Lecture 38: Selection Rules The material in this lecture covers the following in Atkins

Lecture 38: Selection Rules The material in this lecture covers the following in Atkins.15 Molecular Symmetry 15.5 Vanishing integrals and orbital overlaps (a) The criteria for vanishing integrals (b) Orbitals with nonzero overlaps (c) Symmetry-adapted linear combinations 15.6 Vanishing integrals and selection rules

Lecture on-line Selection Rules (PowerPoint) Selection Rules (PDF) Handouts for this lecture

Thus, the three mirror planesshown here are related by threefoldrotations : C3v=v’

Character Table

Symmetry operations in the same class are related to one another by the symmetry operations of the group.

and the two rotations shown here are related by reflection in v.vC3=C3

-1

C3v

The dimension is 6 since we have 6 elements.

We have three different symmetryrepresentations as we have threedifferent classes of symmetry elements

Character Table C3v

The pz orbitaldoes not change

with E, C3, C3-1

σv, σ'v ,σ"vThe symmetryrep. is A1

px pydoes change

with E, C3, C3-1

σv, σ'v ,σ"v

X

Y

X

Y

Character Table C3v

X

Y

X

Y

X

Y

px p'x p''x

Epx =px ; C3px =p'x; C3-1 px =p"x

X

Y

X

Y Y

Character Table C3v

Epy =py ; C3py =p'y; C3-1 py =p"y

py p'y p''y

px py( )D(C3)= px py( )−

12

32

3

2−

1

2

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

px py( )D(C3−1 )= px py( )

−12

−32

−3

2−

1

2

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

The trace is -1 forboth matrices

px py( )D(E)= px py( )1 0

0 1

⎛

⎝ ⎜

⎞

⎠ ⎟

The trace is 2which is also thedimension ofthe representation

Character Table C3v

px py( )D(σv)= px py( )−1 0

0 1

⎛

⎝ ⎜

⎞

⎠ ⎟

px py( )D(σv' )= px py( )

12

−32

−3

2−

1

2

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

px py( )D(σv" )= px py( )

12

32

3

2−

1

2

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

The trace is -1 forboth matrices

Character Table C3v

Typical symmetry-adapted linear combinations of orbitals in aC 3v molecule.

Character Table Constructing Linear combinationsHow are they constructed

a1

a2

ex

ey

C3v Original basis SA SB SC E SA SB SCC3

+ SB SC SA

C3− SC SA SB

σv SA SC SBσv' SB SA SCσv" SC SB SA

Character Table

Construct a table showing the effect of each operation on each orbitalof the original basis

To generate the combination of aSpecific symmetry species, takeEach column in turn and

(I) Multiply each member of theColumn by the character of the Corresponding operator

SA

SB

SC

Constructing Linear combinations

C3v

Original basis SA SB SC E 1SA 1SB 1SCC3

+ 1SB 1SC 1SA

C3− 1SC 1SA 1SB

σv 1SA 1SC 1SBσv' 1SB 1SA 1SCσv" 1SC 1SB 1SA

Character Table

(I) Multiply each member of theColumn by the character of the Corresponding operator

(2) Add and divide by group order

ψa1 =13

(SA +SB +SC)

For A1 :

SA

SB

SC

Constructing Linear combinations

C3v

Original basis SA SB SC E 1SA 1SB 1SCC3

+ 1SB 1SC 1SA

C3− 1SC 1SA 1SB

σv -1SA -1SC -1SBσv' -1SB -1SA -1SCσv" -1SC -1SB -1SA

Character Table

(I) Multiply each member of theColumn by the character of the Corresponding operator

(2) Add and divide by group order

ψa2 =0

For A2 :

SA

SB

SCConstructing Linear combinations

C3v

Original basis SA SB SC E 2SA 2SB 2SCC3

+ -1SB -1SC -1SA

C3− -1SC -1SA -1SB

σv 0 SA 0 SC 0 SBσv' 0 SB 0 SA 0 SCσv" 0 SC 0 SB 0 SA

Character Table

(I) Multiply each member of theColumn by the character of the Corresponding operator

(2) Add and divide by group order

ψ1=16

(2SA −SB −SC)

For E :

SA

SB

SC

ψ2 =16

(2SB −SA −SC)

ψ3 =16

(2SC −SA −SB)

Constructing Linear combinations

Character Table Constructing Linear combinations

ψ1=16

(2SA −SB −SC) ψ2 =16

(2SB −SA −SC)

ψ3 =16

(2SC −SA −SB)

C3v

Any of the functions ψ1,ψ2, and ψ3 can be expressedin terms of the two other (linear dependence)

Linear independent solutions :

1. Any of the functions ψ1=ψI16

(2SA −SB −SC)

2. Difference of other two : ψII =12

(SA −SB)

C3v

Molecular orbitals of NH3

a1

ex ey

Character Table

a1

a1

ex

ex

ey

ey

Orbitals of the same symmetry species may havenon-vanishing overlap.

Character Table Structure of character table

This diagram illustrates the three bonding orbitals that may be constructed from (N2s, H1s) and (N2p, H1s) overlap in a C3v molecule.

(a) a1;

(b) and (c) the two components of the doubly degenerate e orbitals. (There are also three antibonding orbitals of the same species.)

