1 / 93

# MOLECULAR SYMMETRY AND SPECTROSCOPY - PowerPoint PPT Presentation

MOLECULAR SYMMETRY AND SPECTROSCOPY. [email protected] Download ppt file from. http://www.few.vu.nl/~rick. At bottom of page. We began by summarizing. Chapters 1 and 2. Spectroscopy and Quantum Mechanics. f. Absorption can only occur at resonance. h ν if = E f – E i = Δ E if.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'MOLECULAR SYMMETRY AND SPECTROSCOPY' - caleb-bauer

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

AND SPECTROSCOPY

http://www.few.vu.nl/~rick

At bottom of page

Chapters 1 and 2. Spectroscopy and Quantum Mechanics

f

Absorption can only occur at resonance

hνif = Ef – Ei = ΔEif

νif

i

M

Integrated absorption coefficient (i.e. intensity) for a line is:

______

8π3 Na

~

~

~

ε(ν)dν

=

I(f ← i) = ∫

F(Ei )

νif

Rstim(f→i)

S(f ← i)

(4πε0)3hc

line

Use Q. Mech. to calculate:

ODME of H

μfi = ∫ (Ψf )* μAΨi dτ

ODME of μA

Fundamentals of Molecular Symmetry,

Taylor and Francis, 2004.

The first 47 pages:

Chapter 1 (Spectroscopy)

Chapter 2 (Quantum Mechanics) and

Section 3.1 (The breakdown of the BO Approx.)

P. R. Bunker and Per Jensen:

Molecular Symmetry and Spectroscopy,

2nd Edition, 3rd Printing,

NRC Research Press, Ottawa, 2012.

http://www.crcpress.com

www.chem.uni-wuppertal.de/prb

Fundamentals of Molecular Symmetry,

Taylor and Francis, 2004.

The first 47 pages:

Chapter 1 (Spectroscopy)

Chapter 2 (Quantum Mechanics) and

Section 3.1 (The breakdown of the BO Approx.)

P. R. Bunker and Per Jensen:

Molecular Symmetry and Spectroscopy,

2nd Edition, 3rd Printing,

NRC Research Press, Ottawa, 2012.

http://www.crcpress.com

www.chem.uni-wuppertal.de/prb

We then proceeded to discuss Group

Theory and Point Groups

“Group”

A set of operations that is closed wrt “multiplication”

“Point Group”

All rotation, reflection and rotation-reflection operations

that leave the molecule (in its equilibrium configuration)

“looking” the same.

“Matrix group”

A set of matrices that forms a group.

“Representation”

A matrix group having the same shaped multiplication

table as the group it represents.

“Irreducible representation”

A representation that cannot be written as the sum

of smaller dimensioned representations.

“Character table”

A tabulation of the characters of the irreducible representations.

E C3σ1

C32σ2

σ3

Two 1D

irreducible

representations

of the C3v group

The 2D representation M = {M1, M2, M3, ....., M6}

of C3v is the irreducible representation E. In this

table we give the characters of the matrices.

Elements in the same class have the same characters

3 classes and 3 irreducible representations

x

EC2σyzσxy

(+y)

z

4 classes and 4 irreducible representations

E C3σ1

C32σ2

σ3

C3v

3

2

1

Spectroscopy

M

f

hνif = Ef – Ei = ΔEif

MMMM

i

S(f ← i) = ∑A | ∫ Φf* μAΦi dτ |2

Quantum Mechanics

ODME of H and μA

μfi =

∫ Φf* μAΦi dτ

Group Theory and Point Groups

(Character Tables and Irreducible Representations)

PH3

8

the geometrical symmetry of the

equilibrium structure.

Point group symmetry not appropriate

when there is rotation or tunneling

Use energy invariance symmetry

instead. We start by using inversion

symmetry and identical nuclear

permutation symmetry.

Inversion (CNPI) Group

Contains all possible permutations

of identical nuclei including E. It also

contains the inversion operation E*

and all possible products of E* with

the identical nuclear permutations.

GCNPI = GCNP x {E,E*}

(after separating translation)

Vee + VNN + VNe

THE GLUE

In a world of infinitely powerful computers we could solve the Sch. equation numerically and that would be that. However, we usually have to start by making approximations. We then

selectively correct for the approximations made.

