spring semester 2015 vibrational spectroscopy of molecules...

Spring Semester 2015

Ch t 3 I f d S tVibrational Spectroscopy of Molecules

Chapter 3 Infrared Spectroscopy

Sang Kuk LeeDepartment of ChemistryDepartment of ChemistryPusan National University

April 2015April 2015

Vibrational energy corresponds to infrared (IR) region.Wh t i f ti t f IR t ? B d t th

Spring Semester 2015, Molecular Spectroscopy : Chapter 3 Infra-red Spectroscopy1

What informations can we get from IR spectrum? Bond strength

Types of vibrationsWhat is vibration? It is a periodic change of bond length (stretching) and bond angle (bending). Stretching shows higher frequency than bending.

Symmetrical Antisymmetrical Scissoring Rocking Wagging Twistingstretching stretching

Stretching Bending

How many vibrations are possible?

Vib i l d f f d T l d f f d (3N)

Stretching Bending

Vibrational degree of freedom = Total degree of freedom (3N) –Translational degree of freedom (3) – Rotational degree of freedom (2(3))

= 3N 3 2(3) = 3N 5(6) for linear (nonlinear) molecules


= 3N – 3 – 2(3) = 3N - 5(6) for linear (nonlinear) molecules

3 1 THE VIBRATING DIATOMIC MOLECULE3.1 THE VIBRATING DIATOMIC MOLECULEWhat types of potential energy do we have from approaching two atoms?

3.1.1 The Energy of a Diatomic Molecule

There exist attractive and repulsive forces between them. Net potential

= Attractive (-) + Repulsive (+)

n n Repulsive

Attractive at long distance,Nuclear-electronNuclear-nuclear, Electron-electron

Attractive


g ,Repulsive at short distance

Most stable

How to describe the potential curve ?

Similarity in potential curve

Simplest expression

Harmonic oscillator

Real shape : Looks quite different from parabola.

Similar shapes in this region : We can use this simplest curve forcan use this simplest curve for

analysis.


3 1 2 The Simple Harmonic Oscillator3.1.2 The Simple Harmonic OscillatorHarmonic oscillator gives the simplest potential curve.

• ExtensionHooke’s law : Extension is

p p

proportional to the force.

ExtensionOn integrating

Max.Min.

What is harmonic oscillator ?


It obeys Hooke’s law

How to express vibrational energy ?How to express vibrational energy ?

Classical vibrational frequency is given byWe can derive the vibrational frequency equation from classical mechanics.

q y g y

ν λ-1

After solving Schrodinger equation of harmonic oscillator, the solution is given as follows. The only difference is ½. g y

Vibrational energy

The quantum mechanical expression of vibrational energy


The quantum mechanical expression of vibrational energy contains ½ which is very important in understanding the nature.

Vibrational energy from parabola potential of harmonic oscillator: Real molecules belong to anharmonic oscillatorto anharmonic oscillator.

Bond strength (force constant)

zero point energy g ( )

Reduced massWe can see the effect of k, force constant by comparing single, double, and triple bonds. Also see the effect of reduced mass from C-H C-D and C-T bonds


from C-H, C-D, and C-T bonds

Intervals are the same no matter what the v is.

We can calculate the force constant from the above equation

Molecule Frequencyx1013 Hz

Force constantNewton/m

q

x1013 Hz Newton/m

HF 8.72 970 Strong

HCl 8.66 480

HBr 7.68 410All t iti i th

Bond Strength

HI 6.69 320

CO 6.42 1860

All transition energies are the same no matter what the

vibrational quantum number v is. Triple bondWeak


NO 5.63 1530

Selection rule for Harmonic OscillatorSelection rule for Harmonic Oscillator

Selection rule for harmonic oscillator:

∆v = ±1

The nature allows one step

Absorption

movement only.

Emission

Transition energy is independent of

vibrational quantum number, v

We should have the transition energy for Only one transition energy


gyharmonic oscillator no matter what the

vibrational quantum number v is.

