vibrational spectra (infrared spectra)

UNIT-2

VIBRATIONAL SPECTRA (INFRARED SPECTRA)

Lesson Structure

2.0 Objective

2.1 Introduction

2.2 Molecular vibration

2.3 Vibration of a single particle (classical)

2.4 The vibration of two particles system (classical)

2.5 Schrödinger Equation applied to Harmonic oscillator

2.6 Zero point energy

2.7 Selection Rule

2.8 Boltzmann Distribution

2.9 Force constant and Bond strength

2.10 The Anharmonic oscillator

2.11 Fundamental and Overtone bands

2.12 Combination bands

2.13 Vibration of Polyatomic molecules : Normal Modesof vibration

2.14 Group frequency

2.15 Vibration rotation spectroscopy

2.16 Factors affecting the band position & Intensities

2.17 Study of vibrational frequencies of Carbonylcompounds

2.18 Effect of Hydrogen bonding on vibrationalfrequencies

Solved Examples

Model Questions

References

22

Vibrational spectra (Infrared Spectra)

2.0 OBJECTIVE

After studying this unit you will be able to

• Know about different types of molecular vibrations

• Discuss the classical treatment on vibration of single as well as two particles system.

• Apply Schrödinger equation to Harmonic oscillator and derive quantised vibrational

energy levels.

• Derive vibrational energies of diatomic molecules.

• Know about zero point energy, force constant & Bond strength.

• Describe selection rules for vibronic transition and Boltzmann distribution.

• Knew about Anharmonic Oscillator.

• Know about funndamental bands, Overtone bonds and Combination bonds in Infrared

spectra.

• Describe vibrational spectra of linear and non linear triatomic molecules as well as

polyatomic molecules.

• Know about vibration-rotation spectra and P, Q, R Branches.

• Discuss group frequencies & their application.

• Describe factors affecting the band positions and intensities.

• Discuss effect of Hydrogen bonding on Vibrational frequencies.

2.1 INTRODUCTION

Vibrational energy of a molecule corresponds to infrared frequency. The interaction of

infrared radiation with molecular vibration gives infrared spectrum. If the average position

and orientation of a molecule remains constant but the distance between the atoms in a

molecule change, molecular vibrations are said to take place.

A vibrational spectrum is observed experimentally as Infrared as well as Raman Spectra.

But the physical origin of two type of spectra are different. Infrared spectrum is associated

with dipole moment () of the bond whereas Raman spectra are associated with polarizability.

Either the wavelength () or wave number ( -1in cm ) is used to measure the position

of a given infrared absorption. The range of IR spectrum is as :

1 1 1 1

6 8

: 12500 4000 4000 650 650 50

: 0.8 2.5 2.5 15 15 200

1 10 10

Wave number cm cm cm cm

Wave length

where is micron such that m cm

23


Most studied absorption frequency in IR spectroscopy is 200 cm–1 4000 cm–1. It is

fundamental frequency region. Greater than 4000 cm–1 requires very high energy. These are

known as overtone bands. Overtones are multiple of fundamental frequencies (e.g.

1 2 1 22 , 2 , 3 , 3 .etc ). Apart from fundamental & overtone, we also have combination bands

1 2

1 2 Addition band

1 2– Substraction band

2.2. MOLECULAR VIBRATION

A molecule is not a rigid assemble of atoms. A molecule can be considered as a system

of balls of varying masses corresponding to atoms of molecules and spring of varying

strengths, corresponding to the chemical bands of a molecule.

There are two types of fundamental vibration for molecules :

(i) Stretching vibration in which the distance between two atoms increases or

decreases but the atoms remain in the same bond axis.

(ii) Bending or deformation, in which the position of the atom changes relative to

the original bond axis.

The various stretching and bending vibrations of a bond occur at a certain quantized

frequencies. When infrared light of the same frequency is incident on the same molecule

energy is absorbed and amplitude of that vibration is increased. When the molecule reverts

from the excited state to the original ground state, absorbed energy is released as heat.

A nonlinear molecule that contains n atoms has 3n-6 possible fundamental vibrational

modes that can be responsible for the absorption of infrared light. Thus, such simple

molecules as methane (CH4) and Benzene (C6H6) have theoretically, nine and thirty possible

fundamental absorption bands, respectively.

In order for a particular vibration to result in the absorption of infrared energy, that

vibration must cause a change in the dipole moment of the molecule. Thus, molecules that

contain certain symmetry elements will display somewhat simplified spectra. The C=C

stretching vibration of ethylene (H2C = CH2) and the symmetrical C—H stretching of the

few C—H bonds of methane (CH4) don’t result in a absorption band in the infrared region.

The predicted number of peaks will not be observed also if the absorption occurs outside

the region ordinarily examined.

Additional (non-fundamental) absorption bands may occur because of the presence of

overtones (or harmonics that occur with greatly reduced intensity at 12

, 13

... of the

wavelength (twice, thrice times the wave numbers) combination bands (the sum of two or

more different wave numbers), and difference bond (the difference of two or more different

wave numbers)

24


Obviously, there are many possible vibrations in a molecule. However, only those

stretching vibrations which cause a change in dipole moment will show an IR absorption.

Those which show no change in dipole moment may observed by raman spectroscopy. For

example, H2O. It is a bent molecule.

HH

Symmetric stretching

O HH

Asymmetric stretching

O

HH

Scissoring

O

Easy str. > E sym. str. > Esci

or Vasy st. > Vsym. st. > vsci

— CH2 group :

Symmetric stretching

C

H

H

Assymetric stretching

C

H

H

Wagging or out of plane bending

C

H

H

Rocking or Asymmetric inplane bending

C

H

H

Twisting or out of plane bending

C

H

H Scissoring orsymmetic in

plane bending

C

H

H

Fig. (2.1)

25


2.3 VIBRATION OF A SINGLE PARTICLE (CLASSICAL)

Let us consider a particle of mass attached to fixed position through a spring

(tensionless).

Now, we consider the type of vibrational motion, the particle of mass m undergoes.

The spring through which the body is fixed is such that if the particle is removed a distance

from its equilibriums position, it experiences a restoring force (fr) which is proportional to

its displacement from the equilibrium position. A spring which behaves in this manner is

said to obey Hooke’s law.

m

For such behavior we can write df x where fd = driving force, x = displacement from

equilibrium position

df kx

where k = Proportionality constant called force constant

But fd = – fr where fr = restoring force.

rf kx

fr is in opposite direction which tends to keep the particle in equilibrium position.

The force constant (k) which appears in the molecular problem measures the stiffness

of the spring i.e. bond. It gives a restoring force (fr) for unit displacement from equilibrium

position. The negative reign indicates that fr is directed opposite to x.

The potential energy (U) is work that must be done to displace the particle a distance

dx.

Therefore, the potential energy is given by

app ddU f dx f dx

rf dx

rdU

fdx

But from Hooke’s law, we have

26


rf kx

Ud

kx kxdx

dU kxdx

If the equilibrium position is taken as that of zero potential energy

U

0 0

x

dU kxdx 21U

2kx ...(2.1)

This gives the expression for potential energy of vibrating particle.

From the expression for potential energy it is clear that is equation of a parabola. This

is potential energy (U) of particle increases parabolically as the particle moves in either

direction from equilibrium position.

x

k

Fig. (2.2)

The equation describing the motion of the particle can be set up as :

From Newton’s law, we know that

2

2.d x

f ma m mxdt

Also dU

fdx

dUmx

dx

0dU

mxdx

...(2.2)

0kx mx

This gives the equation for vibrational motion.

27


0dU d dx

mdx dt dt

10

dU dm dx

dx dt dt

10

dUm dx

dx dt

0dx

dU m dxdt

0dU mxdx ....(2.3)

This is required form of equation of motion for vibrating particle

Again, since dU

kxdx

dU kxdx

Equation (2.3) becomes as

0kxdx mxdx

Now, we can integrate it to get potential energy (P.E.) and kinetic energy (K.E.) part of

vibrational energy.

0k xdx m xdx

2 2

2 2

kx mxE ; where E is Integration constant which gives total energy

2 21 1

2 2E kx mx ...(2.4)

. . . .E P E K E

where 2 21 1. . & . .

2 2P E kx K E mx

Thus total energy associated with the vibrating particle is the sumn of KE. and P.E.

The expression for the vibrational frequency may be obtained as :

We know that equation for the vibrational motion is

0dU

mxdx

28


–dU

mx kxdx

0mx kx ...(2.5)

It is a differential equation of second degree. It has solution of the form

cos 2x A t ...(2.6)

where A = amplitude of vibration

= vibrational frequency

= Phase angle

2 .sin 2dx

x A tdt

2

2 2& 4 cos 2d x

x A tdt

Putting the value of x & x in equation (2.5), we get

2 24 os 2 2 0mA C t kA Cos t

2 2Cos 2 4 0A t m k

2 24 0m k

2 24 0m k

2

2

1

4

k

m

1

2

k

m...(2.7)

1

2 c

k

c m...(2.8)

This equation is the important classical result for the frequency of vibration. It shows

that a particle with mass m held by a spring with force constant k will vibrate according to

equation (2.2) with frequency given by equation (2.7). Only this frequency is allowed.

The energy with which the particle vibrate can be shown to depend upon the maximum

displacement, i.e. amplitude ‘A’ of the vibration.

