![Relationship between the frequency difference (Δν¯) of spectral lines in rotational spectra
and rotational constant (B) of harmonically rotating molecule giving rise to rotational
spectra.
• The rotational energy of diatomic molecule EJ in
rotational energy level J is given by Schrodinger’s
relation
EJ = h2 . J (J+1)/8π2I ------- (1)
Here h is Planck’s constant; I is the moment of inertia
of a molecule rotating about the axis; J is the rotational
quantum number having the values 1,2,3 ----etc.
The spectral lines are obtained by change (transition)
in rotational energy levels.
The change in energy is related to the frequency (ν)
(pronounced as nu) by the Plancks equation
E = h ν
But ν = c/ λ and ν¯ = 1/λ
Hence ν = c ν¯
and E= h ν¯c --------(2)
Here ν¯ is the frequency in terms of wavenumber
(pronounced as nu bar); c is the velocity of light.
Using eq (1) & (2)
h ν¯c = h2. J (J+1)/8π2I
ν¯ = h . J(J+1)/8 π2I c ------- (3)
The term h /8 π2I c in eq (3) is called rotational
constant (B). Therefore eq (3) will be
ν¯ = B . J(J+1) ------- (4)
Eq (4) gives frequency in terms of wavenumber of
spectral line arising due to different type of rotational
transitions in the molecule.
When the transition take place from J = 0 to J =1
rotational energy level in the molecule, the frequency of
spectral line is calculated by eq (4)
ν¯(0 to 1) = ν¯1 - ν¯0
ν¯(0 to 1) = [B. 1(1+1) ] – [B.0(0+1)]
ν¯(0 to 1) = 2B ------ (5)
eq (5) gives frequency (in terms of wavenumber) of the
spectral line arising due to rotational transition from
J= 0 to J= 1 rotational energy levels.](https://image.slidesharecdn.com/molecularspectroscopy-210212110753/75/Molecular-spectroscopy-11-2048.jpg)
![When the transition take place from J = 1 to J =2
rotational energy level in the molecule, the
frequency of spectral line is calculated by eq (4)
ν¯(1 to 2) = ν¯2 - ν¯1
ν¯(1 to 2) = [B. 2(2+1) ] – [B.1(1+1)]
ν¯(1 to 2) = 4B ------ (6)
eq (6) gives frequency (in terms of wavenumber) of
the spectral line arising due to rotational transition
from J= 1 to J= 2 rotational energy levels.
Similarly when the transition take place from J = 2
to J =3 rotational energy level in the molecule, the
frequency of spectral line as calculated by eq (4)
will be
ν¯(2 to 3) = 6B ------ (7)
eq (7) gives frequency (in terms of wavenumber) of
the spectral line arising due to rotational transition
from J= 2 to J= 3 rotational energy levels.
From eq (5), (6) & (7) it is clear that the
frequency difference between the two
successive spectral lines in the rotational
spectrum will be
Δ ν¯ = ν¯(1 to 2) - ν¯(0 to 1) = 4B-2B = 2B
Δ ν¯ = ν¯(2 to 3) - ν¯(1 to 2) = 6B-4B = 2B
Thus the rotational spectrum of a diatomic
molecule consist of series of equidistance
spectral lines with separation (frequency
difference) of 2B.
In other words, the separation between
successive lines in a rotational spectra are of
equal spacing (2B).](https://image.slidesharecdn.com/molecularspectroscopy-210212110753/75/Molecular-spectroscopy-12-2048.jpg)
![Determination of bond length (internuclear distance) of a diatomic
molecule by using the frequency separation (Δν¯) between the spectral lines
of the rotational spectrum
By measuring the frequency separation (Δν¯) between the successive lines in the rotational spectrum,
the rotational constant (B) can be calculated by eq. (1) as follows
Since Δ ν¯ = 2B
Therefore B = Δ ν¯/2 ------- (1)
From the value of B, the moment of inertia (I) of a diatomic molecule can be calculated by eq. (2) as
follows
I = h /8 π2Bc ------- (2)
From the calculated value of I and by knowing the mass (m1 & m2) of the two atoms in a diatomic
molecule, the internuclear distance (bond length) can be calculated by eq. (3) as follows
Since I = µr2
But µ= (m1m2/m1+m2)
Therefore I = [m1m2/m1+m2]r2
and hence r2 = I [m1+m2/m1m2] -------- (3)
Thus from eq (3) we can calculate the internuclear distance or bond length (r) of a diatomic molecule
by knowing the frequency separation (Δν¯) between the successive lines in the rotational spectrum.](https://image.slidesharecdn.com/molecularspectroscopy-210212110753/75/Molecular-spectroscopy-13-2048.jpg)
![Relation between frequency of fundamental, first and second overtone bands
• For fundamental band arising due to V = 01
ΔE = Ev=1 – Ev =0
By using eq (1)
ΔE = [(1+
1
2
) hω – (1+
1
2
)2 hxω ] - [(0+
1
2
) hω – (0+
1
2
)2 hxω
ΔE = [
3
2
hω – (
3
2
)2 hxω ] - [
1
2
hω – (
1
2
)2 hxω]
ΔE =
3
2
hω –
9
4
hxω -
1
2
hω +
1
4
hxω
ΔE =
3
2
hω -
1
2
hω –
9
4
hxω +
1
4
hxω
ΔE = hω – [
9
4
hxω -
1
4
hxω]
ΔE = hω – 2hxω
ΔE = (1 – 2x)hω ------ (2)
But ω = ω¯ c & ΔE = hν¯c
Therefore eq (2) will be
hν¯c = (1 – 2x)h ω¯ c
ν¯ = (1 – 2x)ω¯ ------- (3)
• For first overtone band arising due to V = 02
ΔE = Ev=2 – Ev =0
By using eq (1)
ΔE = [(2+
1
2
) hω – (2+
1
2
)2 hxω ] - [(0+
1
2
) hω – (0+
1
2
)2 hxω
ΔE = [
5
2
hω – (
5
2
)2 hxω ] - [
1
2
hω – (
1
2
)2 hxω]
ΔE =
5
2
hω –
25
4
hxω -
1
2
hω +
1
4
hxω
ΔE =
5
2
hω -
1
2
hω –
25
4
hxω +
1
4
hxω
ΔE = 2hω – [
25
4
hxω -
1
4
hxω]
ΔE = 2hω – 6hxω
ΔE = (1 – 3x)2hω ------ (4)
But ω = ω¯ c & ΔE = hν¯c
Therefore eq (4) will be
hν1¯c = (1 – 2x)h ω¯ c
ν1¯ = (1 – 3x)2ω¯ ------- (5)
For anharmonic vibrations which are at high energy states having large amplitude of vibration, the vibrational energy will be
Ev = (V+
1
2
) hω – (V+
1
2
)2 hxω --------- (1)
Here x is called anharmonicity constant](https://image.slidesharecdn.com/molecularspectroscopy-210212110753/75/Molecular-spectroscopy-36-2048.jpg)
![Problem 11: In the IR spectra of HCl, the fundamental band appears at 8.658x1013 s-1 and first
overtone band at 1.7004 x1014 s-1. Calculate (i) anharmonicity constant (ii) vibrational frequency of a
molecule (iii) force constant of a bond. Given: H = 1amu; Cl =35.5 amu; 1 amu = 1.66x10-27 Kg;
c =3x108 m.s-1
Given : ν = 8.658x1013 s-1 ν1 = 1.7004 x1014 s-1
Formula: Frequency of fundamental band ν¯ = (1 – 2x)ω¯
Frequency of first overtone band ν1¯ = (1 – 3x)2ω¯
𝑣 = v¯c
Solution: v¯ =
𝑣
𝑐
=
8.658x1013s−1
3x108m.s−1 = 2.886 x105 m-1
v¯1 =
𝑣1
𝑐
=
1.7004 x1014 s−1
3x108m.s−1 = 0.5668 x106 m-1
ν¯
ν1¯
=
(1 – 2x)ω¯
(1 – 3x)2ω¯
2.886 x105
0.5668 x106 =
(1 – 2x)ω¯
(1 – 3x)2ω¯
0.509 =
(1 – 2x)
(1 – 3x)2
X = 0.017 = anharmonicity constant
Frequency of fundamental band ν¯ = (1 – 2x)ω¯
2.886 x105 = [1 – 2(0.017)]ω¯
2.886 x105 = [1 –0.034]ω¯
2.886 x105 = [0.966]ω¯
ω¯ = 2.988 x105 m-1 =vibrational frequency of a
molecule in terms of wave number
µ= (mHmCl) / (mH+ mCl)
= (1x35.5)/(1+35.5) =1.6145x10-27 Kg
K = 4ω¯2π2𝑐2µ
K = 4 (2.988 x105)2 (3.14)2 (3x108)2 (1.6145x10-27)
K = 5.116x102 Nm-1 = force constant](https://image.slidesharecdn.com/molecularspectroscopy-210212110753/75/Molecular-spectroscopy-46-2048.jpg)
