Ch t 2 Mi S tChapter 2 Microwave...

Rotational energy of molecules corresponds to microwave region. Thus, microwave spectroscopy is sometimes expressed as Rotational Spectroscopy

Ch t 2 Mi S tChapter 2 Microwave SpectroscopyInformation from molecular rotation

Sang Kuk LeeDepartment of ChemistryDepartment of ChemistryPusan National University

March 2015March 2015

2–1Spring Semester 2015, Molecular Spectroscopy : Chapter 2 Microwave Spectroscopy

Analyzing molecular motion(degree of freedom)y g ( g )• Molecules may experience a variety of motion depending on their

structure, energy, and environment. These are translation, rotation, and vibration.

• Each atom needs 3 coordinates (x, y, z) to describe the position.

• We need 3N coordinate to describe the motion. However, atoms combine to form molecule, so there is a restriction on coordinate.

• The necessary restrictions are given as follows.

1) Translation always needs 3 coordinates (x, y, z).

2) Rotation needs 2 or 3 rotational axes (x-, y-, z-axis) depending on the structure of molecules. (2 for linear, 3 for nonlinear)

3) Vib ti d 3N(t t l) 3(t ) (2 3)( t) di t3) Vibration needs 3N(total)-3(trans)-(2 or 3)(rot) coordinates,

• We call the necessary number of coordinates as degree of freedom.


• Please check degree of freedom of He, H2, CO2, H2O, C6H6.

Energy of each type of motion: Vibration > Rotation > Translation

Translational motion is almost continuous.

This chapter


Energy difference between two levels:

Electronic (vis) > Vibrational (IR) >> Rotational (MW) >>> Translational

This chapter

How to analyze rotational motion of molecules.

2.1 THE ROTATION OF MOLECULES

Condition for center of mass is

m1r1 = m2 r2

yAssume the simplest diatomic molecules of m1 and m2

2.1 THE ROTATION OF MOLECULES

There are 3 rotational axesy

x

y

z

Short distance for heavy one, long distance for light one.


Rotational motion always has an axis of rotation.

Short Review on Classical Mechanics

Linear Motion :Linear Motion :

Describe the motion in Cartesian coordinate (x, y, z)

Rotational Motion : Need moment of inertia, I

Describe the motion in Polar coordinate (r θ φ)Describe the motion in Polar coordinate (r, θ, φ)

In order to describe the rotational motion, we define the moment of inertia Di t f


the moment of inertia Distance from rotational axis

(x y z) space oriented coordinate

x

(x,y,z) space oriented coordinate(a,b,c) moment of inertia oriented coordinate

y

z

y

Cartesian coordinate : x y zz Cartesian coordinate : x, y, zPolar coordinate : r, θ, φRotational coordinate : a b cRotational coordinate : a, b, c

Molecular rotational motion is described in terms of (a, b, c) axes


rather than (x, y, z) axes.

Definition of molecular rotationDefinition of molecular rotation

How can we discriminate a, b, and c. It is very simple.

• We have moment of inertia for each rotational axis. These I I d I f d I i ilare Ix, Iy, and Iz for x, y, and z axes. In a similar way, we

can have Ia, Ib, and Ic for a, b, and c axes.

• We can assign a, b, and c in a molecular frame to satisfy Ia ≤ Ib ≤Ic Sequence order of magnitude of moment of inertia

We should know how to match (x, y, z) with (a, b, c) in a molecular frame.


Copied from Atkins, “Physical Chemistry’

H2O, three different moment of inertia.


Linear molecules

Relationship among moments of inertia Ia ≤ Ib ≤Ic

r

SameThree axes, a, b, and c

r1

r2

Zero

2

There are three moments of inertia (Ia, Ib, Ic). But, we need only two (Ib, Ic) for linear molecules because Ia is zero due to ri =0.


Symmetric tops : Two types

NHH

H H

Prolate (cigar-type) Oblate (disc-type)( g yp ) ( yp )


Two moments are same, but one is different.

Spherical tops : Simplest one

Three moments of inertia (Ia, Ib, Ic) of the same magnitude.