C3v

Original basis 2pA 2pB 2pC E 2pA 2pB 2pCC3

+ 2pB 2pC 2pAC3

− 2pC 2pA 2pBσv -2pA -2pC -2pBσv' -2pB -2pA -2pCσv" -2pC -2pB -2pA

Character Table

Construct a table showing the effect of each operation on each orbitalof the original basis

To generate the combination of aSpecific symmetry species, takeEach column in turn and

(I) Multiply each member of theColumn by the character of the Corresponding operator

Constructing Linear combinations+

+

+

-

-

-

2pA

2pB

2pC

C3v

Character Table

(I) Multiply each member of theColumn by the character of the Corresponding operator

(2) Add and divide by group order

ψa1=

16(2pA +2pB +2pC −

2pA −2pB −2pC)=0

For A1 :

Constructing Linear combinations+

+

+

-

-

-

2pA

2pB

2pC

Original basis 2pA 2pB 2pC E 2pA 2pB 2pCC3

+ 2pB 2pC 2pAC3

− 2pC 2pA 2pBσv -2pA -2pC -2pBσv' -2pB -2pA -2pCσv" -2pC -2pB -2pA

C3v

Character Table

(I) Multiply each member of theColumn by the character of the Corresponding operator

(2) Add and divide by group order

ψa2=

16(2pA +2pB +2pC +

2pA +2pB +2pC)=

13(2pA +2pB +2pC)

For A2 :

Constructing Linear combinations+

+

+

-

-

-

2pA

2pB

2pC

Original basis 2pA 2pB 2pC E 2pA 2pB 2pCC3

+ 2pB 2pC 2pAC3

− 2pC 2pA 2pBσv 2pA 2pC 2pBσv' 2pB 2pA 2pCσv" 2pC 2pB 2pA

C3v

Character Table

(I) Multiply each member of theColumn by the character of the Corresponding operator

(2) Add and divide by group order

ψe1 =16(2(2pA)−2pB −2pC)

For E :

Constructing Linear combinations+

+

+

-

-

-

2pA

2pB

2pC

Original basis 2pA 2pB 2pC E 2pA 2pB 2pCC3

+ -2pB -2pC -2pAC3

− -2pC -2pA -2pBσv 0 0 0σv' 0 0 0σv" 0 0 0

ψe2 =16(2(2pB)−2pA −2pC)

ψe3 =16(2(2pC)−2pA −2pB)

Character Table Constructing Linear combinations

ψ1=16(2(2pA)−2pB −2pC) ψ2=

16(2(2pB)−2pA −2pC)

ψ3=16(2(2pC)−2pA −2pB)

C3vAny of the functions ψ1,ψ2, and ψ3 can be expressedin terms of the two other (linear dependence)

Linear independent solutions :

1. Any of the functions ψ1=ψI16(2(2pA)−2pB −2pC)

2. Difference of other two : ψII =12(2pA −2pB)

++

--

2pA

2pB

2pC

+

+

+

-

-

-

2pA

2pB

2pC

+

+

+

-

-

-

2pA

2pB

2pC

Character Table Vanishing integrals and selection rules

We have that an expectation value withregard to the operator ˆ O

is given by : < ˆ O > = ψˆ O ψ∫ dv

For ˆ O =1 we have <1> = ψψ∫ dv=1

Thus ψψ must transform as A1

Thus for < ˆ O > = ψˆ O ψ∫ dv to be different from zero

ˆ O must transform as A1

Expectation value

Character Table Vanishing integrals and selection rules The dipole operator is given by:

r μ =e[ x

r e x + y

r e y + z

r e z]

For C1 x,y, and z all belong to totally symmetric rep

I

F

ClBr

μ≠0

For Cs x,y, belong to totally symmetric rep

N

Quinolineμ≠0in plane

For Ci x,y,and z do not belong to totally symmetric rep

C

C

Cl H

H Cl

Trans CHCl=CHClμ=0inversion

Character Table Vanishing integrals and selection rules

The dipole operator is given by: r μ =e[ x

r e x + y

r e y + z

r e z]

For cnh all components different from totally symmetric rep

For Cn and Cnv the z components belongs to totally symmetric rep

Character Table Vanishing integrals and selection rules

The dipole operator is given by: r μ =e[ x

r e x + y

r e y + z

r e z]

For higher symmetries : TdOh Dnd Dnh

No dipole

1a1

1b2

1b1

2a1

2b2

Character Table Vanishing integrals and selection rules Optical transitions

The intensity of the opticaltransition of electron fromorbital φ1 to φ2 isrelated to the transitiondipole integral:

.. and thetransition 1b1→ 2a1

Let us consider H2O

<r μ >= φ1∫

r μ φ1dv

= r e x φ1∫ x φ1dv

+ r e y φ1∫ y φ1dv

+r e z φ1∫ z φ1dv

Character Table Vanishing integrals and selection rules Optical transitions

<r μ >= φ1∫

r μ φ1dv

= r e x φ1∫ x φ1dv

+ r e y φ1∫ y φ1dv

+r e z φ1∫ z φ1dv

X−component

E C2 σv σv'

ψ1 [1b1] 1 -1 1 -1ψ2 [2a1] 1 1 1 1x 1 -1 1 -1ψ1xψ2 1 1 1 1

y−component

E C2 σv σv'

ψ1 [1b1] 1 -1 1 -1ψ2 [2a1] 1 1 1 1y 1 -1 -1 1ψ1yψ2 1 1 -1 -1 y−component

E C2 σv σv'

ψ1 [1b1] 1 -1 1 -1ψ2 [2a1] 1 1 1 1z 1 -1 1 1ψ1yψ2 1 -1 1 -1

The polarizations of the allowed transitions in a C2v molecule.The shading indicates the structure of the orbitals of the specified symmetry species. The perspective view of the molecule makes it look rather like a door-stop; however, from the side, each `door-stop' is in fact an isosceles triangle.

What you should learn from this lecture

1.You should be able with the help of a charactertavle to decide which components are differentfom zero in :<

r μ >= φ1∫

r μ φ1dv

= r e x φ1∫ x φ1dv

+ r e y φ1∫ y φ1dv

+r e z φ1∫ z φ1dv