‫‪Spectroscopy‬‬
‫‪Dr.Abdulhadi Kadhim.‬‬
‫‪Ch.3‬‬
‫اﻟﺠﺎﻣﻌﺔ اﻟﺘﻜﻨﻮﻟﻮﺟﯿﺔ‬
‫ﻗﺴﻢ اﻟﻌﻠﻮم اﻟﺘﻄﺒﯿﻘﯿﺔ‬
‫اﺳﺘﺎذ اﻟﻤﺎدة ‪ -:‬اﻟﺪﻛﺘﻮر ﻋﺒﺪ اﻟﮭﺎدي ﻛﺎﻇﻢ‬
‫ﻓﺮع اﻟﻔﯿﺰﯾﺎء اﻟﺘﻄﺒﯿﻘﯿﺔ‬
‫اﻟﻤﺮﺣﻠﺔ اﻟﺮاﺑﻌﺔ‬
‫‪Chapter three:‬‬
‫‪Infrared Spectroscopy‬‬
‫)‪(IR‬‬
‫‪42‬‬
Spectroscopy
Dr.Abdulhadi Kadhim.
Ch.3
Infrared Spectroscopy (IR)
Fundamental region (2.5-15.4µm)
Near IR (0.75-2.5µm)
Far IR (15.4-microwave)
IR
Vibrational energy of diatomic molecule:We are all familiar with the vertical oscillations of amass (m)
connected to a stretched spring of a force constant k whose ether end
is fixed.
The simple harmonic motion with a fundamental frequency:°
=
If two masses in a diatomic molecule m1 and m2 we used the reduced
mass
∴ =
in quantum mechanically, the vibrational energy is given by
°
=
+
υ = 0,1,2,3 − − − −
°
Where υ is the vibrational quantum No.
The energy in Cm-1
=
=( + )
°
\
°
=( + )
Where ° the freq. in cm-1. These energy levels are equally spaced
and the energy of lowest state ° =
\
°
∶
43
Spectroscopy
Dr.Abdulhadi Kadhim.
Ch.3
An harmonic oscillator
In addition to rotational motion , molecule has vibrational motion
which is an harmonic. Hence a further correction to the centrifugal
distortion to be made to account for the vibrational excitation of the
molecule.
The empirical equation of the potential energy of the diatomic
molecule is given by
V =D 1−e
(
)
This equation called morse equation.
V = morse potential.
This potential represent the actual curve to a very good degree of
approximation except when r=0 → Vm gives a high finite value while
the actual value is infinity.
v At r→ ∞ → Vm→De
De: dissociation energy.
v At
r=re →
Vm→0
V
D
re
r
44
Spectroscopy
Dr.Abdulhadi Kadhim.
Ch.3
Let x=r-re
x<< 1 very small
= 1− +
= 1−
Or
+
(
We take the first two term
=
≅
Where
=
)
!
2!
−
−
(
=1−
[1 − ( − )] =
\
3!
)
!
+ ⋯ … … … … … … ..
+ ⋯ … … … … … … ..
=
1−
[− ( − )] =
≅1−
( − ) = (2
)( − ) =
(
)
( − )
\(
− )
: 2a2De=force constant for small displacement.
\
=
in (Hz)
: freq. for small displacement.
Substitution Vm solving this eq. using pert buation theory we
gets G(v) = ω
v+
−ω x
v+
ω x :an harmonicity correction and constant.
ω >> ω x
The effects of an harmonicity:1- The energy levels are not equally spaced.
2- Selection
rules
for
allowed
∆v=±1,±2,±3,……
v When ∆v=±1→ G(v1)= G(1)- G(0)
transition
45
Spectroscopy
Dr.Abdulhadi Kadhim.
Ch.3
First fundamental freq.= ω -2ω x
v When ∆v=±2 → G(v2)= G(2)- G(0)
First over ton fundamental freq.=2 ω -6ω x
ω
D =
4ω x
We can derive the equation of D
1
1
(
)
G v =ω v+ −ω x
2
2
The maximum energy at
=0
− − − −(1)
Then G(v) =De
( )
= ω -2ω x
v+
2vω x =ω − ω x
∴ =
ω
= ω − 2vω x − ω x
ω
ω
Sub in eq. (1)
∴
( )=
=
=
+
−
2
+
−
−
−
+
−
(
2
+
2
)
46
Spectroscopy
Dr.Abdulhadi Kadhim.
Ch.3
D =
=
−
=
Dissociation energy
Example(1):-
The fundamental vibration spectrum of Co molecule at
the wave number 2138 Cm-1 and the line which represent by
the transition v=2→ v=3 at wave number 2091 Cm-1 calculate
1. The harmonic constant ( ).
2. The unharmonic constant (
3. The Dissociation energy.
).
