity constant for CIF is xe = 12.48 × 10–3 and fundamental vibrational frequency 793.1 cm–1. Calculate the dissociation energy of CI—F molecule in kJ mol–1 and eV. 4. The spectroscopic bond dissociation energy of ClO is 15324 cm–1. Calculate the equilibrium bond dissociation energy of ClO if the fundamental vibrational frequency is 780 cm–1. 5. The fundamental vibration frequency of HCl and HI are 2990 and 2309 cm–1 and anharmonicity constant 0.0174 and .006086 respectively. Calculate the dissociation energy for both the molecules. Also calculate and show which molecule is less stable (1 cm–1 = 1.987 × 10–23 J mol–1). 6. The absorption spectrum of O2 shows coarse vibrational structure ending with a continuum at 56876 cm–1. The upper electronic state dissociate into two O atoms—one of these in its ground state and other in excited state with an excitation energy of 15875 cm–1. Calculate the dissociation energy from ground state. 7. Electronic absorption spectra of some of the diatomic molecules are observed as follows: (all in Hz).

Electronic Spectra

123

CO CO+ CO– 14 14 14.606 × 10 6.2199 × 10 3.63 × 1014 NO CIO Cl2 14 14 13.19 × 10 9.5 × 10 5.15 × 1014 (i) Using Fig. 7.1, assign these bands to various transitions. (ii) Explain the different values obtained in the case of CO and its ionic forms. (iii) Which transitions are forbidden transition? 8. The values of ν in cm–1 and xe in ground state for CaO and CH are as follows: CaO ( ν cm–1) CH ( ν cm–1) CaO (xe) CH (xe) 742 2099 6.56 × 10–3 16.2 × 10–3 Calculate the dissociation energy of these molecules. 9. Using Fig. 7.1 write down the electronic configuration of NF, NF–. Name the first transition from ground state to excited state, and state if they are not allowed. Describe how the bond length varies in this species. 10. Using Fig. 7.1 write down the electronic configuration of CO and CO+ . State the nature of first allowed transition in these molecules. 11. Explain (i) why NO molecules dimerise to N2O2?, (ii) why it at once react with oxygen to give NO2 molecule? 12. Using Fig. 7.2 explain why the color of halogens changes from yellow to purple as we go from fluorine to iodine? Also explain why N2 and O2 are colorless while halogens are colored? 13. Explain how the excess inhalation of CO gas leads to congestion and then death? 14. Explain using Fig. 7.1 why CN molecule dimerises to C2N2 and bond is formed between carbon and carbon but CN– as well as CN+ does not dimerise? 15. Using Fig. 7.2 write down the electronic configuration of C2 molecule. State if it will be paramagnetic or not. 16. Using Fig. 7.1 write down the electronic configuration of BC and BN. Name the first transition from ground state to excited state, and state if they are allowed. Describe how the bond length varies in these species? 17. Following are the values of some molecules for fundamental vibration and anharmonicity constant. Calculate the maximum value of vibrational quantum number and thus dissociation energy of these molecules. νe Fundamental νe xe Molecule Anharmonicity constant, xe vib. in cm–1 27 Al 1 H 1682.56 29.09 17.29 × 10 –3 Ba1 H 1168.31 14.50 12.41 × 10–3 11 B 1 H

2366.9

49.39

20.867 × 10–3

12 C 14 N

2858.5 2068.59

63.0 13.087

22.04 × 10 –3 6.326 × 10 –3

CO HCl

2169.81 2990.99

13.288 52.81

6.124 × 10 –3 17.66 × 10 –3

HF HBr

2998.19 2648.97

45.76 45.217

15.26 × 10 –3 17.07 × 10 –3

1639

19.87

12.1 × 10 –3

3737.76

84.88

22.71 × 10–3

12

C 1H

HI DH

a b

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CHAPTER

8

Electronic Transitions in Organic Molecules 8.1

VARIOUS BONDS AND TRANSITIONS

Visible and ultraviolet spectra that result from electronic transitions in organic molecules are encountered with σ and π orbitals only. Both of which may be classified into bonding and anti-bonding orbitals and the latter may be distinguished by means of an asterisk. The nonbonding orbitals, frequently denoted by n, are localized on a particular atom, while σ and π orbitals are spread over whole molecules. The contour MO diagrams of different molecules are different. However, even then the relative position of these molecular orbitals energy levels is shown in Fig. 8.1.

Fig. 8.1: Types of electronic transitions

In the ground state σ, π and nonbonding orbitals are occupied leaving σ* and π* vacant. Degeneracy of various orbitals has not been shown in this diagram. Inspection of Fig. 8.1 shows: (i) Orbital energies increase in the order σ< ππ→π* ≈ n→σ*> n→π*. Of all the possible transitions, the last three are responsible for absorption in the region 200–800 nm, whereas others require much higher energy.

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Elements of Molecular Spectroscopy

(ii) σ→σ σ→σ* transitions: A σ–σ* transition requires the largest amount of energy. Saturated hydrocarbon e.g., methane, ethane undergo σ–σ* transitions and show absorption maximal at 96 nm in far (vacuum) ultraviolet region, far below 200 nm or above 50000 cm–1. In this region strong absorption is shown by oxygen molecule. Hence the spectra can be obtained only if whole of spectrometer is evacuated of air and recorded in He–Ar atmosphere. The equipment for this work is more expensive and require a highly skilled operator to use this technique. It has been mainly exploited in studying bond energies etc. and is not usually very helpful in structure determination. →σ (iii) n→σ →σ* and n →π →π* transitions: Nonbonding electrons are held least strongly in a molecule. The n→σ* and n→π* transition are quantum mechanically forbidden transition and hence are of weak intensities. This type of transitions are observed in compounds containing such heteroatom as nitrogen, oxygen, sulphur, and halogen. The absorption maximals of some compounds with nonbonding electrons are as follows: CH3NH2

CH3OH

CH3SH

CH3Cl

CH3Br

CH3I

213 nm

183 nm

278 nm

173 nm

204 nm

258 nm

These compounds show absorption at considerable longer wavelength than corresponding saturated compounds without heteroatom. For instance CH3I absorbs at a wavelength 258 nm due to n→σ* transition requiring relatively small excitation energy as compared to methane which has no such transaction but show absorption at 96 nm only. An important criterion employed for distinguishing n→σ* band from other bands is disappearance of band in acid solution. In presence of hydrogen ion the lone pair of electron gets protonated and hence one should not expect n→σ* transitions. n→π* transitions are exhibited by compounds containing nonbonded electrons in an unsaturated molecule. It is never observed alone but accompanies n→σ* and π→π*. For example, carbonyl compounds show four main regions of absorption. Two regions representing promotion of an oxygen 2p lone pair electron to antibonding π* orbital give n→π* triplet and n→π* singlet transitions respectively around 280 nm. The third and fourth bands are of moderately strong and strong intensities respectively. These may be assigned to n→σ* and π→π* transitions. π→π π→π* transition: This transition is an allowed transition and hence of high intensity. The π→π* transitions depend on the length and shape of conjugated system the longer the conjugated chain, the longer the wavelength at which it absorbs. The π→π* absorption band of ethylene is located in the far ultraviolet region at λmax = 162.5 nm, while λmax of butadiene and hexatriene are found at 217 and 260 nm. With increase in number of conjugated double bonds further displacement to longer wavelength showing that π and π* energy levels come closer to one another. Similarly λmax for vitamin A, containing five conjugated double bonds is equal to 326 nm β-carotene contain eleven conjugated double bonds has a λmax = 45l nm. The absorption spectra of cyclic conjugated system also shift to longer wavelength as shown in Fig. 8.2. It shows the absorption bands of the polyacene series correspond to one electron transition from highest occupied molecular orbital (HOMO), to lowest vacant molecular orbitals (LVMO) corresponds to π→π* transition. As the number of rings increase absorption maxima shift to longer wavelength.

Electronic Transitions in Organic Molecules

127

Addition of C O group will further increase the conjugation but will also give n→π* transition. Aldehydes and ketones exhibit weak absorption band, which is not found in the corresponding hydrocarbon at long wavelength. For instance, formaldehyde yields n→π* absorption bands at 290 nm with a strong absorption of π→π* transition at 156 nm while ethylene shows absorption at 162.5 nm. Similarly, acrolein and benzaldehyde show weak absorption at long wavelengths. The carbonyl groups of aldehydes and ketones are responsible for these long wavelength absorption band characterized as n →π* transition but of low intensity. LVMO

cm–1

cm –1

cm –1

HOMO

Fig 8.2 : Absorption spectra and the corresponding electronic transitions for the polyacene series HOMO level; LVMO level

Electron donating substituents, —NH2, —NR2, —OH, —SH, halogens all of which have nonbonding electrons. Substitution of these groups in a conjugated system makes these nonbonded electrons to interact with pie electrons and is partially migrated to conjugate system. This sets in mutual repulsion of electrons, facilitating the π→π* transition at lower wavelength. The π→π* transition of benzene is observed at 260 nm. However, on introduction of electron donating substituent, shifts not only absorption to visible region side but also intensifies the absorption as shown in Table 8.1.

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Table 8.1: Absorption maxima of some substituted benzene

Compound

ν cm–1

∆ ν cm–1

C6H6

38089

–

C6H5CH3

37484

– 605

C6H5F

37818

–271

C6H5Cl

37052

–1037

C6H5Br

36996

–1093

C6H5OH

36350

–1739

C6H5NH2

34034

– 405

C6H4Cl o, m, p

36230 36186 35743

– 1859 – 1903 – 2346

Such a shift of absorption to visible region side is called red shift. Two situations arise when (a) electron spins in the π and π* energy levels are parallel, (b) the electron spins in the π and π* energy levels are anti-parallel. The former transition (a) π → π* gives rise to a triplet state with energy difference ∆E1 while in the latter case, a π → π* transition give rise to singlet state with energy difference ∆E2 is obtained. Note that ∆E2 > ∆E1 i.e., triplet is of lower energy than singlet. On the other hand shift in n →π* transitions by the electron donating group to move toward ultra violet region, and is called blue shift. This is a result of movement of nonbonding energy levels nearer to π-orbitals (Fig. 8.3). The effect of chlorination of pyrazine on the n→π* and π→π* transitions of pyrazine is shown in Fig. 8.3. Substitution of chlorine produces the usual π→π* transitions characterized by red shift. In contrast the n→π* transition undergoes a blue shift and n→π* transition is attributed to the transfer of one of the nonbonding electron into the π electron system, thus increasing

Fig. 8.3: Frequency shifts due to chlorination of pyrazine

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129

its negative charge. In the excited state arising from the n→π* transition of a substitution produce the additional negative charge, will prevent the unshared electrons of the substituent from migrating into the π-electron system. Therefore, the stabilization energy due to electron migration will be reduced in the excited state but not in the ground state. Obviously this phenomenon cannot take place in the parent compound without substituent. Thus, we expect the energy of the n→π* transition to be increased as a result of substitution.

8.2

SOME IMPORTANT TERMS USED IN ELECTRONIC TRANSITIONS

1. λmax value: The value of wavelength at which absorption maximum occurs is called the λmax and is expressed in nanometer (nm) (1 nm = 109 metres). 2. εmax value: It is shown as molar absorption or molar extinction coefficient and is a measure of intensity of absorption. The ε value is characteristic of a particular compound at a given wavelength. Usually along with the wavelength of maximum absorption λmax, molar extinction coefficient εmax is also expressed. The intensity of absorption can be expressed as transmittance (T), which is defined as the ratio of the intensity of the radiation transmitted from the sample (I) to that of the radiation incident on the sample (I0) i.e., ...(8.1) T = I / I0 Intensity of absorption is more conveniently expressed in terms of absorbance (A) i.e., A = log (I/T) = log10 (I0/I)

...(8.2)

Absorbance of a band is related to the sample thickness (l) and concentration (c) of the absorbing species. The relation is expressed in the form of Beer-Lambert law as below: A = log10 I0/I = εcl

...(8.3)

Absorbance is a dimensionless quantity. Concentration c is usually expressed in mol dm–3 and path length (l) in cm, hence ε has the unit of dm3 mol–1 cm–1. If we use SI units of mol dm–3 for concentration and in m for path length, the units of ε will be 2 m mol –1. We can obtain the values of ε in m2 mol–1 units from those in dm3 mol–1 units in the following manner: ε = dm3 mol–1 cm–1 = 10 –3 m3 and md–1 (10–2 m) = 10–1 m2 mol–1. Values of ε in SI units can therefore, easily be obtained from published values in dm3 mol–1 cm–1 by dividing the numerical quantity in latter units by 10. Absorption bands with εmax value > 103 m2 mol–1 are considered to be high intensity or strong bands whereas those with εmax values < 102 m2 mol–1 are known as low intensity bands or weak bands.

Chromophore Functional groups, which exhibit electronic absorption of a characteristic nature in visible and ultraviolet region, are called chromophores. Examples of chromophores are —COOH, —COCl, —CONH2, —N O, NO2, —N N—, —C absorption maxima for some typical chromophores.

C C , —C C—, C O, N and so on. Table 8.2 lists the

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Table 8.2: Absorption data for isolated chromophores

Chromophoric group

System

Transition λ max nm

ε max m2 mol–1

Solvent

Ethylene

π→π*

165

1000

Vapor

C—R

Acetylene

π→π*

173

6000

Vapor

C

O

Acetone

π→π* n→π*

188 279

900 1.5

n-hexane

C

O

Acetaldehyde

n→π*

290

1.6

Heptane

6.0

Water

–

–

5000

Water

–

–

Ethylene

RCH

Acetylenic

R—C

Carbonyl

Example

CHR

R1 R2 R1

Carbonyl H Carbonyl

RCOOH

Acetic acid

n→π*

204

Amido

RCONH2

Acetamide

n→π*

< 208

C

N—

Acetoxime

π→π*

190

Nitrile

—C

N

Acetonitrile

π→π*

< 160

Azo

—N

N—

Azomethane

n→π*

347

0.47

Dioxane

Nitrose

—N

O

Nitroso butane

π→π* n→π*

300 665

10 2.0

Ether

Nitrate

—ONO2

Ethyl nitrate

n→π*

270

1.2

Dioxane

Nitro

—NO2

Nitro methane

n→π*

271

1.86

Alcohol

Nitrite

—ONO

Amyl nitrite

π→π* n→π*

218.5 346

112

Ether

Azomethane

Functional group bearing a π-bond together with or without atoms containing a lone pair of electrons constitutes what has become known as chromophores.

