Lecture 37: Symmetry Orbitals The material in this lecture covers the following in Atkins.15 Molecular Symmetry Character Tables 15.4 Character tables and symmetry labels (a) The structure of character tables (b) Character tables and orbital degeneracy (c) Characters and operators (d) The classification of linear combinations of orbitals 15.5 Vanishing integrals and orbital overlaps (a) The criteria for vanishing integrals (b) Orbitals with nonzero overlaps (c) Symmetry-adapted linear combinationsLecture on-line Symmetry Orbitals (PowerPoint) Symmetry Orbitals (PowerPoint) Handouts for this lecture
C2v
C2
Character Table Structure of character table
One dimensional irreduciblerepresentations have the character 1 for E.
They are termed A or B.
A is used if the character of theprinciple rotation is 1.
B is used if the character of the principle rotation is -1
A1 has the character 1 for alloperations
Character Table Structure of character table
C3
C3
C3
C3 C2
Td
Irreducible representations withdimension 2 are denoted E
Irreducible representations withdimension 3 are denoted T
Number of symmetry species(irreducible representations) = Number of classes
C2vCharacter Table Structure of character table
A px orbital on the central atom of a C2v molecule and thesymmetry elements of the group.
Epx = 1px; C2px = -1 px
vpx = 1px; v’ px = -1 px
The irrep. is B1 andThe symmetry b1
C2vCharacter Table Structure of character table
A py orbital on the central atom of a C2v molecule and thesymmetry elements of the group.
Epy = 1py; C2py = -1 py
vpy = - 1py; v’ py = 1 py
The irrep. is B2 andThe symmetry b2
+ -+ -
C2 σv
σv'
C2vCharacter Table Structure of character table
A pz orbital on the central atom of a C2v molecule and thesymmetry elements of the group.
Epz = 1pz; C2pz = 1 pz
vpz = 1pz; v’ pz = 1 pz
The irrep. is A1 andThe symmetry a1
++
-
+
C2 σv
σv'
C2vCharacter Table Structure of character table
A dxy orbital on the central atom of a C2v molecule and thesymmetry elements of the group.
Edxy = 1dxy ; C2dxy = 1 dxy
vdxy = -1dxy ; v’ dxy = - 1 dxy
The irrep. is A2 andThe symmetry a2
+
C2 v
v'
+
+-
-+
+
C2 v
v'
+
-
+
+
C2vCharacter Table Structure of character table
A 1s+ orbital on the two terminal atoms of a C2v molecule and thesymmetry elements of the group.
E 1s+ = 1 1s+ ; C2 1s+ = 1 1s+
v 1s+ = 1 1s+ ; v’ 1s+ = 1 1s+
The irrep. is A1 andThe symmetry a1
C2vCharacter Table Structure of character table
A 1s- orbital on the two terminal atoms of a C2v molecule and thesymmetry elements of the group.
E 1s- = 1 1s- ; C2 1s- = -1 1s-
v 1s- = -1 1s- ; v’ 1s- = 1 1s-
The irrep. is B2 andThe symmetry b2
+
C2 v
v'
+
-
+
-
+
C2 v
v'
+
-
+
-+
-
C2vCharacter Table Structure of character table
A 2p- orbital on the two terminal atoms of a C2v molecule and thesymmetry elements of the group. E 2p- = 1 2p- ; C2 2p- = 1 2p-
v 2p- = -1 2p- ; v’ 2p- = -1 2p-
The irrep. is A2 andThe symmetry a2
The value of an integral I (for example, an area) is independent of the coordinate system used to evaluate it.
That is, I is a basis of a representation of symmetry species A1 (or its equivalent).
Character Table Structure of character table Structure of character table
I= s1∫ s2dv
Y
z
1s1 +1s21s+=
2py
Character Table Structure of character table
I= 1s+(1)2py(1)dv∫ = I=C2 1s+(1)2py(1)dv∫ =
I= [C21s+(1)][C22py(1)][C2dv]∫ =
I= 1s+(1)[−2py(1)][dv]∫ =
I=− 1s+(1)2py(1)dv∫ =−I
Y
z
1s1 +1s21s+=
2py
This is only possible if I=0
Character Table Structure of character table
I= 1s−(1)2pz(1)dv∫ = I=C2 1s−(1)2pz(1)dv∫ =
I= [C21s−(1)][C22pz(1)][C2dv]∫ =
I= [−1s−(1)]2pz(1)dv∫ =
I=− 1s−(1)2pz(1)dv∫ =−I
This is only possible if I=0
Y
z
1s1 -1s21s-=
2pz
Y
z
1s1 -1s21s-=
2pz
Character Table Structure of character table
We must have in general I= ˆ O f1f2dv=∫(ˆ O f1)(ˆ O f2)dv=∫
c f1f2dv∫with c = 1
If c ≠ 1I= 0
General procedurefor determining thesymmetry of product f1f2
1. Decide on the symetry species of theIndividual functions f1 and f2 by reference to the character table, andwrite their characters in two rows in the same same order as in the table
Character Table Structure of character table
1. Decide on the symmetry species of theindividual functions f1 and f2 by reference to the character table, andwrite their characters in two rows in the same same order as in the table
I= 1s+(1)2py(1)dv∫
Y
z
1s1 +1s21s+=
2pz
2py 1 -1 -1 11s+ 1 1 1 1
2. Multiply the numbewrs in each column,Writing the results in the same order
2py1s+ 1 -1 -1 13. The new character must be A1
For the integral to be non-zero
The symmetry species is B2
Character Table Structure of character table
I= 1s−(1)2py(1)dv∫
2py 1 -1 -1 11s- 1 -1 -1 1
2py1s- 1 1 1 1
The symmetry species is A1
Y
z
1s1 -1s21s-=
2pz
For the integral
I= f1f2dv∫
It should be clear thatthe above procedureonly provide a A1 character if f1 and f2
belongs to the same sym.representation
1s-1s+
pxpy
pz
Character Table Structure of character table
A1
B1B2
B2A1
2pz overlaps (interacts) with 1s+
2py overlaps (interacts) with 1s-
C2v
C2
Character Table Structure of character table
The character table of a group is the list of characters of all its irreducible representations.