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Similar to Molecular spectroscopy
Molecular spectroscopy
- 1. Molecular Spectrsocopy Dr.Pravin U.Singare Department of Chemistry, N.M. Institute of Science, Bhavan’s College, Andheri (West), Mumbai 400 058
- 2. Introduction • It dealswith the interaction of electromagnetic radiations with the matter giving rise to a molecular spectra. • Molecular spectra arises due to absorption or emission of electromagnetic radiation by the molecules. • When a molecules in exited state returns to the ground state by emission of light radiation, then the resulting spectra is called Emission spectra. • When the molecules are exposed to the beam of electromagnetic radiations, the molecules will absorb some wavelength of light radiations and remaining radiations are transmitted. • As a result, certain dark lines or bands are observed at corresponding wavelength when the transmitted radiations are observed in the spectrometer. • This observed spectrum is called absorption spectrum of the molecules.
- 3. • It isthis absorption spectrum which gives valuable information regarding the structure and bonding within the molecules. • As a result the absorption spectra are more informative than emission spectra. Electromagnetic Radiations Wavelength (nm) Energy Radio waves 1012 - 108 Low energy Micro waves 107 - 105 Far IR 15x104 - 3x104 IR 3x104 - 2x104 Near IR 2x104 - 1x104 Visible 800 - 400 U.V. 400 - 1.0 X-rays 1.0 - 0.1 γ -rays 0.1 - 2x10-3 High energy
- 4. Molecular Energy Levels •The molecule consists of number of electronic energy levels E0, E1, E2, E3 etc. • The energy difference between the different electronic energy levels goes on decreasing i.e. E1-E0 > E2-E1 > E3-E2 • Electronic energy levels are further subdivided in to number of vibrational energy levels V0, V1, V2, V3 etc. • The energy difference between the different vibrational energy levels also goes on decreasing i.e. V1-V0 > V2-V1 > V3-V2. • Vibrational energy levels are further subdivided in to number of rotational energy levels J0, J1, J2, J3 etc. • The energy difference between the different rotational energy levels however goes on increasing i.e. J1-J0 < J2-J1 < J3-J2. J1 J2 E3 E2 V1 E0 E1 V2 V3 Energy V0 J0 J3
- 5. • When amolecule is exposed to the electromagnetic radiation, it may result in transition of electrons from lower electronic energy level to higher electronic energy level (electronic transition). vibrations of atoms within the molecule about its mean position (vibrational transition). rotation of the molecule about the axis (rotational transition). • Rotational transition requires least energy (Ej), vibrational transition requires intermediate energy (Ev) while the electronic transition require highest energy (Ee). i.e. Ee > Ev > Ej
- 6. •When a moleculeabsorbs energy from radiations in microwave or Far IR region (less energy radiations), then there will be only rotational energy changes (rotational transition) in the molecule. •When a molecule absorbs radiation from IR region (intermediate energy radiations), then there will be vibrational as well as rotational energy changes in the molecule. •While absorption of radiations from UV region will bring about electronic transitions along with vibrational and rotational energy changes in the molecule. •In other words if energy sufficient for electronic transition is supplied to a molecule, it will also result in simultaneous vibrational and rotational transitions in the molecule.
- 7. Rotational Spectra (MicrowaveSpectra) • Absorption of electromagnetic radiations from Microwave or Far IR region will result in the rotational transitions in the molecule and the resulting spectra is called Microwave spectra or Rotational Spectra. • When a polar diatomic molecules like HCl and non linear polar molecules like CH3Cl rotates about the axis, thereby generating an electric field. • This electric field will interact with the electric field of microwave radiation resulting in absorption of the energy by the molecule. This gives rise to rotational spectra. • So such polar molecules are microwave active. • On the other hand nonpolar molecules like H2, CCl4, CH4, C6H6 when exposed to microwave radiations will rotate about the axis but they do not generate any electric field on rotation. • As a result such molecule do not interact with microwave radiations. Hence no spectra is observed. • So such nonpolar molecules are said to be microwave inactive. • Hence it is necessary that a molecule to undergo change in rotational energy (rotational transition), it must possess a permanent diploe movement. • In other words “To be Microwave active, for rotational spectra a molecule must be polar”
- 8. Relation between momentof Inertia and bond length of a rotating polar diatomic molecule (I =µr2) • Consider a simple rigid polar diatomic molecule AB. Let m1 & m2 be the masses of the two atoms A & B respectively. • Let G be the center of gravity about which the molecule is rotating. • Let r1 and r2 be the distance of the two atoms from the center of gravity. • Let r be the interatomic distance (bond length) such that r = r1 + r2. • For particle of mass m revolving around a fixed point at a distance of r, the moment of inertia will be mr2. • On the same basis, the total moment of inertia (I) of the above rotating diatomic polar molecule will be I = m1r1 2 + m2r2 2 I = m1r1r1 + m2r2r2 ---------(1) As the system is balanced about the center of gravity (G), moments of both the atoms are equal. i.e. m1r1 =m2r2 r1 = m2r2/m1 and r2 = m1r1/m2 -----(2) Substituting the values of r1 & r2 from eq (2) in eq (1) r r2 r1 G B A
- 9. I = m1r1(m2r2/m1) + m2r2 (m1r1/m2) I = r1 (m2r2) + r2 (m1r1) I = r1 r2 (m2) + r2 r1 (m1) I = r1 r2 (m2+ m1) ------ (3) But r = r1 + r2 --------- (4) Substituting the value of r2 from eq (2) in eq (4) r = r1 + m1r1/m2 r = r1 ( 1+ m1/m2) r = r1 (m2+ m1/m2) r1 = r (m2/m2+ m1) ------- (5) Substituting the value of r1 from eq (2) in eq (4) r = m2r2/m1 + r2 r = r2 (m2/m1 +1) r = r2 (m2+ m1/m1) r2 = r (m1/m2+ m1) ------- (6) Substituting eq (5) & eq (6) in eq (3) I = r (m2/m2+ m1) r (m1/m2+ m1) (m2+ m1) I = r2 (m1m2) (m2+ m1) / (m2+ m1)2 I = r2 (m1m2) / (m2+ m1) ----------- (7) The term (m1m2) / (m2+ m1) in eq (7) is known as reduced mass of the system and it is represented by the symbol (µ). Hence eq (7) will become I = µr2 --------- (8) Eq (8) gives the relation between moment of inertia (I) and the internuclear distance (r) of a molecule. From the knowledge of internuclear distance (r) respective masses of the two atoms, the moment of inertia (I) of a simple polar diatomic molecules can be calculated by eq (8).