Auxochrome A saturated group with nonbonded electrons when attached to a chromophore alters both wavelength and intensity of the absorption is called an auxochrome e.g., —OH, —OR, —NH2, —Cl, —SO2NH2. The auxochrome, though by itself is unable to impart colour to a compound but when attached to a chromophore may have a colour. The auxochromic effect depends on the ability of the chemical group to donate electrons into conjugated system. This has been most studied with aromatic systems and the spectral shifts of monosubstituted aromatic compounds have been correlated with the electron donating power of auxochromes. The electrons donating power of some common auxochromes decreases in the order O– > NHCH3 > NH2 > OH > Cl > CH3 > NH+3 = H. Substitution of p-ethoxy in azobenzene, maximum absorption wavelength is shifted to 65 nm higher with double its intensity. Likewise benzene

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131

shows λmax shifts to 254 nm (ε = 230) and aniline has λmax at 280 nm (ε = 1430). In the presence of acid, aniline is converted to anilinium ion, which has no nonbonding electron, and λmax shifts to 250 nm (ε = 160).

Bathchromic, Hypsochromic, Hyperchromic and Hypochromic Shifts

Molar extinction coeff.

Some older established terms should be mentioned here although their use should be progressively discouraged. The terms and effects are shown in Fig. 8.4 for a typical electronic spectrum.

Fig. 8.4: Hypochromic, hyperchromic, bathchromic and hypsochromic shifts, displayed in a typical electronic spectrum

A shift of absorption maximum towards longer wavelength produced by a change in medium or by the presence of an auxochrome is called bathchromic shift or red shift. For example, carbonyl group shows λmax somewhere in ultraviolet region but their 2, 4-dinitrophenyl hydrazone derivatives show λmax in the visible region. A shift towards shorter wavelength caused by a change of medium or by removal of conjugation is referred to as hypsochromic shift or blue shift. For example, the conjugation of the lone pair of electron on the nitrogen atom in aniline with π-bond system of benzene ring is removed by protonation. Aniline exhibit λmax = 280 nm while anilinium ion exhibit λmax at 250 nm a case of blue shift. Substitution of an amine, alkoxy, hydroxy or halogen groups on the carbonyl compounds result in displacement of absorption maximum to shorter wavelength. An effect leading to decreased absorption intensity is called hypochromic effect. An effect leading to increased absorption intensity is called hyperchromic effect, which is opposite to hypsochromic effect. It should be noted that auxochrome generally show bathchromic shift but substitution of alkyl or aryl group to auxochrome NH2 exchanges the bathchromic effect. On the contrary, acetylation of OH or NH2 generally has a hypochromic effect. Sometimes, substitution of a particular functional group in compound does not change the λmax but either lowers or raises the molar extinction coefficient than that of the parent compound.

8.3

ABSORPTION DUE TO ETHYLENE CHROMOPHORE

Ethylene has four σ-bonds one C—C σ-bonds and one C—C π bond with electronic configuration in ground state as (σC–H)8(σC–C)2 (πC–C)2.

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In ethylene four electronic transitions are possible which are π→π*, π→σ*, σ→π*, σ→σ*. Out of these the π→σ* and σ→π* are symmetry forbidden. Of the other two, the π–π* transition would appear at 165 nm (εmax = 1000), σ→σ* transition lies in vacuum UV and is out of the range of normal UV spectrophotometer. Substitution of the alkyl groups to ethylenic compounds moves the absorption to longer wavelengths (red shifts). As the number of alkyl groups increase a progressive red shift is observed. Attachment of a heteroatom like N, S bearing nonbonded electrons to the ethylenic linkage also gives rise to a red or bathchromic shift. For example, methyl vinyl sulphide (CH3—S—CH CH2) absorbs at 228 nm. The absorption of cyclic mono-olefins resemble those of the open chain olefins and the absorption has no relationship to ring size. When there are two or more isolated ethylenic bonds in a molecule, it absorbs at the same position as the single ethylenic chromophore. Conjugation markedly affects the position of absorption due to the C C chromophore giving rise to bathchromic shift. Thus, the λmax value for π→π* transition for the 1, 3-butadiene is 217 nm as compared to 185 nm for the 1, 5-hexadiene. CH2

C—C

CH2

1, 3-Butadiene

λmax = 217 nm, εmax = 2100 m2 mol–1

CH2

CH — CH2

CH2 — CH

CH2

1, 5-hexatriene

λmax = 185 nm, εmax = 2100 m2 mol–1

This relatively large increase in wavelength of absorption due to conjugation can be explained as follows: In ethylene the two 2p atomic orbitals combine to form a set of π and π* molecular orbitals. In conjugated dienes such as 1, 3-butadiene when π and π* molecular orbitals of two ethylenic linkages are close enough, overlap can occur. As a result of combination of two π-MOs give two delocalized orbitals of lower and higher energy (π1 and π2). Similarly, the two π* orbitals give rise to two delocalized π* orbitals of different energies (π*3 and π*4) Fig. 8.5. Thus the lowest energy corresponds the π2→ π*3 transition in 1, 3-butadiene that occurs at a longer wavelength (217 nm) as compared to the lowest energy π→π* transition of 1, 5-hexadiene (185 nm).

Fig. 8.5: Molecular orbital energy relationship between π-orbital of isolated and conjugated dienes

As the extent of conjugation increases, it lowers still further the energy of transition from the highest occupied π orbitals to lowest unoccupied π* orbitals, thereby λmax value increases. Table 8.3 gives the wavelengths of some conjugated polyenes that demonstrate this effect.

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133

Table 8.3: Absorption bands of conjugated polyenes

Number of conjugated π bonds λ max/nm

2

3

4

5

6

7

8

217

268

304

334

364

390

410

The effect of substituent and geometry of the absorption bands of conjugated dienes are fairly consistent. A set of empirical rules has been formulated by Woodward to predict the absorption of open chain (acyclic) and six membered ring dienes. Feiser and Scott have modified these rules. The rules are summarized in Table 8.4. Table 8.4: Woodward rules for predicting π–π* absorption in dienes

Value assigned to parent open chain diene Value assigned to parent heteroannular diene Value assigned to parent homoannular diene Value assigned to parent acyclic trienes Increment for (a) each alkyl substituent or ring residue (b) each exocyclic double bond (c) each double bond extending conjugation (d) auxochrome – O (acyle) —OR alkoxy —S— thioether —Cl, — Br —N (alkyl)2 Solvent shift minimal Calculated λmax Total =

217 nm 214 nm 253 nm 245 nm 5 nm 5 nm 30 nm 0 6 nm 30 nm 5 nm 60 nm

In order to apply these rules you must be able to identify the types of structures referred to in Table 8.4. The basic chromophore unit is, 1, 3-butadiene which is considered the parent acyclic (or non-cyclic) diene:

If saturated alkyl groups are attached to the diene, then an additional contribution for each group is added.

PROBLEM 8.1 Calculate the position of π→π* absorption band in 1, 2, 4-trimethyl butadiene using Woodward rules and Fieser-Scott rules.

SOLUTION

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Elements of Molecular Spectroscopy

Base value for acyclic diene (1) Table 8.4 = 217 nm. For three methyl groups a, b, c, add 3 × 5 = 15 nm; total 232 nm. Predicted λmax for π→π* transition — 232 nm. Observed λmax for π→π* transition — 231 nm. If the diene system is contained in a single ring, it is termed homoannular e.g., compound II. On the other hand, if it is spread over two rings, it is said to be heteroannular e.g., compound (III).

PROBLEM 8.2 Calculate the position of π→π* absorption band in Compound II and III using Woodward rules and Fieser-Scott rules.

SOLUTION For compound (II) base value for homoannular diene (Table 8.4) = 253 nm For three ring residents a, b and c, add 3 × 5 = 15 nm. The lower double bond in A is attached to but is outside ring B i.e., it is acyclic to ring B add 5 nm. Predicted λmax value = 273 nm Observed λmax value = 275 nm Note that the groups marked with a star (*) do not contribute as they are not directly attached to the diene system. For compound (III) base value of heteroannular diene (Table 8.4) = 214 nm. For three ring residues a, b, c, add 3 × 5 = 15 nm. The double bond in ring A is in exocyclic position to ring B: add 5 nm. Predicted λmax = 234 nm. Observed λmax = 235 nm.

PROBLEM 8.3 Calculate the λmax value for the compound B given below. It has been observed that the parent cis diene (A) has a λmax at 253 nm.

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135

SOLUTION In the parent cis diene (A) absorption

= 253 nm 3 Ring residues + 2 methyl groups = 5 × 3 + 5 × 2 = 25 nm 2 double bonds which extend conjugation = 2 × 30 = 60 nm 3 Exocyclic double bonds 3 × 5 = 15 nm = 0 nm CH3 COO— Total = 353 nm The compound B is expected to observe λmax at 353 nm which agree with experimental value. Thus in all the three cases, the predicted λmax values are in very good agreement with the observed values.

8.4

ACETYLENIC AND BENZENOID CHROMOPHORES

The electronic spectra of acetylenic and the benzenoid chromophores are more complex than those of the ethylenic chromophores. These cannot be explained by following the model presented for the ethylenic chromophore. They exhibit three absorption bands which are shown below: Acetylene 152 nm (strong), 182 nm (moderate), 220 nm (weak). Benzene 183.4 nm (strong), 204 nm (strong), 254 nm (weak). In each case, you can see the lowest energy absorption band is weak, which is characteristic of a forbidden transition e.g., n →π*. But there are no nonbonding electrons in these molecules. Thus all the three transitions arise from π→π* transitions. Acetylene and benzene are highly symmetrical molecules having degenerate molecular orbitals. Transition between degenerate orbitals in these cases give rise to the complexity in their electronic spectra.

8.5

CARBONYL CHROMOPHORE

Carbonyl groups contain, in addition to σ electrons, pair of π electrons and two pairs of nonbonding electrons. Saturated aldehydes and ketones exhibit three absorption bands due to π→π*, n→σ* and n→π* transitions. π→π*, n→σ* transition appear in the vacuum ultraviolet region near 150 nm and 190 nm. The third band is due to the forbidden n→π* transition appears as weak band in the region 270–300 nm. Some values are as under: CH3COCH3 HCHO CH3CHO n→π* (Weak band) n→σ* (Intense band) π→π* (V. Intense band)

310 nm

290 nm

279 nm

190 nm

189 nm

190 nm

150 nm

151 nm

150 nm

In contrast to the situation in alkenes, alkyl substitution moves n→π* band to higher energy. Auxochromes such as Cl, OH and NH2 cause a larger shift in the carbonyl absorption to shorter wavelengths. When a carbonyl group of a ketone is conjugated with a carbon-carbon double bond ( C C ) the compound is known as an enone or α, β-unsaturated ketone e.g., methyl vinyl ketone (CH3COCH CH2). Conjugation has an effect on energy of π–π* transition similar to that in alkenes.

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Elements of Molecular Spectroscopy

As the energy of the π* orbitals is lowered by conjugation, the π→π* and n→π* absorption move to longer wavelengths. Thus in some enones results are as Ethylene

C

C

Formaldehyde

H

C

Acroline or Propanal

O

CH2

Dimethyl vinyl ketone CH3

CH—CHO

H C

CH3

H

C—CO—CH3

n→π* 290 nm

n→π* 336 nm

n→π* 315 nm

π→π*

π→π*

π→π*

π→π*

217 nm

156 nm

202 nm

237 nm

As with the conjugated dienes, there are empirical rules to predict the position of the π→π* bands in enones summarized in Table 8.5. These rules were put forth by Woodward and modified by Feiser and Scott. Table 8.5: Fieser and Scott rules for predicting π→π* absorption in α, β-unsaturated ketones (enones and aldehydes)

Value for parent acyclic ketone or 6–ring cyclic 5–ring cyclic

215 nm 202 nm

Value for parent unsaturated aldehyde

207 nm

C

C—CHO

Value for parent unsaturated acids and esters

C

C—COOH(R)

Increments for (a) each double bond extending the conjugation (b) each alkyl group or ring residue

197 nm

α 10

β 12

30 nm γ 18

δ 18

35 6 35 – 15 25 –

30 6 30 80 12 30 95

30 6 17 – 12 25 –

50 6 31 – 12 25 –

Addition for each substituent —OH —OAc —OMe —SR thioether —Cl —Br —NH2, —NHR, NR2 If one double bond is exocyclic to one ring If one exocyclic to two rings Homodiene component

— — —

5 nm 10 nm 39 nm

Electronic Transitions in Organic Molecules

137

Solvents shifts: Above values shifted to longer wavelength in water and to shorter wavelength in less polar solvents. For common solvent the following correction should be applied in computing λmax. Water +8 nm

Methanol 0

CHCl3 –1 nm

Dioxane –5 nm

Diethyl ether –7 nm

Hexane –11 nm

Cyclohexane –11 nm

PROBLEM 8.4 Apply the Fieser and Scott rules to α, β-unsaturated ketone to predict the absorption maximum in compound.

SOLUTION Value for parent α, β-unsaturated ketone For β-substituent (marked a) add 12 nm For γ-substituent (marked b) add 18 nm For two double bonds extending conjugation, add 2 × 30 For homoannular diene component, add 39 nm For exocyclic double bond, add 5 nm Calculated value of λmax

— — — — — — —

215 nm +12 nm +18 nm +60 nm +39 nm +5 nm 349 nm

Three λmax values have been observed for this compound and these are 230 nm, 278 nm and 348 nm. The longest wavelength peak is in excellent agreement with the calculated value.

8.6

SOLVENT EFFECTS ON ELECTRONIC SPECTRA

As with vibrational spectra, the phase of the sample or the solvent used while measuring the spectrum can make a marked difference on electronic spectra. Broadly, there are two extremes; the vapour phase and nonpolar solvents on the one hand and polar and hydroxyl solvents on the other. Let us consider the effect of solvents on π→π* and n→π* transitions one by one.

