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SCYA1101: Engineering Chemistry UNIT 2 : Molecular Spectroscopy
UNIT 2
MOLECULAR SPECTROSCOPY
Electromagnetic Spectrum – Interaction of Radiation with Matter – Energy Levels in
Molecules – Microwave Spectroscopy – Principle – Classification of Molecules Based on
Moment of Inertia – Rotational Energy Expression (J Levels) – Calculation of J for CO
Molecule – Vibrational Spectroscopy – Normal Modes of Vibrations – Vibrations of
Polyatomic Molecules (CO 2
and H 2
O) – Determination of Force Constant – Electronic
Transitions in Organic Molecules – Mathematical Derivation of Beer Lambert’s Law –
Stimulated Emission – Lasers in Action – Excimer Laser, Diode Laser and Gas Laser.
2.1 INTRODUCTION TO MOLECULAR SPECTROSCOPY
2.1.1 Electromagnetic Spectrum
1. Define electromagnetic radiations.
Electromagnetic radiations are composed of radio waves, infrared light, visible light,
ultraviolet ( uv ) light, X–rays and –rays. The electromagnetic radiations characterized by
their wavelength, frequency, wave number and energy.
2. What is electromagnetic spectrum?
Electromagnetic spectrum is the arrangement of various types of electromagnetic
radiations in increasing order of their wavelengths or decreasing order of their
frequencies.
3. Draw electromagnetic spectrum with variable energy.
EMR – rays
X–rays UV Visible Infrared Microwaves Radio waves
Energy (eV) 10
6
10
4
10
2
1 10
‒
10
‒
10
‒ 9
4. What is meant by energy of electromagnetic radiations?
Electromagnetic radiations consist of particles having small packets of energies called
photons. The amount of energy corresponding to each photon is expressed by Planck’s
equation as: 𝐸 = ℎ × 𝛾.
5. Define the terms: wavelength, frequency, velocity, and wave number of a
electromagnetic wave.
Wavelength (λ): The linear distance between successive maxima or minima of a wave
is known as its wavelength. It is expressed in Å or nm or cm or m.
Frequency (γ): The number of vibrations (or) oscillations possible per second is known
as frequency of a wave. Its unit is s
‒
Velocity (c): The product of wavelength and frequency is equal to the velocity of wave.
It is mathematically expressed as: 𝑐 = 𝜆 × 𝛾. Its unit is cms
‒
or ms
‒
SCYA1101: Engineering Chemistry UNIT 2 : Molecular Spectroscopy
Wave Number ( 𝜸
−
) : The number of waves per unit distance is called wavenumber. Its
unit is cm
‒
. The reciprocal of wavelength is also known as wave number. It is
mathematically expressed as: 𝛾
−
𝛾
𝑐
. Its unit is cm
‒
2.1.2 Interaction of Radiation with Matter and Energy Levels in Molecules
1. Define spectroscopy
Spectroscopy is the technique, which deals with transitions (electronic, vibrational,
and/or rotational transitions) of a molecule from one energy level to another by the
absorption of suitable electromagnetic radiations.
2. What is energy level quantization? How is quantization useful for spectroscopic
analysis of molecules?
Energy level quantization is the discrete energy levels present in molecules with respect
to their vibrational, rotational and electronic transitions. Spectroscopy is made possible
only due to quantization of energy levels in molecules.
3. List out the different types of energies contributing to total energy of molecules.
When a photon is absorbed by a molecule, the energy is transferred to the molecule. This
energy causes vibrational, rotational, electronic, or translational transitions in molecule.
E
total
= E
translational
+ E
vibrational
+ E
rotational
+ E
electronic
4. What happens to a molecule when it is irradiated with (i) IR light (b) Microwave
radiation?
The photon of infrared light causes a molecule to undergo vibrational and rotational
transitions. The photon of microwave radiation causes a molecule to undergo rotation
transitions.
5. Describe rotational, vibrational and electronic transitions giving details about
their energy levels and the EMR used.
Electronic transitions: The electronic transition in molecules is caused by the absorption
of high energy photons which can send electrons to high energy level. The electronic
transition occurs within the visible and ultraviolet regions of the electromagnetic
spectrum.