The CNPI Group for the Water Molecule

The Complete Nuclear Permutation Inversion (CNPI) group

for the water molecule is {E, (12)} x {E,E*} = {E, (12), E*, (12)*}

+

+

e

e

H2

H1

O

O

E*

(12)

-

e

O

H2

H1

H2

H1

(12)*

Nuclear permutations permute nuclei (coordinates and spins).

Do not change electron coordinates

E* Inverts coordinates of nuclei and electrons.

Does not change spins.

Same CNPI group for CO2, H2, H2CO, HOOD, HDCCl2,…

H

N1N2N3

1

C1

F

3

C2

2

I

C3

O

F

D

2

O

O

1

3

1

3

H H

12C 13C

D H

2

1

2

H

1

+

H

H

3

2

3

GCNPI = {E, (12), (13), (23), (123), (132)}x {E, E*}

= GCNP x {E, E*}

GCNPI = {E, (12), (13), (23), (123), (132)}x {E, E*}

GCNPI={E, (12), (13), (23), (123), (132),

E*, (12)*, (13)*,(23)*, (123)*, (132)*}

Number of elements = 3! x 2 = 6 x 2 = 12

Number of ways of permuting

three identical nuclei

H5

C1

H4

C2

I

C3

D

The CNPI Group of C3H2ID

GCNPI = {E, (12), (13), (23), (123), (132)}

x{E, (45)}x {E, E*}

= {E, (12), (13), (23), (123), (132),

(45),(12)(45), (13)(45), (23)(45), (123)(45), (132)(45),

E*, (12)*, (13)*, (23)*, (123)*, (132)*,

(45)*,(12)(45)*, (13)(45)*, (23)(45)*, (123)(45)*,

(132)(45)*}

Number of elements = 3! x 2! x 2 = 6 x 2 x 2 = 24

H5

C1

H4

C2

I

C3

D

Number of elements

= 3! x 2! x 2 = 6 x 2 x 2 = 24

If there are n1 nuclei of type 1, n2 of

type 2, n3 of type 3, etc then the total

number of elements in the CNPI group

is n1! x n2! x n3!... x 2.

H5

The

Allene molecule

C1

H4

C3H4

C2

H7

C3

H6

Number of elements = 3! x 4! x 2 = 6 x 24 x 2 = 288

H5

The

Allene molecule

C1

H4

C3H4

C2

H7

C3

H6

Number of elements = 3! x 4! x 2 = 6 x 24 x 2 = 288

Sample elements: (456), (12)(567), (4567), (45)(67)(123)

H5

00H

The

Allene molecule

C1

H4

C3H4

C2

H7

C3

H6

00H

Number of elements = 3! x 4! x 2 = 6 x 24 x 2 = 288

How many elements?

C3H4O4

H5

00H

The

Allene molecule

C1

H4

C3H4

C2

H7

C3

H6

00H

Number of elements = 3! x 4! x 2 = 6 x 24 x 2 = 288

C3H4O4

3! x 4! x 4! x 2 = 6912

only on the chemical formula

Number of elements in the CNPI groups of various

molecules

(C6H6)2 12! x 12! x 2 ≈ 4.6 x 1017

Just need the chemical formula to

determine the CNPI group. Can be BIG

Molecule PG h(PG) h(CNPIG) h(CNPIG)/h(PG)

H2O C2v 4 2!x2=4 1

PH3 C3v 6 3!x2=12 2

Allene D2d 8 4!x3!x2=288 36

C3H4

Benzene D6h 24 6!x6!x2=1036800 43200

C6H6

This number means something!

End of Review of Lecture One

22

Symmetry operations are operations that leave the energy of the system (a molecule in our case) unchanged.

Using quantum mechanics:

A symmetry operation is an operation that

commutes with the Hamiltonian:

RHn = HRn

group of the water molecule

(12) E* 1 1

1 -1

-1 -1

-1 1

(12)*

1

-1

1

-1

E 1

1

1

1

A1

A2

B1

B2

It is called C2v(M)

group of the water molecule

(12) E* 1 1

1 -1

-1 -1

-1 1

(12)*

1

-1

1

-1

E 1

1

1

1

A1

A2

B1

B2

It is called C2v(M)

Now to explain how we label

energy levels using

irreducible representations

For the water molecule (no degeneracies, and R2 = identity for all R) :

H = E

RH = RE

Since RH = HR and E is a number, this leads to HR = ER.

