13 Rotational and Vibrational Raman spectra

epgp books

 

Contents:

 

1.     Raman Scattering

2.     Rotational Raman Spectra

3.     Vibrational Raman Spectra

4.     Vibrational – Rotational Raman Spectra

5.     Intensity Alternation in Raman Spectra

6.     Infrared spectroscopy

7.     Infrared and Raman Spectroscopy – What’s the Difference?

8.     Raman and IR active modes for CO2

9.     Difference between IR-Spectra and Raman Spectra Summary

 

1. Raman Scattering

 

Raman spectroscopy is concerned with radiation scattering from a sample. The Scattering occurs when an incident photon interacts with the electric dipole of a molecule. This scattering process can be either elastic or inelastic. Most incident photons are elastically scattered by the molecule (Rayleigh scattering).In Rayleigh scattering the energy of the incident photons equals the energy of the scattered photons. A small fraction of light is scattered at energies different than that of the incident photons (Raman Effect). The Raman Effect is an inelastic process

The two situations arising with Raman scattering are

 

a.     Scattered photons have a lower energy (Stokes scattering – photon emitted)

b.     Scattered photons have a higher energy (anti-Stokes scattering – photon absorbed)

  • The energy of a vibrational mode depends on the molecule’s structure and environment.
  • Raman spectra of different molecules are unique
  • Raman intensity lines are 0.001% (at most) of the source intensity.
  • The intensity can be increased by 103 – 106 orders of magnitude if the sample is adsorbed on the surface of colloidal metal particles.

 

2. Rotational Raman Spectra

 

The rotational energy levels of diatomic molecules are given by

F (J) = BJ (J+1) – DJ2 (J+1) 2

 

Where J = 0, 1, 2, ……… D is the centrifugal distortion constant and is ignored in Raman spectroscopy, due to the precision of the measurements. Therefore, retaining only the first term, the rotation energy levels are

 

F (J) = BJ (J + 1)

 

∆J = 0, ± 2 are the selection rules for the transition between rotational levels in Raman spectroscopy. There is a change of two units in the rotational quantum number that is connected with the symmetry of polarizability ellipsoid. The ellipsoid presents the same appearance after π radians during end-over-end rotation in case of a diatomic molecule. The magnitude of the frequency shift is given by

∆v̅ = F (J +2) – F(J) = B(J + 2) (J + 3) – BJ(J +1) = 4B(J+ 3/2 )

Rotational Raman spectrum

 

Figure depicts the transitions between the rotational energy levels. The selection rules dictate a series of equally spaced Raman lines on either side of the undisplaced line. The two series of lines are called S branches and both having ∆J = 2. The transition J → J + 2 results in a shift to longer wavelengths and are referred to Stokes lines while transition J + 2 → J results in a shift toward a shorter wavelengths refereed to anti-Stokes lines. The separation between lines of both the series is 4B while the separation between the exciting line (undisplaced line) and first of each of the anti-Stokes and Stokes lines is 6B. The intensities of the Raman lines reflect the population of the rotational levels. As the initial state of anti-Stokes lines is higher in energy than that of the corresponding Stokes lines, the anti-Stokes lines are correspondingly weaker. This effect is small as the rotational energy levels are closely spaced except in light molecules.

 

3. Vibrational Raman Spectra

 

In simple harmonic oscillator model of diatomic molecule, the energy levels are given by

 

The selection rules for vibrational Raman spectra are ∆v = ± 1. The vibrational transitions are observed at

Therefore the Stokes and anti-Stokes components are obtained that are shifted by frequency ∆E/ h from Rayleigh line.

 

Most of the molecules are in the lowest vibrational state at normal temperature, and few are in the excited vibrational states. The intensity of the Stokes lines corresponding to the transition 0 →1 is thus much greater than that of 1→0 transition. The separation between the lowest vibrational levels is so large that only the Stokes lines are observed at at normal temperature in the diatomic molecules. The potential is not that of simple harmonic oscillator. The non-harmonicity of the vibrational potential causes the spacing between the vibrational levels not to be constant. Thus the occurrence of separated Stokes lines originates from different vibrational levels.

 

4. Vibrational – Rotational Raman Spectra

 

For a given vibrational band

v̅ =v̅o + B’vJ’(J’+ 1) – B’’vJ’’(J’’ + 1) ‘

Here v̅ o is the frequency of pure vibrational transition without taking into account the rotation. The selection rules for vibrating rotator are those of anharmonic oscillator and the rotator, i.e.,

 

∆v = ± 1, ± 2, …….