Names of irreducible representations: A1,A2,B1,B2.
Characters of irreduciblerepresentations
The integral of the function f = xy over the tinted region is zero.In this case, the result is obvious by inspection, but group theory can be used to establish similar results in less obvious cases.
Character Table Structure of character table
I= xy∫ f(x,y)dxdyf(x,y)= 1 inside triangle;f(x,y)=0 outside triangle
C3vAppendix 1
The integration of a function over a pentagonal region.
I= xy∫ f(x,y)dxdyf(x,y)= 1 inside pentagon;f(x,y)=0 outside pentagon
Character Table Structure of character table C3v
Appendix 1
Character Table
Typical symmetry-adapted linear combinations of orbitals in a C 3v molecule.
Two symmetry-adapted linear combinations of the p-basis orbitals. The two combinations each span a one-dimensional irreducible representation, and their symmetryspecies are different.
Constructing Linear combinations
a1
a2
ex
ey
b1
a2
How are they constructed
Original basis 1S1 1S2 E 1S1 1S2 C2 1S2 1S1 σv 1S2 1S1 σv' 1S1 1S2
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 C2v
+
C2 v
v'
+
1s1
1s2
For A1 :
Original basis 1S1 1S2 E 1S1 1S2 C2 1S2 1S1 σv 1S2 1S1 σv' 1S1 1S2
Character Table Constructing Linear combinations C2v
+
C2 v
v'
+
1s1
1s2
For A1 :
(I) Multiply each member of theColumn by the character of the Corresponding operator
(2) Add and divide by group order
ψ1=14(1s1+1s2+1s2+1s1)=
12(1s1+1s2)
ψ'1=14(1s2+1s1+1s1+1s2)=
12(1s1+1s2)
Original basis 1S1 1S2 E 1S1 1S2 C2 1S2 1S1 σv 1S2 1S1 σv' 1S1 1S2
Character Table Constructing Linear combinations C2v
+
C2 v
v'
+
1s1
1s2
For A2 :
(I) Multiply each member of theColumn by the character of the Corresponding operator
(2) Add and divide by group order
ψ1=14(1s1+1s2−1s2−1s1)=0
ψ'1=14(1s2+1s1−1s1−1s2)=0
Original basis 1S1 1S2 E 1S1 1S2 C2 1S2 1S1 σv 1S2 1S1 σv' 1S1 1S2
Character Table Constructing Linear combinations C2v
+
C2 v
v'
+
1s1
1s2
For B1 :
(I) Multiply each member of theColumn by the character of the Corresponding operator
(2) Add and divide by group order
ψ1=14(1s1−1s2+1s2−1s1)=0
ψ'1=14(1s2−1s1+1s1−1s2)=0
Original basis 1S1 1S2 E 1S1 1S2 C2 1S2 1S1 σv 1S2 1S1 σv' 1S1 1S2
Character Table Constructing Linear combinations C2v
+
C2 v
v'
+
1s1
1s2
For B2 :
(I) Multiply each member of theColumn by the character of the Corresponding operator
(2) Add and divide by group order
ψ1=14(1s1−1s2−1s2+1s1)=
12(1s1−1s2)
ψ'1=14(1s2−1s1−1s1+1s2)=
12(1s2−1s1)
Original basis
2pxA 2px
B
E 2pxA 2px
B
C2 -2pxB -2px
A
σv 2pxB 2px
A
σv' -2pxA -2px
B
Character Table Constructing Linear combinations C2v
For A2 :
(I) Multiply each member of theColumn by the character of the Corresponding operator
(2) Add and divide by group order
ψ1=14(2px
A −2pxB −2px
B +−2pxA)=
12(2px
A −2pxB)
+
C2 v
v'
+
+-
+
-
2pxA
2pxB
ψ'1=
14(2px
B −2pxA −2px
A +−2pxB)=
12(2px
B −2pxA)
Original basis
2pxA 2px
B
E 2pxA 2px
B
C2 -2pxB -2px
A
σv 2pxB 2px
A
σv' -2pxA -2px
B
Character Table Constructing Linear combinations C2v
For B1 :
(I) Multiply each member of theColumn by the character of the Corresponding operator
(2) Add and divide by group order
ψ1=14(2px
A +2pxB +2px
B +−2pxA)=
12(2px
A +2pxB)
+
C2 v
v'
+
+-
+
-
2pxA
2pxB
ψ'1=
14(2px
B +2pxA +2px
A +2pxB)=
12(2px
B +2pxA)
What you should learn from this course
1. Be able to assign symmetries to orbitals from character tables.
2. Be able to use character tables to determine whether the overlap between two functions might be different from zero.
3. Be able to use character table to construct symmetryorbitals as linear combination of symmetry equivalentatomic orbitals