29


2.4 THE VIBRATION OF THE TWO PARTICLE SYSTEM (CLASSICAL) :

DIATOMIC MOLECULE

m2 at equilibrium

extended

compressed

Restoring force

Original length

Restoring force

m1

x1 x2

x2x1

Fig. (2.3)

Let us consider conservative system of two particles having mass m1 and m2, joined by

a massless perfectly elastic spring. On applying force (f) the particle move only along bond

axis of the system with displacement x1 and x2 from equilibrium position. Such motions are

assumed to be harmonic in nature giving rise to harmonic vibrations. The magnitude of the

force that restores each particle to the equilibrium position is proportional to the extent of

compression or extention of the bond (spring) i.e. Restoring force 2 1f x x

2 1(x x )f k f kx ...(2.8)

where k is characteristics of a bond called the force constant.

2 1x x x = displacement with respect to mean position. The negative sign indicates

that the restoring force (f) acts in a direction opposite to displacement. Also, if x is positive,

it corresponds to extension and compression gives a negative value of x.

The work that must be done to displace the particles by a distance dx is - fdx. This

work is stored in the system as potential energy (P.E.), dU so that

dU = – fdx ...(2.9)

If zero potential energy is taken at the equilibrium position, then

0 0 0

U x x

dU fdx kx dx

222 1

1 1

2 2U kx k x x ...(2.10)

30


It complicates the system as it is no longer an equation of simple parabola. Similarly

kinetic energy (K.E), T of the bond is given by

2 21 2

1 21

2

dx dxT m m

dt dt

...(2.11)

For each particle i, the Langrange equation can be written as

i

d dT dU

dt dx dx

= 0

For particle 1, 2

11 2 1 2 12

d xm K x x K x x

dt

or 1 1 2 1m x k x x ....(2.11(a))

and for particle 2

2

22 2 12

d xm k x x

dt

2 2 2 1.m x k x x ...(2.11(b))

The change in sign of two equation is due to their vibration in opposite

direction.

Equation (2.11(a) has solution, 1 1 os2x A C t

and equation (2.11(b)) has solution, 2 2 os2x A C t

1 1 2 in 2x A S t

and 2 21 1 4 2x A Cos t

2 21 1 2 14 os 2 os 2m A C t K A A C t

2 21 1 2 14m A k A A

2 21 1 1 24 0m A kA kA

31


2 2 21 1 2( 4 )A 0m k KA ...(2.12 (a))

Similarly from equation (2.11 (b)), we get

2 12 2 k x xm x

2 2 os 2x A C t

22 2 2 in 2

dxx A S t

dt

2

2 222 22

–4 os 2d x

x A C tdt

2 22 2 2 14 os 2 os 2m A C t k A A C t

2 22 2 1 24 0m A kA kA

2 21 2 24 0kA m k A

Thus equation (2.12(a)) (2.12(b)) are simultaneous equation of first degree

Solution : (i) A2 = A2 = 0 0

It is a trivial solution and so is meaningless

(ii) For non-trivial solution we construct secular determinant of A1 & A2

2 21

2 22

40

4

m K K

K m K

From this secular equation, we can obtain expression for frequency

2 2 2 21 24 4 0m k m k k

4 4 2 2 2 2 2 22 1 216 4 4 0m m m k m k k k

32


2 2 2 21 2 1 24 4 0m m m k m k

2 21 2 1 24 0m m m m k

2 21 2 1 24 m m m m k

1 22

21 2

1

4

k m m

m m

2

2

1.

4

k; where

1 2

1 2

m m

m m

is reduced mass

1

2

k...(2.13)

It gives the expression for frequency of vibration of two particles and its

relation with force constant (k) and reduced mass () of the system. The frequency of the

vibration is written as osc in Hz and osc in cm–1 as :

then 2 0v and hence 0v

1

2osck

... (2.13 a)

1

2osck

c...(2.13 b)

Also, if 2 24 0

then = 0 and hence = 0

It corresponds to the motion in which both particles are displaced by the same amount

in same direction i.e. x1 = x2.

Thus = 0 corresponds to translational motion

On substituting the osc in equation (2.12 a) & (2.12 b), we get

1 2 1

2 1 2

A m x

A m x ...(2.14)

33


If m2 is lighter than m1, the vibrational amplitude of m2 will be correspondingly

greater than that of m1.

On substituting the value of force constant k = 2 24 osc in equation (2.10),

we get

2 2 21 14

2 2 oscU kx x

2 2 22 oscU x ...(2.15)

It shows that in simple harmonic motion (S.H.M) the potential energy (U) is

proportional to the square the displacement of the centre of gravity of the

molecule. The potential energy is parabolic.

The concept of reduced mass () reduces the vibration of two atoms in a

molecule to the vibration of a single mass point, whose amplitude equal the

amplitude change (A2-A1) of the vibrating atoms in the molecule. An increase in energy

will make the oscillations more vigorous, i.e. the degree of compression or extension will be

greater but the vibrational frequency osc will be the same. Such a model gives a vibrational

frequency independent of the amount of bond distortion. However classical mechanics

allows amplitudes and therefore the energy of vibration to attain any value contrary to the

quantum nature of energy.

(Fig. 2.4): Some of the vibrational energy levels & allowed transition of H.O.

diatomic molecule (NO)

2.5. SCHRÖDINGER EQUATION APPLIED TO HARMONIC OSCILLATION

There are a few simple systems where the potential energy is not constant, yet the

Schordinger equation can be exactly solved. For example, vibration of a diatomic molecule

and motion of an atom in a crystal lattice.

34


Let us consider a particle of mass ‘m’ attached to a weightless spring and restricted in

the same, way so that it can move only in the x-direction. The force acting on this particle is

given by Hooke’s law as :

df x

fd

frm

Since direction of driving force (fd) and restoring force (fr) are opposite to each other,

d rf f

rf x

rf kx where k is proportionality constant known as force constant.

If fr = 1 dyne, x = 1 cm k = 1 dyne / cm.

It is defined as equal to force per unit displacement. This type of force is called harmonic.

Whenever the motion of a particle can be described by the simple law known as Hooke’s

law, then the system is said to be harmonic oscillator.

The potential energy, U is given by dU

fdx

dU fdx kxdx

dU kxdx

On integrating, we get

0 0

U x

dU k xdx

21

2U kx

The kinetic energy, T is given by

221 1

2 2

dxT mv m

dt

22 21 1

2 2xp

m vm m

where xp mv is linear momentum. For two particle system having mass m1 and m2,

the kinetic energy, is given by

35


2 21 2

1 21 1

2 2

dr drT m m

dt dt

2 21 1 2 2

1 1

2 2T m r m r

The reduced mass, is given by

1 2

1 1 1

m m

22 2 21 1

2 2 2xp

T x x

Total energy associated with the system is given as

H T U

221

2 2xp

H kx

This is expression of energy through classical mechanics

Classically the equation of motion is expressed as

2

2.d x

m kxdt

(Newton’s law)

which has the general solution as 0 sink

x x tm

where 0 &x are arbitrary constants, x0 being the amplitude of the oscillation. According

to classical mechanics, the particle oscillates from 0x x to 0x x

sinusoidally with the time at a frequency

1

2

k

m

The energy then changes back and forth from the kinetic energy form to the potential

energy form, being exclusively kinetic when x = 0 and exclusively potential when x = x0. The

total energy is constant given by 20

2

kx. It can have any positive value, there being no limit

36


on the value of x0 . Now let us cansider the properties of such a system according to the law

of quantum mechanics.

Let us first set up Hamiltonian operator for Harmonic oscillator (H.O.)

We know that

221

2 2xp

H U T kxm

2ˆ 21ˆ ˆ

2 2xp

H kxm

2 22

2

1ˆ

2 2kx

x

(

ˆxp ix

22ˆx

x

p i2 2

2 2 2

2 2i

x x

)

The schodinger equation for harmonic oscillator then may be written as : we

know that S.E. in operator form is as

H E

2 22

2

1

2 2kx E

x

22

2 2 2

2 1 2

2kx E

x

22

2 2

2 10

2E kx

x....(2.16)

This is Schrödinger equation for harmonic oscillator. The problem is now to find the

well behaved functions which satisfy the equation (2.16) and the allowed energy levels.

The solution of this equation vanishes at infinite and is single valued and

finite, and the energy is discontinuous but changes by integral value of the

vibrational quantum number v given by

1

2 2osc

h kE v

...(2.17)

where v = 0, 1, 2, 3 .... known as vibrational quantum number (can take only

positive integer values, including zero).

The quantum energy levels with the simple harmonic oscillator as a model are

equidistant and have been represented in figure (2.4)

37


It should be particularly noted that energy at v = 0 is not zero but 1

2 osch and is called

zero point energy. It has no counterpart in the classical approach. This is also in accordance

with the Hiesenberg’s uncertainty principle, i.e. at 0 K (–273ºC) when even translational,

rotational motion have been frozen, uncertainty of the position of the molecules still exist

due to zero point energy and is equal to 1

2 osch per vibrational mode.

For example in case of NO,

11904osc cm

zero point energy = –34 101 16.625 10 1904 3 10

2 2h

= 18923 × 10–24 J

It is conventional to express vibrational modes and energy levels in cm–1 as follows

:

11( )

2osc

osc

EG v cm

hc

where G(v) is called term value, vosc the molecular vibration in wave numbers, and v

is the vibrational quantum number.

Zero point energy in cm–1 = 2oscv

zero point energy of NO in cm–1 = 11904

9522

cm

Again, 1

= +2 2π

h kE v

Thus the quantum mechanics requires that only certain discrete energies are

assumed by the vibrator. The term

2

h k

appears in both classical as well as quantum mechanical treatment

38


Also, from classical mechanics treatment we know that

1 2

1 2

1 1

2 2m

k m mk

m c m m....(2.18)

From equation (2.17) and (2.18), we have

1

2mE v h

....(2.19)

where m is the vibrational frequency of mechanical model.