H(R) = E(R)

E is nondegenerate. Thus RΨ = cΨ.

But R2 = identity. Thus c2 = 1, so c = ±1 and R = ±

R = (12), E* or (12)*

The eigenfunctions have symmetry

+ Parity

- Parity

Ψ1+(x)

Ψ2-(x)

x

x

Ψ-(-x) = -Ψ-(x)

Ψ3+(x)

Eigenfunctions of H

must satisfy

E*Ψ = ±Ψ

x

Ψ+(-x) = Ψ+(x)

E*ψ(xi) = ψE*(xi), a new function.

ψE*(xi) = ψ(E*xi) = ψ(-xi) = ±ψ(xi)

Since E*ψ(xi) can only be ±ψ(xi)

This is different from Wigner’s approach

See PRB and Howard (1983)

- Parity

Ψ1+(x)

Ψ2-(x)

x

x

Ψ-(-x) = -Ψ-(x)

Ψ3+(x)

Eigenfunctions of H

must satisfy

E*Ψ = ±Ψ

x

Ψ+(-x) = Ψ+(x)

and (12)ψ = ±ψ

There are four symmetry types of H2O wavefunction

(12) E* 1 1

1 -1

-1 -1

-1 1

E 1

1

1

1

(12)*

1

-1

1

-1

R = ±

A1

A2

B1

B2

A2 x B1 = B2, B1 x B2 = A2, B1 x A2 x B2 = A1

(12) E* 1 1

1 -1

-1 -1

-1 1

E 1

1

1

1

(12)*

1

-1

1

-1

A1

A2

B1

B2

We are labelling the states using

the irreps of the CNPI group

∫ΨaHΨbdτ = 0 if symmetries of Ψa and Ψb are different.

∫ΨaμΨbdτ = 0 if symmetry of product is not A1

A2 x B1 = B2, B1 x B2 = A2, B1 x A2 x B2 = A1

(12) E* 1 1

1 -1

-1 -1

-1 1

E 1

1

1

1

(12)*

1

-1

1

-1

A1

A2

B1

B2

Thus, for example, a wavefunction of “A2 symmetry”

will “generate” the A2 representation:

∫ΨaHΨbdτ = 0 if symmetries of Ψa and Ψb are different.

∫ΨaμΨbdτ = 0 if symmetry of product is not A1

Eψ=+1ψ(12)ψ=+1ψE*ψ=-1ψ(12)*ψ=-1ψ

(12) E* 1 1

1 -1

-1 -1

-1 1

E 1

1

1

1

(12)*

1

-1

1

-1

A1

A2

B1

B2

Thus, for example, a wavefunction of “A2 symmetry”

will “generate” the A2 representation:

∫ΨaHΨbdτ = 0 if symmetries of Ψa and Ψb are different.

∫ΨaμΨbdτ = 0 if symmetry of product is not A1

Eψ=+1ψ(12)ψ=+1ψE*ψ=-1ψ(12)*ψ=-1ψ

For the water molecule we can, therefore label

the energy levels as being A1, A2, B1 or B2

using the irreps of the CNPI group.

The vibrational wavefunction for the v3 = 1

state of the water molecule can be written

approximately as ψ = N(Δr1 – Δr2).

Eψ=+1ψ(12)ψ=-1ψE*ψ=+1ψ(12)*ψ=-1ψ

This would be labelled as B2.

c = ε, ε2 (=ε*), or ε3 (=1)

If n = 3,

Suppose Rn = E where n > 2.

We still have RΨ = cΨ for nondegenerate Ψ, but now RnΨ = Ψ.

Thus cn = 1 and c = n√1, i.e. c = [ei2π/n]a where a = 1,2,…,n.

where  = ei2/3

C3 C32

eiπ = -1

ei2π = 1

For nondegenerate states we hadthis as the effect of a symmetry operation on an eigenfunction:

For the water molecule ( nondegenerate) :

H = E

RH = RE

HR = ER

Thus R = c since E is nondegenerate.