Asymmetric tops : very complicateWithout

center of mass

center of mass

center of mass

Ocenter of mass

HO

H


Classification of molecules : 4 types of molecules

• Linear molecules: Ib = Ic, Ia = 0

• Symmetric tops

Prolate: I < I = IProlate: Ia < Ib = IcOblate: Ia = Ib < Ic Text error

• Spherical tops: Ia = Ib = Ic• Asymmetric tops: I ≠ I ≠ I• Asymmetric tops: Ia ≠ Ib ≠ Ic


What is the symmetry of water?

x y z

What is the symmetry of water?

Oz

x y z

a b c

Spectroscopic data of waterHH

x a b c

Rotational constants:A: 8332 x102 GHzB: 4347 x102 GHzC 2985 102 GH

y

Molecule lies on the (x,y) plane.z axis is perpendicular to plane C: 2985 x102 GHz

The related mass-dependent (adiabatic) equilibrium bond length

z-axis is perpendicular to plane.

The related mass dependent (adiabatic) equilibrium bond length and bond angle of H2O is req =0.957(85) Å and θeq=104.50°, respectively, while those of D2O are req =0.957(83) Å and θeq=104.49°.


θeq 104.49 .

A, B, and C are inversely proportional to Ia, Ib, and Ic, respectively.

Planar moleculesPlanar molecules

I I I ( t i ll d)Ia + Ib = Ic (geometrically proved)C axis is perpendicular to molecular plane.


2 2 ROTATIONAL SPECTRA2.2 ROTATIONAL SPECTRA

Rotational spectrum is observable in the microwave region and l h l tt f di t i d li l lalso show very regular pattern for diatomic and linear molecules.

MW iMW region

Regular pattern


Absorption/emission of microwave

• To absorb/emit microwave, molecules should have a dipole moment which generates an oscillating electric field with time. That is, only polar molecules can be observed through microwave spectrum.

• We can observe the rotational spectrum of nonpolar molecules through Raman spectrum which will be studied later.

• The possible molecules to be detected through e poss b e o ecu es o be de ec ed ougmicrowave spectrum are HCl, H2O, NH3, CHCl3, CO, .. (all polar molecules).


CO spectrum : 12C16O, 13C16O, 12C18O Trace amount

13C: 1.108%18O: 0.204%

12C16OShow regular pattern


The shape looks like the 2nd derivatives.

CH CH=CHCOOH

cis

transCH3CH=CHCOOH

Two isomers : trans is more stable

transTwo isomers show different moment of inertia (I) even though they have the same molar mass.


2.3 DIATOMIC MOLECULES :Classical analysisTry to obtain spectral information from molecular structure for diatomic molecules

r0 = r1 + r2

molecular structure for diatomic molecules

m1 m2

r0

m1r1 = m2r2

Condition for center of mass

r r Moment of inertia of diatomic

CenterofMass COM

r1 r2

Rotational axis


From the previous two equations,

EJ = hν, νλ=c

= hc/λReduced mass : definition

and

AftersolvingSchrodingerequation1/λ = EJ/hcReduced mass : definition of reduced mass

wavenumberwavenumber

J: Rotational quantum numberB: Rotational constant


Rotational constant B is inversely proportional to moment of inertia.

Addition to presentationp

• What is reduced mass?• What is reduced mass?

• The actual rotation of diatomic molecule is the rotation of two atoms around the center of mass (COM), so we can easily obtain thearound the center of mass (COM), so we can easily obtain the moment of inertia as I=m1r1

2+m2r22=[m1m2/(m1+m2)]r0

2 =μr02

• The same moment of inertia is generated from the rotation of mass μ connected by string of length r0 to center of mass.

• The rotation of one object is mathematically easier than the rotation f t bj t d t fof two objects around center of mass.

• This is the reason why we introduced the reduced mass, μ in rotation of molecules


rotation of molecules.

From the above equation,

Rotational energy increases with rotational quantum number (J)

12B

Rotational energy increases rapidly with rotational quantum number (J)

10B

quantum number (J).

q ( )

MW 8B

6B

4B2B

Difference


Rotational Transition Energy :Rotational Transition Energy : Energy difference between two states.

Scientists discovered that the transition is

S l ti l ∆J ±1

possible between ∆J = ±1 : selection rule

12B

Selection rule: ∆J = ±1+1 : Absorption10B

-1 : Emission

Selection rule: The transition probability can

8B

6B p ybe obtained by integrating two wave functions in quantum mechanics .2B

6B

4B

2B


Important: Interval between two lines is always 2B

2B’∆J = +1 for absorption spectrum

2B2BB=1.92 cm-1

Guess what J number can be assigned to each line.