- 10. Rotational spectra isobtained when molecule absorbs micro wave radiations. By absorbing radiations of different frequency in microwave region there will be rotational transitions in the molecule from one rotational energy level to higher energy level giving rise to series of dark spectral lines in the microwave (rotational) spectra which is recorded on the spectrophotometer. Thus microwave (rotational) spectra consists of series of spectral lines at different frequencies (ν). The frequency of individual spectral lines corresponds to the frequency of microwave radiations absorbed by the molecule. Spectral lines Molecule Incident Microwave/ Far IR radiations Transmitted Microwave / Far IR radiations Spectrophotometer Intensity Frequency in terms of wavenumber (ν¯) ν¯(0to1) ν¯(1to2) ν¯(2to3) ν¯(3to4)
- 11. Relationship between thefrequency difference (Δν¯) of spectral lines in rotational spectra and rotational constant (B) of harmonically rotating molecule giving rise to rotational spectra. • The rotational energy of diatomic molecule EJ in rotational energy level J is given by Schrodinger’s relation EJ = h2 . J (J+1)/8π2I ------- (1) Here h is Planck’s constant; I is the moment of inertia of a molecule rotating about the axis; J is the rotational quantum number having the values 1,2,3 ----etc. The spectral lines are obtained by change (transition) in rotational energy levels. The change in energy is related to the frequency (ν) (pronounced as nu) by the Plancks equation E = h ν But ν = c/ λ and ν¯ = 1/λ Hence ν = c ν¯ and E= h ν¯c --------(2) Here ν¯ is the frequency in terms of wavenumber (pronounced as nu bar); c is the velocity of light. Using eq (1) & (2) h ν¯c = h2. J (J+1)/8π2I ν¯ = h . J(J+1)/8 π2I c ------- (3) The term h /8 π2I c in eq (3) is called rotational constant (B). Therefore eq (3) will be ν¯ = B . J(J+1) ------- (4) Eq (4) gives frequency in terms of wavenumber of spectral line arising due to different type of rotational transitions in the molecule. When the transition take place from J = 0 to J =1 rotational energy level in the molecule, the frequency of spectral line is calculated by eq (4) ν¯(0 to 1) = ν¯1 - ν¯0 ν¯(0 to 1) = [B. 1(1+1) ] – [B.0(0+1)] ν¯(0 to 1) = 2B ------ (5) eq (5) gives frequency (in terms of wavenumber) of the spectral line arising due to rotational transition from J= 0 to J= 1 rotational energy levels.
- 12. When the transitiontake place from J = 1 to J =2 rotational energy level in the molecule, the frequency of spectral line is calculated by eq (4) ν¯(1 to 2) = ν¯2 - ν¯1 ν¯(1 to 2) = [B. 2(2+1) ] – [B.1(1+1)] ν¯(1 to 2) = 4B ------ (6) eq (6) gives frequency (in terms of wavenumber) of the spectral line arising due to rotational transition from J= 1 to J= 2 rotational energy levels. Similarly when the transition take place from J = 2 to J =3 rotational energy level in the molecule, the frequency of spectral line as calculated by eq (4) will be ν¯(2 to 3) = 6B ------ (7) eq (7) gives frequency (in terms of wavenumber) of the spectral line arising due to rotational transition from J= 2 to J= 3 rotational energy levels. From eq (5), (6) & (7) it is clear that the frequency difference between the two successive spectral lines in the rotational spectrum will be Δ ν¯ = ν¯(1 to 2) - ν¯(0 to 1) = 4B-2B = 2B Δ ν¯ = ν¯(2 to 3) - ν¯(1 to 2) = 6B-4B = 2B Thus the rotational spectrum of a diatomic molecule consist of series of equidistance spectral lines with separation (frequency difference) of 2B. In other words, the separation between successive lines in a rotational spectra are of equal spacing (2B).
- 13. Determination of bondlength (internuclear distance) of a diatomic molecule by using the frequency separation (Δν¯) between the spectral lines of the rotational spectrum By measuring the frequency separation (Δν¯) between the successive lines in the rotational spectrum, the rotational constant (B) can be calculated by eq. (1) as follows Since Δ ν¯ = 2B Therefore B = Δ ν¯/2 ------- (1) From the value of B, the moment of inertia (I) of a diatomic molecule can be calculated by eq. (2) as follows I = h /8 π2Bc ------- (2) From the calculated value of I and by knowing the mass (m1 & m2) of the two atoms in a diatomic molecule, the internuclear distance (bond length) can be calculated by eq. (3) as follows Since I = µr2 But µ= (m1m2/m1+m2) Therefore I = [m1m2/m1+m2]r2 and hence r2 = I [m1+m2/m1m2] -------- (3) Thus from eq (3) we can calculate the internuclear distance or bond length (r) of a diatomic molecule by knowing the frequency separation (Δν¯) between the successive lines in the rotational spectrum.
- 14. ISOTOPIC EFFECT ONROTATIONAL SPECTRA The effect of presence of isotope in a molecule on the frequency separation of spectral lines in the rotational spectra. • The presence of isotope in a molecule effect the reduced mass (µ) of a molecule but the internuclear distance (r) and geometry of molecule remain same. • As the atomic mass increases reduced mass (µ) increases. • Increase in the value of µ, increases the moment of inertia (I), • When moment of inertia increases, its rotational constant (B) decreases. • Since 2B value represent the spacing in the rotational spectral line, decrease in B value will further result in decrease in spacing between the spectral lines.