Vibrational transitions: The vibrational transition in molecules is caused by lower
energy photons which causes the changes in vibrational energy levels. The vibrational
transitions occur in near infrared region of electromagnetic spectrum.
Rotational transitions: The rotational transition in molecules is caused by very low
energy photons which can produce the changes in the rotational energy levels in the
molecules. The rotational transitions occur in microwave region of electromagnetic
spectrum.
SCYA1101: Engineering Chemistry UNIT 2 : Molecular Spectroscopy
9. Relate moment of inertia and rotational energy levels.
Moment of Inertia and Energy of the rotational energy levels are related as:
𝐽
2
2
[𝐽(𝐽 + 1 )]
The rotational energy level can also be expressed in terms of wave number 𝛾̅ as:
− 1
) = 𝐵 × [𝐽(𝐽 + 1 )]
Where 𝐵 =
ℎ
8 𝜋
2
𝐼𝑐
is known as rotational constant and J is known as rotational quantum
number.
2.2.1 Classification of Molecules Based on Moment of Inertia
1. Explain in detail about the classification of molecules based on the moment of
inertia.
Based on the moment of inertia, the molecular shapes are classified into: linear
molecules, symmetric top molecules, spherical top molecules and asymmetric top
molecules.
1. Linear Molecules
These are the molecules in which all the atoms are arranged in a straight line. Linear
rotors have one moment of inertia equal to zero [ I A
= 0 and I B
= I
C
].
Examples: HCl, CO 2
, OCS, HCN, C
2
H
2
etc.
Linear molecules can be rotated in 3-directions: (i) About the bond axis (ii) End-over-
end rotation in the plane of the paper, and (iii) End-over-end rotation at right angles to
the plane.
In a linear molecule like HCl, when the atoms (H
and Cl) are rotated about the main principal axis
( x - axis), no need to apply much force. This is
because, the relative positions of atoms (H and Cl)
are not changing. Hence, the moment of inertia
along the bond axis will be zero, I A
=0 (almost
negligible).
When the molecule is rotated along the y - axis
(end-to-end rotation in the plane of paper), the
relative positions of H and Cl atoms are changing
and hence the moment of inertia about y - axis, I B
is
greater.
H
Cl
I
A
I
B
I
C
Similarly, when the molecule is rotated along z - axis (end-to-end rotation perpendicular
to the plane), the relative positions H and Cl atoms are changing to 90°,180°, 270°, 360°,
and hence the moment of inertia about z - axis, I C
is also greater. Since I B
and I C
are
mutually perpendicular to I A
, I
B
= I
C
. Therefore, for a linear molecule, the principal
moments of inertia are I A
= 0 and I B
= I
C
SCYA1101: Engineering Chemistry UNIT 2 : Molecular Spectroscopy
2. Symmetric Top Molecules
Symmetrical tops are molecules with two rotational
axes that have the same moment of inertia and one
unique rotational axis with a different moment of
inertia.
Symmetrical tops can be divided into two
categories based on the relationship between the
inertia of the unique axis and the inertia of the two
axes with equivalent inertia.
I
A
I
B
I
C
C
I
H
H
H
If the unique rotational axis has a lower inertia than
the degenerate axes, I A
< (I
B
= I
C
), the molecule is
called a prolate symmetrical top molecule.
Examples: CH 3
I; CH
3
F; CH
3
CN
If the unique rotational axis has a greater inertia
than the degenerate axes, I C
> (I
A
= I
B
), the
molecule is called an oblate symmetrical top
molecule.
Examples: BF 3
; BCl 3
; NH
3
I
A
I
B
I
C
B
F
F
F
3. Spherical Top Molecules:
Spherical top molecule is a special case of a
symmetric tops with equal moment of inertia about
all the three axes, I A
= I
B
= I
C
. Because of their
spherical symmetry, they have zero dipole moment
and they are microwave inactive.