π→ π π* Transitions When a polar solvent is used, the dipole interaction with the solvent molecules lowers the energy of the excited state more than that of the ground state Fig. 8.6. This is due to the fact that excited states are more polar than ground states. The energy difference between excited and ground state is reduced. This leads to a small red shift of the absorption maximum in polar solvents. Thus the π–π* transition show a red shift of the order of 10–20 nm when solvent is changed from hexane to ethanol.

n→ π π* Transitions Solvent effect for n → π* transitions is opposite to that found for π → π* transitions. Polar solvents cause a shift to lower wavelength (blue shift) relative to nonpolar solvent or the vapour phase. The amount is particularly pronounced in hydroxylic solvents. The lone pair electrons in the nonbonding

138

Elements of Molecular Spectroscopy

orbitals hydrogen bond or otherwise interact strongly with the polar solvent, leading to a lowering in energy of the nonbonding orbital whereas π* orbitals is affected much less (Fig. 8.6). The result is an increase in transition energy on going from a less polar to a more polar solvent. For example, in hexane solution acetone shows absorption maxima of 279 nm whereas in aqueous solution, the absorption maximum is at 264.5 nm.

Fig. 8.6: Solvent effect on n → π* and π → π* transition

PROBLEM 8.5 Two isomeric compounds X and Y have the molecular formula C6H8. Both of them decolourize bromine solution in CCl4 as well as alkaline KMnO4. Both of them give cyclohexane C6H12 on catalytic hydrogenation. X shows an λmax at 256 nm while Y shows no λmax beyond 200 nm. What are the structures of X and Y ? SOLUTION Both X and Y add four hydrogen atoms on catalytic hydrogenation to yield cyclohexane. This shows that both X and Y are cyclic hydrocarbon with two double bonds i.e., each one is a cyclohexatriene. This inference is supported by its decolourization of Br2 dissolved in CCl4 as well as decolourization of alkaline KMnO4 reactions. Their absorption maxima are 256 nm and < 200 nm. From this we can infer that X with higher value of λmax = 256 nm is a conjugated diene (Table 8.4) whereas Y with lower value of λmax is not conjugated. Hence X is 1, 3-cyclohexatriene and Y is 1, 4-hexatriene.

PROBLEM 8.6 Three isomer compounds X, Y and Z with molecular formula (C5H6) absorb three moles of hydrogen on catalytic reduction. Y and Z give a white precipitate when passed through ammoniacal silver nitrate but X does not react with it. X and Y show λmax near 230 nm whereas Z gives no absorption maximum beyond 200 nm. Identify X, Y, and Z.

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139

SOLUTION (i) All the three compounds absorb three moles of hydrogen to give n-pentane. From this we infer that they are unsaturated open chain hydrocarbons each containing one double and one triple bond. (ii) Since Y and Z give a white precipitate with ammoniacal AgNO3, the triple bond is present at C. the end of chain as C — C — C — C (iii) Since X does not give a white precipitate with ammoniacal AgNO3 the triple bond does not lie at the end of chain. C—C C—C—C (iv) Since both X and Y show λmax near 230 nm while Z gives it at λmax < 200 nm, both X and Y have conjugated bonds. (v) Now X has a conjugated system of multiple bonds and the triple bond is not near the end. Its structure should be CH2 CH — C C — CH3 1, Penten-3-yne (vi) Y has a conjugated system of multiple bonds and the triple bond is situated at one end. Its structure should be CH — C CH 3, Penten-1-yne CH3 — CH (vii) In Z the multiple bonds are not conjugated and the triple bond is situated at one end. Its structure should be CH2 CH — CH2 — C CH 1, Penten-4-yne

8.7

COLOUR AND CONSTITUTION

The relationship between colour and constitution was pointed out for the first time in 1876 by the German Chemist Otto Witt. He put forward his chromophore-auxochrome theory of colour and constitution. It may be stated at the outset that electronic structure and behaviour of chromophoreauxochrome as dealt in section 8.2 is the same but only context is different. According to Otto Witt. (i) Colour usually appears in an organic compound when it contained certain unsaturated groups, which should more appropriately be called groups with multiple bonds. These groups with multiple bonds exhibit characteristic absorption in visible and ultraviolet region were called as chromophores. A few examples of important chromophore are:

(ii) The compound containing the chromophoric group is called chromogen. It has been noticed that chromogen containing only one chromophore is usually yellow. Depth of colour increases with number of chromophores.

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Elements of Molecular Spectroscopy

A single C C group as in ethylene does not produce any colour. However, if a number of these groups are present in conjugation a colour develops. For example, CH3—(CH CH)6—CH3 is yellow in colour. (iii) A saturated group with nonbonded electrons when attached to chromophore alters both wavelength and intensity of absorption. Witt described such groups as auxochromes or colour augmenters or deepeners. Some examples of important auxochrome are: —OH —OR —NH2 —NHR and —NR2 Hydroxy

Alkoxy

Amino

Alkylated amino

Auxochromes perform following three functions: (a) They deepen the colour of the chromogen. (b) Their presence is necessary to make the chromogen a dye. (c) Auxochrome though by itself is unable to impart colour but when attached to a chromophore may give a colour. An interesting example of the effect of auxochrome groups may be noted with the following compounds:

The sulphonic and carboxyl group posses little auxochromic property but their presence makes the dye soluble in water. The electron donating power of some common auxochromes decrease in the order. O– > NHCH3 > NH2 > OH > Cl > CH3 > NH+3 = H From experience, several empirical observations have been made. For example, in the case of phenols it has been observed that salts of phenol are more strongly coloured than free phenol from which they are made. Presence of auxochromes in meta position to chromophore does not affect the colour. Thus, according to Witt’s theory of colour and constitution chromogen—A chromophore-bearing compound. Dye-chromogen containing an auxochrome. For example, in p-hydroxyazobenzene a bright red N— is the chromophore and —OH is an auxochrome. Similarly, dye; azobenzene is chromogen, —N

Electronic Transitions in Organic Molecules

141

in picric acid, a yellow dye has trinitrobenzene as chromogen. It has three nitro groups as chromophores and —OH group as auxochrome. Some other terms introduced later are called bathchromic and hypsochromic groups. Bathchromic group: Bathchromic group brings about deepening of colour. This increases absorption intensity of the dye. This leads to absorption on longer wavelength i.e., the colour absorbed will be lying near the lower end or red end of spectrum and shift is called bathchromic shift or red shift. Since this end overlap with infrared radiation, energy levels will be more populated and thus emitting more photons making the colour intense. The substitution of alkyl or aryl group in NH2 has a bathchromic effect. Hypsochromic group: Substitution of an amine, alkoxy hydroxy or halogen groups on carbonyl group brings about lightening of colour. The lighting of colour will help absorption of shorter wavelength i.e., the colour absorbed will be lying near the upper end or blue end therefore, it leads to blue shift. Since blue end overlaps with ultraviolet end, the higher energy of spectrum hence lesser number of electrons will be in the excited state consequently the colour spectrum becomes light. The term deepening of colour in dye chemistry is being used for the following changes in colour: Yellow → Orange → Red → Purple → Blue → Green → Black This is the order of change in complementary colours in the visible spectrum.

8.8

SAMPLE PREPARATION

For most such work, cells of path length 1 cm are used, although 0.1 cm and 10 cm cells are commercially available: for accurate work, the sample and reference cells should be matched in optical path length. Synthetic silica and natural silica are both used, but glass absorbs strongly from 333 cm–1 onwards is not particularly useful for organic work. Matched silica cells are expensive and fragile, and they must be thoroughly cleaned after each use and wiped with soft tissue dipped in methanol. These cells must be stored properly, and the optical surface must never be handled. Electronic spectra are usually measured on very dilute solutions, and the solvent must be transparent within the wavelength range being examined. Even with double-beam operation, solvent absorption will not cancel out between the sample and reference beams. Further, a strongly absorbing solvent will allow so little light to pass through the cell that the photo multiplier will be effectively blind, this will make the amplifier to produce a very noisy background on the recorded spectrum. Standard solutions are prepared in very accurate volumetric flasks, of 0.1 per cent concentration usually. This concentration can be used to record the absorbance of weak bands, but solution may have to be diluted to bring the intense bands on scale. If dilution is necessary it should be done by a known factor (for example dilute 2 ml to 20 ml) so that εmax for strong bands can be calculated. Some compounds do not obey Beer Lambert’s law exactly, so that the spectra recorded at differing concentrations do not match. This effect does indicate the presence of certain molecular absorbing species, which are not being considered. Replotting of the spectrum should always be done to convert the absorbance spectrum into ε or log ε: this involves calculating ε from concentration and absorbance figures manually for a range of wavelengths and then redrawing the spectrum on graph paper. Such a spectrum is free from the noise signals and independent of concentration.

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Elements of Molecular Spectroscopy

PROBLEMS 1. Arrange the following in the increasing order of their UV, λmax. (i) Anthracene (ii) Butadiene (iii) Ethylene (iv) Naphthalene. 2. Compounds A and B have the formation C 5H8 and on hydrogenation yield n-pentane. Their ultraviolet spectra, show the following values of λ max : A 176 nm, B 211 nm, C 215 nm (1 Pentene has λmax 178 nm). (i) What are the likely structures of A, B and C? (ii) What kind of information might enable you to assign specific structures to B and C? 3. For the three diene given below, assign with justification the λ max values from amongst 176 nm, 211 nm and 215 nm. (i)

(ii)

(iii) 4. What wavelength is absorbed by the following chromophores?

O || (i)

C

C

(ii)

C

C—C

C

(iii) —C—C

5. What types of compounds absorb UV radiations? Select compounds in support of your answer from the following list: (i) 1, 3-butadiene (ii) 1-hexane (iii) cyclobutane (iv) nitrobenzene (v) 1, 3-cyclohexadiene (vi) chlorobenzene (vii) chloro-cyclohexane. 6. Identify the two geometrical isomers of stibllene C6C5 CH CH—C6H5 from their absorbance values of 294 nm and 278 nm. 7. Write down the possible electronic transitions of (i) CH2 CH—COCH3 (ii) CH2

CH—CHO

8. Which of the following compounds would be expected to absorb light of the longest and shortest wavelength? Give reasons.

Electronic Transitions in Organic Molecules

143

9. An organic compound X with molecular formula C5H8 has the following isomers: (i) (ii)

(iii)

How UV spectra could be used in establishing the identity of the compound X? Explain with reasons. 10. Use the Woodward rules to predict the expected λmax for the following compounds dissolved in ethanol:

11. Use the Woodward rules in Table 8.5 and predict the expected λmax for π→π* transition in the following compound in ethanol:

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Elements of Molecular Spectroscopy

12. A ketone was known to have one of the following isomeric structures shown and had λ max (in ethanol) at 224 nm. What was it?

13. What is Witt’s theory of colour and constitution? Illustrate its application with reference to the use of phenolphthalein as an indicator in acid base titration. 14. Explain the terms chromophore, chromogen, and auxochrome. What are bathchromic and hypsochromic groups. What does deepening of colour in dye chemistry mean? Why are bathchromic groups said to have a red shift and hypochromic groups a blue shift? 15. Explain why a large number of organic compounds are yellow or orange while only a few are blue or violet. 16. The indicator phenolphthalein is made up by using H2SO4 to condense 2 moles of phenol and 1 mole of sulphobenzoic anhydride.

By eliminating 1 mole of water. What is its structure? How will you account for the fact the phenol sulphaphthalein is yellow in acidic and red in alkaline media (pH 6.8–8.4). 17. Explain why aminoazobenzene is yellow while its acid solution is violet in colour.


❖

CHAPTER

9

Transition Metal Complexes and their Electronic Structures 9.1

CRYSTAL FIELD THEORY

Ligands that can in some way donate electron pairs to metal ions or other acceptors to form the socalled coordinate bond. The crystal field theory (CFT), treats the interaction between the metal ions and the ligands as an electrostatic problem in which ligand atoms are represented as point charges or point dipoles. Bond in CFT is treated as purely ionic in nature. It does provide a very simple and easy way, numerically, of treating many aspects of the electronic structure of complex. The basic difficulty with CFT is that it takes no account of the partly covalent nature of the metal-ligand bonds, and, therefore, whatever effects and phenomenon stem directly from covalency are entirely inexplicable in simple CFT. At the opposite extreme, the metal ligand interaction can be described in terms of molecular orbitals formed by overlap of ligands and metal orbitals. However, it does not provide numerical results in such an easy way. Therefore, a kind of modified CFT was developed, what is often called Ligand Field Theory (LFT). z

– y – –

n+1

M

–

x

–

–

Fig. 9.1: Sketch showing six negative charges arranged octachedrally around a central Mn+1 ion with a set of Cartesian axes for reference

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Elements of Molecular Spectroscopy

Consider a metal ion Mn+1, lying at the centre of an octahedral set of point charges as shown in Fig. 9.1. Suppose this metal ion has a single d electron outside the closed shell, such an ion might be Ti3+ or V4+. In the free ion this d electron would have an equal probability of being in any one of the 5d-orbitals, since all are equivalent. Now in a complex, however, all the d-orbtials are not equivalent. Some are concentrated in regions of space closer to the negative ions than the others, and the electron will obviously prefer to occupy the orbital in which it can get as far as possible from the negative charges. By examining the shapes of the d-orbitals (Fig. 9.2) and comparing them with Fig. 9.1 we see that both dz2 and dx2–y2 orbitals have lobes that are heavily concentrated along z and x-y axis thus in the vicinity of the charges, whereas the dxy, dyz and dzx orbitals have lobes that project between the charges. It can also be seen that each of the three orbitals in the latter group, namely dxy, dyz and dzx are equally favourable for the electron; these three orbitals have entirely equivalent environment in octahedral complex (Fig. 9.2). The two relatively unfavourable dz2, and dx2–y2 are also equivalent in relation to the octahedral distribution of charges.

