chem 373- lecture 37: symmetry orbitals

8/3/2019 Chem 373- Lecture 37: Symmetry Orbitals

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Lecture 37: Symmetry OrbitalsThe material in this lecture covers the following in Atkins.15 Molecular Symmetry

Character Tables15.4 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 combinations

Lecture on-lineSymmetry Orbitals (PowerPoint)Symmetry Orbitals (PowerPoint)

Handouts for this lecture


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v 2

2

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 the

principle rotation is 1.B is used if the character of theprinciple rotation is -1

A1 has the character 1 for alloperations


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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


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v 2Character Table Structure of character table

A p x orbital on the central atomof a C 2v molecule and thesymmetry elements of the group.

Ep x = 1p x; C 2px = -1 p x

vpx = 1px; v px = -1 p x

The irrep. is B 1 andThe symmetry b 1


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A p y orbital on the central atomof a C 2v molecule and thesymmetry elements of the group.

Ep y = 1p y; C 2py = -1 p y

vpy = - 1p y; v py = 1 p y


+-

+-

C 2 v

v'


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A p z orbital on the central atomof a C 2v molecule and the

symmetry elements of the group.Ep z = 1p z; C 2pz = 1 p z

vpz = 1p z; v pz = 1 p z

The irrep. is A 1 andThe symmetry a 1

++

-

+

C 2 vv'


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A d xy orbital on the central atomof a C 2v molecule and the

symmetry elements of the group.Ed xy = 1d xy ; C 2dxy = 1 d xy

vdxy = -1d xy ; v dxy = - 1 d xy


+

C2 vv'

+

+

-

-

+


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+

C2 vv'

+

-

+

+


A 1s + orbital on the twoterminal atomsof a C 2v molecule and the

symmetry elements of the group.

E 1s + = 1 1s + ; C 2 1s+ = 1 1s +

v 1s + = 1 1s + ; v 1s+ = 1 1s +



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A 1s - orbital on the twoterminal atomsof a C 2v molecule and the

symmetry elements of the group.

E 1s - = 1 1s - ; C 2 1s - = -1 1s -

v 1s - = -1 1s - ; v 1s - = 1 1s -


+

C2 vv'

+

-

+

-


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+

C2 vv'

+

-

+

-

+

-


A 2p - orbital on the two

terminal atomsof a C 2v molecule and thesymmetry elements of the group.

E 2p - = 1 2p - ; C 2 2p - = 1 2p -

v 2p - = -1 2p - ; v 2p - = -1 2p -



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The value of an integral I (for example, an area) is independentof the coordinate system used to evaluate it.

That is, I is a basis of a representationof symmetry species A1 (or its equivalent ).

Character Table Structure of character tableStructure of character table

I s s dv= 1 2


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Y

z

1s1 +1s 21s+=

2p y


I s p dvy

=

=+

1 1 2 1( ) ( ) I C s p dvy= =+

2 1 1 2 1( ) ( )

I C s C p C dvy= =+[ ( )][ ( )][ ]2 2 21 1 2 1

I s p dvy= =+1 1 2 1( )[ ( )][ ]

I s p dv Iy= = +1 1 2 1( ) ( )

Y

z

1s1 +1s 21s+=

2p y

This is only possible if I = 0


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I s p dvz= =

1 1 2 1( ) ( ) I C s p dvz

= =2

1 1 2 1( ) ( )

I C s C p C dvz= =[ ( )][ ( )][ ]2 2 21 1 2 1

I s p dvz= =[ ( )] ( )1 1 2 1

I s p dv Iz= = 1 1 2 1( ) ( )

This is only possible if I = 0

Y

z

1s1 -1s 21s -=

2p z

Y

z

1s1 -1s 21s -=

2p z


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We must have in generalI O f f dv

Of Of dvc f f dvwith

= = =

( )( )1 2

1 21 2

c = 1

If c 1I = 0

General

f

procedurefor determining thesymmetry of productf1 2

1. Decide on the symetry species of theIndividual functions f1 and f2 by

reference to the character table, andwrite their characters in two rows inthe same same order as in the table


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1. Decide on the symmetry species of theindividual functions f1 and f2 byreference to the character table, and

write their characters in two rows inthe same same order as in the table

I s p dvy= +1 1 2 1( ) ( )

Y

z

1s1 +1s 21s+=

2p z

2p y 1 -1 -1 11s+ 1 1 1 1

2. Multiply the numbewrs in each column,

Writing the results in the same order

2p y1s+ 1 -1 -1 13. The new character must be A 1For the integral to be non-zero

The symmetry species is B 2


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I s p dvy= 1 1 2 1( ) ( )

2p y 1 -1 -1 11s - 1 -1 -1 1

2p y1s- 1 1 1 1

The symmetry species is A 1

Y

z

1s1 -1s 21s -=

2p z

For the integral

I f f dv= 1 2It

belongs

should be clear thatthe above procedureonly provide a Acharacter if f and f

to the same sym.representation

1

1 2


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1s1s +

px

py

pz


A1

B1B2

B2A1

2p z overlaps (interacts)with 1s +

2p y overlaps (interacts)with 1s

-


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v 2

2


The character table of a groupis the list of characters of all itsirreducible representations.