∆      J = 0, ± 2

 

For J’= J’’ + 2 = J + 2 (S-branch), from the above equation

 

v̅S (J) = v̅0 + B’v (J + 2) (J + 3) – B’’v J (J + 1)

v̅S (J) = v̅0 + 6 B’v + (5B’V – B’’v) J + (B’v – B’’v) J2

Where J = 0, 1, 2, ………..

 

For J’= J’’ – 2 = J – 2 (O – branch),

v̅O(J) = v̅0 + B’v (J – 2) (J – 1) – B’’v J (J + 1)

v̅O (J) = v̅0 + 6 B’v + (3B’V – B’’v) J + (B’v – B’’v) J2 (15.17)

 

Where J = 2, 3, ………….

 

For J’= J’’ = J (Q branch)

v̅O(J) = v̅0 + B’v J(J – 1) – B’’v J (J + 1)

v̅O (J) = v̅0 + (B’v – B’’v)J + (B’V – B’’v) J2

Where J = 0, 1, 2, ……….

 

The figure depicts the rotational transitions accompanying a vibrational transition in a Raman spectrum.

 

 

 

 

For vibrational transition 0 →1 , the difference between??′ and ??”  is very small, therefore??′= ??”=B

 

So

ῡs(J) = ῡ0 + 4BJ + 6B

ῡ0(J) = ῡ0 − 4BJ + 2B

ῡQ(J) = ῡ0

Therefore, lines of Q branch are very close to one another and are not separately. This gives rise to an intense line at v̅0 at shorter wavelength side. The separation of the first S(0) and the first O(2) is 12 B. The S branch occurs at higher frequency and O branch is at lower side of the frequency of Q branch. The lines of S and O branches are very much weak. They form a series similar to R and P branches of infrared bands except that the line separation is twice as large.

 

The method of combination differences is used to obtain the value of Bv’ and Bv’’

∆4” F(J) = ῡs(J − 2) − ῡO(J + 2) = 8??“(? + 1/2)
∆4′ F(J) = ῡs(J) − ῡO(J) = 8?′(? + 1/2)

Graphs of ∆′4F(J) versus (J + ½) and ∆”4F(J) versus (J + ½), gives straight lines with slopes 8B’v and 8B’’v respectively

 

5. Intensity Alternation in Raman Spectra

 

Transitions between only even J rotational levels are observed while transitions between odd J rotational levels are missing for Raman spectra of some molecules,. However, there is an alternation in intensity of the adjacent lines in some molecules. The intensity of transitions between odd J value rotational levels is three times stronger than that between even J value rotational levels in case of Raman spectra of H2 molecule. This is understood on the basis of Pauli principle because for homonuclear molecules, the states that can be occupied and the transitions that are allowed are restricted by the symmetry requirement. In case of the integral nuclear spins (0, 1, 2,…..), the complete wavefunction of the molecule must be symmetric with respect to exchange of labels of two identical nuclei, while For half-integral nuclear spin the wavefunction must be antisymmetric in an exchange of the labels of the two nuclei (identical fermions).

 

The total wavefunction Ψ of the molecule is

 

Ψ = ΨeΨvΨrΨns (neglecting the small interactions between the modes associated with the electronic, vibrational, rotational and nuclear spin behavior of the molecule)

 

Where ΨeΨvand Ψr are the electronic, vibrational and rotational wavefunctions andΨns is nuclear spin wavefunctions. Now, Ψe and Ψv are symmetric in an exchange of the labels of the two nuclei. The symmetry of the wavefunction Ψ is governed by the symmetry properties of the product ofΨrΨns

 

For I having half integral value, Ψ and hence ΨrΨns must be antisymmetric in nuclear label exchange and thus if Ψr is symmetric, Ψns must be antisymmetric.

 

Considering the H2 molecule, spin of hydrogen nuclei is 1/2 (fermion). So Ψ must be antisymmetric with respect to interchange of nuclei. The Ψr may be symmetric or antisymmetric depending upon whether J is even or odd.

