Transcript Physical Chemistry 2nd Edition

Chapter 19
The Vibrational and Rotational Spectroscopy
of Diatomic Molecules
Physical Chemistry 2nd Edition
Thomas Engel, Philip Reid
Objectives
• Describe how light interacts with molecules to
induce transitions between states
• Discuss the absorption of electromagnetic
radiation
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Outline
1. An Introduction to Spectroscopy
2. Absorption, Spontaneous Emission, and
Stimulated Emission
3. An Introduction to Vibrational Spectroscopy
4. The Origin of Selection Rules
5. Infrared Absorption Spectroscopy
6. Rotational Spectroscopy
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
19.1 An Introduction to Spectroscopy
•
•
Spectroscopy are tools chemists have to
probe the species at an atomic and molecular
level.
The frequency at which energy is absorbed or
emitted is related to the energy levels involved
in the transitions by
hv  E2  E1
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
19.1 An Introduction to Spectroscopy
•
19.1 Energy Levels and Emission Spectra
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
19.1 An Introduction to Spectroscopy
•
•
During vibration, oscillator will absorb energy in
both the stretching and compression.
The molecule can absorb energy from the field
during oscillation.
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Band name
subHertz
Abbr
subHz
Frequency
ITU band and
wavelength in air
0
Extremely low frequency
ELF
1
Super low frequency
SLF
2
Ultra low frequency
ULF
3
Very low frequency
VLF
4
Low frequency
LF
5
Medium frequency
MF
6
High frequency
HF
7
Very high frequency
VHF
8
Ultra high frequency
UHF
9
Super high frequency
SHF
10
Extremely high frequency
EHF
11
< 3 Hz
> 100,000 km
Example uses
Natural and man-made electromagnetic waves (millihertz,
microhertz, nanohertz) from earth, ionosphere, sun, planets,
etc[citation needed]
3–30 Hz
100,000 km –
Communication with submarines
10,000 km
30–300 Hz
Communication with submarines
10,000 km – 1000 km
300–3000 Hz
Communication within mines
1000 km – 100 km
3–30 kHz
Submarine communication, avalanche beacons, wireless heart
100 km – 10 km
rate monitors, geophysics
30–300 kHz
Navigation, time signals, AM longwave broadcasting, RFID
10 km – 1 km
300–3000 kHz
AM (medium-wave) broadcasts
1 km – 100 m
3–30 MHz
Shortwave broadcasts, amateur radio and over-the-horizon
100 m – 10 m
aviation communications, RFID
FM, television broadcasts and line-of-sight ground-to-aircraft
30–300 MHz
and aircraft-to-aircraft communications. Land Mobile and
10 m – 1 m
Maritime Mobile communications
Television broadcasts, microwave ovens, mobile phones,
300–3000 MHz
wireless LAN, Bluetooth, GPS and two-way radios such as Land
1 m – 100 mm
Mobile, FRS and GMRS radios
3–30 GHz
Microwave devices, wireless LAN, most modern radars
100 mm – 10 mm
30–300 GHz
Radio astronomy, high-frequency microwave radio relay
10 mm – 1 mm
Terahertz imaging – a potential replacement for X-rays in some
Chapter 19: The Vibrational and Rotational Spectroscopy
Diatomic
Molecules
300–3,000of
GHz
medical
applications, ultrafast molecular dynamics, condensedTerahertz
THz
12
nd
1
mm
–
100
μm
matter
physics,
terahertz time-domain spectroscopy, terahertz
Physical Chemistry 2 Edition
computing/communications
© 2010 Pearson Education South Asia Pte Ltd
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
19.2 Absorption, Spontaneous Emission, and Stimulated Emission
•
The 3 basic processes by which photon-assisted
transitions occur are absorption,
spontaneous emission and stimulated
emission.
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
19.2 Absorption, Spontaneous Emission, and Stimulated Emission
•
•
•
In absorption, the incident photon induces a
transition to a higher level.
In emission, a photon is
emitted as an excited state
relaxes to one of lower energy.
Spontaneous emission is a
random event and its rate
is related to the lifetime of
the excited state.
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
19.2 Absorption, Spontaneous Emission, and Stimulated Emission
•
At equilibrium,
B12  vN1  B21  vN2  A21N2
where
•
= radiation density at frequency ν
= rate coefficient
Einstein concluded that
B12  B21
A21 16 2 hv 3
and

B21
c3
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Example 19.1
Derive the equations
B12  B21 and A21 / B21  16 2 hv 3 / c3
using these two pieces of information: (1) the
overall rate of transition between levels 1 and 2
is zero at equilibrium, and (2) the ratio of N2 to
N1 is governed by the Boltzmann distribution.
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Solution
The rate of transitions from level 1 to level 2 is
equal and opposite to the transitions from level 2
to level 1. This gives the equation
B12   N1  B21  N2  A21N2
.
The Boltzmann distribution function states that
N 2 g 2  hv / kT

e
N1 g1
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Solution
In this case g 2  g1 . These two equations can be
solved for  v  , giving     A21 / B12ehv / kT  B21  .
Planck has showed that
 v 
A21
B12ehv / kT
8 hv3
1