The solution:G(v) = ω
v+
1
−ω x
2
G(v1)= G(1)- G(0)
( )=
1+
−
( )=
3
2
2138 =
1+
−
9
4
( )=
By the same way
−2
−
−
−2
0+
1
2
1
2
−
+
1
4
0+
− − − −(1)
G(v2)= G(3)- G(2)
47
Spectroscopy
Dr.Abdulhadi Kadhim.
G(v ) =
49
7
ω −
ω x
4
2
= ω − 6ω x
∴ 2091 =
−6
2091 =
2138 =
-47 cm-1 = -4
∴ω x =
cm
2138 cm-1 =
(2 × 11.75
Sub in eq.(1)
∴
=
= 9.94 × 10
− − − −(2)
‫ﺑﺎﻟﻄﺮح‬
− − − −(1)
−2
= 11.75 cm
= 2161.5
4
5
25
− ω −
ω x
2
2
− − − −(2)
−6
Ch.3
)
(2161.5)
4.44 × 10
=
=
4 × 11.75 9.94 × 10
1
× 3 × 10
= 29.82 × 10
1
× 6.63 × 10
sec
= 147.7 × 10 Joul
D = 29.82 × 10
147.7 × 10
∴D =
1.6 × 10
1
J. S
= 123.56 × 10 ev = 12.356 ev
48
Spectroscopy
Dr.Abdulhadi Kadhim.
Ch.3
H.W.(1):The absorption spectrum of HF molecule in the IR region ,
the first line at 3961 cm-1 and the second line at 7751 cm-1
calculate
1. The dissociation constant 2f.
2. The Dissociation energy D .
Population of vibration energy levels
According to Boltzman distribution
∆ /
=
/
=
∆E For first three transitions
(1) V=0 → V=1 ( Fundamental frequency ) ∆V=+1
∆E = εv=1 - εv=0
∆E = 1 +
∆E=
− 1+
(1 − 2 )
(2) V=0 → V=2
∆E=2
−6
∆E=2 (1 − 3
(3) V=0 → V=3
∆E=3
∆E=3
− 12
(1 − 3
)
)
∆V=+2
−
−
∆V=+3
49
Spectroscopy
Dr.Abdulhadi Kadhim.
Ch.3
Example(2):HCl molecule is absorb the radiation at 2885.9 cm-1 by
using Boltzman distribution calculate the relative Number
for first vibrational level and the ground level at 25oC
suppose the No. of molecule at ground level is one.
The solution:=
=
∆ /
=
. × . ×
= 9.245 × 10
/
× ×
/ .
×
×
H.W.(2):-
HCl molecule is absorb the IR radiation at fundamental
vibrational frequency =2890 cm-1 calculate the force
constant ?
H.W(3):- Calculate the force constant for OH which can
be the spectrum absorb at IR region .
Infrared selection Rules
The selection Rule of IR if the vibration quantum
Number change
1. ∆V=±1 under harmonic approximation.
2. ∆V=±1, ±2, ±3 for anharmonic oscillator.
50
Spectroscopy
Dr.Abdulhadi Kadhim.
Ch.3
Over tones frequency:On other transitions may be take place furthermore
the fundamental frequency, at 2V1, 3V1, 4V1
1. 2V1, 3V1, 4V1 for transition began from v=0.
2. 2V2, 3V2, 4V2 for transition began from v=1.
3. 2V3, 3V3, 4V3 for transition began from v=2.
And the frequency of the 1st overtone = 2 e(1-3xe)
The absorption due to these transition called (( Over tones
frequencies))
The following figure represent the overtones frequencies for
Hcl molecule in IR region.
Zero point energy =
−
1−
V =0 →1
Fundamental frequencies
First over ton
freq.
2nd over ton
freq.
V =0 →2 V =0 →3
0
5000
10000
3rd over ton
freq.
4th over ton
freq.
V =0 →4
V =0 →5
V (cm1)
51
Spectroscopy
Dr.Abdulhadi Kadhim.
Ch.3
Combination bands and differences bands:The selection rules allowed by other transitions to
produce combination bands and difference bands. The
combination bands
is arise by added two or more fundamentals frequencies or
over tons frequencies such as:+
,2
+
,
+
+
These frequencies become allowed but the intensity is very
low.
The difference bands is similar to combination bands but the
transition appear in complex spectra.