-fold

degenerate energy level with energy En

RΨnk=D[R ]k1Ψn1 +D[R ]k2Ψn2 +D[R ]k3Ψn3 +…+D[R ]kℓΨnℓ

For each relevant symmetry operation R, the constants

D[R ]kp form the elements of an ℓℓ matrix D[R ].

ForT = RS it is straightforward to show that

D[T ] = D[R ] D[S ]

The matrices D[T ],D[R ], D[S ]….. form an ℓ-dimensional representation that is generated by the ℓ functions Ψnk

The ℓ functions Ψnktransform according to this representation

We can label energy levels using

the irreps of the CNPI group for any molecule

∫ΨaHΨbdτ = 0 if symmetries of Ψa and Ψb are different.

∫ΨaμΨbdτ = 0 if symmetry of product is not A1

Pages 143-149

Pages 99-101

34

reducing a representation

Example of using the symmetry operation (12):

(12)

r1´

r2´

´

H1

H2

We have (12) (r1, r2, ) = (r1´, r2´, ´)

We see that (r1´, r2´, ´) = (r2, r1, )

r

r2´

r2

r2

r2

r2

r1

r1

r1

r1

r2´

r1´

´

´

3

3

3

3

3

1

2

2

2

2

2

1

1

1

1

3

1

2

2

1

3

2

1

3

E

(12)

´

E*

r1´

r2´

(12)*

´

r2´

r1´

R a = a´ = D[R] a

E

 = 3

 = 1

(12)

 = 3

E*

(12)*

 = 1

aA1 = ( 13 + 11 + 13 + 11) = 2

aA2 = ( 13 + 11  13  11) = 0

aB1 = ( 13  11  13 + 11) = 0

aB2 = ( 13  11 + 13  11) = 1

Γ = Σ aiΓi

i

A reducible representation

 = 2A1 B2

aA1 = ( 13 + 11 + 13 + 11) = 2

aA2 = ( 13 + 11  13  11) = 0

aB1 = ( 13  11  13 + 11) = 0

aB2 = ( 13  11 + 13  11) = 1

Γ = Σ aiΓi

i

i

A reducible representation

 = 2A1 B2

aA1 = ( 13 + 11 + 13 + 11) = 2

aA2 = ( 13 + 11  13  11) = 0

aB1 = ( 13  11  13 + 11) = 0

aB2 = ( 13  11 + 13  11) = 1

Γ = Σ aiΓi

i

i

A reducible representation

 = 2A1 B2

aA1 = ( 13 + 11 + 13 + 11) = 2

aA2 = ( 13 + 11  13  11) = 0

aB1 = ( 13  11  13 + 11) = 0

aB2 = ( 13  11 + 13  11) = 1

Γ = Σ aiΓi

i

i

A reducible representation

 = 2A1 B2

We know now that r1, r2, andgenerate the

representation 2A1 B2

Consequently, we can generate from r1, r2, and three „symmetrized“ coordinates:

S1 with A1 symmetry

S2 with A1 symmetry

S3 with B2 symmetry

For this, we need projection operators

General for li-dimensional irrep i

Diagonal element of representation matrix

Symmetry operation

Simpler for 1-dimensional irrep i

Character

1

General for li-dimensional irrep i

Simpler for 1-dimensional irrep i

Character

Diagonal element of representation matrix

1

Symmetry operation

General for li-dimensional irrep i

Simpler for 1-dimensional irrep i

Character

Diagonal element of representation matrix

1

Symmetry operation

E (12) E* (12)*

A1 1 1 1 1

PA1 = (1/4) [ E + (12) + E* + (12)* ]

S1 = P11A1r1= [ E + (12) + E*+ (12)* ]r1

S3 = P11B2r1= [ E  (12) + E* (12)*] r1

= [ r1  r2 + r1 r2 ] = [ r1  r2 ]

= [ r1 + r2 + r1 + r2 ] = [ r1 + r2 ]

S2 = P11A1= [ E + (12) + E*+ (12)* ]

P11B2= [ E  (12) + E* (12)* ]

= [  +  + +  ] = 

= [   +   ] = 0

Projection operator for A1 acting on r1

PA1

PA1

PB2

PB2

 Is „annihilated“ by P11B2

PB2

S1 = P11A1r1= [ E + (12) + E*+ (12)* ]r1

S3 = P11B2r1= [ E  (12) + E* (12)*] r1

= [ r1  r2 + r1 r2 ] = [ r1  r2 ]

