Infrared Spectroscopy

University of Babylon Undergraduate Studies
College of Science
Department of Chemistry
Course No. Chsc 424 Physical chemistry
Fourth year - Semester 2
Credit Hour: 3 hrs.
Scholar units: Three units
Lectures of Molecular Spectroscopy
Second Semester, Scholar year 2018-2019
Prof. Dr. Abbas A-Ali Draea
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Lecture No. Third: Infrared Spectroscopy
1-Introudction.
2-Theory of Infrared Absorption.
3-Molecular Vibrations.
4- Stretching Vibrations.
5-Applications.

1-Introudction.
IR techneque is proved information about the structure of a compound and as an analytical tool to assess the purity of a compound.
They found main points in IR- spectroscopy:
Chemical bond is electrical combination between two electrical charged compartments (energetic process).
As soon as chemical bond formation as rotational- vibrational motions occurs spontaneously through center of mass.
Differences in electrical charge distribution along the chemical bonds lead to form a permanent electric dipole moment onto the chemical species.
Interaction between a permanent electric dipole moment of active Species and electrical force field of light produced energy absorption and excitation occurs in higher vibration energy levels.
Rotational transitions occurs simultinously within each two vibrational energy level of transition.

Infrared refers to that part of the electromagnetic spectrum between the visible and microwave regions. The IR region is divided into three regions: the near, mid, and far IR as shown in figure 1. The mid IR region is of greatest practical useing. This is the region of wavelengths between 3 x 10–4and 3 x 10–3cm.

Figure 1. Classification of infrared regions.


2-Theory of Infrared Absorption.
Absorption of IR is restricted to excite vibrational and rotational states of a molecule. Even though the total charge on a molecule is zero, the nature of chemical bonds is such that the positive and negative charges do not necessarily overlap in this case. Such molecules are said to be polar because they possess a permanent dipole moment. For a molecule to absorb IR, the vibrations or rotations within a molecule must cause a net change in the dipole moment of the molecule. The alternating electrical field of the radiation interacts with fluctuations in the dipole moment of the molecule. If the frequency of the radiation matches the vibrational frequency of the molecule then radiation will be absorbed, causing a change in the amplitude of molecular vibration. The result of IR absorption is heating of the matter since it increases molecular vibrational energy. Molecular vibrations give rise to absorption bands throughout most of the IR region of the spectrum. The far IR, lying adjacent to the microwave region, has low energy and may be used for rotational spectroscopy.
In order for IR absorbance to occur, they found two conditions must be met:
Must be found a change in the dipole moment of the molecule as a result of a molecular vibration (or rotation). The change (or oscillation) in the dipole moment allows interaction with the alternating electrical component of the IR radiation wave. Symmetric molecules (or bonds) do not absorb IR radiation since there is no dipole moment.
If the frequency of the radiation matches the natural frequency of the vibration (or rotation), the IR photon energy is absorbed and the amplitude of the vibration increases.

The wave numbers (sometimes referred to as frequencies) at which a molecule absorbs radiation give information on functional groups present in the molecule. Certain groups of atoms absorb energy and therefore, give rise to bands at approximately the same frequencies. The chemist analyzes a spectrum with the help of tables which correlate frequencies with functional groups.

3-Molecular Vibrations
There are three main types of molecular transitions that occur in IR
Rotational transitions
When an asymmetric molecule rotates about its center of mass, the dipole moment seems to fluctuate ((تتقلب. ?E for these transitions correspond to n < 100 cm-1 .Quite low energy, show up as sharp lines that subdivide vibrational peaks in gas phase spectra.
Vibrational-rotational transitions
Complex transitions that arise from changes in the molecular dipole moment due to the combination of a bond vibration and molecular rotation.
Vibrational transitions
The most important transitions observed in qualitative mid-IR spectroscopy.
There are two types of molecular vibrations, stretching and bending. Molecule having rigid bond lengths and bond angles. This is not the actual case, since bond lengths and angles represent the average positions about which atoms vibrate.
A molecule consisting of n atoms has a total of 3ndegrees of freedom, corresponding to the Cartesian coordinates of each atom in the molecule.
A nonlinear molecule, 3 of these degrees are rotational and 3 are translational and the remaining corresponds to fundamental vibrations;
A linear molecule, 2 degrees are rotational and 3 are translational. The net number of fundamental vibrations for nonlinear and linear molecules is therefore:

Molecular degrees of freedom (nonlinear 3n– 6) and (linear 3n– 5)

Calculation reveals that a simple molecule such as propane,C3H8, has 27 fundamental vibrations, and therefore, you might predict 27 bands in an IR spectrum! (The actual number is sometimes different) The fundamental vibrations for water, H2O, are given in Figure .2. Water, which is nonlinear, has three fundamental vibrations.