−
,2
−
,
+
Born-Oppenheimer principle
−
((The electrons movements is high speed compared with
nuclei movements , then can be considered the nuclei is
constant when studies the electronic properties of molecules
and the energy of electrons is independent on nuclei energy
and the total energy is equal to sum of vibration energy))
Plus Rotational energy plus electronic energy
.
=
.
+
.
+
.
Vibration – Rotation spectroscopy of diatomic
molecules
The vibration and rotation motion happen in any molecule
at the same time, and for any vibration level consist from
52
Spectroscopy
Dr.Abdulhadi Kadhim.
Ch.3
many Rotational levels. The following figure explain the first
two vibrational level v=0 and v=1 and the rotational levels.
The vibration – rotation spectrum produce by transitions
from rotational levels of vibration level like v=0 to another
rotational levels of vibration level like v=1 and the selection
rule for vibration- Rotation spectra
∆V=±1 , ∆J =±1
The energy of vibration-Rotation for diatomic molecule
E(J,V) = E(J) + E(V)
= BJ(J+1) +
=ℎ
+
°
( + 1) +
+
+
+
v=0 , v=1 ‫ﻣﺜﻼ ﻟﺪﯾﻨﺎ ﺣﺎﻟﺔ اﻻﻧﺘﻘﺎل ﺑﯿﻦ ﻣﺴﺘﻮﯾﯿﻦ‬
∆ J=+1
∆ J=-1
J
/
4
3
v=1
2
1
0
“
J
4
3
v=0
2
P-branch
2B
vo
4
R-branch
53
2B
1
0
Spectroscopy
Dr.Abdulhadi Kadhim.
Ch.3
\
=
\
+
−
\\
\\
+
\
+
+
\\
±
+
The vibrational-rotational absorption spectrum
Consists of two groups of equally spaced lines(R+P)branches with
a gap between them, the center of this gap is υo But experimentally it
is found that the spacing is not constant , this is due to the vibrationrotation coupling.
When take the distortion effect ( centrifugal distortion ) in account the
=
,
( + 1) −
=
( + 1) +
−
+
+
Consider the vibrational rotational transition υ=0→υ=1
Assuming B and D the same for both υ=0, and υ=1 states and
denoting upper state by single prime and lower state by double prime
\
=
=
(1)
\
(1 − 2
+1 −
+
3
2
)+
\\
−
\
−
9
4
\\
\\
+1
\
−
+
−
\\
1
2
\
−
+1 −
1
4
Use of the selection rule ∆J=+1
=
+2
\\
=
−2
\
Where
=
+1 −4
+1 +4
\\
+1
∆J=-1
\
+1
\
+1
\
\
+1
\\
−
\\
\\
+1
\\
+1
∴
J\-J\\=1 gives
,
= 0,1,2, − − − − −
∴
,
−
-
J\-J\\=-1 gives
= 0,1,2, − − − − −
(1 − 2 ) which is the freq. of the υ=0→υ=1transition
54
Spectroscopy
Dr.Abdulhadi Kadhim.
Ch.3
Lines corresponding to ∆J=-1 called P branch.
Lines corresponding to ∆J=+1 called R branch.
,
=
+
=
+
−
(m=J\\+1
and J\+1)
usually D is extremely small and in such cases
,
Example(3):The normal modes of vibration of Co2 molecule are υ1=1330
cm-1, υ2=667 cm-1, υ3=2349 cm-1 and υ4=2349 cm-1 evaluate the zero
point energy of Co2 molecule.
Solution:The Co2 molecule has 4 normal modes of vibration the
symmetric bending mode υ2 is double degenerate.
The zero point energy Eo of Co2= ℎ ∑
∑
∴
\
°
\
= (1330 + 667 + 667 + 2349) = 5013 × 100
= 6.6 × 10
= 4.98 × 10
× 3 × 10 × 5013 × 100
= 0.311
55
Dr.Abdulhadi Kadhim.
Spectroscopy
Ch.3
H.W.(4) :The equilibrium vibration frequency of the Iodine molecule is υs
cm and the anharmonicity constant
= 0.003
-1
What is the intensity of hot band υ=1→υ=2 relative to that of the
fundamental υ=0→υ=1 , if the temperature is 300 oK ?
Note: - The hot band occurs in the temperature higher than the room
temp.
Example(4):The fundamental and first over tone transition of 14N16O are
centered at 1876.06 cm-1 and 3724.20 cm-1 respectively. Evaluate the
equilibrium vibration frequency, the anharmonicity constant, zero
point energy and force constant of molecule?
Solution:-
over tone = 2
Zero point energy
=
= 7.332 × 10
°
=
\
1−
(1 − 2 ) = 1876.06 cm
(1 − 3 ) = 3724.20 cm
\
= 1903.98
56
Spectroscopy
Dr.Abdulhadi Kadhim.
