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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 - PowerPoint PPT Presentation
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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