Examples: Methane (CH 4
); Phosphorus
tetramer (P 4
); Carbon tetrachloride (CCl 4
I
A
I
C
I
B
C
H
H
H
H
4. Asymmetric top molecules
Asymmetric top molecules are a type of polyatomic
molecules having all principal axes of the moment
of inertia are different from each other
(I
A
≠ I
B
≠ I
C
Examples: HCHO; H 2
O; CH
3
OH
I
A
I
C
I
B
C
H
H
O
SCYA1101: Engineering Chemistry UNIT 2 : Molecular Spectroscopy
2
1
1
2
Substituting r 1
and r 2
in equation (1), we get:
2
1
2
×
1
1
2
×
1
2
1
2
1
2
× 𝑟
2
𝐼 = 𝜇 × 𝑟
2
Where, 𝜇 =
𝑚
1
𝑚
2
( 𝑚
1
+𝑚
2
)
is known as reduced mass of the diatomic molecule.
Determination of Energy of a Linear Diatomic Molecule
The angular momentum ( L ) for a rotating molecule can be calculated using the
expression:
𝐿 = 𝐼 × 𝜔 … ( 5 )
Where, 𝐼 is the moment of inertia and ω is the angular velocity.
According to quantum mechanics, the angular momentum is quantized for a rotating
molecule, and it is given as:
Where J is known as rotational quantum numbers.
The energy of a rotating molecule can be determined by using the expression:
𝐽
× 𝐼𝜔
2
Dividing and multiplying the RHS of equation (7) by I , we have:
𝐽
× 𝐼
2
2
𝐽
× 𝐿
2
… ( 8 ) 𝑠𝑖𝑛𝑐𝑒, 𝐿 = 𝐼 × 𝜔
Substituting L value (equation 6) in equation (8), we get:
𝐽
× [
√𝐽(𝐽 + 1 )]
2
SCYA1101: Engineering Chemistry UNIT 2 : Molecular Spectroscopy
𝐽
×
2
2
× 𝐽
𝐽
2
2
[𝐽(𝐽 + 1 )] … ( 9 )
The rotational energy level can be expressed in terms of wave number 𝛾̅.
Dividing the equation (9) by hc on both the sides, we get:
𝐽
2
2
×
× [𝐽(𝐽 + 1 )]
ℎ𝑐
𝜆
𝐸
ℎ𝑐
𝐽
2
×
[
)]
− 1
) = 𝐵 × [𝐽(𝐽 + 1 )] … ( 11 )
Where 𝐵 =
ℎ
8 𝜋
2
𝐼𝑐
is known as rotational constant and J is known as rotational quantum
number.
The rotational energy in terms of wave number can be calculated for various J values by
using the expression (11).
For J = 0; 𝛾̅
− 1
For J = 1; 𝛾̅
− 1
For J = 2; 𝛾̅ (𝑖𝑛 𝑐𝑚
− 1
For J = 3; 𝛾̅ (𝑖𝑛 𝑐𝑚
− 1
The selection rule represents the change
in quantum numbers for allowed
transitions.
For pure rotational transitions, the
allowed transition is ∆𝐽 = ± 1.
The rotational energy for a diatomic
molecule can be diagrammatically
represented as:
J
(cm
)
0
1
2
3
0
2B
6B
12B
The allowed transition, J = 0 1
J = 1 2
J = 2 3
2.2.3 Calculation of J for CO Molecule
3. Explain the determination of J values for the allowed rotational transitions.
Determination Rotational Quantum Numbers ( J values) for Allowed Transitions
We know that, 𝛾̅ (𝑖𝑛 𝑐𝑚
− 1
) = 𝐵𝐽(𝐽 + 1 ), B is rotational constant.
SCYA1101: Engineering Chemistry UNIT 2 : Molecular Spectroscopy
Determination of moment of inertia, I:
The moment of inertia is given by:
1
2
1
2
× 𝑟
2
Where m 1
is the atomic mass of carbon atom; m 2
is the atomic mass of oxygen atom; and
r is the distance of their separation (bond length of CO).