Ch.3
= × 1903.98 (1 − 0.003666) = 948.5
14
Mass of N =23.25x10
16
Mass of O =26.56x10
=
=4
.
.
×
.
.
\
-27
Kg
-27
× 10
=4
Kg
= 12.3975 × 10
× 12.3975 × 10
= 1597 N/m
× 9 × 10
Kg
× (904 × 100)
H.W.(5):Estimate the position of the band center and B value of the
HCl from the few lines of the P and R branches listed below:-
line
P1
P2
P3
P4
2865.1
2843.62
2821.56
2798.94
line
Ro
R1
R2
R3
2906.24
2925.90
2944.90
2963.29
57
Spectroscopy
Dr.Abdulhadi Kadhim.
Ch.3
Example(5):The fundamental band for HCl is centered at 2886 cm-1 .
Assuming that the inter nuclear distance is 1.276 oA. Calculate the
wave No. of the first two lines of each of the P and R branches of
HCl.
Solution:( .
The reduced mass of HCl = (
=
=
= 10.6129
,
=
+2
× .
×
.
. ×
×( .
)×( .
) ( .
×
)
)
)
= 1.627 × 10
×
= ∓1, ∓2, ∓3 − − − − −
= 2886 − 2 × 10.6129 = 2864.77
= 2886 − 4 × 10.6129 = 2843.55
= 2886 + 2 × 10.6129 = 2907.23
= 2886 + 4 × 10.6129 = 2928.45
58
Dr.Abdulhadi Kadhim.
Spectroscopy
Ch.3
IR Spectrometer:Consist from the following parts:(1) Source of radiation:Rod from silicon carbide heated to 1500 oC.
(2) Two mirrors to produce two beams, the first beam
reference beam and the second beam is sample beam.
(3) Cell of sample: - made from nacl.
(4) Grating or lit row mount prism.
(5)The sample may be solid or liquid ((If the sample is gas
must be used special cells of some millimeters thickness))
IR applications:(1) Diagnosis the structure of organic compounds.
(2) Determine the purity of compounds.
(3) Quantitative analysis for different compounds.
(4) A study of chemical interactions.
(5) A study of Hydrogen bonding.
(6) Analysis of orientation and geometric shapes for small
molecules in gas state.
(7) A study of polymers chemical.
59
Spectroscopy
Dr.Abdulhadi Kadhim.
Ch.3
Orational Infrared spectrum for polyatomic
molecules
1- Linear Polyatomic molecules
When molecule contain N of atoms then the axis (x, y, z) this
leads to 3N of axis this molecule need or called 3N freedom degree.
This means
(3N) to characterized the position or motion of center of
molecule mass, and (2N) for rotational motion then:-
3N-5 No. of degree of freedom for vibrational motion
N:-No. of atoms
For Diatomic molecules 3N-5=1 means the No. of vibrational motion
equal to one.
For Linear molecule have three atoms such as Co2 3N-5=3x3-5=4 No.
of vibrational ( Fundamental vibration) two as parallel of axis bond
and two represented the atom motion outside of axis bond .
The first one called ( stretching ) and the second called (bending)
Starching
Starching
60
Spectroscopy
Dr.Abdulhadi Kadhim.
Ch.3
2- Non Linear Polyatomic molecules:The No. of vibration equal to 3N-6 Three for transition motion
and Three for rotational motion.
Example:Water molecule H2o
The No. of vibration 3N-6=3
H
O
O
O
H
H
H
H
H
The energy of stretching is higher than the energy of bending
vibration.
The rotational energy level occurs as a result of rotation of molecules
around axis perpendicular on has symmetry axis and the axis of
symmetry has been neglected due to the mass of atom insure in
nuclear and radius of nuclear is very small compared with atomic
radius this leads to the moment of inertia due to electron rotation
around the axis.
61
Dr.Abdulhadi Kadhim.
Spectroscopy
Ch.3
The Features of Rotational levels:(1) The separated energy between the spectrum lines about 10-3ev.
(2) The wave length about ( 0.1mm-several cm).
(3) Lie in micro wave region.
(4) The moment of inertia equal to I=µr2 .
(5) The energy = Iω =
ℏ
J(J + 1).
(6) Appear clearly in permanent dipole molecule and does not appear
in symmetrical molecule like H2,N2 .
(7) The selection rules ∆J=±1 and µ≠0 .
(8) The energy in room temp. greater than rotational energy
( KT= 2.6 x 10-2 ev) .
(9) Most rotational spectrum shown in absorption spectrum and
appear in cases and vapor.
62