Explain laterData from vibrational spectrum

tripleweak

Can be determined by analysis ofObservation


Can be determined by analysis of rotational spectrum

What is the vibrational modes? Each has different frequency.Vibrational degree of freedom: 3N-3-2(3)Vibrational degree of freedom : 4 for CO2It has only 3 vibrational modes. If two vibrations have the same frequency, it is doubly degenerate one mode

g ( )

it is doubly degenerate one mode.

Upside InsideUpside down

Inside out

Same frequency

HighestLowest


Vibrational modes of CH3Cl(Methyl Chloride)3N-3-3 = 9 vibrational degrees of freedom, but only 6 vibrational

modes because 3 of them are doubly degenerate.

Nondegenerate

Doubly degenerate


Notation of doubly degenerate


CH3Cl(Methyl Chloride)

Why different bandshapes?

Banshape depends on the vibrational symmetry.

We will study it later.

A total of 6 bands

1 Many bands in IR region


1. Many bands in IR region2. Each band shows a different band shape depending on vibrational symmetry

What is the potential energy of real molecules? 3.1.3 The Anharmonic Oscillator

Morse potential : Most well known potential function Di th h i l i fMost well-known potential function Discuss the physical meaning of

Morse potential when r → 0 and ∞

ParabolaNo dissociation

Dissociation limit

We can describe the real

Dissociation limit

molecules by Morse potential. Dissociation energy from potential minimumfrom potential minimum

The function should explain the physics of molecules at small and


p ylong distance.

Vibrational Energy of Anharmonic Oscillator

The following energy equation is obtained by inserting the Morse potential into Schrodinger equation of quantum mechanics.

Additional term

Real molecules Ideal moleculeWe can get the vibrational energy of anharmonic oscillator by inserting the Morse potential into Schrodinger

i

Same intervalequation.

Energy interval between two l l i tti ll ith

Getting smaller interval

levels is getting smaller with increasing v.

Difference?Different freq


Single freq.Different freq.

( v+1/2)

1. Anharmonic oscillator behaves like the harmonic oscillator.2. Oscillator frequency decreases steadily with increasing v.

( )

How to express the getting smaller interval? Add th t ( i lid )Add another term (previous slide)

GettingGetting smaller

Transition energy depends on vibrational quantum number v


vibrational quantum number, v.Not twice and three times

Selection rule for anharmonic oscillatorSelection rule for anharmonic oscillatorSelection rule for anharmonic oscillator is slightly differentfrom that of harmonic oscillator:from that of harmonic oscillator:

∆v = ±1, ±2, ±3,…Intensity weaker rapidly

Dominant

From these selection rules

• Fundamental band : ∆v = +1, v = 0 → 1

• 1st Overtone band : ∆v = +2 v = 0 → 2 twofold• 1 Overtone band : ∆v = +2, v = 0 → 2

• 2nd Overtone band : ∆v = +3, v = 0 → 3

twofold

threefold

• Hot band : ∆v = +1, v = 1 → 2B lt di t ib ti l N /N { E /kT)

Intensity depends on temperature.


Boltzmann distribution law Nv/N0 = exp{-Ev/kT)Most population is at v=0 state.

Vibrational spectrum of CO

P R

Check the position and intensity

p

P- R-

spectrum > band > branch >line∆v = +1, v = 0 → 1

FundamentalBranch will be explained later.

Overtone bands are weak

∆v = +2, v = 0 → 2Why hot band

Overtone bands are weak due to selection rule.

P- R-,y

shows weak intensity?

1st Overtone :TwofoldHot band


∆v = +1, v = 1 → 2

Characteristics of bandsPosition and Intensity of the bands

Fundamental

1st O ertone

From energy equation.

1st Overtone

Less than twofold

2nd Overtone

Less than threefold

HotL h f d l


Less than fundamental

Summary of vibrational transition energy and intensity

Band energy:

2nd overtone band > 1st overtone band > fundamental band > hot band

Band intensity:

Fundamental band >> overtone band ≈ hot bandFundamental band >> overtone band ≈ hot band (depending on the sample condition)

The intensity of hot band changes with temperature of the sample because of the shift of population to higher energy levels.