If we now assume that transitions in vibrational energy levels can be brought about by

radiation, provided the energy of radiation exactly matches the

difference in energy levels E between the vibrational quantum states and

provided that the vibration causes a change in dipole. This difference is identical between

any pair of adjacent levels, because v in equation (2.17) & (2.19) can

assume only whole numbers. That is ,

2mh k

E h ...(2.20)

At room temperature majority of molecules use in the ground state (= 0). Thus

01

2 mE h

In order to move to the first excited state with energy 13

2 mE h requires radiation of

energy

3 1

2 2m m mh h h

The frequency of radiation that will bring about this change is identical to the classical

vibrational frequency of the bond m . Thus,

2radiation mh k

E h E h ...(2.21)

1

2mk

Hz ...(2.22)

39


This expression can also be written in terms of wave number of radiation.

Thus,

1 2 1

1 2

1 1

2 2

k m mkcm

c c m m....(2.23)

where, = wave number of absorption peak in cm–1

k = force constant (in dynes/cm)

c = velocity of light ( 3 × 1010 in cm/s)

m1 & m2 are masses of two atoms

The allowed vibrational energy levels and transition between them for a diatomic

molecule undergoing simple harmonic motion may be shown as :

=0

r eq.Internuclear distance

osc.

osc.

osc.

osc.

osc.

osc.

osc.

osc.

osc.

osc.

19 2

17 2

15

2

13

2

11

2

9

2

7 2

5

2

3 2

1

2

En

erg

y (

cm)

–1

osc.

cm–1

Fig. (2.5). The allowed vibrational energy levels and transition between them for a

diatomic molecule undergoing simple harmonic motion.

40


2.6 ZERO POINT ENERGY (E0)

According to the old quantum theory, the energy levels of a harmonic

oscillation were given by

En n h

If this were true, the lowest energy level would be that with n = 0, and would therefore

have zero energy. This would be state of complete rest and represent the minimum in

potential energy curve. The uncertainty principle does not allow such a state of complete

defined position and completely defined momentum (in this case zero). As a result wave

mechanical treatment show that the energy levels of the oscillator are given by

01

2vE v h ...(2.24)

where v is the vibrational quantum number which may take on the values,

v=0, 1, 2, 3 ..... The vibratory motion of the nuclei of a diatomic molecule can be represented

as vibration of a simple harmonic oscillator. In such an oscillator the vibrational energy Ev is

related to the fundamental vibrational frequency 0 by the above wave mechanical

relationship. The above equation shows that such an oscillator retains the energy 0 01

2E h

in the lowest vibrational level v = 0. This residual energy, called zero point energy of the

oscillator cannot be removed from the molecule even cooling it to 0 K (–273ºC). The enrgy

E0 = 01

2h must be added to the planck’s expression for the mean energy of an oscillator..

The implication is that the diatomic molecule (and indeed any molecule) can never

have zero vibration energy; the atoms can never be completely at rest

relative to each other. The quantity 01

2h Joules or

10

1

2cm , the zero point energy;

depends only on the classical vibration frequency and hence on the strength of the chemical

bond (k) and atomic masses ().

The prediction of zero point energy is the basic difference between the wave mechanical

and classical approaches to molecular vibrations. Classical mechanics could find no objection

to a molecule possessing no vibrational energy but wave mechanics insists that it must

always vibrate to some extent, the latter conclusion has been amply borne out by experiment.

2.7 SELECTION RULE

Further use of the Schrödinger equation leads to the simple selection rule for the

harmonic oscillator undergoing vibrational changes :

1v

41


To this, we must of course add the condition that vibrational energy changes will only

give rise to an observable spectrum if the vibration can interact with radiation, i.e. if the

vibration involves a change in the dipole moment of the molecule. Thus the vibrational

spectra will be observable only in hetronuclear diatomic molecules since homonuclear

molecules have no dipole moment.

Applying the selection rule we have immediately :

11 1

12 2v v osc oscE v v

1osc cm ...(2.25)

for absorption, whatever the initial value of v.

Such a simple result is also obvious from the figure given above; since the

vibrational levels are equally spaced, transitions between any two neighboring states will

give rise to the same energy change. Further since the difference between energy levels we

expressed in cm–1 gives directly the wave number of the spectral line absorbed or emitted.

spectroscopic osc

This, again, is obvious if one consider the mechanism of a absorbing or emission in

classical terms. In absorption for instance, the vibrating molecule will absorb energy only

from radiation with which it can coherently interact and this must be radiation of its own

oscillation frequency.

For example, NO (nitric oxide) molecule.

The expected vibrational energy levels for NO molecules are equally spaced. Transition

between any two neighbouring states will give rise to the same energy change and thus

only one line. The NO as harmonic oscillator should absorb at 1904 cm–1

or

10 1 10 13 10 1904 5712 10osc osc

cc s s

The energy of this quanta of radiation is

34 10 246.626 10 5712 10 37847 10oscE h J J

It is the energy which must correspond to the energy difference, even for a

0 1v v transition.

2.8 BOLTZMANN DISTRIBUTION

With the help of Boltzman distribution we can calculate the number of molecules in v

= 1 state relative to the ground vibrational state at room temperature , i.e. 27ºC or 300 K

42


241

230

37847 10exp exp

1.38 10 300

Ev KT

v

N

N

9.1428 0.000107e

It shows that less than 1% of molecule are in the v =1 state and still small

number in higher levels. It follows that in experiment at room temperature only transition

from the ground vibrational state v = 0 are of major importance. Also the force constant can

be calculated as :

2 24 osck

with 2712.495 10 kg for NO molecule.

10 15712 10osc Hz or s

22 27 104 3.14 12.495 10 5712 10k

11609.4k Nm SI units

5 116.094 10 (CGS units)k dyne cm

Force constant (k) in a quantitative way is a measure of how strong are force of attraction

between the two atoms of a molecule.

The horizonal lines in figure in the previous section represent some of the vibrational

states. The points of intersection of the potential energy curve and horizontal lines give the

value req + q, where Evib = U, the kinetic energy is zero so that

At req +q, Evib= U

21 1

2 2oscv h kq

where q is amplitude of the vibration

for v = 0, the maximum value of amplitude of vibration (qmax) can be

2max

1 1

2 2oschv kq

12

maxoschv

qk

for NO molecule,

qmax =

122437847

0.004851609.4

nm

43


Stiffness of a chemical bond, which is a measured by the force constants (k) is reflected

in the amplitude of vibration. The greater rigidity and lower amplitude of the HF than those

of HI molecules.

2.9 FORCE CONSTANT (K) AND BOND STRENGTHS

In a harmonic oscillator, the restoring force per unit displacement is called force constant

k by (fr = –kx) and is given by

2 2 2 1 2

1 2

4 4m m

Km m ...(2.26)

where is vibrational frequency (in Hz) and m1 and m2 are masses of the oscillating

atoms.

The force constant for diatomic molecule can be obtained by using above equation

provided the vibrational frequency () is known. Force constant for polyatomic molecule

cannot be determined directly. A method has been suggested by assuming that each valency

bond has a certain definite value for the force constant, which is characteristic of the bond

and independents of the molecule in which it occurs. Force constants of some bonds are

given in the table below.

Table 2.1. Force constants (k) of some bonds in dynes/cm

5 55

tan ( )10 10

10

4.9 — 4.8 — 4.6

12.3 12.1 9.5

18.6 17.5 15.8

Force cons t kBond Bond k Bond k

C O C N C C

C O C N C C

C O C N C C

It is an interesting fact that, the force constant (k) increases approximately in proportion

to the multiplicity of the bond, and so the former can be used to give an indication of the

latter. For example, the force constant for the carbon oxygen bond in carbon dioxide has

been found to be 15.7 × 105 dynes/cm. This value lies between C=O and C O as shown in

the table. This result is in good agreement with the resonance structure of carbon dioxide,

which is

– —O C O O C O O C O

Similarly, the force constant given for C O is that for carbon monoxide (CO). Thus

the value 18.6 × 105 dyne/cm provides support to the triple bond structure of carbon

monoxide.

The bond length varies inversely with the bond order and bond energies vary directly

with bond order. Force constant (k) of the bond is directly related to bond order. Hence, as

bond order increases, bond energy as well as force constant increase. As bond energy

increases, bond length decreases and force constant increases. As bond length increases

44


bond energy decreases and force constant also decreases. As force constant increases, bond

length decreases and bond energy increases.

2.10. THE ANHARMONIC OSCILLATOR

Though a simple harmonic oscillator gives a good picture, it does not explain the

following points :

(i) The potential energy and therefore the restoring force increases infinitely with

increasing distance from the equilibrium position. Therefore it places no limit on how far a

bond can be stretched, while in an actual molecule, when atoms are at a great distance from

one another, the attractive force is zero, the bond will break and correspondingly the potential

energy has a constant value.

(ii) The summary of the potential energy curve (figure in the previous section) reflects

that the restoring force for compression and extension is the same which is not the case.

The bonds strongly resist compressing as revealed by relative incompressibility of solids.

Although for small compressions and extensions the bond may be taken as perfectly

elastic for larger amplitudes say greater than 10% of the bond length, a much more

complicated behaviour must be assumed. Figure given below shows, diagrammatically the

shape of the energy curve for a typical diatomic molecule, together with (dashed) the ideal,

simple harmonic parabola.

0.5 1.0 1.5 2.0 2.5 Å

Deq.

1.5Deq.

Energy

0.5 Deq.Deq.

Internuclear distance

Figure (2.6) : The Morose Curve : the energy if a diatomic molecule

undergoing anharmonic extensions and compression.

