Slide 1 Chapter 4 Rigid-Rotor Models and Angular Momentum Eigenstates.

Slide 1

Chapter 4

Rigid-Rotor Models andAngular Momentum Eigenstates

Slide 2

Outline

• Math Preliminary: Products of Vectors

• Rotational Motion in Classical Physics

• The 3D Quantum Mechanical Rigid Rotor

• Angular Momentum in Quantum Mechanics

• Angular Momentum and the Rigid Rotor


• The 3D Schrödinger Equation: Spherical Polar Coordinates

• Rotational Spectroscopy of Linear Molecules

Not Last Topic

Slide 3

Outline (Cont’d.)

• Application of QM to Molecular Structure: Pyridine

• Statistical Thermodynamics: Rotational contributions to the thermodynamic properties of gases

Slide 4

Mathematical Preliminary: Products of Vectors

x y zA A i A j A k

x y zB B i B j B k

Scalar Product (aka Dot Product)

x x y y z zA B A B A B A B

Note that the productis a scalar quantity(i.e. a number)

c o s ( )A B A B

Magnitude:

Parallel Vectors: 0c o s ( 0 )A B A B

A B

Slide 5

x y zA A i A j A k

x y zB B i B j B k

Cross Product

The cross product of two vectors is also a vector.

Its direction is perpendicular to both A and B and is given by the “right-hand rule”.

s i n ( )A x B A B

Magnitude:

Parallel Vectors: 0s i n ( 0 )A x B A B

BxA

0s i n ( 9 0 )A x B A B

Perpendicular Vectors:

0

A B

Slide 6

x y zA A i A j A k

x y zB B i B j B k

A x B

x y z

x y z

i j k

AxB A A A

B B B

y z x yx z

y z x yx z

A A A AA AA xB i j k

B B B BB B

Expansion byCofactors

( ) ( ) ( )y z z y z x x z x y y xA x B A B A B i A B A B j A B A B k

Slide 7

Outline









Slide 8

Rotational Motion in Classical Physics

s i n ( )L r p Magnitude:

Angular Momentum (L)

m

Circular Motion: 0s i n ( 9 0 )L r p r p

or: 2 2( )v

L rp rm v m r m rr

L I ω where 2 vI m r

r

Energy

2 2

2 2

p mvE

m

2 2 2( )

2 2

m r m r

2

2

I or:

2 2( )

2 2

I LE

I I

Momentof Inertia

AngularFrequency

L rxp

Slide 9

Comparison of Equations for Linear and Circular Motion

Linear Motion Circular Motion

Mass 2I m r Moment of inertiam

Velocityv

r Angular velocityv

Momentum L I Angular momentump=mv

Energy2

2

LE

I Energy

2

2

pE

m

or2

2

IE

Energy

2

2

mvE

Slide 10

Modification: Rotation of two masses about Center of Mass

m

m1 m1r

2I m r 2I r

where 1 2

1 2

m m

m m

Slide 11

Outline









Slide 12

Angular Momentum in Quantum Mechanics

L r x p

Classical Angular Momentum

( ) ( ) ( )z y x z y xL y p z p i z p x p j x p y p k

r x i y j z k

x y zp p i p j p k

x y z

i j k

x y z

p p p

x z yL y p z p

y x zL z p x p

z y xL x p y p

Slide 13

Angular Momentum in Quantum Mechanics

QM Angular Momentum Operators

x z yL y p z p

y x zL z p x p

z y xL x p y p

ˆ xp ii x x

ˆ yp ii y y

ˆ zp ii z z

Classical QM Operator

ˆxL i y z

z y

ˆyL i z x

x z

ˆzL i x y

y x

2 ˆ ˆ ˆ ˆ ˆ ˆx x y y z zL L L L L L L

^

Slide 14

Operator Commutation and Simultaneous Eigenfunctions

It can be shown that: ˆ ˆ, 0x yL L Do not commute

ˆ ˆ, 0y zL L Do not commute

ˆ ˆ, 0x zL L Do not commute

Because the operators for the individual components do not commute,one cannot determine two separate components simultaneously.

i.e. they cannot have simultaneous eigenfunctions.

In contrast, it can be shown that:2ˆ , 0zL L

^ Do commute

Because these operators commute, one can determine Lz and L2

simultaneously; i.e. they can have simultaneous eigenfunctions.

Slide 15

Outline









Slide 16

The 2D Quantum Mechanical Rigid Rotor

Assume that two masses are attached by a rigid rod (i.e. ignorevibrations) at a fixed distance, r,and are free to rotate about the Center of Mass in their x-y plane.

m1

m1

r x

y

The angle represents the angle of rotation relative to the x-axis.