Energy of 1st overtone band is equal to that of sum of fundamental band and hot band.


3 2 THE DIATOMIC VIBRATING ROTATOR3.2 THE DIATOMIC VIBRATING ROTATORVibrational + rotational motion simultaneously

Vibration: 1000 cm-1

(3 x1013 times/sec)Rotation: 1 cm-1

The molecule experiences about 1000 times of vibration during single rotation.

(3x1010 times/sec)

Total energy is f t d ibsum of rot. and vib.

Neglected due to small contributionsNeglected due to small contributions

RotVibVib


Χe : centrifugal distorsion constant, Χe <<1

Two rotation-vibrational structures of v=0 and 1

Excited vib. statestrongest weak

Two selection rules, vib. and rot., should be applied to vibration and

Rot. structure(J= 4→5)

should be applied to vibration and rotation

Much larger separation between two

J= 4→3)

∆v= +1; ∆J= +1 ∆v= +1; ∆J= -1Much larger separation between two vibrational states than shown here.

∆v=±1; + : absorption: emission

Rot. structure

- : emission

Ground vib. state


We assume that both the ground (v=0) and excited (v=1) states have the same rotational constants. B’ = B” That is the bond length

does not change.

Since ᇫV >> ᇫJ, ᇫJ = ±1 belongs to absorption

New notation in SpectroscopyL t t J” U t t J’

Upper state – lower stateLower state: J”, Upper state: J’

Pure vibrational transition

There are two choices for rotational selection rule, ∆J = ±1.

Energy increases with J

Energy decreases


∆J= +1 always shows higher frequency than ∆J =-1

gywith J

∆J=-1 ∆J =+1

There are two branchesbranches corresponding to ∆J=+1 and ∆J=-1.

∆J= 0 is not allowed

∆J = +1 : Start from J=0. ∆J = -1 : Start from J=1.

Pure vibrational transition


New notation in Spectroscopy : Lower state: J”, Upper state: J’

Selection rule and branches

We express ∆J in terms of English alphabets such as O P Q R S rather than the numbersO, P, Q, R, S rather than the numbers..

Spectrum > Band > Branch > LineSpectrum Band Branch Line

Line: Single rotational transition

Branch: A collection of rotational transition of ∆J= +1 or ∆J= -1Branch: A collection of rotational transition of ∆J= +1 or ∆J= -1.

Band: A collection of rovibrational transition.

R t ti Vib ti R ib ti


Rotation + Vibration = Rovibration

3.3 THE VIBRATION-ROTATION SPECTRUM OF CARBON MONOXIDE

P R

What are band, branch, and line ?

P- R-One band consists of two branches. Each branch consists of many lines1 band, consists of many lines corresponding to J values.

,2 branches,and many lines.

ΔJ= +1

ΔJ= -1

P- R-ΔJ= +1ΔJ= -1


Fundamental OvertoneX 2 =

High resolution spectrum of COHigh resolution spectrum of CONotation of transition quantum numbers

R(0) : J”=0 → J’=1R(6) : J”=6 → J’=7

P(1) : J”=1 → J’=0P(7) : J”=7 → J’=6

Increasing J Increasing J

New findings:In P-branch: larger spacingIn R-branch: smaller spacingIn R branch: smaller spacing

Interval is getting smaller. Why?

Hot band


Band origin : J”=0 → J’=0 (Pure vibrational transition) : Forbidden transition

Experimental data of COExperimental data of CO


Larger interval

Smaller intervalWhy?

IR spectrum of HCl, H35Cl (75%) and H37Cl (25%). Two lines due to H35Cl and H37ClH Cl and H Cl

P-branch R-branch

R(0)P(1)

We can see the variation of spacing in the P- and R-branches.


Wh does it ha e t o lines?Why does it have two lines?