A purely empirical expression which fits this curve to a good approximation was derivedby P.M. Morse, and is called Morse function

45


2

( ) De 1 qU q e ...(2.27)

where q measures the distortion of bond from the equilibrium length and is a constantsuch that

272

1.2177 10osc oscc

Deh De...(2.28)

De = Dissociation energy of the molecule measured from equilibrium position,expressed in cm–1. The constant determines the narrowness and curvature of the curve. Aplot of U(q) versus q gives an asymmetric curve, what is called he Mose potential curve.The Mose potential curve for the NO molecule is shown in the figure given below withpotential energy U(q) in electron volts.

When equation U(q) = De (1–e–Bq)2

is used instead of 21U( ) ;

2 eqq kq q r r

in the Schrödinger equation, the pattern of the allowed vibrational energylevels is found to be

211 1

2 2V e e eE v v x cm ...(2.29)

where v = 0, 1, 2, ....

and e is an oscillation frequency (expressed in wave numbers) which we shall define

more closely below, and xe is the corresponding anharmonicity constant which, for bondstretching vibrations, is always small and positive ( +0.01), so that the vibrational levelscrowd more closely together with increasing v. Some of these levels are sketched in figuregiven below

Deq.

1.5Deq.

Energy

0.5 Deq.Deq.

0

1

2

3

4

5

678

r=9

D

Internuclear distance

46


cm–1

~e ~2e ~3e

Fig. 2.7: The allowed vibrational energy levels and some transitions between them

for a diatomic molecule undergoing anharmonic oscillations.

It should be mentioned that equation (2.29) like equation (2.27) is an

approximation only; more precise expression for the energy levels require cubic, quartic

etc. terms in 1

2v

with anharmonicity constant ye, ze etc. rapidly diminishing in magnitude.

These terms are important only at large values of v, and we shall ignore them

If we rewrite the equation (2.29), for anharnomic oscillator as

1 11

2 2v e ex v v ...(2.30)

and compare with the energy levels of the harmonic oscillation given as

11

2v oscv cm ...(2.31)

we see that we can write

11

2osc e ex v ...(2.32)

Thus the anharmonic oscillator behaves like the harmonic oscillator but with an

oscillation frequency which decreases steadily with increasing v. If we now consider the

hypothetical energy state obtained by putting v =1

2 (at which, according to equation (2.30)

=0) the molecule would be at the equilibrium point with zero vibrational energy. Its

oscillation frequency (in cm–1) would be

osc e

Thus we see that e may be defined as the (hypothetical) equilibrium oscillation

frequency of the anharmonic system the frequency for infinitely small vibrations about the

equilibrium point. For any real state specified by a positive integral v the oscillation

frequency will be given by equation (2.32)

Thus in the ground state (v = 0) we would have

47


10

11

2e ex cm

and

10

1 11

2 2e ex cm

we see that zero point energy differs slightly from that of the harmonic oscillator

10

1

2 osc cm

The selection rules for the anharmonic oscillator also change to

v 1 , 2, 3 .....

Thus they are the same as for the harmonic oscillator, with the additional

possibility of large jumps. These, however, are predicted by theory and observed in practice

to be of rapidly of diminishing probability and normally only the lines of 1v ±2 and ±3

at the most, have observable intensity. Further, the spacing between the vibrational levels is,

as we shall shortly see, of the order 10–3 cm–1 and at room temperature we may use the

Boltzmann distribution to show

1

0v

v

N

N

=

34 10 3

23

6.63 10 3 10 10exp exp

1.38 10 300

hckT

4.8exp 0.008

In other words, the population of the v = 1 state is nearly 0.01 or some one percent of

the ground sate population. Thus, to a very good approximation, we may ignore all transitions

originating at v = 1 or more and restrict ourselves to three transitions :

1. 0 1v v , 1v with considerable intensity

1 0v v

2 21 1 1 1

1 12 2 2 2e e e e e ex x

11 2e ex cm ...(2.33 a )

2. 0 2, 2v v v With small intensity

2 21 1 1 1

2 22 2 2 2e e e e e ex x

12 1 3e ex cm ...(2.33 b)

48


3. 0 3v v , 3v , with normally negligible intensity

2 21 1 1 1

3 32 2 2 2e e e e e ex x

13 1 4e ex cm ...(2.33 c)

These three transition are shown in the previous figure. To a good approximation,

since 0.01ex , the three spectral lines lie very close to e , 2 e & 3 e . The line near e is

called fundamental absorption, while those near 2 e and 3 e are called the first and second

overtones respectively. The spectrum of HCl for example, shows a very intense absorption

at 2886 cm–1, a weaker one at 5668 cm–1, and a very weak one at 8347 cm–1. If we wish to

find the equilibrium frequency of the molecule from these data, we must solve any two of

the three equations :

1 2 2886e ex

2 1 3 5668e ex

13 1 4 8347e ex cm

and we find 12990e cm , xe = 0.0174

Thus we see that, whereas for the ideal harmonic oscillator the spectral absorption

occurred exactly at the classical vibrational frequency for real, anharmonic molecules the

observed fundamental absorption frequency and the equilibrium frequency may differ

considerably.

The force constant of the bond in HCl may be calculated directly by inserting the

value of e

2 2 2 14 ek c Nm

22 2 1 8 274 3.14 2990 10 3 10 1.64 10k m

= 516 Nm–1

Although we have ignored transitions from v = 1 to higher states we should note that,

if the temperature is raised if the vibration has a particularly low frequency, the population

of the v = 1 state may becomes appreciable. Thus at, say 600 K (i.e. about 300ºC),

49


1

0

v

v

N

N

= exp–2.4 0.09, and transition from v =1 to v =2 will be same 10 percent the

intensity of those from v = 0 to v = 1. A similar increase in the excited state population would

arise if the vibrational frequency were 500 cm–1 instead of 1000 cm–1. We amy calculate the

wave number of this transition as :

4. 1 2; 1v v v normally very weak.

1 1 1 12 6 2

2 4 2 4e e e e e ex x

11 4e ev x cm ...(2.32 d)

Thus, should this weak absorption arise, it will be found close to and at slightly lower

wavenumber than the fundamental (since xe is small and positive). Such weak absorption

are usually called hot bands since high temperature is one condition for their occurrence.

Their nature may be confirmed by raising the temperature of the sample and a true hot

band will increase in intensity.

Therefore, the anharmonisity of the potential function has introduced the following

points :

1. Earlier calculation had shown that only one absorption band will be observed

corresponding to the oscillation frequency. Introduction of anharmonicity in molecular

vibrations could explain the hot bands and overtones.

2. The zero point energy is not half of the oscillation frequency. As in NO molecule it is

978.66 cm–1 which is not half of osc = 1904.4 cm–1.

3. The oscillation frequency and fundamental absorption band are not the same.

The values for NO molecule are 1876.097 cm–1 and 1876.097 cm–1 respectively

4. The energy levels are not equally spaced but a decrease is observed with the increase

in vibrational quantum number.

5. No vibration energy level is expected for 40v in the ground electronic state of the

NO molecule but is expected to dissociates.

2.11 FUNDAMENTAL AND OVERTONE BANDS

The Infrared spectrum of diatomic molecules exhibit more than one bands. Some

additional bands which occur at higher frequencies are attributed to anharmonisity. As such

it need correction at high value of vibrational quantum number. As the vibrational quantum

number v changes form 0 to 1, 1 to 2 etc. the spacing decreases. The former occurs at higher

frequency than latter. The transition for 0 1 vibrational level is called fundamental

vibration. It corresponds to one unit change 1v

50


Those transitions accompanying 2, 3 etc unit change occur in infrared region which

are termed as overtone bands. The modified selection rules for anharmonic vibration are:

1, 2, 3v ....

The first three possible transition are

1. 0 2 , 2v ; first overtone (second harmonic)

2. 0 3 , 3v ; second overtone (third harmonic)

3. 0 4 , 4v ; Third overtone (fourth harmonic)

The selection rule 1v as discussed in the beginning breaks down and all transitions

with v greater than 1 are classified as overtones. Actually the infrared spectrum comprises

both fundamental and overtone bands. But the overtones are less probable and does out

after 0 3 transition

The energy spacing for fundamental and overtones are given as:

1. 0 ´ 1v , ´ 1v v v

1 0E E E

33 11 1

2 2 2 2e e

e e

x xhc hc

9 1

4 4e e e e ehc hc x hc x

2e e ehc hc x

11 2e ehc x cm

1 2 ;vib e vib ex where hc

2. 0 ´ 2v v , ´ 2v v v

2 0

5 5 11 1

2 2 2 2e

e e e

xE E E hc x hc

25 1

24 4e e e e ehe hc x hc x

2 6e e ehc hc x

51


2 1 3e ehc x

st2 1 3 ;1 overtonevib eE x

nd3 1 4 ; 2 overtonevib eE x

3. 0 to 3, 3

3 0

7 7 11 1

2 2 2 2e

e e e

xE E E hc x hc

3 12e e ehc hc x

3 1 4e ehc x

3 1 4vib ex ; 2nd overtone

These three bands will occur at position 1 2e ehc x , 2 1 3e ehc x and 3 1 4e ehc x .

For approximate work, these bands are located at e , 2 e & 3 e cm–1

respectively. The line which is nearer to e is fundamental band. The lines which occur at

two or three time, the fundamental bands ( 2 e and 3 e ) are overtones. Their intensities

follow the order:

2 3e e e

2.12 COMBINATION BANDS

Vibrational spectra are also complicated by the fact that two different vibrations in a

molecule can interact to give absorption peaks with frequencies that are approximately the

sums or differences of their fundamental frequencies. The combination bands are merely

the sum of two or more fundamental frequencies or overtone ( 1 2 1 2 1 2 3; 2 ,

etc.). While the difference bonds are differences of two or more fundamental frequencies or

overtone (e.g. 1 2 1 2 1 2 3, 2 , etc.) The intensities of combination and

difference peaks are generally low.