 

For I = ½, Ml takes the values + ½ or – ½. The nuclear spin wavefunction Ψns is written as α or β corresponding to M1 = +1/2 or – ½, respectively. Both the H1 nuclei labelled 1 and 2 can have either α or β nuclear wavefunction. There are four possible forms of Ψns for the molecule as a whole

Ψns= α(1) α (2); β(1)β(2); α(1) β (2); β(1)α (2);

 

α(1)α(2) and β(1)β (2) are symmetric with respect to interchange of 1 and 2. α(1) β (2) and β(1)α (2) are neither symmetric nor antisymmetric. The linear combination of α(1)β(2) and α(2)β(1) can be made symmetric or antisymmetric. The linear combinations, are

1⁄√2 [α(1)α(2) – β(1)β (2)]      and     1⁄√2 [α(1)α(2) + β(1)β (2)]

are antisymmetric and symmetric, respectively. Thus there are three symmetric and one antisymmetric Ψns. In order for ΨrΨns to be always antisymmetric for H2 molecule, the single antisymmetric Ψns is associated between even J value while three symmetric Ψns are associated with odd J value. Thus transitions between odd J rotational levels are three times stronger than those between even J rotational levels.

 

Consider an example of CO2 molecule which is linear and carbon atom is at the centre of gravity of the molecule. Rotation about centre of gravity involves only oxygen atoms and C does not move. The nuclei of O16 atom have a nuclear spin I = 0. Particle with integral spins obey Bose- Einstein statistics. According to the Pauli principle, Ψ must be symmetric with respect to the interchange of the particle labeling (in this case oxygen nuclei.) The electronic wavefunction Ψe is symmetric with respect to interchange of oxygen nuclei since nothing changes in the interchange of the nuclei. The Ψr is symmetric with respect to the interchange of oxygen nuclei. The symmetry of Ψr depends on the value of J. As spin I = 0, therefore there is no nuclear spin function and hence ignored. In order for Ψ to be symmetric with respect to interchange of oxygen atoms, the Ψr must be symmetric as Ψe and Ψv are symmetric. Therefore, only the states for even J exist and odd J states are missing, that is, only even J states are populated. Hence, transitions between even J value rotational levels can take place and transitions between odd J value rotational levels are missing. Thus in Raman spectra of CO2 every other line is missing.

 

In general case, the number of antisymmetric spin states to the number of symmetric spin states determines the relative intensity of rotational lines. It is shown that the ratio R of antisymmetric states to symmetric spin states gives the ratio of strong and weak lines if two nuclei of homonuclear molecule each having spin I

Assignments:

 

1.     In the rotational Raman spectrum of a diatomic molecule the separation from Rayleigh line of the first Stokes line is different from the subsequent line spacing. Find the ratio of these spacing when only odd levels are populated.

 

The spacing between the Rayleigh line and the first Stokes line corresponding to transition J = 0 → J = 2 is 6B. Separation between adjacent Stokes lines is 4B. As only odd J levels are populated therefore, only transitions between odd J value levels will be observed. Hence first observed Stokes line would correspond to the transition J = 1 → J = 3. This line is at a separation of 10B to Rayleigh line. The Stokes lines corresponding to transitions between even J values are missing, hence the separation between Stokes lines corresponding to two consecutive odd J value transitions would be 8B.

 

The required ratio is then 10B/8B = 5/4.

 

2. Calculate  in  units  of  B,  the  frequency  of  the  rotational  lines  of  H2

 

resulting from the transitions to the excited state characterized by the quantum number J = 4. If the bond length of H2 is 0.07417 nm, determine the spacing between the lines.

 

The mass of hydrogen atom is 1.673 x 10-27Kg.

 

For Stokes lines transition take place from J = 2 → J = 4 and for anti-Stokes line the transition is from J = 6 →J = 4. The position of the first Stokes and anti-Stokes line is at 6B to the position of Rayleigh line. The spacing between the adjacent Stokes lines is 4B and same is true for anti-Stokes lines. The first anti-Stokes line corresponds to the transition J = 2→    J= 0 and therefore anti-Stokes transition J = 6 →J = 4 transition would be 6B + 16B = 22B away from the Rayleigh line. On the other hand first Stokes line corresponds to the transition J = 0 → J = 2 and J = 2 → J = 4 transition would be 8B away from the first Stokes line or 6B + 8B = 14B away from the Rayleigh line.

 

The rotational constant B is given by

 

B= h/8πIc and I=μr2

 

B=60.8 cm-1

 

6.  Infrared spectroscopy

 

• There are three regions; the near-, mid- and far- infrared in the infrared portion of the electromagnetic spectrum. The higher-energy near-IR ~ 14000–4000 cm−1 (~0.8–2.5 μm  wavelength)  can excite overtone or harmonic vibrations.