Figure 2. Stretching and bending vibrational modes of H2O.

Carbon dioxide, CO2, is linear and hence has four fundamental vibrations (Figure .3). The asymmetrical stretch of CO2gives a strong band in the IR at 2350cm–1. This band may notice in samples which, that run on the instruments in the teaching labs; since CO2is present in the atmosphere. The two scissoring or bending vibrations are equivalent and therefore, have the same frequency and are said to be degenerate, appearing in an IR spectrum at 666 cm–1.

Figure 3. Stretching and bending vibrational modes for CO2.

The symmetrical stretch of CO2 is inactive in the IR because this vibration produces no change in the dipole moment of the molecule. In order to be IR active, a vibration must cause a change in the dipole moment of the molecule.*(The reason for this involves the mechanism by which the photon transfers its energy to the molecule ((Recall that the dipole moment is defined as the product of the charge and the distance of separation. The distance has direction, therefore, dipole moments are vectors.)) Of the following linear molecules, carbon monoxide and iodine chloride absorb IR radiation, while hydrogen, nitrogen, and chlorine do not. In general, the larger the dipole change, the stronger the intensity of the band in an IR spectrum.

C?O I—Cl H2 N2 Cl2

Absorbed in IR do not absorb in IR

Only two IR bands (2350 and 666 cm–1) are seen for carbon dioxide, instead of four corresponding to the four fundamental vibrations. Carbon dioxide is an example of why one does not always see as many bands as implied by our simple calculation. In the case of CO2, two bands are degenerate, and one vibration does not cause a change in dipole moment. Other reasons why fewer than the theoretical number of IR bands are seen include:
An absorption is not in the 4000–400 cm–1range.
An absorption is too weak to be observed.
Absorptions are too close to each other to be resolved on the instrument.
Weak bands which are overtones or combinations of fundamental vibrations are observed.

The stretching and bending vibrations for the important organic group, –CH2, are illustrated in Figure .4 (The 3n–6 rule does not apply since the –CH2group represents only a portion of a molecule.) Note that bending vibrations occur at lower frequencies than corresponding stretching vibrations.

Figure 4. Stretching and bending vibrational modes for a CH2 group.

4- Stretching Vibrations.
The stretching frequency of a bond can be approximated by Hooke’s Law. In this approximation, two atoms and the connecting bond are treated as a simple harmonic oscillator composed of 2 masses (atoms) joined by a spring. The energy curve for a simple harmonic oscillator is illustrated in Figure 5. According to Hooke’s law, the frequency of the vibration of the spring is related to the mass and the force constant of the spring, k, by the following formula
?=1/2?.?(k/µ)
Where k is the force constant µ is the reduced mass ? is the frequency of the vibration. In the classical harmonic oscillator,
E = 1/2kx2= h?
Where x is the displacement of the spring.
The energy or frequency is dependent on how far one stretches or compressed the spring, which can be any value. If this simple model were true, a molecule could absorb energy of any wavelength.

Figure 5. Energy curve for a vibrating spring (left) and energy constrained to quantum mechanical model (right).
However, vibrational motion is quantized, it must follow the rules of quantum mechanics, and the only transitions which are allowed fit the following formula: Ev = (v+ 1/2)h?
Where ? is the frequency of the vibration and v is the vibration quantum number (0, 1, 2, 3, . . .).
Ev = (v+ 1/2)?osc by joule units.
?v= E/hC = (v + 1/2)?-osc by Cm-1
?-osc is oscillator frequency, selection rule is ?v=?1.
The lowest energy level is E0 = 1/2 h?, the next highest is E1 = 3/2 h?. According to the selection rule, only transitions to the next energy level are allowed; therefore molecules will absorb an amount of energy equal to 3/2 – 1/2 h? or h?. This rule is not inflexible, and occasionally transitions of 2 h?, 3 h?, or higher are observed. These correspond to bands called overtones in an IR spectrum. They are of lower intensity than the fundamental vibration bands.

