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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 Modes
of 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 Carbonyl
compounds
2.18 Effect of Hydrogen bonding on vibrational
frequencies
Solved Examples
Model Questions
References
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 ( (^) in cm-1) is used to measure the position
of a given infrared absorption. The range of IR spectrum is as :
1 1 1 1
6 8
Wave number cm cm cm cm Wave length
where is micron such that m cm
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, H 2 O. It is a bent molecule.
H^ H
Symmetric stretching
O H H
Asymmetric stretching
O
H
H
Scissoring
O
E (^) asy str. > E (^) sym. str. > E (^) sci or V (^) asy st. > V (^) sym. st. > v (^) sci
— CH 2 group :
Symmetric stretching
C
H
H
Assym etric stretching
C
H
H
Wagging or out of
plane bending
C
H
H
Rocking or
Asymmetric in
plane bending
C
H
H
Twisting or
out of plane bending
C
H
H
Scissoring or
symmetic in
plane bending
C
H
H
Fig. (2.1)
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 ( f (^) r ) 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 fd x where f (^) d = driving force, x = displacement from
equilibrium position
f d kx
where k = Proportionality constant called force constant But f (^) d = – f (^) r where f (^) r = restoring force.
f^ r kx
f (^) r 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
dU fapp dx fd dx
f dxr
r
dU f dx
But from Hooke’s law, we have
dU d dx m dx dt dt
dU d m dx dx dt dt
dU m dx dx dt
dx dU m dx dt
dU mxdx ^ ^ (^0) ....(2.3)
This is required form of equation of motion for vibrating particle
Again, since
dU kx dx
dU kxdx
Equation (2.3) becomes as
kxdx mxdx ^ 0 Now, we can integrate it to get potential energy (P.E.) and kinetic energy (K.E.) part of vibrational energy.
k (^) xdx m xdx 0
2 2 2 2
kx mx E ; where E is Integration constant which gives total energy
2 ^2
E kx mx (^) ...(2.4)
E P E.. K E..
where ^2 ^ ^2
P 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
^ 0
dU mx dx
dU mx kx dx
mx ^ kx (^0) ...(2.5)
It is a differential equation of second degree. It has solution of the form
x A cos 2 t ...(2.6)
where A = amplitude of vibration = vibrational frequency = Phase angle
2 .sin^ 2
dx x A t dt
^ ^
2 & x d x A 4 2 2 cos 2 t dt
Putting the value of x^ & x in equation (2.5), we get
^ ^ ^ ^
mA 4 2 2 C os 2 t kA Cos 2 t 0
A Cos 2 t m 4 2 2 k 0
4 2 2 m k 0
4 2 2 m k 0
2 2
k m
k m
2 c
k c m
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.
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 2 1 2 1 2
dx dx T m m dt dt
^
For each particle i, the Langrange equation can be written as
i
d dT dU dt dx dx
For particle 1,
2 1 1 2 2 1 2 1
d x m K x x K x x dt
or m x 1 1^ k x 2 (^) x 1 ....(2.11(a))
and for particle 2
2 2 2 2 2 1
d x m k x x dt
m 2 (^). x 2^ k (^) x 2 (^) x 1 ...(2.11(b))
The change in sign of two equation is due to their vibration in opposite direction.
