Molecular Spectroscopy For UG and PG Students part 3.pptx

General Molecular Spectroscopy for UG and PG Students. Part-III

1. Shri Shivaji Education Society Amravati’s Shri Shivaji Arts, Commerce & Science College Motala , Dist. Buldana

2. Molecular Spectroscopy Part-3 By Mr. Bhaskarrao Subhashrao Bhise Assistant Professor Head Department of Chemistry Shri Shivaji Arts, Commerce & Science College Motala Dist. Buldana

3. Content Vibrational ( IR ) Spectroscopy :- Pure Rotational Raman Spectrum of Diatomic molecule Condition of Raman Active Molecules Raman Spectroscopy Relationship between vibrational freq. and force constat Selection rule for Vibrational Transitions Expression of vibrational energy of diatomic molecule

4. Vibrational ( Infra Red ) Spectroscopy Vibrational spectroscopy is one most powerful tools for elucidation of molecular structure. Vibrational transition form by absorbing radiation nearly equal to 0.1 ev get excited from ground level. Vibrational spectrum molecules provide information about oscillational frequency of covalent bond and force constant of bond calculated. Generally absorb radiation and change in dipole moment these are IR active. Heteronuclear diatomic molecules, change diople moment during vibration as a result which are IR active. Homonuclear diatomic molecules, dipole moment does not change during vibration and hence these are inactive. Condition for IR active diatomic molecule:-

5. Examples:- H2, N2, O2, Cl2, F2 ------------------IR Inactive HCl, NO, CO, HF, HI ------------------ IR Active Expression for vibrational energy of a diatomic molecule Consider a diatomic molecule which behaves like harmonic oscillator. According to wave mechanics, vibrational energy is given byequation . E vib = (V+ ɷ osc Since, ω osc = Cɷ osc , we get E vib = (V+ Where, V = Vibrational quantum no. V= 0, 1, 2 3, …….. h= planck constant , c= velocity of light = Vibrational frequency of oscillator in term wave no. in cm -1 E vib = vibrational energy in cm -1  

6. V0 V1 V2 V3 V4 Є 0= 1/2ɷosc x hc Є 0= 3/2ɷosc x hc Є 0= 5/2ɷosc x hc Є 0= 7/2ɷosc x hc Є 0= 9/2ɷosc x hc Vibrational Energy Zero point energy …… Fig. Vibrational Energy Level Diagram Zero point energy :- Lowest vibrational energy possessed by the molecule V=0, Є 0= 1/2ɷosc x hc is called zero point energy Selection Rule for Vibrational Transitions:- The selection rule for vibrational transition is Δ V = The plus sign indicate absorption spectra and minus indicate emission spectra. Vibrational spectra studied in absorption mode i.e. Δ V = +1  

7. Position of vibrational band in Heteronuclear diatomic molecule Vibrational absorptional spectrum of diatomic molecule exhibits one band according to selection rule Δ =+1 The molecule present in V=0 and hence band form V=0 to V=1. However other also form Δ =+ 1 i.e V=1 to V = 2, V=2 to V=3, V=3 to V=4 etc ⊽ = Є 1- Є i.e. 3/2ɷosc - 1/2ɷosc = ɷosc cm -1 In case of anharmonic oscillator , the selection rule are Δ V = , etc. Δ V = give fundamental bands ( strong ) Other band with Δ V = etc are called as overtone ( very weak ).  

8. Raman Spectroscopy In 1928, Sir C. V. Raman discovered that monochromatic beam of light passed through a substance in solid, liquid or gaseous state, the light is scattered containing additional frequency over and above of incident light frequency this phenomenon is known as Raman Effect. The spectrum of scattered light is characteristic property of scattering substance is called as Raman spectrum of molecule. The scattered lines having lower frequency ( High Wavelength) than that of incident frequency are called as S tokes lines . The scattered lines having higher frequency ( Low Wavelength) than that of incident frequency are called as Anti-stokes lines .

9. If ⊽i is wave number of incident radiation and ⊽ s is scattered radiation by given molecule, The Raman shift given by Δ ⊽ = ⊽i - ⊽s For Stokes l ines :- Δ ⊽ = + ve i.e. ⊽i > ⊽ s or ⊽s < ⊽i For Anti-Stokes lines :- Δ ⊽ = - ve i.e. ⊽i < ⊽s or ⊽s > ⊽i

10. Low Frequency ( High Wavelength) Stokes Line High Frequency ( Low Wavelength ) Anti-Stokes Line Rayleigh Line Incident Light Characteristic of Raman Spectrum:- The intensity of stokes lines is always greater than Anti-stoke lines. Raman shift, Δ ⊽ generally lies within far and near Infrared regions. Raman line are symmetrical to Rayleigh line. Raman line are due to change in polarization of the molecules.