2.13 VIBRATION OF POLYATOMIC MOLECULES : NORMAL MODES OF

VIBRATION

In case of polyatomic molecules, there can be a number of vibrational modes. We

do not apply the selection rules regarding dipole moment of the molecule in vigorous

manner. In even a non polar molecule (e.g. CO2) with change in dipole moment during

52


vibration gives the infrared active bands. The force constant values are different in polyatomic

molecule and as a result of which the vibrational spectrum reveals a complicated structure.

Consequently there is a direct way of assigning all their vibration . The only way is to

calculate the degree of freedom associated with various molecular motion which determine

the different types of vibration.

Molecules vibrate in a number of ways, we call them vibrational modes. These

vibrations are attributed to the compression or extension of chemical bonds or bending (or

deformation) of bond angles. Each vibration mode has got a characteristics frequency of

vibration.

Let us consider a molecule containing N-atoms. We can refer to the position of each

atom by specifying three coordinates (e.g. x, y & z certesian co ordinate). Thus the total

number of coordinate values is 3N and we say the molecule has 3N degree of freedom.

However, once all 3N coodinates have been fixed , the bond distances and bond angles of

the molecule are also fixed and no further arbirary specification can be made.

Now, the molecule is free to move in three dimensional space, as a whole

without change of shape. This translational movement uses three of 3N degrees of freedom

leaving 3N -3.

In general, also the rotation of a non linear molecule can be resolved into components

about three perpendicular axes. Specification of these axes also requires three degrees of

freedom, and molecule is left with 3N-6 degree of freedom. The only other motion allowed

to it is internal vibration, so we know immediately that a non linear N atomic molecule can

have 3N-6 different internal vibrations.

Non linear molecule : 3N-6 fundamental vibration.

On the other hand, if the molecule is linear, there is no rotation about the bond axis,

hence only two degrees of rotational freedom are required leaving 3N-5 degrees of vibrational

freedom, one more than in case of non-linear molecule.

Thus, linear molecule : 3N -5 fundamental vibration

In both cases, since N-atomic molecule has N-1 bonds (acyclic) between its

atoms, (N-1) of the vibrations are bond stretching motions.

The other 2N-5 (non linear) or 2N-4 (for linear molecule) are bending motions.

These 3N-5 (for linear molecules) and 3N-6 (for non-linear molecule)

vibrational motion are also referred as normal modes of vibration. In general , a normal

modes, of vibration is defined as molecular motion in which all the atoms move in phase

and with the same frequency.

Now these normal vibrations are further classified as :

(a) Stretching mode :

In this mode, a molecule may vibrate by compression or extension of bonds. We call it

53


bond stretching vibration. The distance between the vibrating atoms are periodically

changing and thus the mean value is called bond length.

The different types of stretching mode are further classified as

(i) Symmetric stretching and

(ii) Asymmetric stretching

In a symmetric stretching bonds one either compressed or elongated, so that the

symmetry of the molecule is preserved.

In symmetric stretching one bond undergoes extension while the other is

compressed.

For example, in the case of methylene group, H—C—H; the two H-atoms move away

from the central carbon atom without change in bond angle in case of symmetric stretching.

In case of asymmetric stretching one H-atom approaches carbon atom while the other H-

atom moves away from the carbon atom.

Symmetricstretching

Asymmetricstretching

Stretching vibrations

(b) Bending mode

Such vibrations may consists of a change in bond angle between bonds with a common

atom or the movement of a group of atoms with respect of the remainder of the molecule

without movement of the atoms in the group with respect to one another. These are of four

types :

(i) Scissoring : In scissoring the two atoms concerned to a atom move towards and

away from each other with deformation of the valency angle (in plane bending)

(ii) Rocking : In rocking, the structural units swings back and forth in the plane of

the molecule (in plane bending)

(iii) Wagging : wagging, the structural unit strong back and forth out of the plane of

the molecular out of plane bonding)

(iv) Twisting : In twisting, the structural unit rotates about the bond which joins it

to the remainder of the molecule fact of plane bending.

54


Scissoring(In plane bending)

Rocking(In plane bending)

Wagging

(Out of planebending)

Twisting(Out of plane bending)

Fig. 2.8. (Bending vibrations)

In a molecule containing more than two atoms, all the four types of vibrations may be

possible.

Let us consider the infrared spectrum of carbon dioxide (CO2) which is a linear triatomic

molecule and thus has a four normal nodes, i.e. 3N-5 = 3 × 3 – 5 = 4 modes of vibration.

Non linear triatomic molecules, such as H2O, SO2, N2O etc have 3N- 6 = 3 × 3-6 =3

vibrational modes.

Examples : (a) H2O

O

C2 axis

H H

Let us consider water (H2O) molecule. it is non linear (bent) and triatomic

N = 3

Normal modes of vibration = 3N — 6 = 3 × 3 – 6 = 3

Each of motion is described as stretching or bending depending upon the

nature of change in molecular shape .

H H

Symmetric

stretching 1

O

(a)

Parallel ( || )

H H

Bending 2

O

(b)

Parallel ( || )

H H

Asymmetric

stretching 3

O

(c)

Perpendicular ( )

55


The IR spectrum of water shows the three bands at 1595 cm–1 3652 cm–1 and 3756

cm–1. The force constant for stretcing vibration is more than that of bending vibration and

so it is easier to deforms or bend than to stretch it.

The stretching vibrations (1 & 3) have frequency considerably larger than bending

vibration (2). Moreover, the asymmetric stretching (3) occurs at larger frequency than the

symmetric stretcing (1). The three modes of vibrations corresponds to

(i) 3652 cm–1 (1) Symmetric stretching

(ii) 3756 cm–1 (3) Asymmetric stretching

(iii) 1595 cm–1 (2) Bending

Thus 2 > 1 > 2

(b) CO2

It is a linear triatomic molecule.

Here N = 3

Normal modes of vibration = 3N– 5 = 3 × 3– 5 = 4

Its normal modes of vibration are shown in the figures given below

(i) Symmetric stretching

~ 1330 cm–1

1

O C O

(ii) Asymmetric stretching

or 2349.3 cm–1

3

O C O

(iii) Bending

or 667.3 cm–1

2

O C O

It is non polar molecule with no permanent dipole moments ( = 0). The two stretching

modes of vibrations are :

(i) Symmetric stretching : Here both covalent bonds are in same phase. The molecule

remains symmetrical in course of vibration with no change in dipole moment. Therefore,

symmetric stretching vibration is IR inactive.

56


O C O

= 0, 0

d

dr IR inactive

11340i cm

(ii) Asymmetric stretching :

here one bond contracts and other expands. This leads to disappearance of initial

symmetry of the molecule giving the formation of instantaneous dipole making the band

IR active

O C O

A bond which occurs at 2349 cm–1 for CO2 is assigned to asymmetric stretching (3)

(iii) Bending vibration : The molecule can bend in two independent directions at

right angles to each other. They occur at the same frequency and so they give a doublet i.e.

a double degenerate vibration. There is a loss of symmetry during bending vibration of the

molecule giving the formation of instantaneous dipole and is IR active.

This absorption bond occurs at 667 cm–1

O C O O C O

If CO2 is a linear molecule, two fundamental bands would be expected in

infrared region. A doubly degenerate band at 667 cm–1 assigned to bending

vibration and a high energy band at 2349 cm–1 is due to asymmetric vibration. The symmetric

stretching (1) is Raman active and it appears near 1340 cm–1.

2.14 GROUP FREQUENCY

Due to 3N-6 and 3N-5 rules it is well evident that a complex molecule is expected to

have an infrared spectrum that can exhibit a large number of normal vibrations. Each normal

mode involves some displacement of all, or almost all the atoms in the molecule. However

in some of the normal modes, all atoms may undergo approximately the same displacement

while in other the displacement of a small group of atoms may be much more vigourous

than the displacement of the remainder of atoms. Therefore the normal nodes can be divided

into two classes.

(i) Skeletal vibration

(ii) Characteristics group vibrations

57


Skeletal vibration :

It involves the displacement of all the atoms to the same extent. A linear or branched

chain structure and molecular moiety gives the skeletal vibration and it usually falls in the

range of 1400-700 cm–1. Thus group such as

—C—O—C—C

C

C

CC

give rise to several skeletal modes of vibration and hence several absorption bonds. It is

difficult to assign a particular band to a specific vibrational mode but the whole complex of

bands observed is highly typical of a molecular structure. If changing a substituent results

in a marked change in the chemcical nature, it is reflected in the change of absorption

bands and vice versa.

These bands are generally known as the finger print bands, because the molecular or

structure can often be recognized simply by the appearance of this part of the spectrum.

Characteristics Group frequency

It involves only a small portion of the molecule, the remainder being more or less

stationary. From the observation of the infrared spectra of a number of

compounds having a common group of atoms, it is found that, regardless of rest of the

molecule, this group absorbs over a narrow range of frequencies called the group frequency.

The group frequencies are usually almost independent of the structure of the molecule

as a whole and, with a few exception fall in the region well above and well below that of the

skeletal modes. We see that the vibrations of light atoms in terminal groups (for example —

CH3, —OH, —C N, > C = O etc.) are of high frequency, while those of heavy atoms (—C—

Cl, —C—Br, metal - metal etc.) are low in frequency. Their frequencies and consequently

their spectra, are lightly characteristics of the group, and can be used for analysis.