 B21
c3 ehv / kT  1
For these two expressions to be equal
B12  B21 and A21 / B21  8hv3 / c3  16 2 hv3 / c3
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
19.3 An Introduction to Vibrational Spectroscopy
•
•
The vibrational frequency depends on two
identity vibrating atoms on both end of the
bond.
This property generates characteristic
frequencies for atoms joined by a bond known
as group frequencies.
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Example 19.2
A strong absorption of infrared radiation is
observed for 1H35Cl at 2991 cm-1.
a. Calculate the force constant, k, for this molecule.
b. By what factor do you expect this frequency to
shift if deuterium is substituted for hydrogen in this
molecule? The force constant is unaffected by this
substitution.
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Solution
a. We first write E  hv  hc /  
Solving for k,
E  hv 
hc


h
2
k

k/
.
and
c
2 1.00834.969 
k  4     4 2 2.998 108 2991100 
1.66110  27  516.3N / m
35.977 
 
2
2


b. The vibrational frequency for DCl is lower by
a substantial amount.
mH mcl  mD  mC 
 1.0078  36.983 


 

  0.717

mD mcl  mH  mC 
 2.0140  35.977 
HCl
DCl
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
19.3 An Introduction to Vibrational Spectroscopy
•
•
•
19.2 The Morse Potential
The bond energy D0 is defined with respect to
the lowest allowed level, rather than to the
bottom of the potential.
 ( x  x ) 2
V
(
x
)

D
[1

e
]
The energy level is
e
e
1   hv  
1

En  hv  n   
n



2  4 De 
2

2
2
D0  Enmax  E0
1   hv  
1

 hv  nmax   
 nmax    De
2  4 De 
2

2
Enmax
1   hv  
1

E0  hv  0   
0 
2  4 De 
2

2
2
2
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
19.3 An Introduction to Vibrational Spectroscopy
•
Parameters for selected model are shown.
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
19.4 The Origin of Selection Rules
•
The transition probability from state n to state
m is only nonzero if the transition dipole
moment
satisfies the following condition:
 xmn   m* x  x x  n x d  0
where x = spatial variable
μx = dipole moment along
the electric field
direction
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
19.5 Infrared Absorption Spectroscopy
•
•
Atoms and molecules possess a discrete energy
spectrum that can only be absorbed or emitted
which correspond to the difference between
two energy levels.
Beer-Lambert law states that
I  
 e   Ml
I 0  
where I(λ) = intensity of light leaving the cell
I0(λ) = intensity of light passing dl distance
l = path length
ε(λ) = molar absorption coefficient
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Example 19.4
The molar absorption coefficient    for ethane is
40 (cm bar)-1 at a wavelength of 12 μm. Calculate
I   / I 0   in a 10-cm-long absorption cell if ethane is
present at a contamination level of 2.0 ppm in one
bar of air. What cell length is required to make
I   / I 0    0.90 ?
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Solution
Using
I  
 e   Ml
I 0  
I  
 exp  40 2.0 10 6 1.0  0.9992  1.0
I 0  
 


This result shows that for this cell length, light
absorption is difficult to detect. Rearranging the
Beer-Lambert equation, we have
 I   
1
3


l
ln 


ln
0
.
90

1
.
3

10
cm
6

40 2.0 10
M    I 0   
1


Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
19.5 Infrared Absorption Spectroscopy
•
•
Coupled system has two vibrational
frequencies: the symmetrical and antisymmetric
modes.
For symmetrical and asymmetrical, the
vibrational frequency is
1
vsymmetric 
2
vqntisymmetric 
1
2
k1

k1  2k2 

Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
19.6 Rotational Spectroscopy
•
•
•
•
19.3
19.4
19.5
19.6
Normal
Normal
Normal
Normal
Modes for H2O
Modes for CO2
Modes for NH3
Modes for Formaldehyde
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Example 19.5
Using the following total energy eigenfunctions for
the three-dimensional rigid rotor, show that the
J=0 → J=1 transition is allowed, and that the J=0
→ J=2 transition is forbidden:
Y00  ,   
1
4 1/ 2
1/ 2
 3 
Y  ,    

 4 
cos 
0
1
1/ 2
 5 
Y  ,    

16



0
2
The notation
functions.
Mj
Yj
3 cos  1
is used for the preceding
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Solution
Assuming the electromagnetic field to lie along the
zaxis, z   cos , and the transition dipole moment
takes the form