11. Condition for Raman Active Molecules When molecule placed in external magnetic field, its electron and nuclei are displaced. Thus, dipole moment in produced in the molecule due to displacement of electron and nuclei is said to be polarized. The scattering of light is explained on the basis of polarizibility of molecules. Raman active molecules to be vibrations/ rotation should be change polarization of molecule. Mononuclear diatomic molecules are Raman active due to change p[ olarization during vibration/ rotation . Raman active molecules but these are inactive in microwave / IR spectroscopy due to no change in dipole moment and no permanent dipole moment.

12. Origin of Stokes and Anti-Stokes lines in Raman spectrum on the basis of quantum theory. Rayleigh Line:- When molecule absorb electromagnetic radiation ( hv ), it is excited from lower energy to higher energy level. The life time of excited state is very short, it de-excite immediately to lower level. In addition with posses vibrational and rotational energy levels. After de-excitation of molecule returns to same vibrational/ rotational level will have exactly same energy ( Frequency) as incident frequency. This transition gives to very intense line in raman spectrum of molecule called as Rayleigh line. 2) Stokes line :- When molecule absorb electromagnetic radiation ( Incident light) get excited additional with vibrational / rotational transitions. It de-excites as same transition then scattered photon has less energy compare to incident energy this transition gives stokes line. In de-excitation loss some energy to the molecule which used for vibrational/ rotational excitation.

13. 3) Antistokes line:- After excitation molecule de-excited to vibrational/ rotational ground level then scattered light ( energy ) has more energy than that of incident light ( energy ). This transition gives antistoke line in Raman spectrum of molecule.

14. Pure Rotational Raman Spectrum of the Diatomic Molecule Selection rule for pure rotational spectrum:- Only those rotational transition are allowed for which Δ j =0 or Δ j = When Δ j =0 , the scattered Raman radiation will be same frequency as incident light. When Δ j = give stokes line and Δ j = give anti-stokes lines.   Raman shift for diatomic molecule The energy of rigid rotator is given by Ej = J(J+1) or ε j = or ε j = BJ(J+1) cm -1 As selection rule for rotational transition is Δ j= For stoke line Δ j = Raman shift is given by  

15. For stokes line Δ j = For Anti-stokes line Δ j = Δ ⊽ =( ε j +2 – ε j ) cm -1 Δ ⊽ =( ε j – ε j +2 ) cm -1 = [B(J+2)(J+3)] – [BJ(J+1)] =B[J 2 + 5J + 6 – J 2 – J] = -2B (2J+3) cm -1 -----------(2) Δ ⊽ = 2B ( 2J+3) cm -1 ----------------(1) Δ ⊽ = B( 2J+3) cm -1 where J=0,1,2,3-- Position of stokes and antistokes line in pure rotational spectrum of diatomic molecule are shown in the following table   Sr. No. Line J Stokes line ⊽ s= ⊽ i -2B(2J+3) cm -1 Antistokes line ⊽ s= ⊽ i +2B(2J+3) cm -1 1 First line ⊽ s= ⊽ i -6B ⊽ s= ⊽ i +6B 2 Second line 1 ⊽ s= ⊽ i -10B ⊽ s= ⊽ i +10B 3 Third line 2 ⊽ s= ⊽ i -14B ⊽ s= ⊽ i +14B 4 Fourth line 3 ⊽ s= ⊽ i -18B ⊽ s= ⊽ i +18B 5 Fifth line 4 ⊽ s= ⊽ i -22B ⊽ s= ⊽ i +22

16. Thus, it is clear that from above table stoke & antistoke lines separated from Rayleigh line by 6B cm-1, where subsequent lines are separated by 4B cm-1. Raman rotational transition spectrum is shown as. Δ j = Δ j = Stokes lines ( Rayleigh line) Antistokes lines 4 6 6 4 4 4 4 4 18B 14B 10B 6B ⊽ I 6B 10B 14B 18B Rotational transition and Rotational R aman spectrum of diatomic molecules  

17. Vibrational Raman Spectrum of Diatomic Molecule Pure vibrational Raman spectrum of diatomic molecule results due to bibrational transition accordind to selection rule Δ ⊽ = For stoke line Δ ⊽ = +1 and Antistoke line Δ ⊽ = -1. According to wave mechanics, vibrational energy is given by E vib = (V+ ɷ osc J OR E vib = (V+ ɷ osc cm -1 3 Є =7/2 ɷ osc 2 Є =5/2 ɷ osc 1 Є =3/2 ɷ os Є =1/2 ɷ osc (stoke line) ( Antistoke line) Thus pure vibrational spectrum a stoke line will appear at ⊽ s = ⊽ i – ɷ osc Where, as Antistoke line will appear ⊽ s= ⊽ I + ɷ osc  

18. T H A N K Y O U