Given:
m 1
= 12.0001 amu = 12.0001×1.66×
‒
kg; m 2
= 15.9994 amu = 15.9994×1.66×
‒
kg; r = 112.8 pm = 112.8×
‒
m
1
2
1
2
12. 0001 × 1. 66 × 10
− 27
×
15. 9994 × 1. 66 × 10
− 27
( 12. 0001 × 1. 66 × 10
− 27
𝑘𝑔) + ( 15. 9994 × 1. 66 × 10
− 27
1
2
1
2
19. 92 × 26. 559 × 10
− 54
2
( 19. 92 + 26. 559 ) × 10
− 27
𝑚 1
𝑚 2
( 𝑚
1
+𝑚
2
)
529 × 10
− 54
𝑘𝑔
2
- 479 × 10
− 27
𝑘𝑔
1
2
1
2
= 11. 38 × 10
− 27
kg
Substituting the above values in equation (2):
𝐼 = 11. 38 × 10
− 27
kg × ( 112. 8 × 10
− 12
2
2
𝐼 = 11. 38 × 10
− 27
kg × 1. 272 × 10
− 20
2
𝐼 = 1. 44 × 10
− 46
kg𝑚
2
Substituting moment of inertia in equation (1), we get:
2
6. 626 × 10
− 34
2
− 1
8 × 3. 142 × 3. 142 × 1. 44 × 10
− 46
2
× 3 × 10
8
− 1
- 626 × 10
− 34
- 42 × 10
− 36
− 1
− 1
−𝟏
SCYA1101: Engineering Chemistry UNIT 2 : Molecular Spectroscopy
J values for first three rotational transition for CO molecule are:
For J = 0, we have:
2B = 2 × 1.935 = 3.87 cm
‒
For J = 1, we have:
4B = 4 × 1.935 = 7.74 cm
‒
For J =2, we have:
6B = 2 × 1.935 = 11.61 cm
‒
2.3 VIBRATIONAL SPECTROSCOPY
1. What is vibrational level transition?
It is the transition from one vibrational energy level to higher vibrational energy levels
by absorbing IR radiation.
2. Give the principle of IR spectroscopy.
When a molecule absorbs the energy of IR light, its atoms and molecules are excited and
undergoes transition from one vibrational energy level to next higher vibrational energy
level. This phenomenon is known as vibrational spectroscopy (or) infrared spectroscopy.
3. What are the conditions (selection rules) for IR absorption by molecules? 1. There should be a change in the dipole moment of the molecules upon IR absorption. 2. The frequency of the incoming radiation should match with the frequency of the
energy level difference between the vibrational states. Vibrational transition occurs
when ∆𝜗 = ± 1.
4. What are the infrared active molecules? Give examples.
Those molecules which possess either a permanent dipole moment or the dipole moment
arise due to vibration of atoms in molecules are infrared active.
Examples: CO, NO, CN, HCl
5. Define degrees of freedom.
The number of degrees of freedom is equal to the sum of coordinates necessary to locate
all the atoms of a molecule in space. Each atom has three degrees of freedom
corresponding to the three Cartesian coordinates (X, Y, Z) which is necessary to describe
its position on relative to other atoms in a molecule.
For linear molecules, the degrees of freedom, 3 N = translational (3 coordinates) +
rotational (2 coordinates) + vibrational.
For non-linear molecules, the degrees of freedom, 3N = translational (3 coordinates) +
rotational (3 coordinates) + vibrational.
SCYA1101: Engineering Chemistry UNIT 2 : Molecular Spectroscopy
S. No. Stretching Vibrations Bending Vibrations
- If two bonded atoms continuously
oscillate between the bonds by
changing their distance without
altering the bond angles, it is
known as stretching vibration.
If two bonded atoms continuously
oscillate between the bonds by
changing the bond angles with respect
to centre atom in and out of bond axis,
it is known as bending vibration.
- There are two types of stretching
vibrations namely symmetrical
stretching and unsymmetrical
stretching.
There are two types of bending
vibrations namely in-plane bending
(scissoring and rocking) and out-of-
plane rocking (wagging and twisting).
2.3.1 Fundamental Modes of Vibrations
1. Explain the types of vibrations with their diagrammatic representations.
Fundamental Modes of Vibrations (The mechanism of Interaction of Vibration of
Molecules and Infrared Radiations)
When a substance absorbs light in IR-regions, its molecules and atoms are excited in turn
leads to bond deformation. This is mainly due to two types of fundamental modes
vibrations namely stretching vibration and bending vibration.
1. Stretching Vibration
If two bonded atoms continuously oscillate between the bonds by changing their distance
without altering the bond angles, it is known as stretching vibration.
Example (i): Diatomic molecules like H–H or H–Cl vibrates in one way by moving
towards or away from each other when they interact with suitable IR radiations.