 

The  mid-infrared  ~  4000–400 cm−1(2.5–25 μm)  is  used  to  study  the fundamental  vibrations  and  associated rotational-vibrational structure. The far-infrared ~ 400–10 cm−1 (25–1000 μm) lying adjacent to the microwave region, has low energy and is used for rotational spectroscopy.

 

• Infrared spectroscopy is based on the fact that molecules absorb frequencies that are characteristic of their structure. These absorptions occur at resonant frequencies which mean the frequency of the absorbed radiation matches the vibrational frequency.

• The energies are affected by the shape of the molecular potential energy surfaces, the masses of the atoms, and the associated vibronic coupling.

•     In Born–Oppenheimer and    harmonic    approximations,    i.e.    when the molecular   Hamiltonian corresponding   to   the   electronic ground state can be approximated by a harmonic oscillator in the neighbourhood of the equilibrium molecular geometry, the resonant frequencies are associated with the normal modes corresponding to the molecular electronic ground state potential energy surface. The resonant frequencies are also related to the strength of the bond and the mass of the atoms at either end of it. Thus, the frequencies of the vibrations are associated with a particular normal mode of motion and a particular bond type.

•  Number of Vibrational modes

• In IR active sample, the vibrational mode must be associated with changes in the dipole moment. However, a permanent dipole is not necessary, as the requirement is only for a change in dipole moment.

• A  molecule   can   vibrate  in   many  ways,  and   each   way  is    called a vibrational mode. Consider a molecule with N number of atoms, linear molecule has 3N – 5 degrees of vibrational modes, while nonlinear molecules  have  3N – 6  degrees  of  vibrational modes  or  vibrational degrees of freedom. For example H2O is a non-linear molecule for which 3 × 3 – 6 = 3 degrees of vibrational freedom, or modes occur.

•  As simple diatomic molecules have only one bond, there is only one vibrational band. In case of the symmetrical molecule like. N2, the band is not observed in the IR spectrum, but found in the Raman spectrum. Asymmetrical diatomic molecules like CO, absorb in the IR spectrum. As more complex molecules have many bonds, so their vibrational spectra are also more complex or the big molecules have many peaks in their IR spectra.

•  The atoms in a CH2X2 group, where X represent any other atom, can vibrate  in  nine  different  ways.  Six  of  these  vibrations  involve  only the CH2 portion: symmetric and anti symmetric stretching, scissoring, rocking, wagging and twisting, Structures that do not have the two additional X groups attached have fewer modes because some modes are defined by specific relationships to those other attached groups. For example, in water, the rocking, wagging, and twisting modes do not exist because these types of motions of the H represent simple rotation of the whole molecule rather than vibrations within it. The simplest and most important IR bands arise from the “normal modes,” corresponding to the simplest distortions of the molecule. In some cases, “overtone bands” are observed. These bands arise from the absorption of a photon that leads to a doubly excited Vibrational state.

• Practical IR spectroscopy

• The infrared spectrum of a sample is recorded by passing a beam of infrared light through the sample. When the frequency of the IR is the same as the Vibrational frequency of a bond or collection of bonds, absorption occurs. Examination of the transmitted light reveals how much energy was absorbed at each frequency (or wavelength). This measurement can be achieved by scanning the wavelength range using a monochromator. Alternatively, the entire wavelength range is measured using a Fourier transform instrument and then a transmittance or absorbance spectrum is generated using a dedicated procedure.

•  This technique is commonly used for analyzing samples with covalent bonds. Simple spectra are obtained from samples with few IR active bonds and high levels of purity. More complex molecular structures lead to more absorption bands and more complex spectra.

• Uses and applications

• Infrared spectroscopy is a simple and reliable technique widely used in both organic and inorganic chemistry, in research and industry. It is used in quality control, dynamic measurement, and monitoring applications such as the long-term unattended measurement of CO2 concentrations in greenhouses and growth chambers by infrared gas analyzers.

• It is also used in forensic analysis in both criminal and civil cases, for example in identifying polymer degradation. It can be used in determining the blood alcohol content of a suspected drunk driver.

•IR-spectroscopy has been successfully used in analysis and identification of pigments in paintings and other art objects such as illuminated manuscripts.

• A useful way of analyzing solid samples without the need for cutting samples uses ATR or attenuated total reflectance spectroscopy. Using this approach, samples are pressed against the face of a single crystal. The infrared radiation passes through the crystal and only interacts with the sample at the interface between the two materials.

• With increasing technology in computer filtering and manipulation of the results, samples in solution can now be measured accurately (water produces a broad absorbance across the range of interest, and thus renders the spectra unreadable without this computer treatment).