8B 10B 12B 14B 16B 18B 20B

Absolute wavenumber


Rotational constant

Moment of inertia

We can calculateWe can calculate reduced mass

We obtain bond length (molecular structure)


We obtain the bond length of CO from the observed spectrum.

2 3 2 The Intensities of Spectral Lines : Already studied2.3.2 The Intensities of Spectral Lines : Already studied

Th b d i t it i MW tThe observed intensity in MW spectrum. Three factors contribute to the intensity.

Transition

Factors affecting the intensity of spectral lines:

1 Transition dipole moment : Selection rule1. Transition dipole moment : Selection ruleDipole moment of molecules

2. Population at the initial state :P l ti i i b B lt di t ib ti lPopulation is given by Boltzmann distribution law

3. Photon energy is proportional to signal intensity of detector


Rotational Boltzmann distributionRotational Boltzmann distribution

0 < exp function < 1

The population decreases with increasing energy state, but the tendency is slow for heavier moleculestendency is slow for heavier molecules.

li ht

heavier

lighter Exponentially decreased


Degeneracy in rotational levelsg y

D (2J+1)Degeneracy = (2J+1)

Orientation of rotationOrientation of rotation


Degeneracy increases with J numberDegeneracy increases with J number

2J+1

J=2 J 3J=2 J=3


Distribution of molecular population over statesDistribution of molecular population over states

• Molecular population are distributed over the states according to their energy.

• The population of states is given by Boltzmann distribution.Boltzmann distribution :• Boltzmann distribution :

NJ/N0 = exp{-(EJ –E0)/kT} Decreasing population with increasing J numberDecreasing population with increasing J number.

• Also we have to consider the degeneracy of states: Degeneracy of rotational states = (2J+1)

Increasing degeneracy with increasing J number.• Thus total distribution is

(Degeneracy) x (Boltzmann distribution)

We have already studied what the degeneracy is in quantum chemistry.


y g y q y

Distribution of molecules over rotational levelsDistribution of molecules over rotational levels

Large moleculeJmax

Distribution is given by Boltzmann distribution law below

Small molecules

Jmax can be obtained from 1st derivative of the function.

J

What J has the maximum population?

It should be an integer number

Jmax

The larger the rotational constant is, the smaller the molecular size is.

We can also calculate the temperature

Jmax increases with temperature, T.


T from the maximum J, Jmax and B.Homework #2

2B’Guess what temperature of sample is from Jmax and B.

2B’

2B

Intensity variation with J.

Jmax


2 3 3 The Effect of Isotopic Substitution : Different mass2.3.3 The Effect of Isotopic Substitution : Different massRotational constant B is inversely proportional to th i f l l

EJ = BJ(J+1)

the size of molecules.Smaller

interval for

Larger interval for lighter one

12CO: B=1.9211813CO: B’=1.83669

heavier isotopes

lighter one

CO: B 1.83669Heavy molecules have small B values.

Large difference at high J.

Small difference at low J


They have different B values due to different mass.Small difference at low J.

2B’

2B

12CO B 1 9211812CO: B =1.9211813CO: B’=1.83669


2 3 4 Th N i id R

The real molecules have a little bit flexible bond length which increases with increasing J number.

2.3.4 The Non-rigid Rotator

Bond stretches with increasing rotational speed

The bigger the molecules, gg ,the smaller the B values.

Experimental data: The B values d ith


decrease with increasing J.

Try to explain the flexibility of the bond length.

2.3.5 The Spectrum of a Non-rigid RotatorAdditional term

B >> D

Vibrational frequencyIf we extend further additional terms, then we have

This equation should be applied for larger J such as J=100.

From a simple equation,ΔJ = +1 for absorption


D, H, K: Centrifugal distortion constants

2 3 5 The Spectrum of a Non-rigid Rotator2.3.5 The Spectrum of a Non-rigid Rotator

Correction term to non-rigid rotator: The contribution increases with J

The larger the rotational quantum number is, the greater the difference between two neighboring lines is.