H H
H H
Normal
IR
H H
interaction
Example (ii): However, triatomic molecules such as CO 2
possesses two different stretching
modes namely symmetrical stretch and unsymmetrical stretch.
In the symmetrical stretch, each O moves towards C or away from C simultaneously.
O C O
Normal
IR O C O
O C O
Symmetrical stretching
interaction
In the unsymmetrical stretch, one O moves towards C while the other O moves away from
C.
O C O
Normal
IR
O C O
Asymmetric stretching
O C O
interaction
SCYA1101: Engineering Chemistry UNIT 2 : Molecular Spectroscopy
2. Bending Vibration
If two bonded atoms continuously oscillate between the bonds by changing the bond angles
with respect to centre atom in and out of bond axis, it is known as bending vibration.
There are two types of bending vibrations:
IN-PLANE bending vibrations: All the atoms are on same plane; Examples: Scissoring
and rocking.
OUT-OF-PLANE bending vibrations: If two atoms are on same plane while the one atom
is on opposite plane; Examples: Twisting and wagging.
Scissoring: If the two atoms joined
to centre atom move towards and
away from each other, it is known as
scissoring.
Centre Atom
1
2
3
Centre Atom
1
2
3
Scissoring
Rockin g: If the two atoms joined to
centre atom move simultaneously in
the same direction, it is known as
rocking.
Centre Atom
1
2
3
Centre Atom
1 2
3
Rocking
Twisting: If one of the atoms moves
up (coming forward) and the other
moves down (going back) with
respect to centre atom and changing
the bond angles, it is known as
twisting.
Twisting
Centre Atom
+
3
+ means coming forward
means going backward
Wagging: If the two atoms move
back and forth in the same direction
with respect to the centre atom and
changing the bond angles, it is
known as wagging. Such vibrations
are generally referred to as
degenerate modes.
Centre Atom
+
3
+ means coming forward
means going backward
Wagging
Centre Atom
+
3
SCYA1101: Engineering Chemistry UNIT 2 : Molecular Spectroscopy
Hence, CO 2
has
four modes of
vibrations.
Since the two
bending modes
have the same
vibrational
energy, they are
known as
degenerates, and
only one
absorption band is
expected for these
two bending
modes of
vibrations.
So, for CO 2
, we
have total of two
absorption bands.
Asymmetrical stretching (2565 cm
‒
O C O
Normal
IR
O C O
Asymmetric stretching
O C O
Exhibits
net
dipole
moment
IR active
In-plane bending vibration (526 cm
‒
O C O
Has net
dipole
moment
IR active
Out-plane bending vibration (526 cm
‒
O C O
Has net
dipole
moment
IR active
2. Mode of Vibrations in H 2
O Molecule
Description Mode of vibrations Change in
dipole
moment
IR
active/
inactive
H
2
O is a non-linear
tri-atomic
molecule.
Vibrational degrees
of freedom =
3N – 6 =
3×3 – 6 = 3.
Hence, H 2
O has
three fundamental
modes of
vibrations a
symmetrical
stretching, an
unsymmetrical
stretching and a
bending mode.
Symmetrical stretching (3686 cm
‒
O
H
H
O
H H
Expansion
Comprssion
Involve in
change in
dipole
moment
IR active
Asymmetrical stretching (3606 cm
‒
O
H
H
O
H
H
E
x
p
a
n
s
i
o
n
E
x
p
a
n
s
i o
n
C
o
m
p
r
s
s
i
o
n C
o
m
p
r
s
s
i
o
n
Involve in
change in
dipole
moment
IR active
Bending vibration (In-plane
scissoring) (1885 cm
‒
O
H H
O
H H
Involve in
change in
dipole
moment
IR active
SCYA1101: Engineering Chemistry UNIT 2 : Molecular Spectroscopy
2.3.2 Determination of Force Constant for a Diatomic Molecule
1. Explain how the force constant of a diatomic molecule is determined?
The molecules are made up of atoms linked by chemical bonds. The movement of atoms
and the chemical bonds are like spring and balls (vibrations). This characteristic vibration
is known as natural frequency of vibration.