= [ r1 + r2 + r1 + r2 ] = [ r1 + r2 ]

S2 = P11A1= [ E + (12) + E*+ (12)* ]

P11B2= [ E  (12) + E* (12)* ]

= [  +  + +  ] = 

= [   +   ] = 0

Projection operators for A1 and B2

PA1

PA1

PB2

PB2

 Is „annihilated“ by P11B2

PB2

S1 = P11A1r1= [ E + (12) + E*+ (12)* ]r1

S3 = P11B2r1= [ E  (12) + E* (12)*] r1

= [ r1  r2 + r1 r2 ] = [ r1  r2 ]

= [ r1 + r2 + r1 + r2 ] = [ r1 + r2 ]

S2 = P11A1= [ E + (12) + E*+ (12)* ]

= [  +  + +  ] = 

Projection operators for A1 and B2

PA1

PA1

PB2

Aside: S1, S2 and S3 have the symmetry and form of the

normal coordinates.

A1

A1

The three

Normal modes

of the water

molecule

B2

The vanishing integral theorem

Pages 136-139

Pages 114-117

But first we look at

The symmetry of a product

Pages 109-114

1 -1

-1 -1

-1 1

THE SYMMETRY OF A PRODUCT

(12)*

1

-1

1

-1

E 1

1

1

1

A1

A2

B1

B2

A2

Eψ=+1ψ(12)ψ=+1ψE*ψ=-1ψ(12)*ψ=-1ψ

∫ΨaHΨbdτ = 0 if symmetries of Ψa and Ψb are different.

B1

Eφ=+1φ(12)φ=-1φE*φ=-1φ(12)*φ=+1φ

∫ΨaμΨbdτ = 0 if symmetry of product is not A1

The symmetry of the product φψis B1 x A2 = B2.

A2 x B1 = B2, B1 x B2 = A2, B1 x A2 x B2 = A1

B1 x B2, A1 x A2, B1 x A2, B2 x A2, B1 x B1,…

A2 A2 B2 B1 A1

A1 A1 = A1

A1 A2 = A2

A2 A2 = A1

A1 E= E

A2 E= E

E E= A1  A2  E

E E: 4 1 0

Reducible representation

Characters of the product representation are the products

of the characters of the representations being multiplied.

Symmetry of triple product is obvious extension

- Parity

Ψ+(x)

Ψ-(x)

x

x

Ψ-(-x) = -Ψ-(x)

Ψ+(x)

∫Ψ+Ψ-Ψ+dx = 0

x

- parity

Ψ+(-x) = Ψ+(x)

- Parity

Ψ+(x)

Ψ-(x)

The vanishing integral theorem

x

x

Ψ-(-x) = -Ψ-(x)

Ψ+(x)

∫f(τ)dτ = 0 if symmetry of f(τ) is not A1

∫Ψ+Ψ-Ψ+dx = 0

x

- parity

Ψ+(-x) = Ψ+(x)

1 -1

-1 -1

-1 1

USING THE VANISHING

INTEGRAL THEOREM

E 1

1

1

1

(12)*

1

-1

1

-1

Symmetry of H

A1

A2

B1

B2

Using symmetry labels and the vanishing

integral theorem we deduce that:

∫Ψa*HΨbdτ = 0 if symmetry of Ψa*HΨb is not A1,

Integral vanishes if Ψa and Ψb have different symmetries

vanishes if symmetries not the same

This means that we can

“block-diagonalize” the

Hamiltonian matrix

A1 A2 B1

ψ1ψ2ψ3ψ4ψ5ψ6ψ7ψ8

0

0

0

0

0

0

0

0

. . .

. . .

. . .

0

Ψ1

Ψ2

Ψ3

Ψ4

Ψ5

Ψ6

Ψ7

ψ8

0

0

A1

A2

B1

0

0

. . .

. . .

. . .

0

0

0

0

0

0

. .

. .

0

0

0

Symmetry is preserved on diagonalization

ODME of μA

μfi = ∫ (Ψf )* μAΨi dτ

Can use symmetry to determine if this ODME = 0

This ODME will vanish if the symmetry

of (Ψf)* μAΨi is not A1

0

0

1 -1

-1 -1

-1 1

What is the symmetry ofμA ?