Always same intervalIdeal molecule

Large difference here

Always same intervalInterval decreases with increasing J

moleculeReal molecule


g

We need n equations to solve n variables.How to determine the 2 bond lengths ?

Isotopic SubstitutionTwo different bond lengths (variables)

SCOrco rcs

Equation 1

Isotope substitution

Equation 2


We need two equations to solve two variables (bond length).

How many independent variables are needed?y p

1 1 variable

2 2 variables

1 1 variable

3 varibales

2 variables


2

1 ariable1 variable

1 variable


J represent the total angular momentum K is the angular

2.4.2 Symmetric Top MoleculesThere are two kinds of rotational quantum numbers, J and K (See next page).

momentum. K is the angular momentum along symmetric axis.

We have two types of axes, symmetry axis and rotational axis. Two axes are always perpendicular to y p peach other for linear molecules.

No rotation about symmetric axis, K=0y ,

In symmetric rotor there are two types of axesIn symmetric rotor, there are two types of axes, symmetry axis and rotational axis. However, two axes may have different angle each other. If they are parallel, then J=K. If they are perpendicular,


p , y p p ,then K=0 like linear molecules.

Quantum numbers J and KQuantum numbers, J and KSince there are three axes, the K component should not be bigger than J.

JKAngular momentum of rotation about the symmetric axis

z

≤Total angular momentum

symmetric axis

Symmetric axis

yy

x Rotation about symmetric axis

K = J, (J – 1), (J – 2), …, 0, …, -(J - 1), -J

Clockwise rotation about the Counter clockwise rotation


axis perpendicular to symmetric axis

about the axis perpendicular to symmetric axis

Calculate rotational energy of symmetric-tops

P: angular momentumgP = Iω

IC = IB ≠ IA→ PX2/2IB + PY

2/2IC(=B) + PZ2/2IA

For prolate tops

C B A X B Y C(=B) Z A

= PX2/2IB + PY

2/2IB + PZ2/2IB + PZ

2/2IA - PZ2/2IB

= P2/2IB + PZ2/2IA - PZ

2/2IB Add and subtract !!

Symmetric topsProlate: IA<IB=IC

= P2/2IB + (1/2IA - 1/2IB) PZ2A-axis

Prolate: IA<IB ICOblate: IA=IB<IC

From Schrodinger equationFrom Schrodinger equation, we get the following energy equations.


Final equation:

positive

Energy equation of prolate symmetric tops.

Symmetric topsWith an exactly same method, we get the equation of Prolate: IA<IB=ICOblate: IA=IB<IC

the oblate. The only difference is that symmetry axis is c-axis for the oblate symmetric-top.

Energy equation of oblate symmetric tops.

negative


Rotational energy of symmetric-tops with J and KProlate Oblate

Energy increases with J and K.

Energy increases with J but decreases with K.dec eases t

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


Define A= 1/IA, etc., then A ≥ B ≥ C

Selection rule of symmetric-tops: ∆J = ±1, ∆K = 0

+1: Absorption1 E i i-1: Emission

What is the physical meaning of ∆K = 0 ?What is the physical meaning of ∆K 0 ?

Rotational quantum number K represents the rotational speed about the symmetric axis while J

indicates the total rotational speed.

Important meaning: The absorption/emission of radiation byImportant meaning: The absorption/emission of radiation by symmetric-top molecules does not change the rotational speed about symmetric axis.


speed about symmetric axis.

Rotational spectrum of symmetric tops

K structure of the same J


2.4.3 Asymmetric Top Moleculesy pThe analysis of asymmetric tops is so complicate that we will not study it at the undergraduate level.

H OH2O


2.5 TECHNIQUES AND INSTRUMENTATIONQ

MW is already monochromatic, so we don’t need to separate pthe frequencies.


InstrumentationInstrumentation

The source and monochromatorElectronically controlled device

Beam directionWaveguide device

Sample and sample spaceGas phase sample vaporized from liquid/solidp p p q

DetectorCrystal detector cooled by liquid helium temp.Crystal detector cooled by liquid helium temp.


2 6 CHEMICAL ANALYSIS BY MICROWAVE2.6 CHEMICAL ANALYSIS BY MICROWAVE SPECTROSCOPY

• Detect ozone in the upper atmosphere.• Useful for detecting freon gas in the upper atmosphere.