Equation (2.11(a) has solution, x 1^ ^ A 1^ C^ os 2 t
and equation (2.11(b)) has solution, x 2^ ^ A 2^ C^ os 2 t
x 1^ ^ A 1^ ^2 ^ S^ in^ 2 t
and ^ ^ ^ 2 2 x 1 (^) A 1 (^) 4 Cos 2 t
^ ^ ^ ^ ^ ^ ^
2 2 m A 1 1 (^) 4 C os 2 t K A 2 (^) A 1 C os 2 t
^
2 2 m A 1 1 (^) 4 k A 2 (^) A 1
4 ^2 2 m 1 (^) A 1 (^) kA 1 (^) kA 2 0
2 2 2 ( 4 m 1 (^) k ) A 1 KA 2 0 ...(2.12 (a))
Similarly from equation (2.11 (b)), we get
m x 2 2 k x 2 x 1
x 2 (^) A 2 (^) C os 2 t
2 2 (^2) (^2) in (^) 2 dx x A S t dt
(^) (^)
2 2 2 2 2 2 2 –4^ os^2
d x x A C t dt
(^) (^) (^) 2 2 m A 2 2 (^) 4 C os 2 t k A 2 (^) A 1 C os 2 t
(^) (^) 2 2 m A 2 2 (^) 4 kA 1 (^) kA 2 0
(^) (^) 2 2 kA 1 (^) 4 m 2 (^) k A 2 0
Thus equation (2.12(a)) (2.12(b)) are simultaneous equation of first degree
Solution : (i) A 2 = A 2 = 0 ^0
It is a trivial solution and so is meaningless
(ii) For non-trivial solution we construct secular determinant of A 1 & A (^2)
2 2 1 2 2 2
m K K
K m K
From this secular equation, we can obtain expression for frequency
^ ^ ^ ^ ^
2 2 2 2 4 m 1 (^) k 4 m 2 k k 0
16 4 m m^4 2 (^) 4 2 2 m k 1 4 2 2 m k 2 k^2^ k^2 0
If m 2 is lighter than m 1 , the vibrational amplitude of m 2 will be correspondingly greater than that of m 1.
On substituting the value of force constant k = (^4) ^2 2 osc in equation (2.10),
we get
(^1 2 1 42 ) 2 2 osc
U kx x
U (^) oscx ...(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 (A 2 -A 1 ) 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.
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 :
f (^) d x
f d
f r
m
Since direction of driving force ( f (^) d ) and restoring force ( f (^) r ) are opposite to each other,
f d fr
f r x
f r kx where k is proportionality constant known as force constant.
If f (^) r = 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 f dx
dU fdx kxdx
dU kxdx
On integrating, we get
0 0
U x dU^ k^ xdx
U kx
The kinetic energy, T is given by
2 1 2 1 2 2
dx T mv m dt
2 1 2 2 1 2 2
m v p^ x m m
where p^ x mv^ is linear momentum. For two particle system having mass m 1 and m 2 ,
the kinetic energy, is given by
on the value of x (^) 0. 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
2 (^12) 2 2
H U T p^ x kx m
2 ˆ ˆ^1 ˆ^2 2 2
H p^ x kx m
2 2 2 2
kx x
p ˆ^ (^) x i x
2 ˆ 2 x x
p i
2 2 2 2 2 i x (^2) (^) x 2
The schodinger equation for harmonic oscillator then may be written as : we know that S.E. in operator form is as
H E
2 2 2 2
kx E x
2 2 2 2 2
kx E x
^
2 2 2 2
E kx x
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
osc
h k
E v
^
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)
It should be particularly noted that energy at v = 0 is not zero but
2 osc
h (^) 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
2 osc
h (^) per vibrational mode.
For example in case of NO,
1904 ^1
osc cm
zero point energy = ^ ^ ^ –34^ ^ ^ ^10
h
= 18923 × 10 –24^ J
It is conventional to express vibrational modes and energy levels in cm –1^ as follows :
osc osc
E
G v cm
hc
^ ^
where G( v ) is called term value, v (^) osc the molecular vibration in wave numbers, and v
is the vibrational quantum number.
Zero point energy in cm –1^ = 2
v osc
zero point energy of NO^ in cm^ –1^ =^
cm
Again, (^)
h k
E v
Thus the quantum mechanics requires that only certain discrete energies are assumed by the vibrator. The term
h k
appears in both classical as well as quantum mechanical treatment
This expression can also be written in terms of wave number of radiation.
Thus,
(^)
1 2 1 1 2
k k m^ m cm c c m m ....(2.23)
where, (^) = wave number of absorption peak in cm – k = force constant (in dynes/cm) c = velocity of light ( 3 × 10 10 in cm/s) m (^) 1 & m (^) 2 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 :
= 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
Energy (cm
)
osc.
cm–
Fig. (2.5). The allowed vibrational energy levels and transition between them for a diatomic molecule undergoing simple harmonic motion.
2.6 ZERO POINT ENERGY (E 0 )
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
(^) (^) 0
v 2 E 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 E (^) v 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 ^ 0
E 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
E 0 = 0
h (^) 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 0
h (^) Joules or 1 0 ^1 2
cm (^) , 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 :
v 1