(12)*

1

-1

1

-1

E 1

1

1

1

A1

A2

B1

B2

μA = ΣCre Ar

r

Charge on

particle r

A coordinate

of particle r

A = space-fixed X, Y or Z

EμA= ?μA (12)μA = ?μA E*μA= ?μA (12)*μA= ?μA

1 -1

-1 -1

-1 1

What is the symmetry ofμZ ?

(12)*

1

-1

1

-1

E 1

1

1

1

A1

A2

B1

B2

μA = ΣCre Ar

r

Charge on

particle r

A coordinate

of particle r

A = space-fixed X, Y or Z

EμA= +1μA (12)μA = +1μA E*μA= -1μA (12)*μA= -1μA

μA has symmetry A2

(12) E* 1 1

1 -1

-1 -1

-1 1

Γ(H) = A1

E 1

1

1

1

(12)*

1

-1

1

-1

Symmetry of H

R = ±

A1

A2

B1

B2

Symmetry of μA

Γ(μA) = A2

Using symmetry labels and the vanishing

integral theorem we deduce that:

∫Ψa*HΨbdτ = 0 if symmetry of Ψa*HΨb is not A1,

∫Ψa*μAΨbdτ = 0 if symmetry of ψa*μAψb is not A1,

(12) E* 1 1

1 -1

-1 -1

-1 1

Γ(H) = A1

E 1

1

1

1

(12)*

1

-1

1

-1

Symmetry of H

R = ±

A1

A2

B1

B2

Symmetry of μA

Γ(μA) = A2

Using symmetry labels and the vanishing

integral theorem we deduce that:

∫Ψa*HΨbdτ = 0 if symmetry of Ψa*HΨb is not A1,

that is, if the symmetry of

Ψa is not the same as Ψb

Pages 113-114

∫Ψa*μAΨbdτ = 0 if symmetry of ψa*μAψb is not A1,

that is, if the symmetry of the product ΨaΨbis not A2

b

JKaKc

a

c

Γrot

KaKc

e e A1

o o A2

e o B1

o e B2

Allowed transitions

65

Ch. 7

Ch. 6

There are problems with the CNPI Group

Number of elements in the CNPI groups of various

molecules

Huge groups. Size bears no relation to geometrical symmetry

C6H6, for example, has a 1036800-element CNPI group,

but a 24-element point group at equilibrium, D6h

Often gives SUPERFLUOUS multiple symmetry labels

PH3

3

2

1

There are two VERSIONS

of this molecule

PH3

TWO VERSIONS: Distinguished

by numbering the identical nuclei

Very, very

high

potential

barrier

2

~12000 cm-1

3

1

1

2

3

Bone et al., Mol. Phys., 72, 33 (1991)

(order of CNPI group)/(order of point group)

For H2O this is 4/4 = 1

H

For H3+ this is 12/12= 1

C1

F

C2

For PH3 or CH3F this is 12/6 = 2

I

C3

For O3 this is 12/4 = 3

D

12/1 = 12

For HN3 this is 12/2 = 6

2 1 3

2 3 1

3 1 2

3 2 1

The six versions of HN3

H

N1 N2 N3

H

C1

F

C2

I

C3

D

12 versions

(order of CNPI group)/(order of point group)

For H2O this is 4/4 = 1

H

For H3+ this is 12/12= 1

C1

F

C2

For PH3 or CH3F this is 12/6 = 2

I

C3

For O3 this is 12/4 = 3

D

12/1 = 12

For HN3 this is 12/2 = 6

(order of CNPI group)/(order of point group)

For H2O this is 4/4 = 1

H

For H3+ this is 12/12= 1

C1

F

C2

For PH3 or CH3F this is 12/6 = 2

I

C3

For O3 this is 12/4 = 3

D

12/1 = 12

For HN3 this is 12/2 = 6

For C6H6 this is (6!x6!x2)/24

= 1036800/24 = 43200

label the energy levels of a molecule

that has more than one version.