Fig. 9.2: Sketches showing the distribution of electron density in the 5d-orbitals with respect to a set of six octahedrally arranged negative charges

Thus, in the octahedral environment of six negative charges, the metal ion has two kinds of d-orbitals: three of one kind, equivalent to one another and conventionally labelled t2g, and two of another kind, equivalent to each other, conventionally labelled eg. Further, result may be expressed in an energy level diagram as shown in Fig. 9.3. The energy difference between eg and the t2g of orbitals as ∆o, where the subscript o stands for octahedral. Some books also represent it as 10Dq. The additional feature of Fig. 9.3—the indication that the e g levels like 3/5∆ o or 6Dq lie above and the t 2g levels like 2/5∆o or 4Dq lie below the energy of the unsplit d-orbitals will now be explained. Let us suppose that a cation containing 10d electrons, two in each of the d-orbitals, is first placed at the centre of hollow sphere whose radius is equal to the M–X internuclear distance, and that charge of total quantity 6e is spread uniformly over the sphere. In this spherically symmetric environment, the d-orbitals are still fivefold degenerate. The entire energy of the system i.e., the metal ion and the charged sphere has a definite value. Now suppose that total charge on sphere is caused to collect into six discrete point charges each of the magnitude e and each lying at the vertex of an octahedron but still on the surface of sphere. Merely redistributing the negative charge over the surface of sphere in this manner cannot alter total energy of the

Transition Metal Complexes and their Electronic Structures

147

system when the metal ion consists entirely of spherically symmetrical electron shells, and yet we have already seen that as a result of these redistribution, electrons in eg orbitals now have higher energy than those in t2g orbitals. It must, therefore, be that the total energy of four electrons equals the total decrease in energy of the six t2g electrons. This then implies that the rise in the energy of the eg orbitals are 3/5 times the drop in energy of the t2g orbitals, which is equivalent to the 3/5: 2/5 ratios shown.

Fig. 9.3: Energy-level diagrams showing the splitting of a set of d-orbitals by octahedral and tetrahedral electrostatic crystal fields.

This pattern of splitting, in which the algebraic sum of all energy shifts of all orbitals is zero is said to “preserve the centre of gravity of the set of levels”. This centre of gravity rule is quite general for any splitting pattern when the forces are purely electrostatic and where the set of levels being split is well removed in energy from all other sets, which they might be able to interact. As Ti (III) ion has only one electron, then ∆o for the ligand can be obtained from its spectra in their presence. The single electron present in the d-orbital will occupy the lowest energy level available giving the ground state configuration t2g . Absorption of light by complex will excite the electron to the next energy level giving the excited state configuration of eg so that electron absorbs radiation corresponding to energy difference between t2g and eg levels (= ∆o). As the absorption spectrum of [Ti (H2O)6]3+ has an absorption maximum at ν = 20400 cm–1, ∆o for water as ligand is given by: Energy = hν = h ν c = (6.624 × 10–34 Js molecule) (20400 cm–1) (3 × 1010 cm s–1) = 4.073 × 10–19 J molecule–1 = (4.073 × 10–19 × J molecule–1) (6.023 × 1023 molecule mol–1) = 244.4 kJ mol–1. Similarly, ∆o for fluoride ions in ReF6 (also a d1 species) corresponds to 325000 cm–1 giving a value of 338 kJ mol–1. These are typical values for ∆o and are of the same order of magnitude as the energy of a chemical bond. The measurement of ∆o is carried by observing electronic spectra of complex in solution; lowest frequency at which the absorption maximum is observed corresponds to t2g to eg transition and thus ∆o. The value of absorption maximum of hydrate complexes of d 4 , d 5 , d 6 and d 7 are given in Table 9.1. However, ∆o cannot be easily obtained from the absorption spectra of the complexes because of the (a) interelectronic repulsion, (b) overlapping of the electronic transitions. Since in these transitions there is no change in number of nodal planes these transitions are not allowed and hence have low intensity.

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Elements of Molecular Spectroscopy

Table 9.1: Absorption maximum of some of the hydrate complexes

Cr2+

Mn3+

Mn2+

Fe3+

Fe2+

Co2+

dn

d4

d4

d5

d5

d6

d7

Configuration

t23g e1g

t23g e1g

t 23g eg2

t 23g eg2

t 23g eg2

t 23g eg2

∆o in cm–1

13900

21000

8600

13700

10400

9300

∆o in kJ mol–1

166.37

251.35

102.92

164.0

124.47

111.31

Ion

9.2 CRYSTAL FIELD STABILIZATION ENERGY (CFSE) FOR WEAK AND STRONG OCTAHEDRAL FIELDS AND PAIRING ENERGIES The electronic configuration of the metal ions, and hence the magnetic properties of the complexes can easily be understood from the d-orbital splitting in the ligand fields. The electronic configuration of the ion will be given by the following considerations: • The electron occupies the orbitals of lowest energy in the ground state.

• Due to reduced interelectronic repulsion in the different orbitals in a degenerate level, Hund’s rule is obeyed. • The quantum mechanical exchange energy for parallel spins is higher than that for opposite spins. • If pairing of electrons take place; the energy of the system will be raised by P, the pairing energy of the system. Now, the energy of triply degenerate t2g level relative to the unperturbed d-obitals is – 4Dq whereas that of the eg orbitals is + 6Dq. The Crystal Field Stabilization Energy (CFSE) for d 1, d 2 and d 3 will therefore be – 4Dq, – 8Dq and – 12Dq respectively as the electrons will remain unpaired (Hund’s rule) and enter different degenerate levels.

For a d4 ion: Two possibilities exist (a) The fourth electron will go to the eg level giving the configuration

t23g e1g 4 t 2g

CFSE = – 4Dq × 3 + 6Dq × 1 = – 6 Dq CFSE = – 4Dq × 4 + P = –16Dq + P

The pairing energy P includes the effect of all the factors involved except the crystal field splitting. Thus the t23g e1g configuration having four unpaired electrons (high spin) will be stabler (at a lower 4 configuration having two unpaired electrons (low spin) if energy) than the t 2g – 6Dq < –16Dq + P or 10Dq < P i.e., if the 10Dq = ∆o for the ligand is lower in magnitude than pairing energy P, pairing will take place. Hence, ligands producing a weak field (< P) will give a high spin complex, whereas the ligands giving strong field (> P) will result in the formation of low spin complexes. Thus, [Mn (H2O)6]3+ is a high spin complex ( t23g e1g ) whereas [Mn (CN)6]3– is a low spin complex t2g as the 10Dq for H2O is less than that for CN– and is more than the pairing energy for d4 Mn3+ ion.

Transition Metal Complexes and their Electronic Structures

149

The high spin states are stabilized by reduced interelectronic repulsions and increased exchange energies from the interchange of positions by the electron with parallel spins.

For a d5 ion: Two possibilities exist (a) The fourth and fifth electrons go to eg level giving configuration t 23g eg2 for weak field ligands. CFSE = – 4Dq × 3 + 6Dq × 2 = 0 with 10Dq < P 5 and (b) Electron pairing may take place in the t2g level giving the configuration t 2g CFSE = – 4Dq × 5 + 2P = 20Dq + 2P having 10Dq > P

For a d6 ion In a weak field case the sixth electron has to undergo pairing with one of the t2g electron as all the d-orbitals are singly occupied with configuration t24g eg2 . CFSE = – 4Dq × 4 + 6Dq × 2 + P = – 4Dq + P 6 In strong field ligands will give t 2g as electronic configuration and CFSE = – 4Dq × 6 + 3P = – 24Dq + 3P It should be kept in mind that P is not a constant but changes from ion to ion and ligand to ligand. For a d7 ion (a) For a high spin, configuration will be t25g eg2 with CFSE = – 4Dq × 5 + 6Dq × 2 + 2P = – 8Dq + 2P with 10Dq > P (b) For a low spin, configuration will be t26g e1g with CFSE = – 4Dq × 5 + 6Dq × 2 + 3P = – 8Dq + 2P with 10Dq < P These results are summarized in Table 9.2. Table 9.2: CFSE and electronic arrangements in octahedral complex

Number of d-electrons

Arrangement in weak field ligands CFSE

t2g

↑

–4Dq

↑

– 4Dq

↑↑

–8Dq

↑↑

–8Dq

↑↑↑

–12Dq

↑↑↑

–12Dq

–6Dq

↑↓↑↑

–16Dq + P

↑↓↑↓↑

–20Dq + 2P

– 4Dq + P

↑↓↑↓↑↓

–24Dq + 3P

↑↑

– 8Dq + 2P

↑↓↑↓↑↓

↑

–18Dq + 3P

↑↓↑↓↑↓

↑↑

–12Dq + 3P

↑↓↑↓↑↓

↑↑

–12Dq + 3P

d9

↑↓↑↓↑↓

↑↓↑

– 6Dq + 4P

↑↓↑↓↑↓

↑↓↑

– 6Dq + 4P

d10

↑↓↑↓↑↓

↑↓↑↓

↑↓↑↓↑↓

↑↓↑↓

t2g

eg

Arrangement in strong field ligands

d1 d2 d3 d4

↑↑↑

↑

d5

↑↑↑

↑↑

d6

↑↓↑↑

↑↑

d7

↑↓↑↓↑

d8

0

0

eg

CFSE

0

150

Elements of Molecular Spectroscopy

For a metal ion, a critical value of 10Dq exists which equals P, the pairing energy of the electrons in the ion. If the 10Dq for a ligand is more than P, low spin complexes will be formed otherwise high spin complexes will result. The pairing energy P depends on inherent interelectronic repulsion between the electrons occupying the same orbitals (these decreases with size of the orbitals) 3d > 4d > 5d due to decreasing electron density but remains same in a period. Approximate theoretical estimates of the mean pairing energies for the relevant ions of the first transition series have been made from spectroscopic data. In Table 9.3 these energies along with ∆o for some complexes are listed. It will be seen that the theory developed above affords correct predictions in all cases. It should be further noted that the mean pairing energies vary irregularly from one metal ion to another as do the values of ∆o for a given set of ligands. Table 9.3: Crystal field splitting, ∆o and electron pairing energies, P for some transition metal ions (energies in cm–1)

Configuration

Ion

P

Ligands

∆o

Spin state Predicted

Observed

d4

Cr2+ Mn3+

23,500 28,000

6H2O 6H2O

13900 21000

High High

High High

d5

Mn2+ Fe3+

25,500 30,000

6H2O 6H2O

7500 13700

High High

High High

d6

Fe2+

17,600

Co3+

21,000

6H2O 6CN– 6F– 6NH3

10400 32850 13000 23000

High Low High Low

High Low High Low

Co2+

22,500

6H2O 6NH3

9200 11000

High High

High High

d7

Table 9.3 shows the d 5 system should be exceptionally stable in their high spin states, whereas the should be exceptionally stable in their low spin states. These exceptions are in excellent agreement with the experimental facts. d 6 system

9.3

FACTORS AFFECTING MAGNITUDE OF 10DG OR ∆ O

Some of the factors that affect the magnitude of 10Dq, the legend field splitting are as follows: (i) Charge on the ion: The increased charge on the metal ion tends to pull the negatively charged ligand closer to itself, making the interaction to become more strong thereby increasing the 10Dq values. Theoretically a change from +2 to +3 state should increase the 10Dq by 50 per cent. Due to the increased interelectronic repulsions at closer distances the actual increase is about 40 per cent only. (ii) As will be discussed in section 9.4, the 10Dq values of tetrahedral complexes are lower than the values of the octahedral complexes, which in turn is less than the values of planar complexes.

Transition Metal Complexes and their Electronic Structures

151

(iii) The spectrochemical series: It has been found by experimental study of spectra of a large number of complexes containing various metal ions and various ligands, that ligands may be arranged in a series according to their capacity to cause d-orbital splitting. This series for the more common ligands is Ligand : f :

I– 0.7 Ac 0.94 phen 1.34

< Br– 0.72 < H2 O 1.00 < NO–2 1.4

< SCN– 0.73 < NCS– 1.02 < CN– 1.7

< Cl– 0.78 < NC– 1.15 < CO 1.7

< NO–3 0.83 < py 1.23

< F– 0.90 < NH3 1.25

< OH– < C2 O–4 < 0.94 0.99 < en < bpy < 1.28 1.33

Jorgensen (1962) has given a field factor f for the ligands taking f for water as 1.00. Standard f values range from 0.7 to 1.7 for the weak and the strong field ligands. The idea of this series is that the d-orbital splitting and hence the relative frequencies of visible absorption bands for two complexes containing the same metal ion but different ligands can be predicted from the above series whatever the particular metal ion may be. Naturally, one cannot expect such a simple and useful rules to be universally applicable. The following qualifications must be remembered in applying it: 1. The series is based upon data for metal ions in common oxidation states. Because the nature of the metal-ligand interaction in an unusually high or unusually low oxidation state of metal may be in certain respects qualitatively different from that of the metal in a normal oxidation state, striking violations of the order shown may occur for complexes in usual oxidation state. 2. Even in metals in their normal oxidation states inversion of order of adjacent or nearly adjacent members of the series are sometimes found. The spectrochemical series is an experimentally determined series and incorporates all factors like the d-orbital splitting, π-bond formation etc. and does not depend on CFSE only. 3. Nature of metal ion: Due to the spin pairing energies, higher coordination number and tendencies toward higher coordination, the d-orbital splitting are in the order 3d < 4d < 5d, the increase being about 40–50% from 3d to 4d and 20–30% from 4d to 5d. In a period, however, the 10Dq values do not change much (7800 cm–1 for Mn2+ and 11000 cm–1 for Cr+2. Jorgensen (1962) has also defined a g factor for the metal ions such that (10Dq)octahedra = f in cm −1 ligand

×

g ×

1000

metal ion

which is in the order Mn2+ g : 8.0

< Ni2+ 8.7

< Co2+ 9.0

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Appendix I

Photochemistry Photochemistry concerns itself with the study of the effect of visible and near ultraviolet radiation on chemical reactions and with the rates and mechanisms by which reactions initiated by light proceed. Ordinary or thermal reactions are initiated by activation brought about through molecular collisions. Its characteristics of all such reactions that can occur only when the reactions are accompanied by a free energy decrease. If a free energy increase is involved, no reaction is possible. However, thermal activation is not the only means by which the energy of atoms and molecules can be raised sufficiently to cause reaction. We have seen that atoms and molecules can absorb radiation. In fact, with absorption of a sufficiently large quantum of radiant energy a molecule may be ruptured. Such absorption of light by an atom or molecule leads to its excitation; and if the activation is sufficiently great, chemical reaction may result. It is in this manner that absorbed light can affect the rate of a chemical reaction and frequently bring about chemical changes under conditions where thermal activation alone would not be effective. The rate of thermal uncatalyzed reactions at any fixed concentration can be varied only by change of temperature. With photochemical reactions, however, the rate can be controlled also by varying the intensity of the light used for irradiation. In the latter reactions the number of molecules activated depends on intensity of the light and hence the concentration of activated molecules will be proportional to the light intensity to which the reactant is exposed. With sufficiently intense light sources it is thus possible to attain reactions at ordinary temperatures, which would not result thermally except at considerably elevated temperatures. Again, since photochemical activation does not depend to any degree of temperature, the rate of activation is usually temperature independent. Any increase in the rate of a photochemical reaction with temperature is primarily due to thermal reactions, which follow the activation process. Further, not only spontaneous reactions can be made to proceed photochemically, but also many reactions are attained by a free energy increase. In the spontaneous reactions the light acts to speed up the thermal reactions, on the other hand, the radiant energy supplied to the system may increase the free energy of the reactants sufficiently to make the ∆G negative. An outstanding example of such a process is photosynthesis. Under the action of sunlight and promoted by chlorophyll, carbon dioxide and water are combined in plants into complex carbohydrate and oxygen. On removal of the light products oxidize slowly back to carbon dioxide and water, releasing at the same time the energy accumulated from the sun’s radiation.