- 15. Since I =µr2 --------(1) B = h/8π2Ic --------(2) Substituting eq (1) in eq (2) we get B = h/8π2 (µr2 )c --------(2) Consider two isotopic molecules x and y. Since the presence of isotope in a molecule effect the reduced mass (µ) of a molecule but the internuclear distance (r) and geometry of molecule remain same then for those two molecules eq (2) will be Bx = h/8π2 (µxr2 )c & By = h/8π2 (µyr2 )c ------ (3) Then Bx / By = µy / µx -----------(4) From eq (4) it is clear that if reduced mass (µ) increases due to the presence of isotope, the spacing between the spectral lines (B) decreases. This is called isotopic effect. Also Δ ν¯x = 2Bx and Δ ν¯y = 2By Therefore Δ ν¯x / Δ ν¯y = Bx / By ----------(5) From eq (4) and eq (5) Δ ν¯x / Δ ν¯y = Bx / By = µy / µx
- 16. Rotational spectra ofcarbon mon oxide (CO) molecule • In the diagram, the spectral lines of carbon monoxide are shown. • It is seen that in case of 13CO, due to relatively large mass of 13C isotope, the spacing between the spectral lines is less. • While in case of 12CO, due to small mass of 12C isotope, the spacing between the spectral lines is relatively large. • This difference in the spacing between the spectral lines of the two spectra is due to the isotopic effect. 12CO 13CO ν¯(0 to 1) ν¯(1to 2) ν¯(2 to 3) ν¯(3 to 4) ν¯(0 to 1) ν¯(1 to 2) ν¯(2 to 3) ν¯(3 to 4)
- 17. Limitations of RotationalSpectra 1. Molecules having permanent dipole moment like HCl, HCN can be studied by microwave spectroscopy. However homonuclear diatomic molecules like H2, N2, O2 which are non polar are microwave inactive and hence can not be studied by rotational spectroscopy. 2. Rotational spectra of a molecule can be recorded only if the molecule is in gaseous phase. This is because only in gaseous phase molecules have free rotational motion. But in case of solids and liquids due to the strong intramolecular forces of attraction, the free molecular motion is restricted. As a result, substances in solid and liquid phase can not be studied by rotational spectroscopy.
- 18. Important Formulas ofRotational spectroscopy I = µr2 µ= (m1m2) / (m2+ m1) Δ ν¯ = 2B B = h /8 π2I c Δ ν¯ = h /4π2I c For same molecule having two isotopes x & y Δ ν¯x / Δ ν¯y = Bx / By = µy / µx
- 19. Problem 1: Thepure rotational spectrum of HCl has spectral lines at 2.12 x103, 4.24x103 and 6.36x103 m-1. Calculate the bond length of HCl. (Given: H = 1amu; Cl = 35.46 amu; h = 6.626x10-34 J.s; c =3x108m.s-1; 1amu = 1.66x10-27Kg Given: ν¯1 = 2.12 x103 m-1, ν¯2 = 4.24x103 m-1 and ν¯3 = 6.36x103 m-1 Formulas: Δ ν¯ = 2B I = h /8 π2Bc µ= (m1m2) / (m2+ m1) r2 = I / µ Solution: Δ ν¯ = ν¯2 - ν¯1 = ν¯2 - ν¯1 = 2.12x103 m-1 B = Δ ν¯ / 2 = 2.12x103 / 2 = 1.06x 103 m-1 I = h /8 π2Bc = 6.626x10-34 / 8(3.14)2(1.06x103) (3x108) I = 6.626x10-34 / 250.828x1011 = 2.6417x10-47 J.s = 2.6417x10-47 Kg.m2 µ = mH.mCl/mH + mCl =1x35.46/1+35.46 = 0.9726 amu = 0.9726 x1.66x10-27 = 1.6145x 10-27 Kg r2 = I/µ = 2.6417x10-47 / 1.6145x 10-27 r = 1.6362 𝑥 10 − 20 = 1.279 x 10-10m
- 20. Problem 2: Arotational spectral line appears in the spectrum of C16O at 2.135x105m-1. What will be the frequency of the spectral line if C18O is used? Solution: Let µx be the reduced mass of C16O and µy be the reduced mass of C18O. µx = mC.mO/mC+mO = 12x16/12+16 = 6.857 amu µy = mC.mO/mC+mO = 12x18/12+18 = 7.200 amu µy/ µx = 7.200/6.857 = 1.050 Let ν¯x be the reduced mass of C16O and ν¯y be the reduced mass of C18O. But ν¯x = 2.135x105m-1 (given) ν¯x / ν¯y = µy / µx 2.135x105 / ν¯y = 1.050 2.135x105 / 1.050 = ν¯y ν¯y = 2.033x105 m-1 Therefore if C18O is used the frequency of the spectral line will be 2.033x105 m-1
- 21. Problem 3: Thefrequency separation of the successive lines in the rotational spectrum of H35Cl is 2.120x103m-1 while that of *H35Cl is 1.089x103m-1. Calculate the isotopic mass of *H atom. Solution: Let the frequency separation of H35Cl = 2Bx = 2.120x103m-1 Let the frequency separation of *H35Cl = 2By = 1.089x103m-1 Therefore 2Bx/2By = 2.120x103 / 1.089x103 = 1.947 Let µx be the reduced mass of H35Cl = (1)(35)/(1+35) = 35/36 = 0.9722 amu Let µy be the reduced mass of *H35Cl = (mH*)35/(mH*+35) But Bx / By = µy / µx Therefore 1.947 = (mH*)35/(mH*+35)/ 0.9722 (mH*)35/(mH*+35) = (1.947) (0.9722) (mH*)35/mH*+35 = 1.893 35mH* = 1.893(mH*+35) 35mH* = 1.893mH* + 66.255 35mH* = 1.893mH* + 66.255 35mH* - 1.893mH* = 66.255 33.107mH* = 66.255 mH* = 2.00amu
- 22. Problem 4: Calculatethe rotational constant of CO molecule, if its moment of inertia is 1.46x10-46 Kgm2. Solution: B = h /8 π2I c B = 6.625x10-34 /8(3.14)2(1.46x10-46) (3x108) B = 6.625x10-34 / 345.921x10-38 B = 0.01915 x10-34 x1038 B = 0.01915 x 104 B = 1.915 x 102 m-1
- 23. Problem 5: Thefrequency separation of successive lines in the rotational spectrum of HF is 4.18x103m-1. Calculate the rotational constant and moment of inertia of the molecule. Solution: Δ ν¯ = 4.18x103m-1 (given) But Δ ν¯ = 2B B = Δ ν¯/2 = 4.18x103/ 2 = 2.09x103m-1 Also I = h /8 π2B c I = (6.625x10-34)/8 (3.14)2(2.09x103) (3x108) I = (6.625x10-34)/ 494.56 x1011 I = 0.0134x10-45 I =1.34x10-47 J.s2 I =1.34x10-47 Kg.m2
- 24. Problem 6: Thefrequency separation of the successive lines in the rotational spectrum of 12C16O is 3842 m-1 while that of *C16O is 3673 m-1. Calculate the isotopic mass of *C atom. Answer: The isotopic mass of *C atom in *C16O is 13 amu
- 25. Vibrational spectra (IRSpectra) • Absorption of IR radiation by the molecule will bring about the change in vibrational energy level with in the molecule giving rise to closely packed absorption bands (spectral lines) in vibrational spectrum. • When a molecule having permanent dipole moment is placed in the path of IR radiation, the electric field of radiation will exert force on the charged atoms in the molecule as a result the molecule begins to vibrate. • If the rate of vibration of charged atoms in the molecule is high, then the molecule will strongly absorb IR radiations giving very intense absorption bands (spectral lines). • On the other hand if the rate of vibrations are low, then the absorption bands (spectral lines) will be of less intensity. • Thus polar molecules are IR active (i.e. such molecules will show lines in IR spectra). • While homonuclear non polar molecules like H2, N2, O2 are IR inactive (i.e. such molecules will not show lines in IR spectra). • Different vibrations in the molecule are expressed as fundamental modesor normal modes of vibrations.