35Cl

H35Cl (75%) and H37Cl (25%)

37Cl With better resolution

Two isotopes exist in the nature.


p

P- R-


Decreasing interval

Low resolution spectrumLow resolution spectrumWe cannot see each lines from low resolution spectrum, but we can

calculate the Jmax from equation J

+ for R - for P-branches

Jmax

+ for R, - for P-branchesWe have already studied how to calculate Jmax.

Jmax

Many lines are overlapped.


What looks like the spectrum at lower resolution?

J = 3Jmax 3Jmax = 3

Jmax depends on temperature of sample.

Smaller interval

Larger interval

1. H35Cl and H37Cl show different vibrational transition due to zero point energy.2. R-branch shows narrow interval compared to P-branch due to Morse potential.3 Intensity of each J transition depends on the population at the initial state


3. Intensity of each J transition depends on the population at the initial state.

3.4 BREAKDOWN OF THE BORN-OPPENHEIMER APPROXIMATION : THE INTERACTION OF ROTATIONS AND VIBRATIONS Vibration affects rotational speed.

What is the Born-Oppenheimer approximation?

Max Born was the Nobel prize winner in Physics in 1954

The nuclei are much more massive than the electrons, and thus move slowly relative to the electron Thereforeand thus move slowly relative to the electron. Therefore, the electrons can be considered to move in a field produced by nuclei fixed at some internuclear separation.

We assumed that there is no interaction between ib ti d t ti B t t ll th i i t tivibration and rotation. But actually there is an interaction

between them. This means that the bond length (r) depends on vibrational quantum numbers v


depends on vibrational quantum numbers, v.

An increase in the vibrational energy is accompanied by an increase in the vibrational amplitude and hence the value B will depend on the quantum number v.

Vib. rot. constant

Vibration affects rotational speed.

v=2r2re r0 r1< < <

Definition of bond length at a given v

Center pointv=2

v=1

0

r2

r1B B B B> > >

Morse potential is unsymmetric, steep slope at v=0

r0re

Be B0 B1 B2> > >

Real molecules have a different B with v. B d l th i ith ib ti l

steep slope at short distance, but slow at long distance.


0e Bond length increases with vibrational quantum number v.

distance.

The transition energy is given by the following equations.The transition energy is given by the following equations.

Use of different rotational constant B for different vibrational states

B’ (B1) ≠ B” (B0)Generally,

With B O h i i tiR-branch With Born-Oppenheimer approximation

We can make a similar equation for

For P-branchsimilar equation for R-branch.


Different B values

By combining P- and R-branch equations, we obtain the following equation.

M = + (J+1): R branch-+

M = + (J+1): R-branchM = - (J+1): P-branch

because Be B0 B1 B2> > >

GettingGetting smaller interval For CO molecule

Increase of bond length


Second order function has a minimum or maximum.

Interval difference in P- and R-Branches

Interval between adjacent lines is getting smaller with higher frequency region.

… > P(10) > P(9) > … > P(1) > R(0) > R(1) > …> R(7) > R(8) >…

N l i th i r > r B < BNow we can explain the region. r1 > r0 → B1 < B0

m= + : R-branchm= - : P-branch

P

y = ax2 + bx +ca > 0 minimuma < 0 maximum

negative

-ν

P-

R-

Maximum ν

+

Overlap of many lines : bandheadR-


m Now, we can explain why the spacing is getting smaller with J number in R-branch.

Interval difference in P- and R-Branches

m= +: R branch

−

+ m= +: R-branchm= -: P-branch

+

because Be B0 B1 B2> > >

νParabola curve with maximum.Bandhead at R-branch

νParabola curve minimum.Bandhead at P-branch

-

r1 > r0 → B1 < B0 r1 < r0 → B1 > B0Smaller interval and back again.

Smaller intervalPure vib. energy -

P-branch R-branch

-

P-branch R-branch

Smaller interval and back again.

m+-

m+-


Now, we can explain why the spacing is getting smaller with J number in R-branch.

Pure vibration: Band origing

bandheadbandhead

Smaller interval and back again

Blue : R-branchRed : P-branch


We have a

RP bandhead around here, but cannot see that due to intensity


3.5 THE VIBRATIONS OF POLTATOMIC MOLECULES

The spectrum is complicate because there are many different kinds of bands fundamental overtone hot combinationkinds of bands, fundamental, overtone, hot, combination, difference, etc.

