lecture 37: symmetry orbitals the material in this lecture covers the following in atkins. 15...

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


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


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


+ -+ -

C2 σv

σv'


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'


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


+

C2 v

v'

+

+-

-+

+

C2 v

v'

+

-

+

+


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+



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-


+

C2 v

v'

+

-

+

-

+

C2 v

v'

+

-

+

-+

-


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 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


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


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


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


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


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


A1

B1B2

B2A1

2pz overlaps (interacts) with 1s+

2py overlaps (interacts) with 1s-

C2v

C2


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.


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 :


Character Table Constructing Linear combinations C2v

+

C2 v

v'

+

1s1

1s2

For A1 :


(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)



+

C2 v

v'

+

1s1

1s2

For A2 :



ψ1=14(1s1+1s2−1s2−1s1)=0

ψ'1=14(1s2+1s1−1s1−1s2)=0



+

C2 v

v'

+

1s1

1s2

For B1 :



ψ1=14(1s1−1s2+1s2−1s1)=0

ψ'1=14(1s2−1s1+1s1−1s2)=0



+

C2 v

v'

+

1s1

1s2

For B2 :



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


For A2 :



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


For B1 :



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

lecture 37: symmetry orbitals the material in this lecture covers the following in atkins. 15...

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