For example, the —CH3 group gives rise to a symmetric C—H stretching

absorption invariably falling between 2850 & 2890 cm–1, an asymmetric stretching frequency

at 2940 -2980 cm–1, a symmetric deformation (i.e. the opening and closing of the C

H H

I

H

‘umbrella’) at about 1375 cm–1 and an asymmetic deformation at about 1470 cm–1. Again,

the > C = 0 group shows a very sharp and intense absorption between 1600 and 1750 cm–1,

depending largely on the other substituents of the group. For example, let us deduce the

58


structure of thioacetic acid-acetic acid in which one oxygen atom is replaced by sulphur. It

may be CH3CO. SH or CH3CS.OH. The infrared spectrum gives a very clear picture. It

shows a very sharp absorption at about 1730 cm–1, and one at about 2600 cm–1 . These are

consistent with the presence of > C=O and —SH groups respectively. Also there is little or

no absorption at 1100 cm–1 (apart from the general background caused by the skeletal

vibrations), thus indicating the absence of > C = S.

The table given below collects some of the data of characteristics streching frequency

of some molecular groups :

Table 2.2

Group Approximate Group Approximate

frequency (cm–1) frequency (cm–1)

—OH 3600 C=O 1750-1600

—NH2 3400 C=C 1650

CH 3300 C=N 1600

H

3060 C—C

=CH2 3030 C—N 1200-100

CH3 2970 (asym. stretch) C—O

2870 (sym. stretch) C = S 1100

1460 (asym. deform.) C—F 1050

C—Cl 725

1375 (sym. deform) C—Br 650

59


—CH2— 2930 (asym. stretch) C—I 550

2860 (sym. stretch)

1470 (deformation)

—SH 2580

—C N 2250

—C C— 2220

2.15 VIBRATIONAL-ROTATIONAL SPECTROSCOPY

Let us consider a diatomic molecule which has rotational energy separations of 1—10

cm–1, while the vibrational energy separation (of HCl say) were nearly 3000 cm–1. Since the

energies of the two motions are so different we may, as a first approximation, consider that

a diatomic molecule can execute rotations and vibrations quite independently. This is called

B o rn-Oppenheimer approximation and is tantamount to assuming that the combined

rotational vibrational energy is simply the sum of the separate energies :

total rot vibE E E (in joules)

total rot vib ( in cm–1) ...(2.34)

Taking the separate expressions for

rot & vib , we have

,J V J vE E E

2 32 31 1 1 ...BJ J DJ J HJ J 2

–11 1+

2 2e e ex cm

...(2.35)

Initially we ignore the small centrifugal distortion constants D, H etc, and hence write.

2

,1 1

12 2total J v e e eBJ J v x v ...(2.35 a)

The rotational levels are sketched in figure (2.9) given below for the two lowest

vibrational levels, v = 0 and v = 1. There is however no attempt at scale in this diagram since

60


the separation between neighboring J values, is in fact, only some 1

1000 of that between the

v values. Note that since the rotational constant B in equation (2.35 a) is taken to be the

same for all J and v, the separation between two levels of given J is the same in the v = 0 and

v =1 states.

It may be shown that the selection rules for the combined motions are the same as

those for each separately, therefore we have

1, 2v etc , 1J ...(2.36)

Strictly speaking we may also have v =0, but this corresponds to the pure rotational

transitions. However, a diatomic molecule except under very special and rare circumstances,

may not have J = 0, in other words a vibrational change must be accompanied by a

simultaneous rotational change.

In figure (2.10) we have drawn some of the relevant energy levels and tansitions,

designating rotational quantum numbers in the v=0 state as J´´ and in the v =1 as J´.

0

Internuclear axis

Fig. (2.9). The rotational energy levels for two different vibrational

states of a diatomic molecule

The rotational level J” are filled to varying degrees in any molecular population, so the

transitions shown will occur with varying intensities. This is indicated schematically in the

spectrum at the foot of fig. (2.10).

61


Fig. (2.10). Some transitions between the rotational-vibrational levels

of a diatomic molecule together with spectrum arising from them.

An analytical expression for the spectrum may be obtained by applying the selection

rules (Eq. 2.36) to the energy levels (Eq. 2.35 a). Considering only the v =0 v=1 transition,

we have in general

, ´v 1 ", 0J v J J v

1 1 1 1´(J´ 1) 1 2 " " 1

2 4 2 4e e e e e eBJ x BJ J x

62


10 ´ " ´ " 1B J J J J cm

(where for simplicity we write 0 for e (1-2xe)

we should note that taking B to be indentical in the upper and lower

vibrational states a direct consequence of the Born Oppenheimer approximations rotation

is unaffected by vibrational changes

Now we can have

1. 1J i.e. ´ " 1J J or ´– " 1J J

Hence, 1, 2 " 1J v o B J cm , where J” = 0, 1, 2 ...2.37 (a)

2. 1J , i.e. " ´ 1, ´ " 1J J or J J ;

and 1, 0 2 ´ 1 , ´ 0, 1, 2....J v B J cm where J ....2.37 (b)

These two expressions may conveniently may be combined into

1, . 0 2J v spect B m cm ; m = ± 1, ± 2 ...2.37 (c)

where m replacing J”+1 in eqn (2.37 a) and J´ +1 in eq. (2.37 c) has postive values for

J = +1 and is negative if J = –1. Note particularly that m cannot be zero since this would

imply values of J´ or J” to be –1.

The frequency 0 is usually called the band origin or band centre.

Equation (2.37 c), then represents the combined vibration rotation spectrum. Evidently

it will consists of equally spaced lines (spacing = 2B) on each side of the band origin 0 , but

since 0m , the line at 0 itself will not appear. Lines to the low frequency side of 0 ,

corresponding to negative m (i.e. J = –1) are referred to as the P branch, while those to the

high frequency side (m positive, J = +1) are called the R branch. This apparently arbitrary

notation may become clearer if we state here that latter, in other contexts, we shall be

concerned with J values of 0 and ±2, in addition to ±1 considered here, the labelling of the

line series is then quite consistent : Lines arising from

J = – 2 – 1 0 +1 +2

called: O P Q R S

Let us consider the fundamental vibration rotation band of carbon monoxide (CO)

under high resolution with some lines in the P and R branches. The band centre is at about

2143 cm–1 while the average line separation near the centre is 3.83 cm–1. This immediately

gives

63


2B = 3.83 cm–1

B = 1.915 cm–1

This is in good agreement with the value B = 1.92118 cm–1 obtained by microwave

studies. Therefore, we can obtain quite good values for the rotational constant(B) and hence

moment of inertia (I) and bond length from infrared data alone. Historically, of course the

infrared values came first , the more precise

microwave values following much latter.

The bond origin at the mid point of P(1) ( 12139.43 cm ) and R(O) (

12147 .80 cm ) is

at 2143.26 cm–1. This, then is the fundamental vibration frequency of carbon monoxide, if

anharmonisity is ignored. The latter can be taken into account, however, since the first

overtone is found to have its origin at 4260.04 cm–1.

we have

01 2 2143.26e ex

2 1 3 4260.04e ex

From which 12169.74e cm , xe = 0.0061

2.16 FACTORS AFFECTING THE BAND POSITIONS AND INTENSITIES

The isolated multiple bond such as > C = C < or —C C— have group frequencies

which are hightly characteristics. When two such groups which in

isolation, have comparable frequencies occur together in a molecule, resonance

occurs and the group frequencies may be shifted considerably from the expected value.

Thus the isolated carcarboyl in a ketone C=OR

R and the > C = C < double bond, have

group frequencies of 1715 & 1650 cm–1 respectively. However, when the grouping

C = C—C = O occurs, their separate frequencies are shifted to 1675 and about 1600 cm–

1 respectively and the intensity of the >C = C< absorption increases to become comparable

with that of the inherently strong > C = O bond. This is due to Fermi resonance.

Closer coupling of the two groups, as in the ketene radical, >C=C=O, gives, rise to

absorption at about 2100 and 1100 cm–1, which are very far removed from the characteristics

frequencies of the separate groups

Shifts in group frequencies can arise in other ways too, particularly as the

result of interactions between different molecules. Thus the —OH stretching

64


frequency of alcohols is very dependent on the degree of hydrogen bonding, which lengthens

and weakens the —OH bond, and hence lower its vibrational frequency. If the hydrogen

bond is formed between the —OH and say, a carbony group (> C =O), the latter frequency

is also lowered, though to a less extent than the —OH, since hydrogen bonding weakens the

> C=O linkage also. The shifts position of group frequency caused by resonance or

intermolecular effects are in themselves highly characteristic and so are very useful for

diagnostic purposes.

In a similar way, a change of physical state may cause a shift in the frequency of a

vibration. Particularly if the molecule is rather polar. In general a more condensed phase

gives a lower frequency

gas liquid solution solid

For example a shift of about 100 cm–1 is obtained in polar molecule like HCl in passing

from vapour to liquid and a further decrease of 20 cm–1 on solidification.

Non polar CO2 molecule shows almost negligible shifts in symmetric vibration (1=1330

cm–1) but a decrease of about 60 cm–1 on solidification in asymmetic vibration (v3 = 2349.3

cm–1).

In general increasing the mass of atom undergoing, oscillation within the group (i.e. -

increasing ) tends to decrease the frequency e.g., the series CH, CF, CCl CBr, or the values

of >C = O & >C = S. Also increasing the strength of the band and hence increasing the force

constant (k) tends to increase the frequency; e.g. the series —C—X, —C = X, —C X, where

X = C, N or O (in case of first two fragments)

Now let us consider very briefly the intensities of infrared bands we have seen that an

infrared spectrum appears only when the vibration produces a change sin the permanent

electric dipole of the molecule. It is reasonable to suppose, then, that the more polar a

bond, the more intense will be the infrared spectrum arising from vibration of that bond.