- 26. Vibrational modes inpolyatomic molecules • The molecular vibrations are classified into two types as (1) stretching vibrations (2) bending vibrations. • Stretching vibration is a rhythmical movement of atoms along the bond axis such that bond length or interatomic distance will increase or decrease. • If both the bond length increases or decreases at the same time, then it is known as symmetrical stretching vibration. • If one bond length increases and at the same time other bond length decreases, then it is called asymmetrical stretching vibration. Vibrations stretching bending Symmetrical stretching asymmetrical stretching A B C A B C
- 27. • Bending vibrationsconsists of change in bond angle between the bonds with common atom. • When the two bonds come close to each other, the bond angle decreases and when they move away from each other the bond angle increases. • Hence bond angle is taken as an average value of angle between the two bonds in a molecule. • Bending vibrations are also called deforming vibrations. • Such vibrations take place at high energy (lower wavelength). • Bending vibrations are of two types 1. In plane bending vibrations : in which the bonds vibrate in the same plane 2. Out plane bending vibrations: in which the bonds vibrate outside the original plane. In plane bending Out plane bending Vibrations stretching bending
- 28. • In planebending vibrations are further subclassified in to two type 1. scissoring : in which two bonds come close to each other or move away from each other. 2. rocking: in which the two bonds move in the same direction. • Out plane bending vibrations are further subclassified in to two types 1. wagging : in which the two bonds move out of the original plane but on the same side. 2. twisting: in which the two bonds move out of the original plane but in opposite direction. In plane bending Out plane bending Vibrations stretching bending rocking scissoring wagging twisting A B C A B C
- 29. Vibrations in themolecule • https://www.youtube.com/watch?v=ZWwLCnuYRys • https://www.youtube.com/watch?v=G6G3RKU5xPY
- 30. Dependance of frequencyof vibrational spectral lines on force constant and masses of the constituent atoms in a molecule • The diatomic molecular system is stable and the atoms are separated by a distance known as bond length or internuclear distance. • When a molecule is exposed to IR radiations, the electric field of radiations will exert force on the atoms in the molecule as a result the molecule begins to vibrate and the equilibrium internuclear distance get disturbed. • This vibrational energy changes are observed in the form of vibrational spectra.
- 31. • Consider twohypothetical atoms A & B having masses m1 & m2 connected by an elastic spring. A is fixed at one end and B is kept hanging at its normal equilibrium distance r. • When a spring is stretched by an external force and the released, the equilibrium distance get disturbed. • If x is the displacement of an atom from the normal position, then the opposing force (f) will restore the spring back to normal position. • According to the Hook’s law, f α –x f = -Kx --------- (1) • The negative sign indicates that the restoring force (f) and displacement (x) are acting in opposite directions. • In eq (1) K is called force constant (K = -f/x), which is defined as the restoring force (f) per unit displacement and having unit N.m-1. X compressed normal stretched f x f r
- 32. • As aresult of this displacement, the atom B performs simple harmonic motion along the axis of the bond. • If the atoms A & B vibrates simultaneously, then the vibrational frequency (ω) is given by ω = 1 2π √ 𝐾 µ ------------ (2) Here µ is the reduced mass given by the formula m1 x m2/m1+m2 But ω¯ = ω 𝑐 -------- (3) Substituting eq(2) in eq (3) we get ω¯ = 1 2π𝑐 √ 𝐾 µ ------------ (4) Squaring both side of eq (4) we get ω¯ 2 = 𝐾 4π2 𝑐2 µ K = 4ω¯ 2π2𝑐2µ -------------- (5) • From eq (5) it is clear that the force constant is directly proportional to the vibrational frequency in terms of wave number (ω¯) and reduced mass (µ) of the vibrating molecule.
- 33. Significance of forceconstant • Force constant is the measure of strength of the bond. • Higher the value of force constant, stronger is the bond and greater is the bond energy. • The force constant of HF is about 3 times greater than that of HI. • Thus HF bond energy is greater than HI. • As a result HF can not be easily oxidized. • Molecules with double or triple bonds have greater stability than those molecules having single bond. • Hence molecules like CO, NO, C2H2 are more stable.
- 34. Zero point energy •By application of Schrodinger wave equation and quantum mechanical treatment, the vibrational energy of a molecule is given by an equation Ev = (V + 1 2 ) hω ---------- (1) here V is the vibrational quantum number having values 1, 2, 3, 4 ---etc; h is the Planck’s constant & is ω the vibrational frequency. • For the lowest vibrational energy level V = 0,eq (1) will be Eo = ℎω 2 ---------- (2) But ω = 1 2π √ 𝐾 µ Therefore eq (2) will be Eo = ℎ 4π 𝐾/µ -------- (3) • This energy possessed by the molecule at V = 0 vibrational energy level is called zero point energy. • This implies that at any temperature, atoms in the molecule can never be completely at rest relative to each other. • Even at absolute zero temperature, when rotational and translational motion have ceased, the molecule will still have vibrational energy. • From eq (2) it is clear that the zero point energy depends on the vibrational frequency (ω) of a molecule. • From eq (3) it is clear that the zero point energy depends on the force constant which determines the strength of the bond and also depends on the atomic masses.