The 2D Schrödinger equation for therelative motion of two masses is:

22

2V E

1 2

1 2

m m

m m

2 22

2 2x y

Two Dimensional Laplacianin Cartesian Coordinates

Slide 17

22

2V E

If one (a) converts the Laplacian to polar coordinates

(b) assumes that the potential energy is constant (arbitrarily 0)

(c) holds r fixed (i.e. neglects derivatives with respect to r)

It can be shown that the Schrödinger Equation for a 2D RigidRotor becomes: 2 2

2 22E

r

or2 2

22E

I

where 2I r is the moment of inertia

Slide 18

The Solution2 2

22E

I

2

2 2

2 IE

C o n s t a n t

22

2IEm or

2 2

2

mE

I

Note: So far, m can have any value;

i.e. there is no energy quantization

imA e

Assume

( ) imdim Ae

d

2

22

( ) imdim Ae

d

2

22

dm

d

22

2 IEm

Slide 19

2 2

0, 1, 2, 3,2

mE m

I

Application of the Boundary Conditions:Quantization of Energy

imA e

To be a physically realistic

solution, one must have: ( 2 ) ( )

Therefore: 2i m m i i mA e e A e

or 2 1m ie

c o s ( 2 ) s i n ( 2 ) 1m i m

This is valid only for: 0 , 1 , 2 , 3 ,m

Therefore, only certain values for the energy are allowed;i.e. the energy is quantized:

Slide 20

Zero Point Energy

2 2

0, 1, 2, 3,2

mE m

I0 0E

There is no minimum Zero Point Energy.

One encounters a ZPE only when the particle is bound (e.g. PIB,Harmonic Oscillator, Hydrogen Atom), but not in freely movingsystems (e.g. 2D and 3D Rigid Rotor, free particle)

Slide 21

Application of the 2D Rigid Rotor

We have solved the 2D Rigid Rotor primarily as a learning exercise,in order to demonstrate the application of angular Boundary Conditions.

However, the model has a real world application, in that it can be used to characterize the rotation of molecules adsorbed on surfaces.

Example

When an H2 molecule is chemisorbed on a crystalline surface, itsrotation can be approximated as that of a 2D rigid rotor.

The H2 bond length is 0.74 Å Calculate the frequency (in cm-1) of the lowest energy rotational transition of chemisorbed H2.

Slide 22

2H

H H

m

m m

2 2

0, 1, 2, 3,2

mE m

Ir = 0.74 Å = 0.74x10-10 m1 amu = 1.66x10-27 kgħ = 1.05x10-34 J•sh = 6.63x10-34 J•sc = 3.00x1010 cm/s

2(1 )0 .50

1 1

amuamu

amu amu

2 72 81 .6 6 1 0

8 .3 0 1 0x kg

x kgam u

22 2 8 1 08 .3 0 1 0 0 .7 4 1 0I r x k g x m 4 8 24 . 5 5 1 0x k g m

2 2 2 22 1

2 2

m mE

I I

m1 = 0

m2 = 12

2I

234

48 2

1.05 10

2 4.55 10

x J s

x kg m

2 11 . 2 1 1 0x J

E

hc

21

34 10

1.21 10

6.63 10 3.00 10 /

x J

x J s x cm s

16 0 . 9 c m

Slide 23

Outline









Slide 24

The Three Dimensional Schrödinger Equation

In Cartesian Coordinates, the 3D Schrödinger Equation is:

2 2 2 2 2 2

2 2 2( , , )

2 2 2V x y z E

m x m y m z

( , , )x y z

The Laplacian in Cartesian Coordinates is: 22 2 2x y z

T(x) T(y) T(z) V(x,y,z)

Therefore:2

2 ( , , )2

V x y z Em

It is sometimes not possible to solve the Schrödinger exactly inCartesian Coordinates (e.g. the Hydrogen Atom), whereas itcan be solved in another coordinate system.

The “Rigid Rotor” and the Hydrogen Atom can be solved exactlyin Spherical Polar Coordinates.

Slide 25

Spherical Polar Coordinates

To specify a point in space requires three coordinates.In the spherical polar coordinate system, they are:

r 0 r < Distance of point from origin (OP)

0 < Angle of vector (OP) from z-axis

0 < 2 Angle of x-y projection (OQ) from x-axis

Slide 26

Relation of Cartesian to Spherical Polar Coordinates

rz

cos( )z

r

c o s ( )z r

x-axis

y-axis

O

Q

x

y

cos( )x

OQ

c o s ( )x O Q

s i n ( ) c o s ( )x r

sin( )y

OQ

s i n ( )y O Q

s i n ( ) s i n ( )y r

OQ=rsin()

OQ

Slide 27

The Volume Element in Spherical Polar Coordinates

In Cartesian Coordinates,

the volume element is: d V d x d y d z

In spherical polar coordinates,

the volume element is: d V d r r d O Q d

s i n ( )d V d r r d r d

2 s i n ( )d V r d r d d

Slide 28

The Laplacian in Spherical Polar Coordinates

22 2 2x y z

Cartesian Coordinates:

One example of a chain rule formula connecting a derivative withrespect to x, y, z to derivatives with respect to r, , is:

r

x r x x x

It may be shown that by repeated application of chain rule formulaeof this type (with 2-3 hours of tedious algebra), the Laplacian inspherical polar coordinates is given by:

22 2

2 2 2 2 2

1 1 1sin ( )

sin ( ) sin ( )r

r r r r r

Slide 29

Angular Momentum Operators in Spherical Polar Coordinates

ˆzL i x y

y x

It may beshown that

ˆzL i

i

2 ˆ ˆ ˆ ˆ ˆ ˆx x y y z zL L L L L L L

^

It may beshown that

22 2

2 2

1 1sin( )

sin( ) sin ( )L

^

Slide 30

Outline









Slide 31

The 3D Quantum Mechanical Rigid Rotor

3D Schrödinger Equation for a particle (Sph. Pol. Coords.)