This is generally borne out in practice. Thus the intensities of the bonds of the groups given

below decreases in the order

C = O > C = N — > C = C

Similarly, intensities of the bands of the following groups given below decreases in the

order,

— OH > NH > — CH

Because of this reason also, the vibration of ionic crystals lattices often give rise to very

strong absorption.

2.17 STUDY OF VIBRATIONAL FREQUENCIES OF CARBONYL COMPOUNDS

The organic compounds containing carbon oxygen double bond ( C = O) are called

carbonyl compounds.

65


In aldehydes, the carbonyl group is bonded to a carbon and hydrogen while in ketones,

it is bonded to two carbon atoms The carbonyl compounds in which carbonyl group is

bonded to oxygen are known us carboxylic acid, and their derivatives (e.g. esters, anhydrides)

while in compounds where carbonyl carbon is attached to nitrogen and to halogens are

called amides and acyl halides respectively. The general formula of these carbonyl compounds

are given below

C

O

RH

Aldehyde

C

O

R R´

Ketone

C

O

R OH

Carboxylic acid

C

O

R X

Acycl halide

C

O

R O

Acid anhydride

C

O

R´

C

O

R OR´

Ester

C

O

R NH2

Amide

Infrared spectroscopy is by far the best way to detect the presence of a

carbonyl group in a molecule. The strong band due to C = O stretching appears at about

1700 cm–1, where it is seldom obscured by other strong absorption. It is one of the most

useful bands in the infrared spectrum

C = O stretching, strong

RCHO 1725 cm–1 R2CO 1710 cm–1

ArCHO 1700 cm–1 ArCOR 1690 cm–1

–C = C–CHO 1685 cm–1 – C = C – C = O 1675 cm–1

Cyclobutanones 1780 cm–1 Cyclopenanones 1740 cm–1

– C = C – C –

OH O

(enols)

1540-1640 cm–1

The carbonyl band is given not only by aldehydes and ketones, but also by carboxylic

acids and their derivatives. Once identified as arising from an aldehyde or ketone, its exact

frequency can give a great deal of information about the structure of the molecule.

The —CHO group of an aldehyde has characteristics C—H stretching band at 2720

cm–1; this in conjuction with the carbonyl band is fairly certain evidence for an aldehyde.

66


Carboxylic acids and esters also show carbonyl absorption, and in the same general

region as aldehydes and ketones. Acids, however, also show the broad

O—H band. For hydrogen bonded (dimeric) acids O—H stretching gives a strong, broad

band in the 2500 -3000 cm–1 range.

O—H stretching strong broad

—COOH and enols 2500 — 3000 cm–1

R—OH & ArOH 3200 — 3600 cm–1

For hydrogen bonded acids, the C = O band is at about 1700 cm–1.


R—C—O—H

O

1700-1775 cm–1

—C = C —C —OH

O

1680-1700 cm–1

Ar—C—OH

O

1680 – 1700 cm–1

—C = CH —C —

OH O

(enols)

1540 - 1640 cm–1

Acids also show a C—O stretching band at about 1250 cm–1 and bands for

O—H bending near 1400 cm–1 and 920 cm–1 (broad)

Enols, too, show both O—H & C = O absorption; these can be distinguished by the

particular frequency of the C = O band. Aldehydes ketones & esters show the carbonyl

absorption, but O—H band is missing.

Esters usually show the carbonyl band at somewhat higher frequencies than ketones

of the same general structure, furthermore esters show characteristics C—O stretching

bands.

The infrared spectrum of an acyl compound show the strong band in the

neighbourhood of 1700 cm–1 that we have come to expect of C=O stretching. The exact

frequency depends on the family the compound belongs to and, for a member of a particular

family on its exact structure. For ester, for example

67



RCOOR 1740 cm–1

RCOOAror

R–COO–C=C–

1770 cm–1

ArCOORor

– C = C – COOR1715–1730 cm–1

Esters are distinguished from acids by the absence of the O—H band. They are

distinguished from ketones by two strong C—O stretching bands in the

1050-1300 cm–1 region, the exact position of these bands, too, depends upon the structure

of the esters structure.

Besides the carbonyl band, amides (RCONH2) show absorption due to

N—H streching in the 3050-3550 cm–1. region and absorption due to N—H bending in the

1600 -1640 cm–1 region.

Table 2.3. (Infrared spectrum of some carbonyl compounds)

Compounds O—H C—O C = O

Alcohols 3200-3600 cm–1 1000–1200 cm–1

Phenols 3200–3600 cm–1 1140-1230

Ether, (Aliphatic — 1060-1150

& Aromatic)

1200-1275

1020-1075

Aldehdes ketones — — 1675-1725 cm–1

Carboxylic acids 2500-300 1250 1680-1725

Esters — 1050-1300 1715-1740

(two bands)

Acid Chloride — — 1750-1810

Amides (N-H 3050-3550) — 1650-1690

(R CONH2)

68


2.18. EFFECT OF HYDROGEN BONDING ON VIBRATIONAL FREQUENCIES

Hydrogen bonding can occur in any system containing a proton dono group (X —H)

and a proton acceptor (Y) if the s-orbital of the proton can effectively overlap the p or -

orbital of the acceptor group. Atoms X and Y are electronegative with Y possessing lone pair

of electrons.

In organic molecules, the common proton donor groups are carboxylic acid, hydroxyl

amine or amide groups and common proton acceptor atoms are oxygen, nitrogen and

halogens.

The strength of the bond is inversely proportional to the distance between X & Y. The

force constant of both the groups X & Y is altered as a result of hydrogen bonding. Hence

frequenices of both stretching as well as bending vibration are altered because of hydrogen

bonding. The X–H stretching bands move to lower frequencies usually with increased

intensity and band widening. The stretching frequency of the acceptor group, e.g. C = O, is

also reduced but to a lesser degree than the proton donor group. The X—H bond bending

vibration usually shifts to higher frequencies or shorter, wave length when bonding occurs.

Thus hydrogen bonding changes the position and shape of infrared absorption band.

For example spectra of pure alcohols show a wide band for the O—H stretching

vibrations as a result of extensive hydrogen bonding. In case of cyclohexanol, the O—H

stretching vibration occurs ground 3330 cm–1 (lower frequency) when hyrogen bonding is

less extensive a sharper and less intense band is observed at higher frequency at about 3600

cm–1.

In the case of intermolecular hydrogen bonding some OH bonds are bonded and some

are non bonded. So both peaks may be obtained. The sharp hydrogen bonded O—H bond

around 3600 cm–1 is expected to occur in the vapour phase, in dilute solution or if steric

hinderance prevents hydrogen bonding. Generally the infrared spectra of pure solids, liquids

(e.g. cyclohexanol) and many solutions show only the broad hydrogen bonded band. The

bonded O—H stretching appears at lower frequency than the O—H stretch because of

lengthening of the original O—H bond on hydrogen bonding.

R—O—H + O — R R — O + H — O — R R—O ----H----O—R

H H H

(I) (II) (III)

(Lenthening of O—H bond in hydrogen bonding)

Hydrogen bonding may be considered as resonance hybrid as a result of which the

bond gets weakened, its force constant (k) is reduced and hence streching frequency ( ) is

decreased. Thus the value of O—H stretching gives the measure of the strength of the

hydrogen bond. The stronger the hydrogen bond, the longer the O-H band, the lower the

vibration frequency and broader and more intense will be the absorption band.

69


Intermolecular hydrogen bonding involves association of two or more molecules of

the same or different compounds, and it may result in dimer molecules as in carboxylic

acids or in polymer molecules, which exists in pure samples or concentrated solution of

monohydric alcohols. In very dilute solution, formation of intermolecular hydrogen bonds

does not occur because the molecules are much separated. Increasing the concentration of

alcohol or phenol causes the sharp band around 3600 cm–1 to be replaced by a broad band

of lower frequency, which is due to —OH groups that are associated through intermolecular

hydrogen bonding. If we consider the spectrum of tertiary butyl alcohol as pure liquid and

as its solution in CCl4 in high frequency region, we find that in spectrum of pure liquid a

strong bonded O—H stretch around 3360 cm–1 is observed, whereas in its solution with

CCl4, the 3360 cm–1 O—H absorption is also accompanied by a sharp band at 3620 cm–1.

On further dilution with CCl4, the 3620 cm–1 band becomes more intense relative to 3360

cm–1 band. Both these bands are due to O—H stretch. The band at higher frequency is due

to the streching mode of free hydroxy. The streching mode of hydrogen bonded or associated

O—H occurs at lower frequency of 3360 cm–1.

An intramolecular hydrogen bonding remains unaffected on dilution and as a result

the absorption band also remains unaffected. Intermolecular hydrogen bonds are broken

on dilution and as a result there is a decrease in the bonded O—H absorption and an

increase in or the appearance of free O—H absorption. Hydrogen bonding in chelates and

enols is very strong and so the observed O—H stretching frequency may be as low as 2800

cm–1 since these bands are not easily broken on dilution by an inert solvent, free O—H

streching may not be seen at low concentration.

O

H

O

C

OCH3

O

C

CH

C

O

H

CH3 CH3

Chelate of mettylsalicylate Enol of Acetyl acetone

In general, intermolecular hydrogen bonds gives rise to broad bands, while

intramolecular hydrogen bonds give sharp and well defined bands.

Solved Problems

Ex. 2.1 What will be the force constant (k) for the bond in HCl if the fundamental

vibrational frequency is 8.667 × 1013 s–1 ?