J0
z
2
2
0
0
   d  YJ0  ,  cos   ,  Y00 cos  sin d
For the J=0 → J=1 transition,
 
10
z
3
4
2
2
 d  cos
0
0
2
 sin d 
 3  cos  

2 
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
3
3
 
 
 0
 3
3
0
Solution
For the J=0 → J=2 transition,
 z20  
5
8
2
2
0
0
 
 5  3 cos 4  cos 2  
 5  1 1
2
3 cos   1 cos  sin d 



  0



8 
4
2  0
8  4 4 
 d  

The preceding calculations show that the J=0 →
J=1 transition is allowed and that the J=0 → J=2
transition is forbidden. You can also show that is
also zero unless MJ=0 .
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
19.6 Rotational Spectroscopy
•
•
For vibrational spectroscopy, we have to
change the symbol for the angular
momentum quantum number from l to J.
Thus the dependence of the rotational
energy on the quantum
number is given by
h2
h2
E
J J  1 
J J  1  hcBJ J  1
2
2
2ro
8ro
where rotational constant is
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
19.6 Rotational Spectroscopy
•
We can calculate the energy corresponding to
rotational transitions
E  E J final   E J initial  for J  1
h2
h2
J  1J  2  2 J J  1  2hcBJ  1 and for J  1
E 
2
2r0
2r0
h2
h2
J  1J  2 J J  1  2hcBJ
E 
2
2r0
2r0
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Example 19.5
Because of the very high precision of frequency
measurements, bond lengths can be determined
with a correspondingly high precision, as illustrated
in this example. From the rotational microwave
spectrum of 1H35Cl, we find that B=10.59342cm-1.
Given that the masses of 1H and 35Cl are 1.0078250
and 34.9688527 amu, respectively, determine the
bond length of the 1H35Cl molecule.
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Solution
We have
B
h
8 2  cr02
6.6260755 1034
r0 

2
8  cB
2  1.0078250  34.9688527  
27
8 c 
 1.66054 10  10.59342 
 1.0078250  34.9688527 
h
 1.274553 1010 m
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
19.6 Rotational Spectroscopy
•
To excite various transitions consistent with the
selection rule J  J final  J initial  1 , we have
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
19_16fig_PChem.jpg
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
, R:
, Q:
(vibrational
).
ΔJ  1
ΔJ  0
ΔJ  1
Δn  1
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
P,Q,R
branches
of rotational spectrum
© 2010 Pearson Education South
Asia Pte
Ltd
19.6 Rotational Spectroscopy
•
•
•
19.7 Rotational Spectroscopy of Diatomic
Molecules
19.8 Rotational-Vibrational Spectroscopy
of Diatomic Molecules
The ratio for value of J relative to the number
in the ground state (J=0) can be calculated
using the Boltzmann distribution:
nJ g J  J  0 / kT
 h 2 J  J 1 / 2 kT

e
 2 J  1e
n0 g 0
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Rotational Raman Spectra
  E
The molecule can be made
anisotropically polarized and
Raman active.
Selection Rules:
J  0,2
K  0
Linear rotors
J  0,1,2
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Symmetrical rotors
Proof of Rotational Raman Selection
Rules
  E( t )  E cos  t
   0   cos 2R t
ind
i
 ind  ( 0   cos2R t )  (E cosi t )
  0 E cosi t  E cos2R t cosi t
1
  0 E cosi t  Ecos(i  2R ) t  cos(i  2R ) t
2
 ind ,x   x sin  cos
E x  E sin  cos 
ind ,y   y sin  sin 
ind ,z   z cos
E y  E sin  sin 
E z  E cos 
ind    E x sin  cos    E y sin  sin    // E z cos    E sin 2    // E cos2 
 ind
1


2
4  2
1

   //       2,0 (, )E
3
3
3 5 




1
2
(


YJi ,MRotational

YJf ,MJ ,f YJi ,Mof
//
 ) YJ f ,M J ,f and
Chapter 19: TheVibrational
Spectroscopy
J ,i
J ,i Diatomic Molecules
3
3
nd
Physical Chemistry 2 Edition
© 2010 Pearson Education South Asia Pte Ltd
Selection rules
A Typical Rotational Raman
Spectrum (Linear rotors)
J  0, 2
(Linear rotors)
~
 (J  1  J)  ~
i  F(J  2)  F(J)  ~
i  2B(2J  3)
Stokes lines
~
 (J  1  J)  ~
 i  F(J)  F(J  2)  ~
 i  2B(2J  1)
Anti-Stokes lines
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
19_21fig_PChem.jpg
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Vibrational Raman effect, Δ n=+1,-1
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
© 2010 Pearson Education South Asia Pte Ltd
Chapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules
Physical Chemistry 2nd Edition
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