- 35. • When thetransition take place from V= 0 to V =1 vibrational energy level, the spectral line thus obtained is called fundamental band or first harmonic band. • When the transition take place from V= 0 to V =2 vibrational energy level, the spectral line thus obtained is called first overtone band or second harmonic band. • When the transition take place from V= 0 to V =3 vibrational energy level, the spectral line thus obtained is called second overtone band or third harmonic band and so on
- 36. Relation between frequencyof fundamental, first and second overtone bands • For fundamental band arising due to V = 01 ΔE = Ev=1 – Ev =0 By using eq (1) ΔE = [(1+ 1 2 ) hω – (1+ 1 2 )2 hxω ] - [(0+ 1 2 ) hω – (0+ 1 2 )2 hxω ΔE = [ 3 2 hω – ( 3 2 )2 hxω ] - [ 1 2 hω – ( 1 2 )2 hxω] ΔE = 3 2 hω – 9 4 hxω - 1 2 hω + 1 4 hxω ΔE = 3 2 hω - 1 2 hω – 9 4 hxω + 1 4 hxω ΔE = hω – [ 9 4 hxω - 1 4 hxω] ΔE = hω – 2hxω ΔE = (1 – 2x)hω ------ (2) But ω = ω¯ c & ΔE = hν¯c Therefore eq (2) will be hν¯c = (1 – 2x)h ω¯ c ν¯ = (1 – 2x)ω¯ ------- (3) • For first overtone band arising due to V = 02 ΔE = Ev=2 – Ev =0 By using eq (1) ΔE = [(2+ 1 2 ) hω – (2+ 1 2 )2 hxω ] - [(0+ 1 2 ) hω – (0+ 1 2 )2 hxω ΔE = [ 5 2 hω – ( 5 2 )2 hxω ] - [ 1 2 hω – ( 1 2 )2 hxω] ΔE = 5 2 hω – 25 4 hxω - 1 2 hω + 1 4 hxω ΔE = 5 2 hω - 1 2 hω – 25 4 hxω + 1 4 hxω ΔE = 2hω – [ 25 4 hxω - 1 4 hxω] ΔE = 2hω – 6hxω ΔE = (1 – 3x)2hω ------ (4) But ω = ω¯ c & ΔE = hν¯c Therefore eq (4) will be hν1¯c = (1 – 2x)h ω¯ c ν1¯ = (1 – 3x)2ω¯ ------- (5) For anharmonic vibrations which are at high energy states having large amplitude of vibration, the vibrational energy will be Ev = (V+ 1 2 ) hω – (V+ 1 2 )2 hxω --------- (1) Here x is called anharmonicity constant
- 37. Similar to eq(3) & (5) when transition take place from V =0 to V = 3 vibrational energy level, the frequency of second overtone band will be ν2¯ = (1 – 4x)3ω¯ ------- (6) Taking the ratio of eq(3),(5) & (6) ν¯ : ν¯1 : ν¯2 = (1 – 2x)ω¯ : (1 – 3x)2ω¯ : (1 – 4x)3ω¯ Since x the anharmonicity constant is negligibly small as compared to 1, the terms in the bracket will be nearly equal to 1. Hence ν¯ : ν¯1 : ν¯2 = ω¯ : 2ω¯ : 3ω¯ = 1:2:3 Thus the frequencies of fundamental, first and second overtone bands are in the ratio 1:2:3 and in terms of wavelengths they will be in the ratio 1: 1 2 : 1 3
- 38. Finger print regionin IR spectra • The region of IR spectrum between 1500 cm-1 to 500cm-1 consists of series of absorption bands (spectral lines) which are characteristics of a molecule. • The spectra in this region is used to identify the molecule. • This region is known as finger print region. • The absorption bands (spectral lines) in the finger print region are due to stretching and bending vibrations of C-C, C-O-C bonds. • This region is known as finger print region because no two different molecules will have same spectrum in this region. • The absorption bands in finger print region are used to identify the chain and ring structures in the molecules. • It is possible to identify the unknown substance by showing its IR spectrum in finger print region is identical with the IR spectrum of known compound.
- 39. Group frequencies inIR spectrum • In a molecule, the functional group plays an important role in its identification. • The atoms present in the functional group are attached to each other by the chemical bond will vibrate with a frequencies which are characteristics of a functional group in a molecule. • This frequencies are know as group frequencies. • The group frequencies generally lies above or below the finger print region. • Functional groups with light atoms absorbs above the finger print region. eg. >C=O group frequency is 1750-1600cm-1; -OH group frequency is 3700 – 3500cm-1; -CN (aromatic) group frequency 3100-3050 cm-1. • Functional groups with heavy atoms will absorb below finger print region. eg. –I =550cm-1; -Br = 650cm-1; -Cl = 725 cm-1. • The group frequencies will remain same for a give functional group even though the compounds may be different.
- 40. About Finger printRegion & Group Frequencies in IR spectra and Spectral Interference • https://www.youtube.com/watch?v=0S_bt3JI150 • https://www.youtube.com/watch?v=nxZ8WBboqVk • https://www.youtube.com/watch?v=Bxo_DQpFh1c • https://www.youtube.com/watch?v=u2IBdtINsrQ • https://www.youtube.com/watch?v=0WujH72I-bM
- 41. Important formulas onIR spectroscopy K = 4ω¯ 2π2𝑐2µ µ= (m1m2) / (m2+ m1) Eo = ℎω 2 = ℎω¯c 2 ω = ω¯c ω¯ = 1 2π𝑐 √ 𝐾 µ Frequency of fundamental band ν¯ = (1 – 2x)ω¯ Frequency of first overtone band ν1¯ = (1 – 3x)2ω¯ Frequency of second overtone band will be ν2¯ = (1 – 4x)3ω¯
- 42. Problem 7. Thevibrational frequency of H-Cl bond is 8.662x1013 s-1. Calculate the force constant of the bond. Given: H = 1amu; Cl =35 amu; c =3x108 m.s-1; 1 amu = 1.66x10-27 Kg Solution: Given: ω = 8.662x1013 s-1 Formula: µ= (mHmCl) / (mH+ mCl) ω = ω¯c K = 4ω¯ 2π2𝑐2µ Solution: µ= (mHmCl) / (mH+ mCl) = (1x35)/(1+35) = 35/36 = 0.973 amu = 0.973 x1.66x10-27 =1.614x10-27 Kg ω¯ = ω /c = 8.662x1013 s-1 / 3x108 m.s-1 = 2.887 x105 m-1 K = 4ω¯2π2𝑐2µ K = 4 (2.887 x105)2 (3.14)2 (3x108)2 (1.614x10-27) K = 4774.84x10-1 K = 477.484 kg.s-2 K = 4.775 x102 kg.s-2 K = 4.775 x102 N.m-1
- 43. Problem 8. Inthe IR spectra of HBr, the fundamental band appears at 2.242x105 m-1. Calculate the vibrational frequency of a molecule. Given: c =3x108 m.s-1 Given: ν¯ = 2.242x105 m-1 Formula: Frequency of fundamental band ν¯ = (1 – 2x)ω¯ ω = ω¯c Solution: Frequency of fundamental band ν¯ = (1 – 2x)ω¯ since x is very much small than 1 we get ν¯ = ω¯ = 2.242x105 m-1 ω = ω¯c ω = 2.242x105 m-1 x 3x108 m.s-1 ω = 6.726 x1013 s-1
- 44. Problem 9:The vibrationalfrequency of a molecule is 2.5x105m-1. Calculate the zero point energy of a molecule. Given: h = 6.626 x10-34 J.s; c =3x108 m.s-1 Given: ω¯ = 2.5x105m-1 Formula: Eo = ℎω¯c 2 Solution: Eo = (6.626 x10−34) 2.5x105 (3x108) 2 Eo = 24.8475 x10-21 Eo = 2.48475 x10-20 J