3.5.1 Fundamental Vibrations and Their Symmetry

For N-atomic molecules, Total degree of freedom : 3N

Translational degree of freedom : 3Rotational degree of freedom : 2(3)

Vib ti l d f f d 3N 5(6)Vibrational degree of freedom : 3N-5(6)We should observe many bands originating from vibrational mode.


For Non-linear, 3N-6 degrees of vibrational motion

F Li 3N 5 d f ib ti l tiFor Linear, 3N-5 degrees of vibrational motion

Normal modes ofNormal modes of vibration (normal vibration)

3 vibrational degrees of freedom3 vibrational modes


Direction of dipole moment change and symmetry axis

Main vibrations of water isotopologuesp g

G 1 1 1Gas v1, cm-1 v2, cm-1 v3, cm-1

H216O 3657.05 1594.75 3755.93

H217O 3653.15 1591.32 3748.32

H218O 3649.69 1588.26 3741.572

HD16O 2723.68 1403.48 3707.47

D 16O 2669 40 1178 38 2787 92D216O 2669.40 1178.38 2787.92

T216O 2233.9 995.37 2366.61

Vibrational frequencies depend on the mass (reduced mass) and


( )strength of bond.

4 vibrational degrees of freedom4 vibrational degrees of freedom3 vibrational modes

No observation in IR


Spectrum of CO2

Symmetric bending : Di l t

Symmetric stret vib :

Dipole momentAntisymmetric stret. vib : Dipole moment

Symmetric stret. vib : No dipole moment

1. Stretch > Bending

2 Antisymm > symm2. Antisymm > symm stretch

3. Sym. Stretch = No dipole moment = Nodipole moment No obs.

4. Different bandshape


3 5 2 Overtone and Combination Frequencies3.5.2 Overtone and Combination FrequenciesThe actual observed spectrum is more complicate than what we have expected from the simple explanations.

There exist many different kinds of bands. Theses are combination bands and difference bands. There are 3 modes in water molecule, in

hi h th t t i d b ( )which the state is expressed by (v1, v2, v3).

There are many possible combinations.

ν1+ν2, ν1+ν2+ν3, v1+v2+2v3, etc

(v1,v2,v3=0,0,1)( 1 0 0)

Fundamental bands

(v1,v2,v3=0,2,0)

( 1, 2, 3 , , )(v1,v2,v3=1,0,0) (v1,v2,v3=1,1,1)

(v1,v2,v3=1,1,0)

Combination bands(v v v =0 0 0)

(v1,v2,v3=0,1,0)


(v1,v2,v3=0,0,0)Change of two (three) modes at the same timeThere are 3 fundamentals

Difference bands ν1-ν2, 2ν1-ν2, ν1+ν2-ν3, etc

The vib. level of one mode increases while that of other mode decreases

(2 1 0)(2,1,0)The vibrational quantum number of one mode increases.

(1,1,0)The vibrational quantum number of other mode decreases.

(0,2,0)


Only 3 modes

The actual observation is much more complicate than the prediction because of many combination bands.


3.6 THE INFLUENCE OF ROTATION ON THE SPECTRA O O A O C O C SOF POLYATOMIC MOLECULES

3.6.1 Linear Molecules Vibrational mode changes the rotational selection rules.

In diatomic molecules, the direction of vibrational motion is always parallel to molecular axis.

g

y p

Now, try to learn vibrational symmetry. Always parallel for diatomic molecules

Direction of vibrational motion Ⅱ symmetry axis

Symmetry axis

COM

Linear molecules: Always parallel band


How to affect the vibrational transition to rotational spectrum?

But situation is different for polyatomic molecules.

In linear polyatomic molecules, there are two directions of vibrational motion: Parallel and Perpendicular direction

p y

vibrational motion: Parallel and Perpendicular direction

Direction of vibrational motion

Stretching: Parallel (Ⅱ)

COM

Direction of

Bending: Perpendicular (⊥)

vibrational motion

COM

p ( )


What are the difference in spectrum between parallel and perpendicular bands?