200

Elements of Molecular Spectroscopy

1. BEER-LAMBERT’S LAW Beer-Lambert’s law gives the variation of intensity of light absorbed by molecules at a particular wavelength Absorbance = A = log

Io = εcl I

...(1)

where Io = intensity of light of wavelength λ incident on the front of a column of length l in cm of a single absorbing species of concentration c in moles per litre, I is intensity of the light transmitted through the column of material, ε is called molar extinction coefficient in L Mole–1. It is constant for a given species at wavelength λ, log Io/I is called the absorbance (or molar absorbance) to the base 10. Equation (1) can also be written in terms of the natural logarithms. ...(2) ln Io/I = αcl where α = 2.303ε. Unfortunately, in literature the symbol ε and α often appear without specifying that they are to the base 10 or e. The fraction of light absorbed Ia/Io is given by

I Ia − εcl = 1 − = 1 − 10 Io I

...(3)

which hold good for all concentrations. At low concentrations of absorber the fraction is directly proportional to concentrations of absorbing species since ey = 1 – y i.e., Ia = 1 − 10 − εcl = 1 − exp ( −2.303 εcl ) I

...(4) ≈ 2.303 εcl This relationships cannot hold at higher concentrations when it approaches 1, indicating that at higher concentrations all of the light is absorbed independent of absorber concentrations. The most common units used in gas phase photochemistry for concentration are number of molecules cm–3 symbolized by N with natural logarithms. The gas phase absorption coefficient designated, σ, is then in units of cm2 molecule–1, σ is also known as absorption cross-section. The Beer-Lambert law becomes ...(5) ln Io / I = σ Nl A second widely used set of unit for gaseous species is concentrations as pressure, in atmosphere, so that ln Io / I = k pl ...(6) Since pressure depends upon temperature, a reference temperature 270 or 298 K must be specified. For a mixture of absorbing molecules, where each molecule has an extinction coefficient ε at concentration ci (i = 1, 2,......) the Beer-Lambert law becomes

I − ( ε c + ε c + ε c ...)l = 10 1 1 2 2 3 3 Io

...(7)

Appendix I

201

2. GROTTUS-DRAPER LAW The first law of photochemistry, called the Grottus Draper law, states: Only the light, which is absorbed by a molecule, can be effective in producing photochemical changes in the molecule. When light falls over a system one of the following three changes may happen: 1. The system is perfectly transparent to light of wavelength λ, means this light does not interact with molecules and cannot lead to a photochemical reaction. 2. The system is translucent to the light of wavelength λ, means this light cannot pass through the medium, rather it is scattered thus interaction between light and molecules is quite weak so photochemical reactions may not be possible. 3. The system absorbs the light of wavelength λ, means this light interacts with molecules and thus a photochemical reaction may take place. Therefore, to assess the potential for photochemically induced changes, it is essential to know the absorption spectra of reactants. In order to fully characterize the reaction mechanism, one must also know the absorption spectra of the intermediate reaction products.

3. STARK-EINSTEIN LAW A second important law of photochemistry, known as the Stark-Einstein law, follows directly from the particle behaviour of electromagnetic radiations. Absorption of radiation is a one photon process. Absorption of one photon excites one atom or molecule in primary (initialing) step and all subsequent physical and chemical reactions follow from this excited species i.e.,

hc λ when hν is a quantum of light called photon and h is Planck’s constant 6.62 × 10–34 Js. Since chemists generally deal experimentally with moles of material, a convenient unit is a mole of photons defined as l Einstein. The energy of Einstein of wavelength λ in nm is ∈ = hν =

E = 6.02 × 1023 hν = If λ is in angstrom

1.196 × 10 5 kJ einstein . λ

1.196 × 10 6 kJ einstein . E= λ

Another unit used in photochemistry to express the energy is electron volt; 1 eV = 96.46 kJ mole–1 and λ nm thus E=

1.24 × 10 3 eV λ

Since absorption of photon of wavelength λ leads to dissociation, the bond dissociation energy may correspond to absorption peaks. Consider NO molecule, which shows absorption peak at 191 nm, and has bond dissociation energy of 6.496 eV. This corresponds to a wavelength λ i.e., 6.496 =

1.24 × 10 3 or λ = 191 nm λ

202

Elements of Molecular Spectroscopy

Thus, absorption of one photon of light of wavelength 191 nm by one molecule of nitric oxide will give it sufficient energy to dissociate. In other words, NO molecule will be photochemically active to light of wavelength 191 nm.

4. PRIMARY AND SECONDARY EFFECTS OF LIGHT ABSORPTION Absorption of radiation energy may lead to dissociation of the absorbing molecule. In fact, in most of photochemical reactions involving molecules, the primary step is usually dissociation of some molecules into atoms, simple molecules or free radicals, which by further interaction either with each other or with different molecules continue the reaction sequence. The primary photochemical stage is dissociation. The secondary reaction proceeds by thermal means. Consider oxygen molecule show absorption, known as Herzberg continuum or Schumann Range lies in the range 147–130 nm. With the absorption of radiation of this wavelength, oxygen molecule dissociates to give oxygen atoms. Thus, the primary photochemical reaction is hν O2 → O+O

These oxygen atoms so formed react with other molecules to give products through thermal reactions. These are secondary effect

→ O 3 (Secondary effect) O2 + O  Since UV radiation of wavelength 147–130 nm is absent in sunlight available on the earth’s surface, hence no formation of ozone takes place. However, at a height of 32 km, this UV radiation is available that leads to formation of ozone. The ozone molecules so formed absorbs this UV radiation and thus is not available on the surface of earth. Another example is that of chlorofluorocarbons (CFC). All chlorofluorocarbons show absorption in the range 160–280 nm. This UV radiation is available near ozone layer of earth atmosphere. Light spectrum available on the surface of earth is perfectly transparent to chlorofluorocarbons and thus behaves as photochemically inert. These do not dissolve in water and thus are not removed by rain. Slowly these molecules rise and reach the ozone layer. Their photons of 160–280 nm are available, absorption of which lead to primary photochemical reaction. hν CFC—Cl → CFC + Cl This chlorine atom reacts with ozone molecule as follows:

 Secondary Cl + O 3  → ClO + O 2  O |→ Cl + O  reaction 2 The net reaction

Cl O3 + O → 2O2

1 atom of chlorine decomposes 10000 ozone molecules that lead to ozone hole. Why in CFC, chlorine is removed and not fluorine? The C—Cl bond length = 18 nm, C—F bond length = 12–14 nm. Larger the bond length lesser is dissociation energy. Since C—Cl bond energy is less than C—F bond energy thus C—Cl bond breaks. The atmosphere becomes thinner and colder with increasing altitudes for the first nine to sixteen km above ground. Almost all of the water vapors are removed at lower altitudes. The gases undergo

Appendix I

203

rapid vertical mixing and are periodically cleaned by rains including all oxides of nitrogen. However, as a result of denitrification of chemical fertilizers a small amount of N2O formation takes place. N2O molecule is not a very reactive molecule, neither it is removed from atmosphere by rains. Hence, it can survive unchanged in the atmosphere for decades till it reaches near ozone layer. N2O is a linear asymmetrical molecule. It shows absorption starting at 306 nm with flat maxima near 290 nm. It is followed by second and third continuum in 280–260 nm. The photons of this wavelength are available near ozone layer and thus N2O become photochemically active. Absorption of photons of this wavelength triggers the primary photochemical process hν N 2 O → N2 + O

(95%)

(5%) → NO + N Primary photochemical reactions convert most of N2O to Nitrogen but about 5% becomes NO. These NO molecules so formed attacks the ozone molecule very efficiently to produce molecular oxygen and NO2 which then reacts with atomic oxygen to form NO again i.e., NO + O 3  → NO 2 + O 2 NO 2 + O  → NO + O 2

}

Secondary reactions

Both reactions have occurred but NO molecule is still there. This cycle continue till NO molecule is removed photochemically by a photon of wavelength 191 nm. Another interesting molecule is NO2. Ignition of petrol to run vehicles also produces NO2 as pollutant. Presence of even a small amount of NO2 in atmosphere is sufficient to trigger the complex series of reactions, to produce photochemical smog through the visible light in the wavelength 300–420 nm. NO2 is a light brown gas and the primary photochemical reaction that takes place is hν

NO 2  → NO + O λ < 420 nm

The oxygen atom so formed reacts with oxygen molecule O + O2  → O 3

Secondary reaction

Rate of formation of ozone is widely used as a measure of smog formation. A fraction of the NO emitted in the air parcel reacts with ozone to regenerate NO2. → NO 2 + O 2 O3 + NO 

As a consequence of photochemical reaction peak value of ozone is reached in the early afternoon. As ozone is a very reactive molecule, it reacts with many other pollutants producing a variety of eye and throat irritants, aerosols. These molecules reduce visibility by producing smog and other irritating or destructive molecules. Pollution of atmosphere with NO2 has now spread to all metropolitan cities like Delhi and Mumbai. However, near the exhaust of vehicles if catalytic converter is fixed, NO2 gets reduced to nitrogen and oxygen and thus removed from the exhaust gases. Until now we had been talking of gas reactions. Many substances undergo photochemical reactions in liquid state. Again the reaction is initiated by Stark-Einstein law by direct light absorption on the part of reactants. However, it may be anticipated that quantum efficiency of these reactions will be less than

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for the same reaction in the gas phase. The reason for this is that in the liquid state an active molecule may readily be deactivated by frequent collisions with other molecules. Furthermore, because of the very short mean free path in the liquid phase free radicals or atoms when formed photochemically will tend to recombine before they have a chance to get very far from each other. The net effect of these processes will be to keep the quantum yield relatively low. In fact, only those reactions may be expected to proceed to any extent for which the primary products of the photochemical act are relatively stable particles. Otherwise the active intermediates will tend to recombine with the solvent and thereby keep the yield low.

5. QUANTUM EFFICIENCY The photochemical equivalence law applies only to the absorption or primary photochemical process. As a result of the primary absorption only one molecule decomposes, and the products enter no further reaction, the number of molecules reacting will be equal to the number of photons absorbed. More frequently, however, a molecule activated photochemically initiates a sequence of thermal reactions as a result of which several or many reactant molecules may undergo chemical change. Under such conditions there will be no reaction between reacting molecules and the number of energy quanta absorbed. To express the relation between the number of molecules entering reaction and the number of quanta absorbed, the quantum yield or efficiency of a process φ is introduced. Quantum yield, φ is defined as φ=

=

Number of molecules reacting in a given time Number of quanta of light absorbed in the same time Number of molecules reacting in a given time Number of Einstein’ s of light absorbed in the same time

This gives the number of molecules observed to undergo chemical transformation per quantum of absorbed energy. The quantum efficiency of a reaction may vary from almost zero to about 106. Nevertheless, no matter how large or small φ may be, the Einstein equivalence law holds good.

6. EXPERIMENTAL STUDY OF PHOTOCHEMICAL REACTION To measure the rate of a photochemical reaction, it is necessary to irradiate a reaction mixture with light of a selected wavelength and to observe the manner in which the concentration of reactants or products varies with time. For this purpose some arrangement such as that shown in Fig. A-I.1 is required. In this diagram A is a light source emitting radiation of suitable intensity in the desired spectral range. To select radiation of only a single wavelength or to confine the radiation to a narrow band, the light is passed through the lens B into a monochromator or filter at C. From C the light enters the cell D immersed in a thermostat and containing the reaction mixture. Finally, the light transmitted through D falls on some suitable recorder E, where its intensity is determined. The light sources used depend on the spectral range in which radiation is desired and include filament lamps, carbon and metal arcs, and various gas discharge tubes. The reaction cells may be glass or quartz vessels usually with optically plane windows for entrance and exit of light. In some instances, metal cells have been used with windows cemented to the ends. Glass can be used only in

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the visible spectral range. Below 350 nm not only the cells but also any other optical parts through which the light passes must be of quartz. With gases no provision need be made for stirring; solutions, however, must be agitated.