2 7 THE MICROWAVE OVEN2.7 THE MICROWAVE OVEN High J

High rotation: High friction: Raising temp.

Low J

A microwave oven works by passing non-ionizing microwave radiationmicrowave radiation, usually at a frequency of 2.45 GHz (a wavelength

f 12 24 ) h h hof 12.24 cm), through the food.


Stark and Zeeman effectStark and Zeeman effect

• Electromagnetic radiation consists of oscillating electric and magnetic field.

• Molecules have also properties of electric and magnetic dipole moment due to unequal distribution of electrons and spin momentmoment due to unequal distribution of electrons and spin moment.

• Stark effect: Interaction between external electric field and electric dipole moment (Energy = electric dipole moment x electric field)dipole moment. (Energy electric dipole moment x electric field)

• Zeeman effect: Interaction between external magnetic field and magnetic dipole moment (Energy = magnetic dipole moment x magnetic field)

• Stark effect is important in neutral polar molecules but Zeeman ff t i t i l l di l f i d l t Zeffect is strong in molecular radicals of unpaired electrons. Zeeman

effect is the only method of detecting molecular radicals.

St k d Z ff t i il Th diff i St k ff t i


Stark and Zeeman effects are very similar. The difference is Stark effect in electric field, but Zeeman effect in magnetic field.

Stark effect Interaction between dipole moment of molecules and

Suppose there are many boats of different direction in the lake

external electric fieldAll boats have the same energy in a lake no matter what

the lake.

the orientation is.

But they will have different energy

Different energy:

different energy with water flow.

Different energy: Splitting

(Energy = electric dipole moment (boat shape) x electric field

E(boat shape) x electric field (strength of water flow),

E = μΕ cosθ (orientation angle)

θ Net effect : E cosθ

Orientation of molecule with respect to electric field

Field onField offMolecules have different energy with external electric field.

All molecules have the same energy without external electric field.


Energy splitting with fieldEnergy splitting with field

Interaction of external electric field with molecular electric dipole t k th t t b litti ( ti )moment makes the energy state be splitting (separation).

NMR is evaluated in terms of the maximum strength of magnetic field (900 MHz in world, 400 MHz in chemistry)

Better separation

Energy Splitting depends on the quantum number M, orientational quantum number

p

orientational quantum number.

Splitting increases with field strength. E = μΕ cosθ

Field strength We now studied the quantum numbers, J, K, and M.

J: total angular momentum


gK: angular momentum along symmetric axisM: angular momentum along field axis

Obtained the following equation using quantum mechanics

M (MJ) ≤ J as like K ≤ J

J=3

MJ=3( J)

Parallelorientation to field

MJ=0

With field : Several transitions

Perpendicular orientation to field

Without field : One transition

transitions

MJ=2What is the selection rule ?

J=2MJ=0


Introduction of new quantum number, M : Orientation along the field

We need a plane polarized light

Field off : Selection rule depends on the

Electric field of radiation

Electric field

trans. angle between two electric field.

Field on : 3 transitions

For ∆MJ = 0

Field on : 3 transitionsWe can see the separation.


We can determine the electric dipole moment, μ from Stark effect.

Observation of MW from the spaceObservation of MW from the spaceMicrowave spectroscopy is very useful for observing the space because it is transparent to air (no absorption by air).transparent to air (no absorption by air).


Only microwave and visible radiation are through air without absorption.

Radio telescopep

Monitoring microwave from the space to identify what molecules are in the space.

Only microwave and visible radiation are transmitted through air.


The environment of the space is quite different from earth. We can observe very stranger molecules from the space which do not exist on earth.


Astronomers can observe the spectrum using radio telescope, but chemists can identify what molecules exist in the universe by using spectroscopic knowledge

HCCCCCN

identify what molecules exist in the universe by using spectroscopic knowledge.

How to identify the species from observation?

Observation from laboratory

Spectrum looks like very noisy butSpectrum looks like very noisy, but we can see the transitions clearly.


Identification of molecules in the spaceIdentification of molecules in the space

Spectrum from the Space(Taurus Molecular Cloud)

Both spectra agree very wellCH2N2

Spectrum from Lab.

Both spectra agree very well, confirming what the molecule is.


Ch t 2 Mi S tChapter 2 Microwave...

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