S l ti l f P ll l b dSelection rule of Parallel band

Vibrational type changes the rotational selection rule!!!Vibrational type changes the rotational selection rule!!!


It is the same as that of diatomic molecules. P- and R- branches

Only P- and R-branchesParallel band

P-branch R-branch

Only P and R branchesParallel band

No origin band observed


polyatomic linear

Different from diatomic

Gives Q-branch of ∆J=0


Perpendic lar band

Q-branch appears

Perpendicular band

Q branch appears

W b thWe can observe the origin band (pure vibrational transition).

Q-band: energy difference of each J is so small that they show narrow structure.


Q-band Shows very narrow shape.Perpendicular band: QPerpendicular band:3 bands appear

Shows broad shape Shows broad shape.Shows broad shape.

R-band

P-band


P-, Q-, and R- branches : Bending vibrationPerpendicular band

Spectrum observed from HCN. The appearance of Q branch indicates thatQ-branch indicates that this belongs to bending vibration.


P- and R- branches: Stretching vibration

P-branch R-branch

g

No origin band observed


Vibrational modes of acetylene (C2H2)b at o a odes o acety e e (C2 2)

H HC CP-,R-

Stretching :

InactiveNo observation

Stretching : Parallel band

Stretching has higher frequency than bending vibration.Active

InactiveNo observationP-,Q-,R-R

Bending : Perpendicular band

Active


The molecule has 12-3-2=7 vibrational degrees of freedom, but only 5 modes.

band

From this spectrum, we can see the intensity alternation with J number.Repetition of strong/weak intensity

P-Q-

R-We could not resolve Q-resolve Qbranch.


Combination band : ν1 + ν5 (3374+729 cm-1)

H belongs to Fermion.

Intensity alternation

g

Odd number

Even number


I=0 for 16O (Boson)

I=5/2 for 17O

I=0 for 18OI=0 for O

Odd Js disappear for I=0 (Boson) symmetry. Only even Js will be shown.


Thus, the interval between two lines is twice of the molecules with I≠0

Intensity alternationIntensity alternation

Center of No center of inversion inversion

N t l i li d t 14N15N b th h diff t l i


No symmetry law is applied to 14N15N because they have different nuclei.

3.6.2 The Influence of Nuclear Spin3.6.2 The Influence of Nuclear SpinScientists introduced the symmetry property into molecules. The intensity alternation is the results of symmetry consideration.

Symmetry: The nature has her own symmetry (Scientific law).Macroscopic world: Boltzmann law only Already studiedMacroscopic world: Boltzmann law only. Already studiedMicroscopic world:

With symmetry: Bose-Einstein law, Fermi-Dirac lawy y ,Without symmetry: Boltzmann law

When two identical particles with center of symmetry changed its position, Bose-Einstein law: total symmetry is maintained.Fermi-Dirac law: total symmetry is changed.

Ã Ψ = (+1) Ψ : Bose-Einstein (Bosons)

Ã Ψ = (-1) Ψ : Fermi-Dirac (Fermions)Ã :Exchange of two particles


Ã Ψ (-1) Ψ : Fermi-Dirac (Fermions)

particles

Symmetry propertiesSymmetry properties

Total symmetry= translational symmetry x rotational symmetry x vibrational symmetry x electronic symmetry x nuclear spin symmetry

If the vibrational and electronic state have symmetric properties, then total symmetry is the rotational(J) x nuclear spin(I).