- 45. Problem 10: Theforce constant of H-I bond is 2.9x102 N.m-1. Calculate the vibrational frequency of H- I bond. Given: H = 1amu; I =127 amu; c =3x108 m.s-1; 1 amu = 1.66x10-27 Kg Given: K = 2.9x102 N.m-1 Formula: µ= 𝑚𝐻 .𝑚𝐼 𝑚𝐻 +𝑚𝐼 ω¯ = 1 2π𝑐 √ 𝐾 µ Solution: µ= 𝑚𝐻 .𝑚𝐼 𝑚𝐻 +𝑚𝐼 = 1 (127) 1+127 = 127 128 = 0.9922 amu = 0.9922 x 1.66x10-27 = 1.647 x10-27 Kg ω¯ = 1 2 3.14 (3x108) √ 2.9x102 1.647 x10−27 ω¯ = 0.0531 x10-8 1.761𝑥1029 = 0.0531 x10-8 x 0.420 x1015 = 0.0223 x107 ω¯ = 2.23 x105 m-1
- 46. Problem 11: Inthe IR spectra of HCl, the fundamental band appears at 8.658x1013 s-1 and first overtone band at 1.7004 x1014 s-1. Calculate (i) anharmonicity constant (ii) vibrational frequency of a molecule (iii) force constant of a bond. Given: H = 1amu; Cl =35.5 amu; 1 amu = 1.66x10-27 Kg; c =3x108 m.s-1 Given : ν = 8.658x1013 s-1 ν1 = 1.7004 x1014 s-1 Formula: Frequency of fundamental band ν¯ = (1 – 2x)ω¯ Frequency of first overtone band ν1¯ = (1 – 3x)2ω¯ 𝑣 = v¯c Solution: v¯ = 𝑣 𝑐 = 8.658x1013s−1 3x108m.s−1 = 2.886 x105 m-1 v¯1 = 𝑣1 𝑐 = 1.7004 x1014 s−1 3x108m.s−1 = 0.5668 x106 m-1 ν¯ ν1¯ = (1 – 2x)ω¯ (1 – 3x)2ω¯ 2.886 x105 0.5668 x106 = (1 – 2x)ω¯ (1 – 3x)2ω¯ 0.509 = (1 – 2x) (1 – 3x)2 X = 0.017 = anharmonicity constant Frequency of fundamental band ν¯ = (1 – 2x)ω¯ 2.886 x105 = [1 – 2(0.017)]ω¯ 2.886 x105 = [1 –0.034]ω¯ 2.886 x105 = [0.966]ω¯ ω¯ = 2.988 x105 m-1 =vibrational frequency of a molecule in terms of wave number µ= (mHmCl) / (mH+ mCl) = (1x35.5)/(1+35.5) =1.6145x10-27 Kg K = 4ω¯2π2𝑐2µ K = 4 (2.988 x105)2 (3.14)2 (3x108)2 (1.6145x10-27) K = 5.116x102 Nm-1 = force constant
- 47. IR Spectrophotometer (Instrumentation) •In IR spectrophotometer, the absorption of IR radiation by a substance is measured. Two types of instruments 1. Single beam IR spectrophotometer 2. Double beam IR spectrophotometer • Double beam IR spectrophotometer is having greater accuracy as compared to single beam instrument. • In double beam instrument, the beam of IR radiation from the source is split by a mirror(M1) in to two half beams- sample beam & reference beam. • The reference beam passes through the reference cell, in which blank solution is placed which is usually the solvent used for sample preparation. • The sample beam passes through the sample cell in which the sample solution is placed. Sample cell Reference cell Beam Collector Source of IR radiations Detector & Recorder Prism M1 M2 M3 M4 Slit S1 Slit S2 M4 M5
- 48. • Radiation source:two types 1. Nernst glower which is a bar made up of Cerium oxide, Zirconium oxide and thorium oxide. Such bar is electrically heated to a high temperature between 1000oC to 1800oC. 2. Globar which is a bar of sintered silicon carbide. When both the sources are heated to same high temperature, IR radiations of uniform intensity are given out for a long period. • Prism monochromators: is made of materials like CaF2, KBr, NaCl as this materials are transparent to IR radiations. • Detectors: used is Bolometer which consists of thin metal conductor. When radiations fall on the conductors, its temperature changes, which will result in change in resistance of conductor. The change in resistance is measure of the intensity of radiation. The signal developed by the detector is amplified and recorded by the recorder. • Sample and reference cells: made of NaCl or KBr are generally used. • Samples: 1. For liquid samples, the sample is dissolved in non polar solvent like CCl4 or CHCl3. The solution prepared in non polar solvent must be diluted so that the intramolecular forces of attraction which are usually strong in solids will get reduced to minimum. (See the video ahead https://www.youtube.com/watch?v=uuAjA9UbA7Y) 2. For solid samples, the sample is intimately mixed with KBr and the mixture is pressed in to a pallets using a special mould through a hydraulic press. (See the video ahead https://www.youtube.com/watch?v=ntjHg6BT1E0) Some times solid sample is mixed with Nujol to prepare the fine paste or slurry and the paste is applied on KBr disc. (see the video ahead https://www.youtube.com/watch?v=uuAjA9UbA7Y ). Note: Nujol is a brand of mineral oil having density 0.838 g/mL at 25 °C, used in infrared spectroscopy. It is a heavy paraffin oil so it is chemically inert and has a relatively uncomplicated IR spectrum.
- 49. IR Instrumentation Below isthe video showing IR Instrumentation • https://www.youtube.com/watch?v=OiukFtC8E04
- 50. Solid Sample preparationfor IR analysis Below is the video showing IR Sample preparation • https://www.youtube.com/watch?v=ntjHg6BT1E0
- 51. Solid Sample preparationfor IR analysis Below is the video showing Solid sample preparation for IR analysis • https://www.youtube.com/watch?v=lTAHqg_Q_5I
- 52. Liquid Sample preparationfor IR analysis Below is the video showing liquid sample preparation for IR analysis • https://www.youtube.com/watch?v=uuAjA9UbA7Y
- 53. Raman Spectroscopy • Whena light radiations of visible region of definite frequency is passed through a gas. liquid or solid samples and the scattered light is observed at a angle to incident light it is observed that the scattered light is having the same frequency as that of incident light. • Such type of scattering of light radiation without change in frequency is called Rayleigh scattering. • Due to Rayleigh scattering the spectral lines which are obtained are called Rayleigh lines and the corresponding frequency is called Rayleigh frequency. • In addition to Rayleigh lines, the scattered light radiations also shows some additional lines of modified frequencies which lies over and below the Rayleigh lines. • The lines with frequencies lower than main Rayleigh line are called Stokes lines. • The lines with frequency higher than that of main Rayleigh line are called anti-stoke lines. • This effect is called Raman effect and the lines whose frequencies are modified due to Raman effect are called Raman Lines. • Thus the resulting Raman spectra consist of Raman lines which are observed on both sides of main Rayleigh lines. • This shifting of Raman lines frequency from the main Rayleigh line is called Raman shift which generally lies in the range of 100x102 to 3000x102 m-1. • The Raman shift does not depend upon the frequency of incident light radiation but it is regarded as characteristics of a substance causing Raman effect.