Fig. A-I.1: Apparatus for study of photochemical reactions (schematic)

The radiant energy is measured with some form of thermopile or actinometer. The thermopile is essentially a multijunction thermocouple consisting usually of silver and bismuth soldered to metal strips blackened with lamp, platinum or bismuth black. The radiation falling on the blackened strips is absorbed almost completely and nonselective, and is converted into heat. The heat thus generated raises the temperature of the hot junctions above the cold, and the current generated thereby is measured. The current produced is proportional to the energy absorbed and this, in turn, depends on the intensity of the incident light and the area exposed to it. Thermopiles are calibrated with standard light sources. Photocells may be used in place of thermopiles. Still other devices are chemical actinometers, which are merely gas mixtures or solutions sensitive to light. When radiation impinges upon these, a chemical reaction ensures whose extent is determined by the amount of energy absorbed. The most common of these is the uranyl oxalate actinometer, consisting of 0.05 molar oxalic acid and 0.01 molar uranyl sulphate (UO2SO4) in water. Under the action of light following reaction takes place: UO 22 + + hν  →[UO 22+ ] * Uranyl ion

[UO 22+ ] * +

COOH

 →[UO 22+ ] + H 2O + CO 2 + CO

COOH Extent of reaction can be ascertained by titrating the remaining oxalic acid with permanganate solution. For best results this actinometer should be calibrated against a thermopile in the spectral range in which it is to be employed. When this is not possible, the quantum yields recorded in the literature may be utilized. This actinometer is applicable to radiations lying between 200 and 500 nm. To obtain the total radiant energy in unit time the procedure is as follows: First, the empty cell or the cell filled with solvent alone in the case of solutions, is interposed in the light beam, and a reading is taken. This gives the total energy incident in a given time upon the system. Next, the cell with the reactants is substituted and the reading is again taken, which gives now the total energy transmitted. The difference between these readings is the total energy absorbed by the reacting mixture in the given time.

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The rate of the chemical reaction taking place in the system is ascertained in the usual manner. For this purpose the change in some physical property can be followed or samples can be removed periodically from the cell and analyzed. It is thus possible to collect data on the rate of the chemical reaction and the light intensity, from which the rate law and quantum yield may be reduced.

PROBLEM The quantum efficiency after the decomposition of uranyl oxalate in an experiment was found to be 0.510 at 300 nm. When light was passed through an empty cell, decomposition of 6.201 × 10–3 mole of the oxalate took place in 2 hours. When the cell contained acetone, and irradiation continued for 10 hours, 1.4 × 10–3 mole of acetone were decomposed and light not absorbed by acetone decomposed 2.631 × 10–2 mole of the oxalate. What is the quantum efficiency for the acetone decomposition? SOLUTION Amount of oxalate decomposed in 2 hrs in the empty cell = 6.201 × 10–3 mole. Amount of oxalate decomposed in 10 hrs in the empty cell = 5 × 6.201 × 103 = 3.1005 × 10–2 mole. The number of photons incident in 10 hrs =

(3.1005 × 10 −2 mole )(6.022 × 10 23 mol −1 ) = 3.276 × 10 22 0.570

Amount of oxalate decomposed after decomposing acetone = 2.631 × 10–2 mole Number of photons not absorbed by acetone in 10 hours =

(2.631× 10−2 )(6.022 × 10 23 ) = 2.78 × 1022 0.570

Hence, number of photons absorbed by acetone = 3.276 × 1022 – 2.78 × 1022 = 4.96 × 1021 Number of molecules of acetone decomposed = (1.4 × 10–3 mole) (6.022 × 1023 mole–1) = 8.43 × 1020 Hence, the quantum efficiency = φ =

8.43 × 10 20 4.91 × 10 21

= 0.17

7. HIGH AND LOW QUANTUM YIELDS Case I: High Quantum Efficiency, φ > 1 In reactions where number of reactant molecules undergoing chemical change per photon of radiation absorbed is more than one, extra energy absorbed by the reactant molecules is transferred to its product, which may take part in secondary reactions. For example, decomposition of two molecules of HI by absorption of only one photon of light has the mechanism as follows: → H + I (i) [Primary reaction] HI + hν  → H2 + I (ii) [Secondary reaction] H + HI  → I2 I+I  or

→ H2 + I2 2HI + hν 

(iii)

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In the primary process only one molecule of HI is decomposed to its elements by absorbing one photon of light and this is in accordance to Einstein law of photochemical equivalence. Subsequently, the obtained H atom reacts with another HI molecule yielding an atom of iodine. The iodine atoms obtained in steps (i) and (ii) combine to give a molecule of iodine. Thus, the overall effect is that 2 molecules of HI undergo chemical change by absorption of one photon of light. Hence φ = 2. There are many photochemical reactions in which a large number of molecules may undergo chemical change by the absorption of one photon of light. In such reaction the quantum yield has been found to vary from 104 to 105. At moderate gas pressures and with comparable proportion of chlorine and hydrogen photochemical reaction takes place.

Cl 2 + hí  → 2Cl

[Primary reaction]

 Cl + H 2  → HCl + H   H + Cl 2  → HCl + Cl Secondary reactions  1 Cl (at walls)  → Cl 2  2  Here the chains acquire great length and as many as 106 molecules of hydrogen chloride are obtained per quanta of light. In all combinations of atoms to form stable molecules considerable energy is evolved. Unless this energy is removed, the resulting molecule cannot be stable. Consequently, in all such reactions it is necessary to assume either that a third body is involved in the collisions which carries off the energy, or that the reaction occurs at the walls. The third body may be a molecule of the reactants, the products, or some other neutral impurity present in the system does this job.

Case II: Low Quantum Efficiency, φ < 1. In case when φ < 1 (i) The primary products may combine to form the reactants. (ii) The activated molecules may be deactivated by collisions, by fluorescence or by internal rearrangement before they start to react. (iii) Some of the activated molecules may not acquire sufficient energy for the chemical change. For example, in the formation of HBr from H2 and Br2, the quantum efficiency is 0.02 only.

Br2 + hí  → 2Br

[Primary reaction]

Br + H 2  → HBr + H   → HBr + Br  H + Br2   Secondary reactions H + HBr  → H 2 Br   Br + Br  → Br2  Such reactions show the variation of the quantum yield with temperature and are accountable by changes in the magnitude of reaction rate constants of secondary reaction with temperature. The quantum yield increased with temperature up to 2 near 200°C, and was still higher at higher temperatures.

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The photochemical decomposition of ammonia proceeds quantitatively according to reaction.

→ N2 (g) + 3H2 (g) 2NH3 (g)  With an average quantum yield of 0.25. The following mechanism has been proposed to explain their results → NH2H NH3 + hν 

[Primary reaction]

   H+H → H 2  NH 2 + NH 2  → N 2 H 4  Secondary reactions  N 2H4 + H  → NH 3 + NH 2   NH 2 + NH 2  → N 2 + 3H 2   NH 2 + H  → NH 3

From this mechanism it can be shown that quantum yield should be small and that it should vary with pressure.

8. FLUORESCENE See Chapter 7, Section 7.7—Dissipation of Energy by Excited Molecules.

9. PHOSPHORESCENCE See Chapter 7, Section 7.8—Phosphorescence.

10. POSSIBLE FATES OF AN ELECTRONICALLY EXCITED MOLECULE— JABLONSKI DIAGRAM Once a molecule is excited into an electronically excited state by absorption of photon, it can undergo a number of primary processes. Photochemical processes are those in which the excited species dissociates, isomerizes, rearranges or react with another molecules. Photophysical processes include radiative transition in which excited molecule emits light in the form of fluorescence or phosphorescence and returns to the ground state, intramolecular nonradiative transitions in which some of all the energy of the absorbed photon ultimately gets converted to heat. These photophysical processes are often displayed in the form of the Jablonski type energy level diagram shown in Fig. A-I.2. Common singlet states are labelled as S0, S1, and S2 and so on and triplets are labelled as T1, T2, T3, and so on in order of increasing energy. Vibrational states are shown as being approximately equally spaced horizontal lines. Radiative transitions, for example, fluorescence (F) and phosphorescene (P) are shown as solid lines and nonradiative transitions as wavy lines. Vertical distances between the vibrational levels of the singlet ground state S0 and the two electronically excited states, the first excited singlet S1, and its triplet T1 corresponding to their energy gaps.

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lst triplet T1

Fig A-I.2: Jablonski diagram illustrating photophysical radiative and nonradiative transitions. S0 = ground singlet state, S1 = 1st excited singlet state. T1 = 1st triplet state. A = absorption of light, F = fluorescence, P = phosphorescence, IC = internal conversion, ISC = intersystem crossing. Radiative transitions are shown by solid lines, nonradiative by wavy lines. Photochemical processes are not indicated.

The excited molecule can undergo the following process in addition to direct photo ionization: (a) If the excited state is completely repulsive i.e., there are no attractive forces strong enough to lead to a potential energy minimum, then dissociation into two or more molecular fragments, free radicals ions will occur within approximately one vibration (10–12–10–13). (b) Even if the excited electronic state is ‘bound’ i.e., it possesses a potential energy minimum, the absorbed photon may be of sufficient energy to raise the vibrational energy to a dissociative limit. This type of direct photodissociation may also be extremely rapid for diatomic molecules. An example of photodissociation is a spin allowed dissociation of O 3 at

λ < 320 nm. O3 (B2 ) + hν(λ < 320 nm)  → O(1D) + O2 (1∧ g ) . Term symbols of the electronic states are indicated in brackets. Both O2 and O are in their electronic excited states. With more complex species, large number itself may make its dissociative lifetime very large and either of the competitive process de-excitation or chemical reaction may take place. (c) Fluorescence (F) is defined as emission of light due to a transition between states of like multiplicity, i.e., S1 → S0 + hν. The molecule may return to the stable ground state by radiative emission of light from the lowest vibrational level. This is an allowed transition and hence the life time of upper state with respect to fluorescence is usually short, typically. (d) 10–6–10–9 s. However, this emission will be of lower energy than the excitation energy for molecule because of the nonradiative dissipation of vibrational energy in the excited state (F). For example, the fluorescence life of OH in the electronically excited 2Σ+ state is 0.7 µs.

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(e) Phosphorescence (P) is defined as emission light due to a transition between states of different spin multiplicities. Because this is not an allowed transition for an ideal unperturbed molecule, phosphorescence lifetimes tend to be relatively long 10–3 – 10–2 s. (f) Intersystem crossing (ISC) is the intramolecular crossing from one state to another of different multiplicity without emission of radiation. In Fig. A-I.2, (ISC)1 shows the transfer from first excited singlet state S1 to the first excited triplet state. Since the process is horizontal, the total energy remains the same and the molecule initially produced in upper vibrational level of T1 from which it is deactivated is shown by the vertical wavy line. Since this is a nonradiative process, it is not limited by the ∆S = 0. The resulting new excitation can no longer undergo a rapid radiative decay to ground state because it is forbidden. Similarly, (ISC)2 shows the intersystem crossing from T1 to upper vibrational states of the ground state, S0 from which vibrational deactivation to V" = 0 then occurs. (g) Internal conversion (IC) process is the intramolecular crossing of an excited molecule from one state to another of the same multiplicity without the emission of radiation. As seen in Fig. A-I.2, the horizontal wavy line (IC) represents the internal conversion of S1 → S0, this is generally followed by vibrational deactivation to V" = 0 but no chemical change. Such process is quite rare but found in aromatic molecules and cyanine dyes. (h) The excited (donor D) molecular may be deactivated by radiationless energy transferred to another (acceptor A) molecule. Because of the forbidden character of triplet state in phosphorescence, triplet state is long lived than the excited singlet state. It can, therefore, readily undergo collision energy transfer at normal encounter of distances by the spin allowed reaction.

→ D(Singlet) +A(Triplet) D(Triplet) + A(Singlet)  Provided the transfer is exothermic (∆Eacceptor < ∆Edonor) or at least less than approximately 10 kJ endothermic. Triplet states are quenched in most fluid solutions by this type of energy transfer to dissolved oxygen and solvent impurities, even in the absence of added quenchers.

11. CHEMILUMINESCENCE In photochemical reactions chemical change result from absorption of light. The converse of this process is emission of light by a system at ordinary temperature as a result of a chemical reaction. Such emission of cold light is called chemiluminescence. It is a well-known phenomenon. When yellow-phosphorus is oxidized in oxygen or air at low pressures and at temperatures between –10 and 40°C, the phosphorus is converted to P2O5 with the emission of a visible greenishwhite luminescence. Similarly, the oxidation of certain Grignard reagents, silicon compounds and 3-aminophthalic acid and hydrazine in alkaline solution is accompanied by light emission. In all these instances part or all of the energy change of the reaction, instead of appearing as heat, goes to activate electronically some molecule. The activated molecule emits the excitation energy as radiation and reverts and resorts to a normal state. The chemiluminescences, which is possibly best understood, is that emitted in the reaction of alkali metal vapors with halogens. When streams of alkali metal vapors and halogens at pressures of 10–2 to 10–3 nm Hg are mixed, alkali halides are formed with the emission of luminescence. The probable mechanism suggested is

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211

(i) Na + Cl2

 →

NaCl + Cl

(ii) Na2 + Cl

 →

NaCl + Na

(iii) Na + Cl

 →

NaCl

It is believed that in the (ii) step sufficient energy is acquired by the NaCl molecule to excite a Na atom on collision as

NaCl * + Na  → NaCl + Na * Na* atom on returning to ground state emits characteristics spectrum or exhibit chemiluminescences. Grignard reagent produces greenish blue luminescence in its reaction with chloropicrin or on its oxidation by air. Only aryl magnesium halides in which magnesium atom is attached to one unsaturated carbon atom exhibit chemiluminescences in solution while in solid state, both alkyl and aryl magnesium halides show this phenomenon.

PROBLEMS 1. State and explain Einstein law of photochemical equivalence and give example of its application in photochemical reaction. Mention some reactions, which appear to deviate from this law. 2. The quantum yield in the combination of hydrogen and chlorine is 103 whereas it is only 0.01 when hydrogen combines with bromine. Discuss the reasons for the difference. 3. Distinguish between the primary and secondary process in a photochemical reaction. How does the distinction permit the explanation of quantum yield of 2 in the dissociation of HI? 4. Write short notes on: (i) Quantum yield (ii) Fluorescence (iii) Phosphorescence (iv) Chemiluminescences (v) Jablonski diagram. 5. Fluorescence is a fast process while phosphorescence is a slow process. Explain. 6. Elaborate the reasons for obtaining low and high quantum field. 7. In the photochemical decomposition of ethylene iodine

→ C2H4 + I2 C2H4I2 + hν  with radiation of 424 nm, the iodine formed after 20 minutes required 41.14 cm3 of 0.0025 mol dm–3 solutions of Na 2 S 2 O 3 for neutralization. The intensity of the light sources was 9.15 × 10–4 Js–1. Calculate the quantum yield assuming that the absorption of energy was complete. 8. A certain system absorbs 3 × 1016 quantas of light per second. On irradiation for 10 minutes 0.002 mole of the reactant was found to have reacted. Calculate the quantum yield of the process. 9. An actinometer contained 10.0 ml solution which was 0.01 M with respect to uranyl sulphate and 0.05 M with respect to oxalic acid. This mixture was exposed to a wavelength of 260 nm. A radiation equivalent to 7.95 × 1022 kJ mol–1 was absorbed. After the experiment the concentration of oxalic acid was found to be 0.0332 M. Calculate the quantum yield for the reaction of decomposition of oxalic acid taking place in the actinometer.