Character Table of CNPI group of PH3

E (123) (23) E* (123)* (23)*

(132) (31) (132)* (31)*

(23) (23)*

GCNPI

GCNPI

12 elements

6 classes

6 irred. reps

PH3 (or CH3F)

Using the CNPI Group

A1’

E’ + E’’

A1’’

A1’’ + A2’

A2’

A2’’

A1’ + A2’’

E’

E’’

PH3 (or CH3F)

Using the CNPI Group

A1’

E’ + E’’

A1’’

A1’’ + A2’

A2’

A2’’

A1’ + A2’’

E’

Why this

Degeneracy?

E’’

are two versions of the PH3 molecule and the

tunneling splitting between these versions is not

observed.

For understanding the spectrum

this is a “superfluous” degeneracy

(particularly for CH3F)

(order of CNPI group)/(order of point group)

For H2O this is 4/4 = 1

For PH3 or CH3F this is 12/6 = 2

For O3 this is 12/4 = 3

For C3H4 this is 288/8 = 36

For C6H6 this is 1036800/24 = 43200,

and using the CNPI group each energy level would

get as symmetry label the sum of 43200 irreps.

Clearly using the CNPI group gives

very unwieldy symmetry labels.

77

works, in principle, and can be used

to determine which ODME vanish.

BUT IT IS OFTEN HOPELESSLY UNWIELDY

In 1963 Longuet-Higgins figured out how to set up a sub-group of the CNPI Group that achieves the same result

without superfluous degeneracies.

This subgroup is called

The Molecular Symmetry (MS) group.

PH3 (or CH3F)

Ab initio calc with neglect of tunneling

Ab Initio CALC IN HERE

Very, very

high

potential

barrier

2

3

1

1

2

3

It would be

superfluous

to calc points

in other min

No observed tunneling through barrier

PH3 (or CH3F)

Only NPI OPERATIONS FROM IN HERE

Very, very

high

potential

barrier

2

3

1

1

2

3

No observed tunneling through barrier

PH3 (or CH3F)

Only NPI OPERATIONS FROM IN HERE

Very, very

high

potential

barrier

2

3

1

1

2

3

(12) superfluous

E* superfluous

(123), (12)* useful

No observed tunneling through barrier

GMS ={E, (123), (132), (12)*, (13)*,(23)*}

If we cannot see any effects

of the tunneling through the

barrier then we only need

NPI operations for one

version. Omit NPI elements

that connect versions since

they are not useful; they are

superfluous.

superfluous

useful

superfluous

GCNPI={E,(12), (13), (23), (123), (132),

E*, (12)*, (13)*,(23)*, (123)*, (132)*}

For CH3F:

useful

GMS ={E, (123), (132), (12)*, (13)*,(23)*}

If we cannot see any effects

of the tunneling through the

barrier then we only need

NPI operations or one

version. Omit NPI elements

that take us between

versions since they are not

useful; they are unfeasible.

If we cannot see any effects

of the tunneling through the

barrier then we only need

NPI operations for one

version. Omit NPI elements

that connect versions since

they are not useful; they are

unfeasible.

superfluous

For PH3

or CH3F:

GCNPI={E,(12), (13), (23), (123), (132),

E*, (12)*, (13)*,(23)*, (123)*, (132)*}

Character Table of CNPI group of PH3 or CH3F

GCNPI={E, (12), (13), (23), (123), (132),

E*, (12)*, (13)*,(23)*, (123)*, (132)*}

H

A1’

E (123) (23) E* (123)* (23)*

(132) (31) (132)* (31)*

(23) (23)*

GCNPI

GCNPI

12 elements

μA

6 irred. reps

A1”

E’ + E’’

A1’

E’ + E’’

A1’’

A1’’ + A2’

A2’

A2’’

A1’ + A2’’

E’

E’’

2 versions → sum or two irrep labels on each level

E (123) (12)*

(132) (13)*

(23)*

H

A1

μA

A2

E

E

A1

A2

A2

E

A1

The MS group gives all the information we need

as long as there is no observable tunneling, i.e.,

as long as the barrier is insuperable.

E’ + E’’

E

E’ + E’’

E

A1’’ + A2’

A2

A1’ + A2’’

A1

CNPIG

MSG

Can use either to determine if an ODME vanishes.

But clearly it is easier to use the MSG.

Unfeasible elements of the CNPI group

interconvert versions that are separated

by an insuperable energy barrier

useful

The subgroup of feasible elements forms a group called

THE MOLECULAR SYMMETRY GROUP

(MS GROUP)

89