2 22

2 2 2 2 2

1 1 1sin( )

2 sin( ) sin ( )r V E

m r r r r r

22 ( , , ) ( , , )

2V r E r

m

Modification: Two masses moving relative to their CM

m1 m1r

2 22

2 2 2 2 2

1 1 1sin( )


r r r r r

1 2

1 2

m m

m m

Slide 32

2 22

2 2 2 2 2

1 1 1sin( )


r r r r r

The Schrödinger Equation in terms of the L2 operator^

22 2

2 2

1 1sin( )

sin( ) sin ( )L

^

The L2 operator is:^

2 2 22

2 2 2 2

1 1 1sin( )

2 2 sin( ) sin ( )r V E

r r r r

22 2

2

1 1

2 2r L V E

r r r I

2I rwhere^

Radial KE Rotational KE

PE

Slide 33

The Quantum Mechanical Rigid Rotor

2 2 22

2 2 2 2

1 1 1sin( )

2 2 sin( ) sin ( )r V E

r r r r

The Rigid Rotor model is used to characterize the rotation ofdiatomic molecules (and is easily extended to linear polyatomic molecules)

It is assumed that: (1) The distance between atoms (r) does not change.

(2) The potential energy is independent of angle [i.e. V(,) = Const. = 0]

22 2

2

1 1

2 2r L V E

r r r I

^

Therefore:

2 2

2 2

1 1sin( )

2 sin( ) sin ( )E

I

21

2L E

I

^

Slide 34

2 2

2 2

1 1sin( )

2 sin( ) sin ( )E

I

( , )

This equation can be separated into two equations, one containing only and the second containing only .

Assume: ( , ) ( ) ( )

Solution of the Rigid Rotor Schrödinger Equation

Algebra + Separation of Variables

221

sin( ) sin( ) sin ( )2

E CI

2 2

2

1

2C

I

and

We will only outline the method of solution.

Slide 35

Solution of the equation is rather simple.

However, solution of the equation most definitely is NOT.

Therefore, we will just present the results for the quantum numbers,energies and wavefunctions that result when the two equations are solved and boundary conditions are applied.

221

sin( ) sin( ) sin ( )2

E CI

2 2

2

1

2C

I

and

Slide 36

The Rigid Rotor Quantum Numbers and Energies

The Quantum Numbers:

Note that because this is a two dimensional problem, thereare two quantum numbers.

0 , 1 , 2 , 3 , 0 , 1 , 2 , ,m

The Energy:2

( 1)2

EI

Note that the energy is a function of l only. However, there are2 l + 1 values of m for each value of l . Therefore, the degeneracyof the energy level is 2 l + 1

2 1g

Remember that for a classical Rigid Rotor:2

2

LE

I

Comparing the expressions, one finds for the

angular momentum, that: ( 1)L

Slide 37

An Alternate Notation

The Quantum Numbers: 0 , 1 , 2 , 3 ,J

0 , 1 , 2 , ,M J

The Energy:2

( 1)2JE J JI

2 1Jg J

When using the Rigid Rotor molecule to describe the rotationalspectra of linear molecules, it is common to denote the twoquantum numbers as J and M, rather than l and m.

With this notation, one has:

Slide 38

The Wavefunctions

When both the and differential equations have been solved,the resulting wavefunctions are of the form:

,( , ) ( ) ( ) ( )mimmN e P

The are known as the associated Legendre polynomials.( )mP

)(mP

The first few of these functions are given by:

00 1P

01 c o s ( )P

11 s in ( )P

0 22

13 cos ( ) 1

2P

12

1sin( ) cos( )

2P

2 22 s in ( )P

We will defer any visualization of these wavefunctions untilwe get to Chapter 6: The Hydrogen Atom

Slide 39

Spherical Harmonics

( , ) ( ) ( ) ( )mimlm mY N e P

The product functions of and are called “Spherical Harmonics”,Ylm(, ):

They are the angular solutions to the Schrödinger Equation for anyspherically symmetric potential; i.e. one in which V(r) is independentof the angles and .Some examples are:

0 01 0 1 0 1 1 0( , ) ( ) c o s ( )iY N e P N

11 1 1 1 1 1 1( , ) ( ) s i n ( )i iY N e P N e

0 0 22 0 2 0 2 2 0( , ) ( 0 ) 3 c o s ( ) 1iY N e P N

12 1 2 1 2 2 1( , ) ( ) s i n ( ) c o s ( )i iY N e P N e

2 2 2 22 2 2 2 2 2 2( , ) ( ) s i n ( )i iY N e P N e

11 1 1 1 1 1 1( , ) ( ) s i n ( )i iY N e P N e

12 1 2 1 2 2 1( , ) ( ) s i n ( ) c o s ( )i iY N e P N e

2 2 2 22 2 2 2 2 2 2( , ) ( ) s i n ( )i iY N e P N e

Slide 40

( , ) sin( )iY Ne One of the Spherical Harmonics is:

Show that this function is an eigenfunction of the Rigid RotorHamiltonian and determine the eigenvalue (i.e. the energy).

cos( )iYNe

sin( ) sin( ) cos( )iYNe

2 2s in ( ) co s ( )iN e

21 2 s in ( )iN e

1sin( ) 2 sin( )

sin( ) sin( )

iiY Ne

Ne

2 2

2 2

1 1sin( )

2 sin( ) sin ( )

Y YEY

I

21

2HY L Y EY

I

^or

Slide 41

( , ) s i n ( )iY N e

2 2

2 2

1 1sin( )

2 sin( ) sin ( )

Y YEY

I

1sin( ) 2 sin( )

sin( ) sin( )

iiY Ne

Ne

sin( )iYiNe

2

2sin( )iY

Ne

2

2 2

1

sin ( ) sin( )

iY Ne

2 2

2 2

1 1sin( )

2 sin( ) sin ( )

Y YHY

I

2

2 sin( )2 sin( ) sin( )

i iiNe Ne

NeI

2

2 sin( )2

iNeI

2

22

YI

Slide 42

2

22

HY YI

Therefore:2

22

EI

Note: Comparing to:2

( 1)2

EI

1we see that:

Slide 43

Outline









Slide 44

Angular Momentum and the Rigid Rotor

The Spherical Harmonics, Ylm(, ), are eigenfunctions of theangular momentum operators:

22 2

2 2

1 1sin( )

sin( ) sin ( )L

^

ˆzL i

Note: It is straightforward to show that L2 and Lz commute;

i.e. [L2,Lz] = 0.

Because of this, it is possible to find simultaneouseigenfunctions of the two operators which are, as shown above,the Spherical Harmonics.

^ ^

^^

2 2( , ) ( 1 ) ( , )l m lmL Y Y ^

The eigenvalues are given by the equations:

ˆ ( , ) ( , )z lm lmL Y m Y

Slide 45

As discussed earlier, the restrictions on the quantum numbersare given by:

0 , 1 , 2 , 3 , 0 , 1 , 2 , ,m

Therefore, both the magnitude, |L|, and the z-component, Lz, of theangular momentum are quantized to the values:

( 1 0 , 1, 2 , 3 ,L

0 , 1 , 2 , ,zL m m z-axis

Lz |L|

If a magnetic field is applied, its directiondefines the z-axis.

If there is no magnetic field, the z-directionis arbitrary.

Slide 46

( , ) s i n ( )iY N e One of the Spherical Harmonics is:

22 2

2 2

1 1sin( )

sin( ) sin ( )L

^ ˆ

zL i

Show that this function is an eigenfunction of L2 and Lz and determinethe eigenvalues.

^^

We’ve actually done basically the first part a short while ago.

Remember: 21

2HY L Y EY

I

^ 2

22

Y

I

Therefore: 2 22L Y Y ^ 2( 1 ) Y 1

z

YL Y

i

sin( )ii N ei

Y m Y

1m

Slide 47

Preview: The Hydrogen Atom Schrödinger Equation

2 22

2 2 2 2 2

1 1 1sin( )


m r r r r r

3D Schrödinger Equation in Spherical Polar Coordinates

The Hydrogen atom is an example of a “centrosymmetric” system,which is one in which the potential energy is a function of only r, V(r).

In this case, the Schrödinger equation can be rearranged to:

2 2 22

2 2 2 2 2

1 1 1( ) sin( )

2 2 sin( ) sin ( )r V r E

m r r r m r r

Radial Part Angular Part

Note that the Angular part of the Hydrogen atom Schrödinger equationis the same as Rigid Rotor equation, for which the radial part vanishes.

Therefore, the angular parts of the Hydrogen atom wavefunctions arethe same as those of the Rigid Rotor

Slide 48

Outline









Slide 49

Rotational Spectroscopy of Linear Molecules

Energy Levels2

( 1)2JE J JI

2 1Jg J

Equivalent Form:2

2( 1)

8J

hE J J

I

2( 1)

8J

hE hc J J

Ic

( 1 )JE h c B J J

RotationalConstant (cm-1):

28

hB

Ic

Note: You must use c in cm/s, even when using MKS units.

0 0 g0=1

1 g1=3B~

2

E /

hc

[cm

-1]

J EJ gJ

2 g2=5B~

6

3 g3=7B~

1 2

4 g4=9B~

2 0

Slide 50

Diatomic versus Linear Polyatomic Molecules

28

hB

Ic

In general, for linear molecules, the moment of inertia is given by:

2

1

N

i ii

I m r

N is the number of atomsmi is the mass of the atom iri is the distance of atom i from the Center of Mass.

If N=2 (diatomic molecule) the moment of inertia reduces to:

2I r1 2

1 2

m m

m m

where r is the interatomic distance

Slide 51

Selection Rules

Absorption (Microwave) Spectroscopy

For a rotating molecule to absorb light, it must have a permanentdipole moment, which changes direction with respect to the electric vector of the light as the molecule rotates.

J = 1 (J = +1 for absorption)

e.g. HCl, OH (radical) and O=C=S will absorb microwave radiation.

O=C=O and H-CC-H will not absorb microwave radiation.

Rotational Raman SpectroscopyFor a rotating molecule to have a Rotation Raman spectrum, thepolarizability with respect to the electric field direction must change as the molecule rotates. All linear molecules have Rotational Raman spectra.

J = 2

J = +2: Excitation (Stokes line)

J = -2: Deexcitation (Anti-Stokes line)

Slide 52

Intensity of Rotational Transitions

The intensity of a transition in the absorption (microwave) orRotational Raman spectrum is proportional to the number of moleculesin the initial state (J’’); i.e. Int. NJ’’

Boltzmann Distribution:''

'' ''

JE

kTJ JN g e

''( '' 1)

'' (2 '' 1)hcBJ J

kTJN J e

Slide 53

Rotational Spectra

0 0 g0=1

1 g1=3B~

2

E /

hc

[cm

-1]

J EJ gJ

2 g2=5B~

6

3 g3=7B~

1 2

4 g4=9B~

2 0Absorption (Microwave) Spectra

' ''J JE E E

J’’ J’

'( ' 1 ) ''( '' 1 )E h c B J J h c B J J

J’ = J’’+1

( '' 1 ) ( '' 2 ) ''( '' 1 )E h c B J J h c B J J

( 2 '' 2 ) '' 0 , 1 , 2 , 3 ,E h c B J J

(2 '' 2 ) '' 0 , 1, 2 , 3,E

B J Jhc

B~

2 B~

4 B~

6 B~

8~0

Slide 54

0 0 g0=1

1 g1=3B~

2

E /

hc

[cm

-1]

J EJ gJ

2 g2=5B~

6

3 g3=7B~

1 2

4 g4=9B~

2 0Rotational Raman Spectra

' ''J JE E E

J’’ J’

'( ' 1 ) ''( '' 1 )E h c B J J h c B J J

J’ = J’’+2

( '' 2 ) ( '' 3 ) ''( '' 1 )E h c B J J h c B J J

( 4 '' 6 ) '' 0 , 1 , 2 , 3 ,E h c B J J

(4 '' 6 ) '' 0 , 1, 2 , 3,E

B J Jhc

B~

1 4B~

6 B~

1 0~

0

Slide 55

The HCl bond length is 0.127 nm.

Calculate the spacing between lines in the rotational microwave

absorption spectrum of H-35Cl, in cm-1.h = 6.63x10-34 J•sc = 3.00x108 m/sc = 3.00x1010 cm/sNA = 6.02x1023 mol-1

k = 1.38x10-23 J/K1 amu = 1.66x10-27

kg

H Cl

H Cl

m m

m m

1 35

0.9721 35

amu amuamu

amu amu

2I r 22 7 91 .6 1 1 0 0 .1 2 7 1 0x k g x m 4 7 22 . 6 0 1 0x k g m

28

hB

Ic

34

2 47 2 10

6.63 10

8 3.14 2.60 10 3.00 10 /

x J s

x kg m x cm s

As discussed above, microwave absorption lines occur at 2B, 4B, 6B, ...

Therefore, the spacing is 2B

~ ~ ~

~

0 . 9 7 2 a m u 27

271.66 101.61 10

1

x kgx kg

amu

1 11 0 . 7 8 1 0 . 8c m c m

1S p a c i n g 2 2 1 0 . 8 2 1 . 6B x c m

Slide 56

h = 6.63x10-34 J•sc = 3.00x1010 cm/sk = 1.38x10-23 J/K

B = 10.8 cm-1~

Calculate the ratio of intensities (at 250C): 3 4

1 2

I

I

0 0 g0=11 g1=3B

~2

E /

hc

[cm

-1]

J EJ gJ

2 g2=5B~

6

3 g3=7B~

1 2

4 g4=9B~

2 0

3 4 3

1 2 1

I N

I N

3

1

/3

/1

E kT

E kT

g e

g e

12 /

2 /

7

3

hcB kT

hcB kT

e

e

10 /7

3hcB kTe

34 10 1

23

10 6.63 10 3.00 10 / 10.810

(1.38 10 / ) 298

x J s x cm s cmhcB

kT x J K K

= 0.52

0.523 4

1 2

7

3

Ie

I

1 .4

Note: This is equivalent to asking for the ratio of intensites offourth line to the first line in the rotational microwave spectrum.

Slide 57

The first 3 Stokes lines in the rotational Raman spectrum of 12C16O2 are found at 2.34 cm-1, 3.90 cm-1 and 5.46 cm-1.

Calculate the C=O bond length in CO2, in nm.

h = 6.63x10-34 J•sc = 3.00x1010 cm/sk = 1.38x10-23 J/K1 amu = 1.66x10-27 kg

0 0 g0=11 g1=3B

~2

E /

hc

[cm

-1]

J EJ gJ

2 g2=5B~

6

3 g3=7B~

12

4 g4=9B~

2012 . 3 4 6 0 6c m B B

13 . 9 0 1 2 2 1 0c m B B B

15 . 4 6 2 0 6 1 4c m B B B

10 .3 9B c m

10 .3 9B c m

10 .3 9B c m

28

hB

Ic

34 2

2 1 10

6.63 10 /

8 3.14 0.39 3.00 10 /

x kg m sI

cm x cm s

28

hI

Bc

4 6 27 . 1 8 1 0x k g m

Slide 58

CM

rCO rCO

The first 3 Stokes lines in the rotational Raman spectrum of 12C16O2 are found at 2.34 cm-1, 3.90 cm-1 and 5.46 cm-1.

Calculate the C=O bond length in CO2, in nm.

h = 6.63x10-34 J•sc = 3.00x1010 cm/sk = 1.38x10-23 J/K1 amu = 1.66x10-27 kg

4 6 27 . 1 8 1 0I x k g m

2i i

i

I m r C OO 22 20O C O C O C Om r m m r 22 O C Om r

2COO

Ir

m 27

26

16 1.66 10 /

2.66 10

Om amu x kg amu

x kg

46 2

26

7.18 10

2 2.66 10CO

x kg mr

x kg

1 01 .1 6 1 0x m 0 . 1 1 6 1 . 1 6n m A n g s t r o m s

Slide 59

h = 6.63x10-34 J•sc = 3.00x1010 cm/sk = 1.38x10-23 J/K

Calculate the initial state (i.e. J’’) corresponding to the mostintense line in the rotational Raman spectrum of 12C16O2 at 25oC.

Hint: Rather than calculating the intensity of individualtransitions, assume that the intensity is a continuousfunction of J’’ and use basic calculus.

B = 0.39 cm-1~

'' '' 2 ''J J JI N '' '' 1

2 '' 1J J hcB

kTJ e

2'' ''2 '' 1

J J hcBJ e

kT

NJ'' is at a maximum for dNJ''/dJ''=0.

2 2'' '' '' '''' 0 2 '' 1 2 '' 1

'' '' ''

J J J JJdN d dJ e e J

dJ dJ dJ

2 2'' '' '' ''0 2 '' 1 2 '' 1 2

J J J JJ e J e

2'' '' 2

0 2 '' 1 2J J

e J

Slide 60

Therefore: 22 '' 1 2 0J

22 '' 1J

2'' '' 2

0 2 '' 1 2J J

e J

hcB

kT

h = 6.63x10-34 J•sc = 3.00x1010 cm/sk = 1.38x10-23 J/K

B = 0.39 cm-1~

23

34 10 1

2 1.38 10 / 2982 '' 1

6.63 10 3.00 10 / 0.39

x J K KJ

x J s x cm s cm

1 0 6 0 3 2 .6

32.6 1''

2J

1 5 . 8 1 6

Slide 61

Consider the linear molecule, H-CC-Cl.

There are two major isotopes of chlorine, 35Cl (~75%) and37Cl (~25%). Therefore, one will observe two series of linesin the rotational spectrum, resulting from transitions ofH-CC-35Cl and H-CC-37Cl.

Can the structure of H-CC-Cl be determined from these two series?

No. There are 3 bond distances to be determined, but only 2 moments of inertia.

What additional information could be used to determine all threebond distances?

The spectrum of D-CC35Cl and D-CC37Cl

Slide 62

Iz

Ix

Iy

Non-Linear Molecules

Non-linear molecules will generally have up to3 independent moments of inertia, Ix, Iy, Iz.

The Hamiltonian will depend upon the angularmomentum about each of the 3 axes.

22 2

2 2 2yx z

x y z

LL LH

I I I

The Schrödinger Equation for non-linear rotors is more difficult to solve,but can be done using somewhat more advanced methods, and therotational spectra can be analyzed to determine the structure(sometimes requiring isotopic species).

For small to moderate sized molecules (I would guess 10-15 atoms),rotational microwave spectroscopy is the most accurate method fordetermining molecular structure.

Slide 63

Outline (Cont’d.)



Slide 64

N 1.35 Å1.34

1.40 Å1.38

1.39 Å1.38

117o

117

124o

124

119o

119

118o

118

The Structure of Pyridine

Calculated: MP2/6-31G(d) – 4 minutes

Experimental: Crystal Structure

Slide 65

#MP2/6-31G(d) opt freq Pyridine 0 1 C -1.236603 1.240189 0.000458 C -1.236603 -0.179794 0.000458 C -0.006866 -0.889786 0.000458 C 1.222870 -0.179794 0.000458 C 1.053696 1.280197 0.000458 N -0.104187 1.989731 0.000458 H 1.980804 1.872116 -0.008194 H 2.205566 -0.673935 -0.009628 H -0.006866 -1.989731 -0.009064 H -2.189194 -0.729767 -0.009064 H -2.205551 1.760818 0.009628

The Command File for the Structure of Pyridine

Slide 66

The Frontier Orbitals of Pyridine

HOMO LUMO

Calculated: HF/6-31G(d) – <3 minutes

Experimental: Huh???

Slide 67

N 1.35 Å1.40

1.40 Å1.33

1.39 Å1.44

117o

116

124o

120

119o

119

118o

120

The Structure of Excited State Pyridine

S0: Calculated: MP2/6-31G(d) – 4 minutes

T1: Calculated: MP2/6-31G(d) – 11 minutes

Experimental: None

Slide 68

Outline (Cont’d.)



Slide 69

Statistical Thermodynamics: Rotational Contributions to Thermodynamic Properties of Gases

A Blast from the Past

2

,

ln

V N

QU kT

T

2

, ,

ln ln

lnV N T N

Q QH kT kT

T V

,V

V N

UC

T

,P

P N

HC

T

lnU

S k QT

l n ( )A U T S k T Q

,

lnln

ln T N

QG H TS kT Q kT

V

Slide 70

The Rotational Partition Function: Linear Molecules

The Energy:2

( 1)2J J JI

2 1Jg J

0

Jrot kT

JJ

q g e

The Partition Function: 2( 1) / 2

0

2 1J J I

kT

J

J e

( 1)

0

2 1RJ Jrot T

J

q J e

2

2R Ik

It can be shown that for most molecules at medium to hightemperatures:

1J J k T

Thus, the exponent (and hence successive terms in thesummation) change very slowly.

Therefore, the summation in qrot can be replaced by an integral.

Slide 71

( 1)

0

2 1RJ Jrot T

J

q J e

2

28R

h

Ik

( 1)

0

2 1RJ Jrot Tq J e dJ

This integral can be solved analytically by a simple substitution.

( 1) Ru J JT

(2 1) Rdu J dJ

T

Therefore:( 1)

0

(2 1)RJ Jrot RT

R

Tq e J dJ

T

0

u

R

Te du

0

1rot u

R

Tq e

1 0 1

R

T

rot

R

Tq

Slide 72

rot

R

Tq

A Correction

For homonuclear diatomic molecules, one must account for the factthat rotation by 180o interchanges two equivalent nuclei.

Since the new orientation is indistinguishable from the original one, onemust divide by 2 so that indistinguishable orientations are counted once.

For heteronuclear diatomic molecules, rotation by 180o produces adistinguishable orientation. No correction is necessary.

rot

R

Tq

is the "symmetry number"

Homonuclear Diatomic Molecule: = 2

Heteronuclear Diatomic Molecule: = 1

Slide 73

How good is the approximate formula?

( 1)

0

12 1

RJ Jrot T

J

q J e

Exact:rot

R

Tq

Approx:

H2: I = 4.61x10-48 kg•m2 R = 87.5 K

O2: I = 1.92x10-46 kg•m2 R = 2.08 K

Compd. T qrot(ex) qrot(app) Error

O2 298 K 71.8 71.7 0.2%

H2 298 1.88 1.70 10%

H2 100 0.77 0.57 26%

H2 50 0.55 0.29 47%

For molecules like H2, HCl, …, with small moments of inertia, the integral approximation gives poor results, particularly at lowertemperatures.

Slide 74

The Total Partition Function for N Molecules (Qrot)

rot

R

Tq

and Nro t ro tQ q

N

rot

R

TQ

where

2

28R

h

Ik

Slide 75

Internal Energy

,

lnln ln(

rot

R

V N

Q dN T

T dT

N

rot

R

TQ

2

,

ln rotrot

V N

QU kT

T

N

T

Therefore: 2rot NU kT

T N k T An N k T

r o tU n R T or r o tU R T

This illustrates Equipartition of Rotational Internal Energy.

[(1/2)RT per rotation].

ln lnN

rot

R

TQ

ln

R

TN

ln ln ( )RN T

Slide 76

2

, ,

ln ln

ln

rot rotrot

V N T N

Q QH kT kT

T V

Enthalpy

,

ln

ln

rotrot

T N

QU kT

V

Qrot independent of V

r o t r o tH U R T

Heat Capacities

,

rot rotrot VV

V N

C UC R

n T

,

rot rotrot PP

P N

C HC R

n T

Remember that these results are for Diatomic and LinearPolyatomic Molecules.

Slide 77

5

2tranPC R

Chap. 3

Experimental Heat Capacities at 298.15 K

Compd. CP (exp)

H2 29.10 J/mol-K

O2 29.36

I2 36.88

r o tPC R

7 78 .3 1 4 / 2 9 .1 0 /

2 2tra n ro tP PC C R J m o l K J m o l K

We can see that vibrational (and/or electronic) contributionsto CP become more important in the heavier diatomic molecules.