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10. A beam of light was passed through a coloured solution when 85% of the incident light was transmitted. Calculate (i) Optical density of solution, (ii) The percentage transmission, if the column of solution was 8/5 times long as the previous column. 11. Explain why in a pollution free atmosphere, photochemical formation of O3 does not take on the surface of earth. 12. Why is it that chlorofluorocarbon become active only near the ozone layer of atmosphere? How does it act to deplete ozone layer? 13. Discuss the statement that even a small amount of NO2 is sufficient to trigger photochemical reactions that leads to formation of smog.

D❖E

Appendix II

Time Management (MAKING EVERY SECOND COUNT)

The ability to work faster and get more done in less time is not slavery; it is freedom. You are going to have the same big pile of stuff to do everyday whether you want it or not. If you can be more efficient, you can get it done and still have sometime left over for yourself—whether it is to read the paper, jog, and go to disco or play the piano. Productive workers have everyday schedules and they stick with them. Break your day into segments. I suggest using hour increments, although quarter and half day can also work. Write down on a piece of paper the project you will work on during each of these segments. Last thing in a day one should prepare a schedule for next day. As you go through the day, consult your schedule and keep on track. If priorities change, you can change the schedule, but do this in writing. Why does our increment works so well, the reason is if you have all day to task X, you will take all day? If you have only an hour you will work that much more quickly and efficiently. It is okay to redo the schedule as long as you do not miss deadlines. As long as you are organized, keep track of deadlines, and allow enough time to finish each job, you will increase your productivity by working on things you feel in the mood to work on. As the story goes, Charles Schwab, president of Bethlehem Steel in the early 1900 could not seem to get enough work done. Details and minor matters were crowding the time he urgently needed to consider than more important matters. He asked Ivy Lee, a management consultant, what to do about it. Lee handed Schwab a blank sheet of paper. "Write down the most important things you have to do tomorrow. The first thing tomorrow morning, start working on item number one and stay with it until completed. The item number two in the same way. Then number three and so on. This way you will accomplish the most important projects before getting to the less important one". The steel executive tried the idea and recommended it to his associates because the method worked so well. When Schwab asked Lee what his fee was. Lee replied pay me what you think the idea is worth. Schwab sent Lee a check of $25000—a fortune in those days.

1. OVERCOME PROCRASTINATION Procrastination is opportunity assassin. It is the single biggest factor causing people to fall behind in their works. It is advised never defer for a single hour that which can be done now. Having a daily to do list and the assigning you various tasks throughout the day in one-hour increments helps you stay on track and avoid putting things off. As long as you have your short-term

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deadlines and long-term goals in mind, you can be somewhat flexible in your daily schedule, adjusting task and time slots to match your enthusiasm for each project. To overcome procrastination, identify any bad habits you have that wastes your time. For me, it was sleeping an hour after I first woke up in the morning. Because the morning is the most productive work time, by forcing myself to get dressed and go to the office when I wake up, instead of falling back into bed, I increase my productivity tremendously. For you it may be watching a soap opera in the middle of day, spending too much time in surfing the web, talking in chat rooms or on the phone, doing housework or staying up too late at night to read or watch television. After you have identified the bad habit make a list of the ones you must avoid. Another time killer is outside distraction. Any outside distraction is an unscheduled activity that interrupts the task you are currently working on. The key is to physically block out disturbances as much as possible, whether by shutting your door, turning your desk away from passersby and so on. Most of us have certain times during the day when we are most alert and perform better. By handling mentally demanding jobs during your peak energy periods you can get more done in less time. If you increase this period, your efficiency will increase. This period is a function of energy. Energy is a function of many factors, one of it being enthusiasm. When you are enthusiastic, your energy can remain high, even if you are physically tired. When you are bored, your energy drains, and you become lethargic and unproductive. To have peak productivity and energy, maintain peak enthusiasm and avoid boredom. The main cause of boredom is not doing what you want, when you want to do it. Therefore, you should structure your work time in such a way that you are spending most of your time doing what you want, when you feel like doing it. Obviously you should shy away from assignments that bore you. Forcing yourself to work on things you dislike will drain your energy. Of course, when a deadline is looming, you may have no choice but to put aside work you want to do and focus on what has to be done to meet that deadline. Everyone finds something that can help revitalize him throughout the day. When you find what works for you, do it. One of my friends has a private bathroom, and when he feels his concentration and energy waning, he washes his hair. It refreshes him immediately, perhaps the wet hairs of head cools his overhead brain. You can take a break and energize by washing your face, taking a walk, running errands, meeting a friend for coffee, or chatting on the phone.

2. TEN TIPS TO HELP YOU WORK BETTER AND FASTER 1. Have a PC: Anyone in any business who wants to be productive should use a modern PC with latest software. Doing so can double, triple or even quadruple your output. Apparently to have PC means one should know how to use PC effectively. 2. Do not be a perfectionist: I am a non-perfectionist said Isaac Asimov, author of 475 books. I do not look back to regret or worry at what I have written. Be a careful worker, but do not agonize over your work beyond the point where the extra effort no longer produces a proportionately worthwhile improvement in your final product. Be excellent but not perfect. That does not mean you deliberately make errors or give less than your best. It means you stop polishing and fiddling with the job when it looks good to you. Create it, check it, then let it go. “Perfection does not exit,” wrote Alfred de Musset, “To understand this is the triumph of human intelligence; to expect to possess it is the most dangerous kind of madness.”

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215

3. Free yourself from the pressure to be an innovator: As publisher Cameron Foote observes, “Clients are looking for good, not great”. Do your best to the clients requirement. They will be happy. Do not feel pressure to reinvent the wheel or create a masterpiece on every project you take on. Do not beheld up by the false notion that you must uncover some great truth or present a revolutionary ideas and concepts. Just eliminate performance pressure. Do not worry about whether what you are doing is different or better than what others have done before you. Just do the best you can. That will be enough. 4. Make deadlines firm but adequate: Productive people set and meet deadlines. Without a deadline, the motivation to do a task is small to nonexistent. Task without assigned deadline automatically go to the bottom of your priority list. At the same time do not make deadlines too tight. Try to build in some extra time for the unexpected, such as missing piece of information, a last minute change etc. 5. Protect and value your time: Productive people guard their time more heavily than the gold in Fort Knox. They do not waste time. They get right to the point. They may come off as abrupt or dismissive to some people. But they realize they cannot give everyone who contacts them all the time each person wants. They determine how much time to spend with each person. They make quick decisions. They say what needs to be said, do what needs to be done and then move on. Respect hour power. 6. Stay focused: Successful people apply themselves to the task at hand. They work until the work gets done. They concentrate on one or two things at a time. They do not go in a hundred different directions. My experience is that people who are big talkers, constantly spouting ideas or proposing deals and ventures are spreading out in too many different directions to be effective. Efficient people have a vision and focus their activities to achieve that vision. Rolling stone do not gather weights. 7. Be productive: If you want to be productive, there are certain things you will have to give up. These things include the extravagant luxuries of sloth, inertia, laziness and wasted idle time. If you are not willing to give these up, you must seriously question whether being more productive is truly a priority in your life. If it is not that is okay. However, do not complain that there is never enough time and then watch 25 hours of sports on TV each weekend. The primary reason most workers are not productive is that they really do not desire it. In a sense, to be productive at anything, some sacrifice must be made. Everything we want to do, learn, achieve or create has a price. And that price is time. Productive workers pay this price to be productive. 8. Know the value of time and money: Unless you are independently wealthy or have second source of income, most workers need to earn a paycheck to pay rent, health insurance premium, utility bills, grocery, doctors bills and for video rentals. The happiest possible life ideally rests on a balance between four elements: health, career, personal relationship and money. Productive workers that I have met tend to be partially money driven or at least money conscious. They are not content with the meager income of the average worker. They may not aspire to great wealth, but they all want to be comfortable. The accepted definition of success is “Doing what you want, when and where you want to do it, and getting paid well for it—something very very well”. This is what you achieve when you are productive, and put your nose to grindstone, your back into your hammer swings, or your fingertips to computer keys.

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A wise person once remarked, “If you do not know where you are going, you are certainly never going to get there”. Think about where you are going. How much money do you want? If your answer is ‘enough’ or ‘a lot’, you have not clearly defined your income goals. Without knowing where you are going, how will you get there? An important first step toward increasing your income is to set a specific goal. The goal should be difficult, because success requires hard work and a bit of ambition. If you set easy goals, you will always achieve them but you will always achieve below your potential as well. 9. Become more active in your own field: Somewhere along the way you may have lost the zest for science, engineering, business management, a photographer or whatever it is you, do in the first place. You can escape job burnout by rekindling your interest in your profession. Join your professional society, if you have not already. Become active: attend meetings, read journals, present papers. You can even run for office in your local chapter. Take a course or teach one. Take responsibility for training one of the new employees in your department. The people who are active in their field are usually the most successful and the most satisfied with their careers. Would not you contribute more and get noticed in a more favorable light. 10. Mastering the time management see-saw: Time and workload never seem to get in synch. In fact, often the opposite is true. You have heard the old saying, ‘when it rains, it pours’. When we are already too busy, even more work hits in our basket. One of the important filler is to be social with your colleagues. I seldom attend association or professional society activities because I simply do not have the time. But this type of networking is worthwhile. It can stimulate your thinking, expose you to new ideas, and broaden your network of contacts. So if you can fit it in, join a local chapter of a professional group and go to meetings. Get to know fellow specialists in your area. Once they get to know you, your phone may start ringing more frequently than it does now. Good performance always start with clear goals, without clear goals you will quickly become a victim of having too many commitments. You will have no framework in which to make decisions about where you should or should not focus your energy. Everyone has gone through confusion, failures but one has to give direction and set goals for himself. Follow what has been written above and you will be a productive person. But then do not forget P.S. Sindhu who is guiding you.

= >

Appendix III

References A. 1. 2. 3. 4.

ATOMIC STRUCTURE AND ATOMIC SPECTRA Herzberg G., "Atomic Spectra and Atomic Structure", Dover, 1994. Shore B.W., & Menzel D.H. "Principles of Atomic Spectra", Wiley, 1968. Richards W.G., & Scott P.R. "Structure and Spectra of Atoms", Wiley, London, 1976. Condon E.U., & Shortley G.H. "Theory of Atomic Spectra", Cambridge University Press, New York, 1934.

B. 1. 2. 3. 4. 5.

NATURE OF CHEMICAL BONDS Murrell J.N., Kettle S.F.A., & Tedder J.M. "The Chemical Bond", Chihester, 1978. Coulson C.A. "Valence", Oxford University Press, London, 1961. McWeeny "Coulson Valence", Oxford University Press, 1974. Ballhausen C.J., & Gray H.B. "Molecular Orbital Theory", Benjamin, New York, 1965. Robert G. Parr "Quantum Theory of Molecule Electronic Structure, A lecture-note and reprint volume". W.A. Benjamin. Inc., Reading, Massachusetts, 1972. 6. Keniti Higasi, Hiroaki Baba and Alan Rembaum "Quantum Organic Chemistry", Interscience Publisher, New York, 1965. 7. Hoffman, B. "The Strange Story of Quantum", Dover, New York, 1959. 8. Atkins P.W. "Molecular Quantum Mechanics", Clarendono Press, Oxford, 1970.

C. 1. 2. 3. 4.

MOLECULAR SPECTROSCOPY Barrow G. "Introduction to Molecular Spectroscopy", McGraw Hill, New York, 1962. Sindhu P.S. "Molecular Spectroscopy. "Tata-McGraw Hill, New Delhi, 1985. Sindhu P.S. "Fundamentals of Molecular Spectroscopy", New Age International (P) Ltd., 2006. Herzberg G. "Molecular Spectra & Molecular Structure: I" "Spectra of Diatomic Molecules II" "Infrared and Raman Spectra of Polyatomic Molecules", Van Nostrand, 1945, 1950. 5. Herzberg G. "Electronic Spectra & Electronic Structure of Polyatomic Molecules", Van Nostrand, New York, 1966. 6. Holas J. Michael. "Molecular Spectroscopy", John Wiley and Sons Ltd., 1996. 7. Townes C.H. & Schawlow A.L. "Microwave Spectroscopy", Dover Publication, New York, 1975.