A molecule is not just two atoms joined on a spring, of course. A bond can come apart, and it cannot be compressed beyond a certain point. A molecule is actually an inharmonic oscillator. As inter atomic distance increases, the energy reaches a maximum, as seen in Figure 6. Note how the energy levels become more closely spaced with increasing inter atomic distance in the inharmonic oscillator. The allowed transitions, h?, become smaller in energy. Therefore, overtones can be lower in energy than predicted by the harmonic oscillator theory. The following formula has been derived from Hooke’s law. For the case of a diatomic molecule, (? has been substituted for ?, recall that ? = c? from equations).
The energy value for this modulation is
?v= E/hC = (v+ 1/2)?-e - (v + 1/2)2?-e .?e - (v + 1/2)3?-e .?e…..exc by Cm-1
Since ?-e is oscillator frequency, and ?e is a harmonist constant with value did not exceeded 0.01. Selection rule for corrected state is involved long transition, selection rule is ?v=?1,?2,?3,?4…..

Figure 6. Energy curve of vibration levels for vibrating bond of inharmonic oscillator model.

Final Equation shows the relationship of bond strength and atomic mass to the wave number at which a molecule will absorb IR radiation. As the force constant increases, the vibration frequency (wave number) also increases. The force constants for bonds are:

Single bond 5 x 105 dyne/cm
Double bond 10 x 105 dyne/cm
Triple bond 15 x 105 dyne/cm

As the mass of the atoms increases, the vibration frequency decreases. Using the following mass values: C, carbon 12/6.02 x 1023. H, hydrogen 1/6.02 x 1023.

The frequency (?) for a C–H bond is calculated to be 3032 cm–1. The actual range for C–H absorptions is 2850–3000 cm–1. The regions of an IR spectrum where bond stretching vibrations are seen depends primarily on whether the bonds are single, double, or triple or bonds to hydrogen. The following table shows where absorption by single, double, and triple bonds are observed in an IR spectrum.

Bond absorption region, cm–1

C–C, C–O, C–N 800–1300
C=C, C=O, C=N, N=O 1500–1900
C?C, C?N 2000–2300
C–H, N–H, O–H 2700–3800


6-Applications.
Example 1: Calculate the fundmental frequency of hydrogen molecules if you know thats the atomic mass of hydrogen atom is 1.0079AMU and the force constant of hydrogen molecule equal to573.4 Nm-1 .
Solution:-


the frequency of the vibration is related to the mass and the force constant of the spring, k,
?=1/2?.?(k/µ)


Example2: the infrared spectrum of 39K35Cl has a single intense line at 378.0 cm-1. What is the force constant?
Solution:-
Our first step is to convert the wavenumber to frequency, using. This yields ?=cv-
v = 378 cm-1 x 3.00 x 1010 cm/s = 1.134 x 1013 s-1.

The reduced mass of 39K35Cl is
? KCl =( 35AMUx39AMU) / (35AMU +39AMU) x1.67x 10-27kg /AMU = 3.08x 10-26kg
Rewriting our frequency equation to solve for force constant yields
k = (2??)2 ? = (2 ? 1.134 x 1013) 2 x 3.08 x 10-26 = 156 N/m.
Example3:- Determine the differences between two isotopes (16O-16O) and (18O-18O) in infrared spectroscopy. Through the bound O-O stretching frequency of oxyhemocyanin.
Solution:-
The different isotopes in a particular species may give fine detail in infrared spectroscopy. For example, the O-O stretching frequency of oxyhemocyanin is experimentally determined to be 832 and 788 cm-1 for ?(16O-16O) and ?(18O-18O) respectively.
By considering the O-O as a spring, the wavelength of absorbance, ? can be calculated:

where k is the spring constant for the bond, and ? is the reduced mass of the A-B system:

(mi is the mass of atom i).
The reduced masses for 16O-16O and 18O-18O can be approximated as 8 and 9 respectively. Thus: