Physical Chemistry 2Physical Chemistry 2ndnd Edition EditionThomas Engel, Philip Reid

Chapter 27 Chapter 27 Molecular Symmetry

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

ObjectivesObjectives

• Discussion of molecular symmetry• Understanding of spectroscopic selection

rules• Identifying the normal modes of vibration for

a molecule• Determining if a molecular vibration is

infrared active and/or Raman active

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

OutlineOutline

1.Symmetry Elements, Symmetry Operations, and Point Groups

2.Assigning Molecules to Point Groups3.The H2O Molecule and the C2v Point Group

4.Representations of Symmetry Operators, Bases for Representations, and the Character Table

5.The Dimension of a Representation

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

OutlineOutline

6.Using the C2v Representations to Construct Molecular Orbitals for H2O

7.The Symmetries of the Normal Modes of Vibration of Molecules

8.Selection Rules and Infrared versus Raman Activity

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

27.1 Symmetry Elements, Symmetry Operations, 27.1 Symmetry Elements, Symmetry Operations, and and Point Groups Point Groups

• Individual molecule has an inherent symmetry based on the spatial arrangement of its atoms.

• Symmetry of a molecule determines its properties.

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

27.1 Symmetry Elements, Symmetry Operations, 27.1 Symmetry Elements, Symmetry Operations, and and Point Groups Point Groups

• Symmetry elements are geometric entities with respect to which operations can be carried out.

• Symmetry operations are actions with respect to the symmetry elements that leave the molecule in a new configuration.

• A set of symmetry elements forms a group.

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

27.2 Assigning Molecules to Point Groups27.2 Assigning Molecules to Point Groups

• The assignment is made using the logic diagram.

• It is useful to first identify the major symmetry elements.

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

27.2 Assigning Molecules to Point Groups27.2 Assigning Molecules to Point Groups

• A number of point groups are applicable to small molecules.

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

27.3 The H27.3 The H22O Molecule and the O Molecule and the CC2v2v Point Group Point Group

• We would express the symmetry operators mathematically and show that the symmetry elements form a group.

• Convince yourself that the 2 mirror planes belong to different classes and have different symbols.

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

27.3 The H27.3 The H22O Molecule and the O Molecule and the CC2v2v Point Point GroupGroup

• Elements that belong to the same class can be transformed into one another by other symmetry operations of the group.

• For example, the operators belong to the same class.

nnnn CCC ,...,, 2

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

Example 27.1Example 27.1

a. Are the three mirror planes for the NF3 molecule in the same or in different classes?

b. Are the two mirror planes for H2O in the same or in different classes?

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

SolutionSolution

a. NF3 belongs to the C3v group, which contains the rotation operators and the vertical mirror planes . These operations and elements are illustrated by this figure:

ECCCC 3

3

1

3233

ˆ and ,ˆˆ,ˆ

3 and ,2,1 vvv

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

SolutionSolution

We see that converts , and converts . Therefore, all three mirror planes belong to the same class.

b. The figure shows that neither the nor the operation converts . Therefore, these two mirror planes are in different classes.

1

323

ˆˆ CC

' to vv

3C 3 to1 vv 2 to1 vv

2C E

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

27.3 The H27.3 The H22O Molecule and the O Molecule and the CC2v2v Point Point GroupGroup

• The effect of the four operators can also be deduced from the following:

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

27.3 The H27.3 The H22O Molecule and the O Molecule and the CC2v2v Point Point GroupGroup

• The operators can be described by the following 3 × 3 matrices:

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

Example 27.3Example 27.3

Evaluate . What operation is equivalent to the two sequential operations?

Solution:

222ˆˆ and ˆˆ CCC v

ECC

C vv

ˆ

100

010

001

100

010

001

100

010

001ˆˆ

'ˆ

100

010

001

100

010

001

100

010

001

ˆˆ

22

2

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

27.4 Representations of Symmetry27.4 Representations of Symmetry Operators, Bases for Representations, and Operators, Bases for Representations, and the the Character Table Character Table

• The matrices derived are called representations where the multiplication table of the group can be reproduced with the matrices.

• Each operation can be represented by +1 or -1 (represented by ) and multiplication table still satisfied.

41 through

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

27.4 Representations of Symmetry27.4 Representations of Symmetry Operators, Bases for Representations, and Operators, Bases for Representations, and the the Character Table Character Table

• The irreducible representations are the matrices of smallest dimension that obey the multiplication table of the group.

• A group has as many irreducible representations as it has classes of symmetry elements.

• AOs forms a basis for one of the representations.

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

27.4 Representations of Symmetry27.4 Representations of Symmetry Operators, Bases for Representations, and Operators, Bases for Representations, and the the Character Table Character Table

• The information on representations can be assembled in a character table.

• The representation in which all entries are +1 is called the totally symmetric representation.

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

27.4 Representations of Symmetry27.4 Representations of Symmetry Operators, Bases for Representations, and Operators, Bases for Representations, and the the Character Table Character Table

• Columns 2 through 5 has an entry for each operation of the group in each representation called characters.

• Columns 6 through 8 shows many possible bases for each representation, called basis functions.

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

27.5 The Dimension of a Representation27.5 The Dimension of a Representation

• The dimension of a representation is defined as the size of the matrix used to represent the symmetry operations.

• 3×3 matrix operations can be reduced to three 1×1 matrix operations, which consist of the numbers +1 and -1.

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

27.5 The Dimension of a Representation27.5 The Dimension of a Representation

• The dimension of the different irreducible representations, dj, and the order of the group, h, defined as the number of symmetry elements in the group, are related by the equation

• This sum is over each class of symmetry elements in the group, rather than over the elements.

N

jj hd

1

2

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

Example 27.4Example 27.4

The C3v group has the elements and three σv mirror planes. How many different irreducible representations does this group have, and what is the dimensionality of each irreducible representation?

233

ˆ and ˆ,ˆ CCE

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

SolutionSolution

The order of the group is the number of elements, so h=6. The number of representations is the number of classes. belong to one class, and the same is true of the three σv reflections. Although the group has six elements, it has only three classes. Therefore, the group has three irreducible representations. The equation is solved to find the dimension of the representations, and one of the values must be 1. The only possible solution is . C3v group contains one two-dimensional representation and two one-dimensional representations.

233

ˆ and ˆ CC

623

22

21 lll

2 and 321 lll

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

27.5 The Dimension of a Representation27.5 The Dimension of a Representation

• To transform x–y coordinate system by a mirror plane,σ

2cos2sin'

2sin2cos'

yxy

yxx

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

27.5 The Dimension of a Representation27.5 The Dimension of a Representation

• The character for an operator in a representation of dimension higher than one is given by the sum of the diagonal elements of the matrix.

• If the set of characters associated with a representation of the group is viewed as a vector, , with one component for each element of the group, the following condition holds:

jii R

kikihRR ikikjkj

h

jiji

and if 0 where,ˆˆ1

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

27.5 The Dimension of a Representation27.5 The Dimension of a Representation

• Equivalently, . The sum is over all elements of the group.

ikjkjiji hRR ˆˆ

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

Example 27.5Example 27.5

Determine the unknown coefficients a, b, and c for the preceding partially completed character table and assign the appropriate symbol to the irreducible representation.

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

SolutionSolution

We know that the unknown representation is one dimensional and for different values of the index i are orthogonal. Therefore,

We can take the sum over classes and multiplied each term by the number of elements in the class, because all elements in a class have the same character.

0222

032

?

? 1

babcba

cbacccbba

E

A

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

SolutionSolution

We also know that a=1 because it is the character of the identity operator. Solving the equations gives the results of b=1 and c=-1. Because the character of C3 is +1, and the character of σv is -1, the unknown representation is designated A2.

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

27.6 Using the 27.6 Using the CC2v2v Representations to Construct Representations to Construct Molecular Orbitals for H Molecular Orbitals for H22OO

• The character for an operator of the direct

product of two representations is given by

• Overlap integral between two combinations of AOs is nonzero only if the combinations belong to the same representation.

group for the 'ˆor ,ˆ,ˆ,ˆˆ22 vvv CCER

RRR jiproductˆˆˆ

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

Example 27.6Example 27.6

Which of the oxygen AOs will participate in forming symmetry-adapted water MOs with the antisymmetric combination of hydrogen AOs defined by ?sBHsAH 11

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

SolutionSolution

The antisymmetric combination of the H AOs is given by shown in the margin. By considering the C2v operations, convince yourself that the characters for the different operations are . Therefore, belongs to the B2 representation. Of the valence oxygen AOs,

only the 2py orbital belongs to the B2 representation. Therefore, the only MOs formed from and the 2s and 2p orbitals that have the symmetry of the H2O molecule and also have a nonzero overlap among the AOs are the MOs denoted 1b2 and 2b2.

sBHsAH 11

1:'ˆ and 1:ˆ ,1:ˆ ,1:ˆ2 vvCE

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

27.7 The Symmetries of the Normal Modes of 27.7 The Symmetries of the Normal Modes of Vibration Vibration of Molecules of Molecules

• All normal modes are independent in the harmonic approximation.

• Random motion of the atoms in a molecule can be expressed as a linear combination of the normal modes of that molecule.

• For harmonic approximation,

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

27.7 The Symmetries of the Normal Modes of 27.7 The Symmetries of the Normal Modes of Vibration Vibration of Molecules of Molecules

The general method for decomposing a reducible representation into its irreducible representations utilizes the vector properties of the representations Take the scalar product between the reducible representation and each of the irreducible representations in turn, and divide by the order of the group.

jreducible R jreducible R

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

27.7 The Symmetries of the Normal Modes of 27.7 The Symmetries of the Normal Modes of Vibration Vibration of Molecules of Molecules

The result of this procedure is a positive integer ni that is the number of times each representation appears in the irreducible representation. This statement is expressed by the equation

Ni

RRh

Rh

Rhh

n jreducible

h

jjijreduciblejireductibleii

,...,2,1for

ˆˆ1ˆ1ˆ11

1

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Physical Chemistry 2nd EditionChapter 27: Molecular Symmetry

27.8 The Symmetries of the Normal Modes of 27.8 The Symmetries of the Normal Modes of Vibration Vibration of Molecules of Molecules

• Normal modes of a molecule are Raman active if the bases of the representation to which the normal mode belongs are the x2, y2, z2, xy, yz, or xz functions.

• All normal modes that belong to a particular representation have the same frequency.