218

Elements of Molecular Spectroscopy

8. Gordy W., Smith W.V. & Trambarula R. "Microwave Spectroscopy", John Wiley, New York, 1960. 9. Townes C.H., & Schawlow A.L. "Microwave Spectroscopy", McGraw Hill, New York, 1955. 10. Wilson E.B., Docius J.C. and Cross P.C. "Molecular Vibrations", McGraw Hill, 1955. 11. Allan H.C., & Cross P.C. "Molecular Vib-Rotator", John Wiley & Sons, New York, 1985. 12. Nakamoto Kazas "Infrared Spectra of Inorganic and Coordination Compounds", John Wiley & Sons, New York. 13. Gilson T.R., Hendra P.J."Laser Raman Spectroscopy", John Wiley & Sons, New York, 1970. 14. Tobin M.C. "Laser Raman Spectroscopy", Wiley Interscience, New York, 1971. 15. Long D.A. "Raman Spectroscopy", McGraw Hill, New York, 1971. 16. Auram M., & Mateexu Gh.D. "Infrared Spectroscopy Application to Organic Chemistry", Wiley Interscience, New York, 1970. 17. Bararisha H., Labudzinska A. & Terpinki J." Laser Raman Spectroscopy-Analytical Application", Ellis Horwood Ltd, Sussex, England, 1970. 18. Silverstein R.M., Bassler G.C. and Morrill T.C. "Spectroscopy Identification of Organic Compounds", Wiley, New York, 1981. 19. Rao C.N.R. "Ultraviolet & Visible Spectroscopy", Butterworth, London, 1967. 20. Daniel C. Harris, Michael D. Betollcei "Symmetry and Spectroscopy, An Introduction to Vibrational and Electronic Spectroscopy", Oxford University Press, New York, 1978. 21. Jaffe H.H., & Orchin M. "Theory and Application of Ultraviolet Spectroscopy," Wiley, New York, 1962. 22. Turner D.W., Baker C., Arthur D. Baker and Brundle C.R. "Molecular Photoelectron Spectroscopy", Wiley Interscience, New York, 1970. 23. Murrell J. N. "The Theory of Electronic Spectra of Organic Molecules", Chapmann and Hall, London, 1971. 24. Gillam A.E., & Stern E.S. "An Introduction to Electronic Absorption Spectroscopy in Organic Chemistry", 2nd Ed., Arnold, London, 1957. 25. Lever A.B.P. "Inorganic Electronic Spectroscopy", Elsevier, Amsterdam, 1968. 26. Carrington A., & Mclachlan A.B. "Introduction to Magnetic Resonance", Chapmann & Hall Ltd., London, 1979. 27. Pople J.A., Schneider W.G., & Bernstein H.J. "High Resolution Nuclear Magnetic Resonance" McGraw Hill, New York, 1959. 28. Abraham R.J., & Loftus P. "Proton and Carbon-13 NMR-Spectroscopy", Heyden & Sons Ltd., London, 1978. 29. Adman D., Becker "High Resolution NMR, Theory and Chemical Application", 1999. 30. Abraham R.J., & Bleaney B. "Electron Paramagnetic Resonance of Transition Ions", Oxford University Press, 1970. 31. Druelt S., & Taranin J.P. "Chemical and Biochemical Applications of Laser", C.B. Moore Ed., Academic Press, 1979. 32. McLafferly F.W. "Interpretation of Mass Spectrum", Benjamin, New York, 1966. 33. Hill H.C. "Introduction to Mass Spectroscopy", Heyden, London, 1966. 34. Duley W.W. "Laser Processing and Analysis of Materials", Plenum Press, New York, 1983. 35. Ready J.F. "Industrial Applications of Lasers", Academic Press, New York, 1978.

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Appendix IV

Some Very Good Internet Sites On Spectroscopy 1. Spectroscopy now is a good site for UV, Vis, IR, Raman, NMR, and atomic spectra

2. Internet based service for Instant access to chemical data is provided by this site bases. This site connects you to more than 1000 sites on spectroscopy. http://www.2acd labs.com/download/lab_addon.html 3. University of Illinois at Chicago has started a site for preparing org.chem. lecture slides. Go to site http://tigger uic.edu? Enter (i) organic chem 234 lecture slides (ii) Molecular Spectroscopy (iii) Chromatography and prepare the slides. 4. Interpretation of spectra is a technique that requires practice. This site is established to provide chemistry students with a library spectroscopy problems and helps to solve them < www.chem.ucla.edu/web.spectra> 5. In this site animated IR spectra for a variety of organic compounds are given.

6. This is a IR spectroscopy interactive visualization site for organic molecules. It shows how a particular group in a molecule vibrates for a frequency. Go to 8. IR spectroscopy for organic chemists Web Resources Data bases SOBS NIST Chemistry Web book < www.dq.fct.ul.pt/qoa/jas/irhtml> 9. A NMR site that teaches how to interpret a PMR spectra

10. Spectroscopy tutorials in IR, NMR and Mass spectroscopy with interactive problems www.chem ulc.edu/web1/OCOL-I/WIN/HOME.HTP


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Appendix V

Some Relationships Between Commonly Used Units CONCENTRATION 1 molal solution = 1 mol/litre = 1 mol dm–1 1 milligram per litre in water = 1.0 ppm = 1000 ppb = 1.0 g m–3 (Sp. Gravity of water = 1)

ENERGY

Energy Energy

J 1 eV 1J –1 kcal mol × 4.184 × 0.04336 × 349.8 × 489 × 109 cm–1 × 1.196 × 10–2 × 1.24 × 10–4 × 2.859 × 10–3 kJ mol–1 × 0.2390 × 0.0104 × 83.59 × 116.87 × 100 eV × 96.49 × 23.06 × 8.066 × 103 × 1.4887 kcal mol–1 kJ mol–1

= Nm = 1.602 × 10–19 J = 6.238 × 1012 MeV = kJ mol–1 = eV = cm–1 =K = kJ mol–1 = eV = kcal mol–1 = kcal mol–1 = eV = cm–1 = K = kJ mol–1 = kcal mol–1 = cm–1 =K = 2.859 × 104 / λ (nm) = 1.196 × 103 / λ (nm)

Appendix V

221

MULTIPLES OF THE BASE UNITS ARE ILLUSTRATED BY LENGTH 109 giga Gm

Fraction Prefix Unit

104 mega Mm

103 kilo km

10–2 centi cm

1 metre* m

10–3 milli mm

10–6 10–8 10–9 micro angstrom nano µm Å nm

10–12 pico pm

* This is not a prefix.

PHYSICAL CONSTANTS Constants

Symbol

Value

Velocity of light

c

2.9979 × 108 ms–1

Avogadro number

N

6.02 × 1023 mol–1

Planck’s constant

h

6.626 × 10–34 Js

h

=h

2π

1.602 × 10–34 Js

Electron rest mass

me

9.1083 × 10–28 gm

Proton rest mass

mp

1.67239 × 10–24 gm

Neutron rest mass

m

1.67470 × 10–24 gm

Charge of proton

e

1.602 × 10–19 C

Gas constant

R

8.3144 Jk–1 mol–1 R

Boltzmann constant

k=

Faraday constant

F = Ne

N

1.38 × 10–23 JK–1 9.65846 × 104 C mol–1

Permeability of vacuum

µ0

4π × 10–7 NA–2

Permittivity of vacuum

ε0

8.854 × 10–12 F m–1

ε0µ0 = 1/c2

4πε 0

D E

1.112 × 10–10 J–1 C2 m–1

222

Periodic Table 1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

1H 1.008

18 2 He 4.003

3Li 6.641

4 Be 9.012

5B 10.80

6C 12.01

7N 14.01

8O 16.00

9F 19.00

10 Ne 20.18

11 Na 22.99

12 Mg

13 AI 26.98

14 Si 28.09

15 P 30.97

16 S 32.06

17 CI 35.45

18 Ar 39.95

19 K 39.10

20Ca 40.08

21 Sc 44.96

22Ti 47.90

23V 50.94

24 Cr 52.00

25 Mn

37 Rb 85.47

38 Sr 87.62

39Y 88.91

40 Zr 91.22

41 Nb

42 Mo

43 Tc

92.91

95.94

55 Cs

56 Ba

71Lu

72Hf

73 Ta

74W

24.31 26 Fe 55.93

27 Co 58.93

28Ni 58.71

29 Cu 63.55

30 Zn 63.37

31 Ga 69.72

32 Ge 72.60

3 3A s

74.92

34 Se 78.96

35 Br 79.90

36Kr 83.80

44 Ru 101.1

45 Rh 102.9

46 Pf 106.4

47 Ag

(99)

107.9

48 Cd 112.4

49 In 114.8

50 Sn 118.7

51 Sb 121.8

52 Te 127.6

53 I 126.9

54 Xe 131.3

75 Re

76 Os

77Ir

78 Pt

79 Au

80 Hg

81Tl

82 Pb

83 Bc

84 Po

85 At

86 Rn

210

210

222

116Uuh

117Uus

118 Uuo

54.94

132.9 137.3 175.9 178.5 180.9 183.9 186.9 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 87 Fr (223)

88Ra (226)

Lanthanides

104Rf 262.1

105 Db 106 Sg 107 Bh 108 Hs 109 Mt 110Uun 111Uuu 112Uub 262.1 (263) (262) (265) (266) (281) (272) (285)

113Uut

114Unq

115Uup

57La 58Ce 59Pr 60 Nd 138.9 140.1 140.9 144.2

61Pm 62Sm 63Eu 64 Gd 65 Tb 66 Dy 67 Ho 68Er 69 Tm 70 Yb (145) 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.0

89 Ac

90 Th

93 Np

(227)

(232)

91 Pa (231)

92 U (238)

(237)

94 Pu (244)

95 Am

96Cm

(243)

(247)

97Bk (247)

98Cf (251)

99 Es (252)

100 Fm 101 Md 102 No

(257)

(258)

(259)

Elements of Molecular Spectroscopy

Actinides

103Lr (257)

Index A Absorption due to ethylene chromophore 131 Absorption frequency of NO 73 Absorption spectra 30 Acetylenic chromophores 135 Adjusted crystal field theory (ACFT) 153 Alkali halide disc 104 Anharmonicity in vibrations 78 Anharmonicity constant 115-116 Anti-stokes 63 Anti-stokes lines 38 Atomic term symbols 17 Atomic theory 2 Auxochrome 130, 140

B Bathchromic group 141 Bathchromic, hypsochromic, hyperchromic and hypochromic shifts 131 Beer-Lambert law 34, 129, 200 Black body radiation 1 Bohr 2 Boltzmann distribution 25, 60, 78, 117 Born-Oppenheimer approximation 27, 29, 107

C Carbon dioxide 83 Carbonyl chromophore 135 CFSE for tetrahedral symmetry 152 C—H deformation 92 Characteristic group vibrations 87 Chemical shift 174, 177

Chemiluminescence 210 Chromogen 139 Chromophore 129 Chromophore-auxochrome theory 139 Chromophoric group 139 CN molecular 114 CN molecule 110 CO molecule 109, 115 Collision broadening 36 Colour and constitution 139 Continuous Wave (CW) 172 Crystal Field Stabilization Energy (CFSE) 148 Crystal field theory 145, 153

D deBroglie wave 8 Degeneracy 48 Degenerate 60 Dissipation of energy by excited molecule 118 Dissociation energy 115 Dissociation energy and temperature 117 Doppler broadening 35

E Effect of hydrogen bonding 94 Effect of isotopic substitution 56, 75 Eigen functions 13 Electron diffraction 11 Electronic spectra 107 Electronic spectra of complexes 154 Electronic states of atomic hydrogen 18 Electronic transitions 125

224

Electronic transitions in heteronuclear diatomic molecules 107 Emission spectra 30 Energy level diagrams 159 ESR spectra 190 ESR spectra of aromatic ions 191

F Factors affecting magnitude of 10Dq or ∆o 150 Fieser and Scott rules 133, 136-137 Fingerprint region 92 Flame 25 Fluorescence 118, 120 Fluorolube 104 Franck-Condon principle 113, 118 Functional groups 98-100 Fundamental band 78

G Grottus-Draper law 201 Group frequency concept charts 88

H Halogen molecules 112 Hamiltonian operator 12 Heisenberg uncertainty principle 35 HF molecule 111 High and low quantum yields 206 Homonuclear diatomic molcules 111 Hot bands 78-79 Hydrocarbon skeleton 98-100 Hyperconjugation 24 Hypsochromic group 141

I I2 molecule 115 Intensity of spectral lines 33, 58–62

J Jablonski diagram 208 Jahn-Teller effect 161 Jahn-Teller tetragonal distortion 162, 164 Jorgensen g factor 151

Index

L Larmor precession 170, 188 Larmor precession frequency 188 Ligand field theory 153 Limitations of Bohr’s theory 7 Linear harmonic oscillator 21, 71

M Magnetic moments 169 Markownikoff’s rule 24 Metal ions 25 Metallic waves 8 Microwave spectra of the molecule CH 55 Molecular spectroscopy importance 40 Moment of interia 52 Morse potential 77

N n→π* transitions 137 n→σ* and n→π* transitions 126 Nature of radiation 26 Nitrogen molecule 112 NO molecule 110 Nuclear zeeman 175 Nujol 104

O Oxygen molecule 112

P π→π* transition 126, 137 Pairing energy 150 Particle in a box 43 Particle in a one-dimensional box 20 Particle in a three-dimensional box 47 Pascal triangle 181, 191 Phosphorescence 121 Photons 8 Plane-polarized light 26 Polarizability 39 Polyatomic molecules 81 Primary and secondary effects of light absorption 202

Index

Probability 10 Proton magnetic resonance 169

Q Quantization of energy 44 Quantum efficiency 204 Quantum theory of radiation 1

R Radius of the orbit 3 Raman spectroscopy 30, 38 Raman spectrum of HCl 67 Recah parameters 153-154 Regions of electromagnetic radiations 32 Rigid rotator 20, 52 Rigid rotator diatomic 64 Rotational Raman spectrum 62, 64-65, 67 Rotational spectra 51, 79 Rydberg constant RH 5

S σ→σ* transitions 126 Sample preparation 103, 141 Scattering spectra 30, 38 Schrodinger wave equation 11, 12-13 Selection rules 53, 72 Shielding constant 175 Significance of wave function 14 Simple harmonic oscillator 77 Skeletal vibrations 87 Slater integrals 153 Solvent effects on electronic spectra 137 Spectra of hydrogen atoms 5, 17 Spectrometer 172 Spectrophotometers 30 Spectroscopy 24 Spectrochemical series 151 Spectrum of carbon monoxide 55 Spin-spin interaction 179 Splitting 175

225

Square planar coordination 165 Stark-Einstein law 201 Stokes 63 Stokes lines 38 Structure determination 97 Study of photochemical reaction 204

T Tetramethyl silane 177 Thermal energy 74 Thermochemical dissociation energy 116 Transition metal complexes 145 Transition probability 34 Trimethyl silyl propane sulphonic acid (DSS) 177

U Uncertainty principle 9

V Velocity of electron in hydrogen atom 4 Vertical transitions 113 Vibrating diatomic molecule 69 Vibrational energy 74 Vibrational frequency 116 Vibrational Raman spectra 80 Vibrational spectra 69 Vibrational spectra of alkali halides 80

W Water 84 Wave equation for hydrogen atom 14 Wave equations for some systems 20 Wave-particle duality 7 Width of spectral lines 35 Woodward rules 133

Z Zeeman splitting 187 Zero point energy 45, 72-73, 116

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