J: Rotational quantum numberI N l i t bI: Nuclear spin quantum number

Total symmetry = (J) x (I)

(anti) = J(symmetry) x I(anti) = J(anti) x I(symmetry) for fermions

(symmerty) = J(anti) x I(anti) = J(symmetry) x I(symmetry) for bosons


anti = antisymmetry

Rotational symmetry: J=0, even number: symmetry, J= odd number: anti-symmetry

Nuclear spin symmetry: I= 0(12C), 1(D): Bose-Einstein, I= ½ (H), ½(13C): Femi-Dirac14N(I=1), 15N(I=1/2), 16O(I=0), 17O(I=5/2), 18O(I=0), 19F(I=1/2)N(I 1), N(I 1/2), O(I 0), O(I 5/2), O(I 0), F(I 1/2)

Explain laterI: nuclear spin quantum number

For H-C≡C-H (H, I=1/2) : Exchange of Hydrogen

Ratio: symmetry(J-even)/anti-symmetry(J-odd)=1/3

Explain later :Not shown here

For CO2 (O-C-O) (O,I=0) : Exchange of Oxygen

Ratio: anti-symmetry(J=odd)/symmetry(J=even)=0/1

The transition of J=odd number will disappearThe transition of J=odd number will disappear.


I (2I 1)(I 1)I sym: (2I+1)(I+1)

I antisym: (2I+1)I

Nuclear Statistics (weight)For I=1/2, odd J / even J = 3/1, intensity alternation of 3 to 1

For Fermions, Jodd * Isym or Jeven * Iantisym

I antisym: (2I+1)INuclear Statistics (weight) I sym: (2I+1)(I+1)

For Bosons, Jodd * Iantisym or Jeven* Isym

For I=1, odd J / even J = 1/2, , intensity alternation of 3 to 6 For I=0, odd J / even J = 0/1, all Jodds disappear.


I=1/2 (fermion), I+1=3/2, I=1/2 I=1 (boson), I=1, I+1=2,

I+1=3/2, I=1/2 I=1, I=1+1=2 I=0, I+1=1I 1 3/2, I 1/2 I 1, I 1 1 2 I 0, I 1 1

Jodd

Jeven

Jodd


There exists a symmetric property for molecules of center of symmetry.For example) HCl, HF, HD, HCCD, 14N15N : No symmetry properties

H2, HCCH, OCO, 14N14N : Symmetry propertiesThere are two types of symmetry for molecules of center of symmetry.

Fermion: • It obeys the Fermi-Dirac law• It obeys the Fermi-Dirac law.• The total wave function should be antisymmetric upon exchange of two identical nuclei. • The nuclear spin quantum number, I, of the exchangeable nuclei should be half-integer numbers, ½, 3/2, 5/2 ….

For example) H(I=1/2) 13C 15N 19F (I=1/2) 17O(I=3/2)For example) H(I=1/2), 13C, 15N, 19F (I=1/2), 17O(I=3/2)• The different I numbers indicate that they have different inner structure inside nucleus.


There exists a symmetric property for molecules of center of symmetryThere exists a symmetric property for molecules of center of symmetry.For example) HCl, HF, HD, HCCD, 14N15N : No symmetry properties

H2, HCCH, OCO, 14N14N : Symmetry propertiesThere are two types of symmetry for molecules of center of symmetry.

Boson: • It obeys the Bose-Einstein law.

The total a e f nction sho ld be s mmetric pon e change of t o identical• The total wave function should be symmetric upon exchange of two identical nuclei. • The nuclear spin quantum number, I, of the exchangeable nuclei should be

d i b 0 1 2zero and integer numbers, 0, 1, 2….For example) 12C, 14N, 16O (I=0), D(I=1)…


The total wave function is a product of translational, rotational, vibrational, electronic, and nuclear spin wave functions.and nuclear spin wave functions.

Among these wave functions, the affecting ones in the ground electronic and ib ti l t t ( l t t ) t ti l d l ivibrational states (normal state) are rotational and nuclear spin.

Thus, total symmetry is a product of rotational symmetry and nuclear spin symmetry., y y p y y p y y

For rotational symmetry, J(even)=symmetry, J(odd)=antisymmetryFor nuclear spin symmetry, I(0,1,2,..)=symmetry, I(1/2,…)=antisymmetry.