- 54. • If 𝑣iis the frequency of incident light radiations and 𝑣s is frequency of scattered radiations by the molecular species then Raman shift in terms of Raman frequency will be Δ 𝑣 = 𝑣i - 𝑣s . • For stoke lines Δ 𝑣 = is + while for anti stoke lines Δ 𝑣 = is - . • According to quantum theory, Raman effect is considered as an outcome of collisions taking between the photons of light and the molecules of the substance. • Consider a molecule of mass ‘m’ which is present in the energy state Ei moving with the velocity V . • Let this molecule collide with a photon of light having energy given by h𝑣i where 𝑣i is the frequency of incident photon. • Due to collision the molecule undergoes change in energy state . • Let the new energy be represented as Ef and the energy of scattered photon be h𝑣s where 𝑣s is the frequency of scattered photon. • If the collisions are elastic, then by principle of conservation of energy Total initial energy of system = Total final energy of system Ei +1/2mv2 +h𝑣i = Ef +1/2mv2 +h𝑣s Ei +h𝑣i = Ef +h𝑣s h𝑣i - h𝑣s = Ef – Ei h(𝑣i - 𝑣s)= Ef – Ei Δ 𝑣 = Ef – Ei/h ----- (1)
- 55. Case 1: WhenEi = Ef then eq(1) will be Δ 𝑣 = 0 i.e. 𝑣i = 𝑣s. It means that the molecule – photon collisions produces no change in frequency of photon and hence the scattered light will be of same frequency giving rise to Rayleigh lines. Case 2: When Ei >Ef then the molecule after collision with photon loose energy to the photon. Thus the scattered photon will be of high energy h𝑣s and high frequency 𝑣s i.e. 𝑣s > 𝑣i. Hence from eq (1) Δ 𝑣 will be negative and the resulting spectral lines will appear at high frequency above the Rayleigh line frequency giving rise to Anti stoke lines. Case 3: When Ei < Ef then the molecule after collision with photon loose energy to the photon. Thus the scattered photon will be of low energy h𝑣s and less frequency 𝑣s i.e. 𝑣s < 𝑣i. Hence from eq (1) Δ 𝑣 will be positive and the resulting spectral lines will appear at lower frequency below the Rayleigh line frequency giving rise to Stoke lines.
- 56. Important formulas inRaman Spectroscopy Δv¯ = v¯𝑖 - v¯s Δ E= Δ𝑣.h since 𝑣 = 𝑣¯𝑐 Δ E= Δ𝑣¯𝑐h For Rayleigh lines, 𝑣s = 𝑣𝑖 and Δ 𝑣 will be Zero. For anti stoke lines, 𝑣s > 𝑣𝑖 and Δ 𝑣 will be negative. For stoke lines, 𝑣s < 𝑣𝑖 and Δ 𝑣 will be positive.
- 57. Problem 12: Asubstance was exited with radiations of wavelength 4250x10-10 m. A Raman line appear at a wavelength of 4465x10-10 m. (i) is it a stoke line or anti stoke line? (ii) calculate the Raman shift. Solution: λi = 4250x10-10 m λs = 4465x10-10 m v¯𝑖 = 1/ λi = 1/ 4250x10-10 = 2.352x106m-1 v¯s = 1/ λs = 1/ 4465x10-10 = 2.24x106 m-1 Since v¯𝑖 > v¯s it is stoke line Raman shift Δv¯ = v¯𝑖 - v¯s Δ v¯ = 2.352x106 − 2.24x106 = 1.13x105 m-1
- 58. Problem 13: Asubstance was exposed to radiation of 4 x10-7m. The first stoke line appeared at wavelength of 5x10-7m. Calculate the energy change of a molecule. Solution: λi = 4 x10-7m λs = 5x10-7m v¯𝑖 = 1/ λi = 1/ 4 x10-7= 2.5 x106 m-1 v¯s = 1/ λs = 1/ 5x10-7 = 2.0x106 m-1 Raman shift Δv¯ = v¯𝑖 - v¯s Δv¯ = 2.5 x106 − 2.0x106 = 5x105 m-1 Since Δ 𝑣 = ΔE/h Δ E= Δ𝑣.h = Δ𝑣¯𝑐h Δ E = 5x105 x 3x108x6.625x10-34 Δ E = 9.9375x10-20 J
- 59. Application of RamanSpectroscopy • Many molecules and functional groups which scatter visible light will give specific Raman shift. Such groups can be identified by studying the Raman shift. • It is generally utilized to analyze aliphatic as well as aromatic hydrocarbons, alcohols, ethers, aldehydes, ketones. • Homo nuclear molecules like H2, N2, O2 etc will give no information in IR spectra could be readily studied by Raman spectra. Even existence of single, double and triple bonds can be established by Raman spectroscopy. • In organic chemistry, Raman effect help to identify various types of isomerism's and isomers. • It can be used to determine specific linkages in a molecule. • Transition from crystalline state to amorphous state can also be studied. • Since Raman intensity is directly proportional to the concentration of the sample, this principle can be used for quantitative analysis.
- 60. Comparison of IR& Raman spectroscopy IR Spectroscopy • Absorption of IR radiations by vibrating molecule gives IR spectra. • Crystals like CaF2, NaBr are used for liquid samples while for solid samples KBr is used for sample preparation. • Since aqueous solution transmit IR radiations to poor extent, aqueous samples can not be studied by IR spectroscopy. • For IR analysis substance in any condition can be used for analysis. • Homonuclear diatomic molecules having no permanent dipole moment are IR inactive. Hence can not be studied by IR spectroscopy. • IR spectroscopy is more sensitive Raman Spectroscopy • Raman spectra is obtained by scattering of light radiations by vibrating molecule. • Glass or quartz crystals are used as a sample holder or sample cell. • Aqueous samples can be studied by Raman spectroscopy. • For Raman spectroscopy, the substance used for analysis should be pure and colourless. • Homonuclear diatomic molecules are found to be Raman active. Hence can be studied by Raman spectroscopy. • The sensitivity of Raman spectroscopy is less in comparison to IR spectroscopy.
- 61. How IR spectroscopy& Raman Spectroscopy are complimentary to each other? • Information about the molecular structure which could not be obtained alone by IR is provided by Raman spectra. • In fact two spectra's are used together extensively to find complete molecular structure of a substance. • Homonuclear diatomic molecules which are IR inactive can be studied well by Raman spectroscopy as the are Raman active. • Since aqueous solution transmit IR radiations to poor extent, aqueous samples can not be studied by IR spectroscopy but the same aqueous samples can be studied by Raman spectroscopy. • In case of CO2 molecule, there are two vibrations 1. symmetrical stretching 2. asymmetrical stretching vibrations. In this symmetrical stretching vibrations are IR inactive but Raman active, while asymmetrical stretching vibrations are Raman inactive but IR active. • Thus complete information regarding the CO2 molecule can be obtained by using the two spectra’s. • Usually the findings of one spectra are complimented by other to obtain complete information about the molecule. • Thus it can be said that the IR & Raman spectroscopy are complimentary to each other.
- 62. Rule of Mutualexclusion in IR & Raman spectra of a molecules • The rule of mutual exclusion state that for a molecule having center of symmetry, the Raman active vibrations are IR inactive and vice versa. • In other words, for a molecules having center of symmetry, a vibrations can not be simultaneously IR and Raman active. • On the other hand for a molecule having no center of symmetry then not necessarily all vibrations may be both Raman and IR active. • For example, in case of CO2 molecule, there are two vibrations 1. symmetrical stretching 2. asymmetrical stretching vibrations. • In this symmetrical stretching vibrations are IR inactive but Raman active, while asymmetrical stretching vibrations are Raman inactive but IR active. • Thus it can be said that CO2 molecule is having center of symmetry. • Conversely it can also be stated that for a same molecule if both IR and Raman spectra don’t have common spectral lines, then the molecule has a center of symmetry. • If a molecule have common lines in both IR and Raman spectra then that molecule has no center of symmetry.