Solution :

m1 ( for H-atom)= 1.008 amu = 1.008 × 1.66 × 10–24 g =1.6739 × 10–24 g

= 1.6739 × 10–27 kg

m2 (for Cl atom) = 35.5 amu = 35.5 × 1.66 × 10–27 kg = 5.8951 × 10–26 kg

70


1

2

k

c

Reduced mass, = 1 2

1 2

m m

m m

= 1.6277 × 10–27 kg

Force constant, k = 2 22

22 13 27 14 3.14 8.667 10 1.6277 10 Nm

= 483.1 Nm–1 Ans

Ex.2.2. Calculate the vibrational absorption frequency of the carbonyl, > C = O

group if force constant of the double bond is 1.0 ×106 dyne/cm2

Soln. Mass (m1) of oxygen atom = 16 × 1.66 × 10–24 g = 2.65 × 10–23 g

mass (m2) of carbon atom = 12 × 1.66 × 10–24 g = 2.0 × 10–23 g

Reduced Reduced mass () = 1 2

1 2

m m

m m =

23 23

23

2.65 10 2.0 10

2.65 2.0 10

g g

g

235.3 10

4.65

= 1.14 × 10–23 g

We know that

6

10 23

1 1 1.0 10

2 2 3.14 3 10 1.14 10

k

14

10

2.96 10

2 3.14 3 10

=

14

11

2.96 10

1.88 10 = 1.574 × 103 cm–1 = 1574 cm–1 Ans

Ex.2.3 Calculate the approximate wavelength of absorption associated with

C—H bond in streching vibration of methyl group. The force constant for single

bond is 5 × 105 dynes cm–1. The mass of carbon and hydrogen atoms are 2 × 10–23

g and 0.167 × 10–23 g respectively.

Soln. 10

1 1

2 2 3.14 3 10

k

c

1/25 23

23 23

5 10 2 0.16 10

0.167 10 2 10 = 3023 cm–1

11,Now cm

41 13.31 10

3023cm cm

71


4 2 63.31 10 10 3.31 10 3.31m m m

1 41: 1 10Note cm

m

Ex.2.4 Calculate the theoretical number of vibrational degree of freedom in (a)

Benzene (b) carbon dioxide (c) water (d) sulphur dioxide (e) N2O

Soln. (a) Benzene (C6H6) is a non linear molecule.

Thus vibrational degrees of freedom = 3N - 6

= 3 12 6 30 (since N 6C 6H 12)

(b) Carbondioxide (CO2) is a linear triatomic molecule. Thus vibrational degrees of

freedom = 3N —5 = 3 × 3 – 5 = 4

(c) Water (H2O) is a non linear triatomic molecule. Thus, vibrational degrees of

freedom = 3N— 6 = 3 × 3 — 6 =3

(d) Sulphur dioxide (SO2) also non linear triatomic molecule.

Thus, vibrational degree of freedom = 3N–6 = 3 × 3 – 6 = 3

Ex.-2.5 Calculate the theoretical number of vibrational degrees of freedom in (a) HCN

(b) C6H5CH3 (c) CH4 (d) NH3 (e) CCl4 (f) CH3 F

Soln. (a) For HCN, H—C N : Linear triatomic molecule; vibrational degrees of fredom =

3 × 3 – 5 = 4.

(b) For C6H5—CH3, Planar (linear), N = 15 vibrational degrees of freedom = 3N– 5 =

3 × 15 – 5 = 40 .

(c) CH4, non-planar pentaatomic molecule vibrational degrees of freedom = 3N – 6 =

3 × 5 – 6 = 9.

(d) CCl4 , N = 5, vibrational degrees of freedom = 3N – 6 = 3 × 5 – 6 = 9.

(e) CHF3, N =5, vibrational degrees of freedom = 3N – 6 = 3 × 5 – 6 = 9.

It is well evident from above discussion that infrared spectroscopy is very helpful

in deciding the shapes of molecules.

Ex.2.6 Following characteristics absorption peaks have been observed, in the

infrared spectrum of an organic compound having formula, C2H6O

(a) Strong band at 3300 cm–1 (b) band at 2965 cm–1 (c) Band at 2920 cm–1

(d) Bond at 1050 cm–1

Name the compound

Soln: A single broad peak at 3300 cm–1 suggest the presence of an —OH group

(hydrogen bonded ) in the compound

A strong bond at 1050 cm–1 conform the presence of primary alcoholic group

(C—O structure). The bond at 2965 cm–1 and 2920 cm–1 may be due to C—H streching. The

presence of primary alcoholic group (—CH2OH) and the molecular formula C2H6O suggests

that the given compound is CH3—CH2—OH, ethylalcohol.

72


Ex.-2.7 Indicate whether following vibrations will be active or inactive in infra

red region ?

Molecule Motion

(a) CH3—CH3 C—C stretching

(b) CH3—CCl3 C—C stretching

(c) SO2 symmetric streching

(d) H2C = CH2 C—H stretching C = C

H

HH

H

(e) H2C = CH2 C—H stretching C = C

H

HH

H

(f ) H2C = CH2 CH2 wagging C = C

H

HH

H

(g) H2C = CH2 CH2 twisting C = C

H

HH

H

Soln: (a) Inactive

(b) Active

(c) Active

(d) Active

(e) Active

(f) Active

(g) Inactive

73


Ex.2.8 Give the characteritics absorption band in the infrared spectrum of

n-octane, CH3(CH2)6CH3

Soln: (a) A band at 2960 cm–1 is due to C-H streching in CH3 and a band at 2900 cm–1 is

due to C-H streching in CH2. Both these bands indicate the presence of CH3 and

CH2 grouping in the molecule.

(b) A band at about 1470 cm–1 is due to C-H bending in the —CH3 group and a band

at about 1375 cm–1 is due to C-H bending in the —CH2 group.

Ex.2.9 Give approximate positions of characteristics infrared bands in the

following

(a) CH2—C—CH3

O

(b) CH3—CH2—OH

(c) CH3—C—OH

O

(d) CH3—C—CH2—CH =CH2

O

Soln: (a) C—H stretcing band at 2960-2850 cm–1

C = O stretching peak at about 1700 cm–1

(b) C-H stretching peak at 2960-2850 cm–1 and O-H stretching peaks at 3630 cm–1.

(c) O-H stretching peak at about 3030 cm–1 and a band at about 1700-1750 cm–1 due

to C = O gr.

(d) C-H stretching peak at 2960–2850 cm–1; C=O stretching peak at about 1700 cm–

1, C = C stretching peak at 1650 cm–1 and = C—H stretching peak at 3040–3010

cm–1.

Ex.2.10 Give characteristics absorption bands of the carbonyl group in the IR spctra of

the following

(a) CH3—C = O

H

(b) CH3—C —CH3

O

(c) C6H5—C = O

H

(d) CH3—C = O

OH

Soln: (a) CH3—C = O

H

~ 1740 cm–1

74


(b) CH3—C = O

CH3

~ 1700 cm–1

(c) C6H5—C = O

H

~ 1700 cm–1 (d) CH3—C = O

OH

~ 1700-1750 cm–1

Ex.-2.11 Give the approximate characteristics absorption bands of C-N group in

the IR spectra of following compounds

(a) C6H5—NH2 (b) (C6H5)2NH

(c) (C6H5)3N (d) CH3—NH2

Soln: (a) 1340- 1250 cm–1

(b) 1340 – 1280 cm–1

(c) 1360 – 1310 cm–1

(d) 1220 – 1020 cm–1

Ex.2.12. How does you differentiate the following each pair using IR spectra ?

(a) Aniline & N-methylaniline

(b) Acetaldehyde from Ethanol

Soln: (a) Aniline (C6H5-NH2) gives N-H group stretching at about 3400 cm–1

but N-methyl aniline (C6H5-NH-CH3) gives N-H stretching at about

3450 cm–1.

(b) Acetaldehyde (CH3—C = O

H

) gives C = O group stretching at about 1730 cm–1,

while ethanol gives O—H stretching at about 3630 cm–1.

Model Qestions

Q.2.1. Write down the Schödinger equation for Harmonic oscillator and show the

quantization of vibrational energy levels. What is zero point energy?

Q.2.2. (a) What is force constant and how does it related with band strength.

(b) Calculate the difference in frequency expected for the v=0 to v=1 vibrational

transition of HCl35 and HCl37 assuming that the force constant of the two

molecule are identical and equal to 4.84 × 105 dynes/cm.

75


Q.2.3 (a) How will you show using infrared spectrum of o-hydroxy benzaldehyde the

presence of intramolecular hydrogen bonding in the molecule

(b) How many normal modes of vibration are possible for (a) HBr (b) OCS (linear)

(c) SO2 (bent) and (d) C6H6?

Q.2.4 (a) What are the characteristics frequencies that are made use of in establishing keto

enol tautomerism in ethylacetoacetate? How do you show with the help of infrared

spectrum the predominance of enol form?

(b) Why water is not used as a solvent in infrared spectroscopy ?

(c) Why is methanol a good solvent for ultraviolet spectroscopy but not infrared

spectra ?

Q.2.5 (a) What is vibration rotation spectroscopy ?

(b) Explain P, Q, R branches in vibration-rotation spectroscopy.

Q.2.6 (a) Discuss the vibrational frequencies of carbonyl compounds.

(b) Explain the solvent effect on vibrational frequencies.

References

1. Fundamentals of molecular spectroscopy; C.N. Banwell, Tata McGraw Hill, New Delhi.

2. Infrafed spectra of Inorganic and Coordination Compounds; K. Nakamoto; (John Wiley

& Sons, New York)

3. Molecular Spectroscopy, Principles and Chemical applications; P,R singh & S.K. Dikshit

(S.Chand & Co. Ltd., New Delhi)

4. Molecular spectroscopy, P.S. Sindhu; (Tata McGraw Hill, New Dehil)

5. Physical methods in Inorganic Chemistry; R.S. Drago, East West Press Pvt. Ltd., New

Delhi.

vibrational spectra (infrared spectra)

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