• Some instruments will also automatically tell you what substance is being measured from a store of thousands of reference spectra held in storage.

•Infrared spectroscopy is also useful in measuring the degree of polymerization in polymer manufacture. Changes in the character or quantity of a particular bond are assessed by measuring at a specific frequency over time. Modern research instruments can take infrared measurements across the range of interest as frequently as 32 times a second. This can be done whilst simultaneous measurements are made using other techniques. This makes the observations of chemical reactions and processes quicker and more accurate.

•Infrared spectroscopy has also been successfully utilized in the field of semiconductor microelectronics: for example, infrared spectroscopy can be applied to semiconductors like silicon, gallium arsenide, gallium nitride, zinc selenide, amorphous silicon, silicon nitride, etc.

• Another important application of Infrared Spectroscopy is in the food industry to measure the concentration of various compounds in different food products

•The instruments are now small, and can be transported, even for use in field trials.

• Infrared Spectroscopy is also used in gas leak detection devices such as the DP-IR and EyeCGAs. These devices detect hydrocarbon gas leaks in the transportation of natural gas and crude oil.

•  In February 2014, NASA announced a greatly upgraded database, based on IR spectroscopy, for tracking polycyclic aromatic hydrocarbons (PAHs) in the universe. According to scientists, more than 20% of the carbon in the universe may be associated with PAHs, possible starting materials for the formation of life. PAHs seem to have been formed shortly after the Big Bang, are widespread throughout the universe, and are associated with new stars and exoplanets.

• Recent developments include a miniature IR-spectrometer that’s linked to a cloud based database and suitable for personal everyday use and NIR-spectroscopic chips that can be embedded in smart phones and various gadgets.

 

7.  Infrared and Raman Spectroscopy – What’s the Difference?

 

Though Infrared and Raman are two similar spectroscopic techniques but they differ. As the photons interact with molecules and induce transitions among the available energy levels, these transitions result in the emission of photons with various wavelengths.

 

The photons emitted due to transitions between vibrational energy levels of molecules are detected with two spectroscopic techniques, the Infra-red (IR) and Raman Spectroscopy.

  • The difference lies in the nature of the molecular transitions taking place. For a transition to be Raman active there is a change in the polarizability of the molecule during the vibration. This means that the electron cloud of the molecule undergoes positional change. While, for an IR detectable transition, the molecule undergoes dipole moment change during vibration. Therefore, one cannot observe any IR absorption lines in case of symmetrical molecule e.g. O2, as the molecule cannot change its dipole moment. The molecules with a strong dipole moment are hard to polarize.
  • The Raman technique uses a monochromatic beam or laser, in the visible, near-infrared, or near ultraviolet range of the electromagnetic spectrum as an excitation source. In IR spectroscopy, a monochromatic beam is used in the infrared region of the electromagnetic spectrum. In order to observe all the absorption lines within a specific range of the infrared region, the wavelength increases or decreases over time,
  • Another difference is observed in the resulting spectra. The IR technique shows irregular absorbance (or transmittance) lines, depending on the material under test. The Raman spectrum mainly comprises the elastic scattered light line (Rayleigh) and two equally distanced lines Stokes and anti-Stokes, with the second being quite weak and difficult to detect.

It is to note that the Raman technique requires high-stability laser sources and sensitive amplification equipment to detect the weak signal.

 

8. Raman and IR active modes for CO2:

 

As it is known that vibrational Raman shift in Raman Spectrum of the diatomic molecule is same as that of the wavenumber of the fundamental vibrational band in NIR spectrum. It is because of the fact that fundamental infra-red absorption band of diatomic molecule is observed as result of vibrational transition which is ν=0→ν=1 that is same as of vibrational Stokes Raman band.

There are certain Similarities

 

1.    Both are common vibrational spectroscopy useful for assessing molecular motion and fingerprinting species.

 

2.   Both work on the basis of inelastic scattering of a monochromatic excitation of source

 

3.  Routine energy range 200-4000cm-1

 

 

9. Difference between IR-Spectra and Raman Spectra

 

 

Summary:

 

Raman Scattering

Rotational Raman Spectra

Vibrational Raman Spectra

Vibrational – Rotational Raman Spectra

 

Intensity Alternation in Raman Spectra

 

Infrared spectroscopy

Infrared and Raman Spectroscopy – What’s the Difference?

Raman and IR active modes for CO2

Difference between IR-Spectra and Raman Spectra

you can view video on Rotational and Vibrational Raman spectra