If the applied IR frequency is equal to the natural frequency of vibration, then absorption
of IR radiation takes place. So, we can observe a corresponding peak in the spectrum.
Force constant is a measure of bond strength between the two atoms.
Mathematically, force constant is defined as the restoring force per unit displacement.
Force constant , −𝑭 = 𝒌𝒙
Where F is the restoring force, x is the displacement and the negative sign indicates the
force that has direction opposite of x.
The relationship between force constant ( k ) and equilibrium vibrational frequency ( ʋ e
) is
given by:
𝑒
Where μ is the reduced mass of a diatomic molecule, which can be calculated using the
expression 𝜇 =
𝑚
1
𝑚
2
( 𝑚
1
+𝑚
2
)
𝑒
Where γ is the frequency of electromagnetic radiation. Substituting equation (2) in
equation (1), we gete:
… ( 4 ) since, γ =
since, 𝛾̅ =
SCYA1101: Engineering Chemistry UNIT 2 : Molecular Spectroscopy
2.4 ELECTRONIC SPECTROSCOPY
1. Give the principle of uv - visible spectroscopy.
When a molecule absorbs energy in uv - visible region (180-780 nm), its outermost
electrons undergo transitions from the lower electronic energy level to the next higher
electronic energy level. This phenomenon is known as electronic spectroscopy (or) uv-
visible spectroscopy.
2. Give the regions of uv - visible spectrum.
The near ultraviolet region: It ranges from 200 to 380 nm.
The visible region: It ranges from 380 to 780 nm.
3. Reason out why the uv bands are broad?
Electronic level transitions are accompanied by vibrational and rotational level
transitions; hence uv spectral bands are broad.
4. What are the various types of electronic transitions?
There are four types of electronic transitions involved ultraviolet and visible
spectroscopy. They are: –; n–; – and n–.
5. Give the order of energy required for electronic transitions in organic molecules.
The order of required energy of various electronic transitions in organic molecules is:
– > n– > – > n–
**6. Give the different types of electronic transitions with one example each.
- Mention the type of electronic transition taking place in methane, ethylene,**
methanol.
S. No. Types of electronic transitions Examples
- σ → σ * Methane; Cyclohexane
- n → σ * Methanol; Methyl amine
- π → π * Ethylene; Acetylene
- n → π * Acetone 8. List out the differences between π → π * and n → π * electronic transitions.
**S. No. π → π * n → π ***
1 Allowed transition Forbidden transition
2 High energy transition Lower energy transition
3 Molar extinction coefficient (ε) value
lies between 100 to 10000
Molar extinction coefficient (ε)
value is < 100
4 More intense than n → π * Less intense than π → π *
SCYA1101: Engineering Chemistry UNIT 2 : Molecular Spectroscopy
**9. Give Beer-Lambert’s Law.
- Give the equation relating absorbance and concentration in** uv - vis spectroscopy.
Beer-Lambert’s law is stated as “when a beam of monochromatic light passes through a
transparent absorbing solution, the amount of light absorbed ( A ) is proportional to the
concentration of the solution ( c ) and the thickness ( l , path length) of the solution.”
Where, 𝑙𝑜𝑔
𝐼
𝜊
𝐼
is known as absorbance ( A ); l is thickness or path length of the cell; c is
the molar concentration of the solution and ε is the molar extinction coefficient or molar
absorptivity.
11. What are the limitations of Beer-Lambert’s Law?
Beer-Lambert’s law fails:
✓ when the absorbing solute dissociates or associates in solution.
✓ at higher concentrations.
✓ in scattering of light due to particulates in the sample.
12. Define the term transmittance in uv - visible spectroscopy.
Transmittance is the fraction of the incident light transmitted from the total intensity of
incident light.
The transmittance and absorbance are related as:
13. What is molar extinction coefficient?
Molar extinction coefficient is defined as “the absorbance of a one molar solution placed
in a cell of one cm path length”.
14. If the transmittance of a solution is 19.4%, what is its absorbance or optical
density?
Given: Percentage of transmittance (T) = 19.4%
(or) T =
- 4
100
Absorbance, A = log
1
𝑇
= log
1
- 194
= log 1 ‒ log (0.194) = 0 ‒ (‒0.712) = 0.
Absorbance, A = 0.712 × 100 = 71.2%