Names of irreduciblerepresentations: A 1,A 2,B 1,B 2.

Characters of irreduciblerepresentations


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The integral of the function f = xyover the tinted region is zero.In this case, the result is obviousby inspection, but grouptheory can be used to establish

similar results in less obvious

cases.


I xyf x y dxdy= ( , )f x y

f x y

( , )

( , )

=

=

1 inside triangle;

outside triangle0

v 3Appendix 1


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The integration of a function

over a pentagonal region.

I xyf x y dxdy= ( , )f x y

f x y

( , )

( , )

=

=

1 inside pentagon;

outside pentagon0

Character Table Structure of character tablev 3

Appendix 1


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Character Table

Typical symmetry-adaptedlinear combinations

of orbitals in a

C 3v molecule.

Two symmetry-adapted linearcombinations of the p-basis

orbitals.The two combinations each span aone-dimensional irreducible

representation, and their symmetryspecies are different.

Constructing Linear combinations

a1

a 2

ex

ey

b1

a 2

How are they constructed


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Original basis

1S 1SE 1S 1S

C 1S 1S1S 1S1S 1S

1 21 2

2 2 1v 2 1

1 2

v'

Character Table Constructing Linear combinations C v2

+

C 2 vv'

+

1 1s

1 2s

For A :1

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

Corresponding operator(2) Add and divide by group order

1 1 2 2 1 1 214

1 1 1 112

1 1= + + + = +( ) ( )s s s s s s

' ( ) ( )1 2 1 1 2 1 2

1

41 1 1 1

1

21 1= + + + = +s s s s s s


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Original basis

1S 1SE 1S 1S

C 1S 1S1S 1S1S 1S

1 21 2

2 2 1v 2 1

1 2

v'


+

C 2 vv'

+

1 1s

1 2s

For A :2



1 1 2 2 114

1 1 1 1 0= + =( )s s s s

' ( )1 2 1 1 2

1

41 1 1 1 0= + =s s s s


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Original basis

1S 1SE 1S 1S

C 1S 1S1S 1S1S 1S

1 21 2

2 2 1v 2 1

1 2

v'


+

C 2 vv'

+

1 1s

1 2s

For B :1



1 1 2 2 114

1 1 1 1 0= + =( )s s s s

' ( )1 2 1 1 2

1

41 1 1 1 0= + =s s s s

h bl b


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Original basis

1S 1SE 1S 1S

C 1S 1S1S 1S1S 1S

1 21 2

2 2 1v 2 1

1 2

v'


+

C 2 vv'

+

1 1s

1 2s

For B :2



1 1 2 2 1 1 214

1 1 1 112

1 1= + = ( ) ( )s s s s s s

' ( ) ( )1 2 1 1 2 2 1

1

41 1 1 1

1

21 1= + = s s s s s s

Ch T bl C i Li bi i


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Original basis

2p 2p

E 2p 2p

C - 2p - 2p

2p 2p

- 2p - 2p

xA

xB

x

A

x

B

2 xB

xA

v xB

xA

xA

xB

v'


For A :2



114

2 2 2 212

2 2= + = ( ) ( )p p p p p pxA xB xB xA xA xB

+

C 2 vv'

+

+-

+

-

2p xA

2p xB

' ( ) ( )114 2 2 2 2

12 2 2= + = p p p p p pxB xA xA xB xB xA

Ch t T bl C t ti Li bi ti


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Original basis

2p 2p

E 2p 2p

C - 2p - 2p

2p 2p

- 2p - 2p

xA

xB

x

A

x

B

2 xB

xA

v xB

xA

xA

xB

v'


For B :1



114

2 2 2 212

2 2= + + + = +( ) ( )p p p p p pxA xB xB xA xA xB

+

C2 vv'

+

+-

+

-

2p xA

2p xB

' ( ) ( )114 2 2 2 2

12 2 2= + + + = +p p p p p pxB xA xA xB xB xA


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What you should learn from this course

1. Be able to assign symmetries to orbitals from charactertables.

2. Be able to use character tables to determine whetherthe overlap between two functions might be differentfrom zero.

3. Be able to use character table to construct symmetryorbitals as linear combination of symmetry equivalentatomic orbitals

chem 373- lecture 37: symmetry orbitals

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