Thus,Thus, For fermions, J(even) x I(1/2) or J(odd) x I(0,1,2) gives a total of antisymmetryFor bosons, J(even) x I(0,1,2) or J(odd) x I(1/2..) gives a total of symmetry

On the other hand, the degeneracy of nuclear spin is given byI(0,1,2..) = (2I+1)(I+1) for symmetric


( , , ) ( )( ) yI(1/2,..) = (2I+1)(I) for antisymmetric

For HCCH belonging to Fermion due to H (I=½),

J(even) x I (antisymmetry) = J(even) x (2I+1)(I) = J(even)

J(odd) x I (symmetry) = J(odd) x (2I+1)(I+1) = 3J(odd)

Thus, the intensity of J(odd) has three times stronger than that of J(even).

For D2 belonging to Boson due to D (I=1),

J(even) x I (symmetry) = J(even) x (2I+1)(I+1) = 6J(even)

J( dd) I ( i ) J( dd) (2I 1)(I) 3J( dd)J(odd) x I (antisymmetry) = J(odd) x (2I+1)(I) = 3J(odd)

Thus, the intensity of J(even) has twice stronger than that of J(odd).

For CO2 belonging to Boson due to 16O (I=0),

J(even) x I (symmetry) = J(even) x (2I+1)(I+1) = J(even)J(even) x I (symmetry) = J(even) x (2I+1)(I+1) = J(even)

J(odd) x I (antisymmetry) = J(odd) x (2I+1)(I) = 0 J(odd)

Thus we cannot observe the J(odd) lines so the interval between two


Thus, we cannot observe the J(odd) lines, so the interval between two

neighboring lines J(evene) is 4B rather than 2B, as we already studied.

For N2 belonging to Fermion due to 15N (I=½), 2 g g ( ),J(even) x I (antisymmetry) = J(even) x (2I+1)(I) = J(even)J(odd) x I (symmetry) = J(odd) x (2I+1)(I+1) = 3J(odd) Thus, the intensity of J(odd) has three times stronger than that of J(even).

For 14N2 belonging to Boson due to N (I=1),For N2 belonging to Boson due to N (I 1),J(even) x I (symmetry) = J(even) x (2I+1)(I+1) = 6J(even)J(odd) x I (antisymmetry) = J(odd) x (2I+1)(I) = 3J(odd) Thus, the intensity of J(even) has twice stronger than that of J(odd).

For 14N15N2 there is no center of symmetry We cannot see intensity alternationFor N N2 there is no center of symmetry. We cannot see intensity alternation.


3.6.3 Symmetric Top Molecules We have already studied the rotational energy of symmetric top

Vibrational energy including anharmonicity

energy of symmetric-top.

Rotational energy of prolate symmetric top molecules

The rovibrational energy is given by


Selection Rule of Symmetric TopSelection Rule of Symmetric Top

There are two types of transition;There are two types of transition;

Parallel and Perpendicular transitions

Parallel vibrations: Transition dipole moment is parallelto symmetric axis to give the selection rule of

Perpendicular vibrations: Transition dipole moment is perpendicular to symmetric axis to give the selection rule of

It k th t

Additi l t iti i K t b

It makes the spectrum very complicate


Additional transitions in K quantum number

Parallel Band of Symmetric TopParallel Band of Symmetric Top

Q-

P-

Q

R-Similar to the bending mode of linear moleculesP- mode of linear molecules

Simple P-, Q-, and R- branches


Perpendicular Band of Symmetric TopPerpendicular Band of Symmetric Top

3Broad and complicate shape

1Very complicate and broad


Rotational energy of symmetric-tops with J and KProlate Oblate

Two transitions

One transitionOne transition

E(J,K) = BJ(J+1) + (A-B)K2 E(J,K) = BJ(J+1) + (C-B)K2


CH Cl(Methyl Chloride)CH3Cl(Methyl Chloride)

Parallel band

Perpendicular band


CH3Cl(Methyl Chloride) : symmetric-top3 ( y ) y p

Parallel: sharp 3-branches

broad

Parallel: sharp 3-branchesPerpendicular: broad shapenarrow


spring semester 2015 vibrational spectroscopy of molecules...

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