We'll discuss this further in Chap. 5

Slide 78

Entropy

lnrot

rot rot US k Q

T

N

rot

R

TQ

lnN rot

rot

R

T US k

T

ln

R

T RTNk

T

ln

R

TnR R

O2: r = 1.202 Å = 1.202x10-10 m (from QM - QCISD/6-311G*)

8.002

O O O

O O

m m mamu

m m

2 7 2 61 . 6 6 1 0 / 1 . 3 2 8 1 0x x k g a m u x k g

2 2 6 1 0 2 4 6 2(1 . 3 2 8 1 0 ) (1 . 2 0 2 1 0 ) 1 . 9 2 1 0I r x k g x m x k g m

= 2 (Homonuclear Diatomic)

2342

2 2 46 2 23

6.63 102.08

8 8 (3.1416) 1.92 10 1.38 10 /R

x J shK

Ik x kg m x J K

Slide 79

O2 (Cont'd)

lnrot

R

TS nR R

2 .0 8R K

For one mole of O2 at 298 K:

Stran = 151.9 J/mol-K (from Chap. 3)

Stran + Srot = 195.7 J/mol-K

O2: Smol(exp) = 205.1 J/mol-K at 298.15 K

Thus, there is a small, but finite, vibrational (and/or) electronic contribution to the entropy at room temperature.

2981(8.31 / ) ln 8.31 /

(2)(2.08 )

43.8 /

rot KS J mol K J mol K

K

J mol K

Slide 80

E (Thermal) CV S KCAL/MOL CAL/MOL-K CAL/MOL-K TOTAL 3.750 5.023 48.972 ELECTRONIC 0.000 0.000 2.183 TRANSLATIONAL 0.889 2.981 36.321 ROTATIONAL 0.592 1.987 10.459 VIBRATIONAL 2.269 0.055 0.008 Q LOG10(Q) LN(Q) TOTAL BOT 0.330741D+08 7.519488 17.314260 TOTAL V=0 0.151654D+10 9.180853 21.139696 VIB (BOT) 0.218193D-01 -1.661159 -3.824960 VIB (V=0) 0.100048D+01 0.000207 0.000476 ELECTRONIC 0.300000D+01 0.477121 1.098612 TRANSLATIONAL 0.711178D+07 6.851978 15.777263 ROTATIONAL 0.710472D+02 1.851547 4.263345

Output from G-98 geom. opt. and frequency calculation on O2 (at 298 K)

QCISD/6-311G(d)

)2 9 8(/4 8.2/5 9 2.0 KR Tm o lk Jm o lk c a lUU r o tt h e r m

r o t

RKm o lJKm o lc a lC r o tV /3 1 4.8/9 8 7.1

Km o lJKm o lc a lS r o t /8.4 3/4 5 9.1 0 (same as our result)

Slide 81

Translational + Rotational Contributions to O2 Entropy

Note that the other (vibration and/or electronic) contributionsto S are even greater at higher temperature.

0 1000 2000 3000 4000 5000100

150

200

250

300

Expt T TR

S

[J/m

ol-K

]

Temperature [K]

Slide 82

Translational + Rotational Contributions to O2 Enthalpy

There are also significant additional contributionsto the Enthalpy.

0 1000 2000 3000 4000 5000

0

50

100

150

200

Expt T TR

H

[kJ/

mol

]

Temperature [K]

Slide 83

Translational + Rotational Contributions to O2 Heat Capacity

0 1000 2000 3000 4000 500015

20

25

30

35

40

45

Expt T TR

CP

[J/m

ol-K

]

Temperature [K]

Note that the additional (vibration and/or electronic) contributionsto CP are not important at room temperature, but very significantat elevated temperatures.

Slide 84

Helmholtz and Gibbs Energies

l nr o t r o tA k T Q ,

lnln

ln

rotrot rot

T N

QG kT Q kT

V

Qrot independent of V

r o tA

lnN

rot rot

R

TA G kT

ln

R

TNkT

ln

R

TnRT

For O2 at 298 K: R = 2.08 K

298ln (8.314 / ) ln 35.5 /

2 2.08rot rot

R

T KA G RT J mol K J mol K

K

Slide 85

Non-Linear Polyatomic Molecules

It can be shown that:

1/ 21/ 2 3rot

a b c

Tq

2

2

2

2

2

2

8

8

8

aa

bb

cc

hI k

hI k

hI k

The symmetry number is defined as the"number of pure rotational elements(including the identity) in the molecule's point group:NO2: = 2NH3: = 3CH4: = 12

I will always give you the value of for non-linear polyatomic molecules.

One can simply use the expression for qrot above in the sameway as for linear molecules to determine the rotational contributionsto the thermodynamic properties of non-linear polyatomic molecules(as illustrated in one of the homework problems).

Slide 1 Chapter 4 Rigid-Rotor Models and Angular Momentum Eigenstates.

Documents

topic slide

angular momentum eigenstates

rigidrotor models

mv energy orenergy slide

d schrdinger equation

rotational contributions

parallel vectors

perpendicular vectors