(1) Teori Grup Dan Vibrasi-revisi

8/2/2019 (1) Teori Grup Dan Vibrasi-revisi

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SYMMETRY ANDGROUP THEORY


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Facial symmetry


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Invariance to transformation as anindicator of facial symmetry:

Mirror image


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Symmetry And

Group TheoryThe symmetry properties of molecules they can

be used to :predict vibrational spectra,hybridization, optical activity, NMR, UV-

Visible (electronic spectra), XRD, etc.


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Symmetry Elements and

Opertaions

A molecule has a given symmetry element if the

operation leaves the molecule appearing as ifnothing has changed (even though atoms andbonds may have been moved.)


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Symmetry Elements

Element Symmetry Operation SymbolEntire object Identity E

n-fold axis Rotation by 2/n Cn

Mirror plane Reflection

Center of in- Inversion i

version

n-fold axis of Rotation by 2/n Sn

improper rotation followed by reflectionperpendicular to the

axis of rotation


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Identity, E

All molecules have Identity. This operationleaves the entire molecule unchanged. A highlyasymmetric molecule such as a tetrahedral

carbon with 4 different groups attached has onlyidentity, and no other symmetry elements.


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n-fold Rotation

Water has a 2-fold axisof rotation. When

rotated by 180o

, thehydrogen atoms tradeplaces, but the moleculewill look exactly the same.


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n-fold Axis of Rotation

Ammonia has a C3 axis. Note that there are twooperations associated with the C

3axis. Rotation by

120o in a clockwise or a counterclockwise directionprovide two different orientations of the molecule.


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Mirror Planes

The reflection of thewater molecule in either ofits two mirror planes results

in a molecule that looksunchanged.


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Mirror Planes

The subscript v in v,indicates a vertical plane ofsymmetry. This indicates

that the mirror planeincludes the principal axis ofrotation (C2).


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Mirror Planes

The vertical planes, v,go through the carbonatoms, and include the C6

axis.The planes that bisect

the bonds are called dihedralplanes, d.

C6.


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Mirror Planes

The benzene ring has aC6 axis as its principal axis ofrotation.

The molecular plane isperpendicular to the C6 axis,and is designated as ahorizontal plane, h.

C6.


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Inversion

The inversion operation projects each atomthrough the center of inversion, and across tothe other side of the molecule.


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Improper Rotation

An improper rotation is rotation, followedby reflection in the plane perpendicular to theaxis of rotation.


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Improper Rotation

The staggeredconformation ofethane has an S6 axis

that goes throughboth carbon atoms.


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Improper Rotation

Likewise, an S2axis is a center of

inversion.


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Point Groups

Molecules with the same symmetry elementsare placed intopoint groups.


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Identifying point groups


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Point Groups Definitions:

1. Point Group = the set of symmetry operations for a molecule

2. Group Theory = mathematical treatment of the properties of the group whichcan be used to find properties of the molecule

Assigning the Point Group of a Molecule

1. Determine if the molecule is of high or low symmetry by inspection

a. Low Symmetry Groups

b High Symmetry Groups


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b. High Symmetry Groups


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2. If not, find the principle axis

3. If there are C2 axes perpendicular

to Cn the molecule is in D

If not, the molecule will be in C or S

a. Ifsh perpendicular to Cn then Dnh or CnhIf not, go to the next step

b. Ifs contains Cn then Cnvor DndIf not, Dn or Cn or S2n

c. If S2n along Cn then S2nIf not Cn

Id if i i


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We can use a flow chart such as this one todetermine the point group of any object.

The steps in this process are:

1. Determine the symmetry is special (e.g.octahedral).

2. Determine if there is a principal rotationaxis.

3. Determine if there are rotation axesperpendicular to the principal axis.

4. Determine if there are mirror planes.

5. Assign point group.



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Point Groups

Molecules are classified and groupedbased on their symmetry. A point group

contains all objects that have the samesymmetry elements.


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Point Groups

Water andammonia both belongto the Cnvclass of

molecules. Thesehave vertical planes ofreflection, but no

horizontal planes.


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Point Groups

The Dnh groupshave a horizontalplane in addition to

vertical planes. Manyinorganic complexesbelong to these

symmetry groups.

X

X

X

X

Y

Y


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C. Examples: Assign point groups of molecules in Fig 4.8


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p g p g p g


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Rotation axes of normal symmetry molecules


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Perpendicular C2 axes

Horizontal Mirror Planes


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Vertical or Dihedral Mirror Planes and S2n Axes

Examples: XeF4

, SF4

, IOF3

, Table 4-4, Exercise 4-3

f


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D. Properties of Point Groups1. Symmetry operation of NH3

a. Ammonia has E, 2C3

(C3 and C23) and 3sv

b. Point group = C3v

2. Properties of C3v(any group)

a. Must contain E

b. Each operation must

have an inverse; doing bothgives E (right to left)

c. Any product equalsanother group member

d. Associative property


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REPRESENTATIONS OF

POINT GROUPS


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Matrices Why Matrices? The matrix representations of the point groups operations will

generate a character table. We can use this table to predict properties.

Definitions and Rules Matrix = ordered array of numbers

Multiplying Matrices

The number of columns of matrix #1 must = number of rows of matrix#2

Fill in answer matrix from left to right and top to bottom

The first answer number comes from the sum of [(row 1 elements ofmatrix #1) X (column 1 elements of matrix #2)]

The answer matrix has same number of rows as matrix #1

The answer matrix has same number of columns as matrix #2

5243or17

23

5438

4327

4862414

403207

84

37

62

51


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e) Relevant example:

321

100

010

001

321

II.Representations of Point Groups

Matrix Representations of C2v

Choose set of x,y,z axes

z is usually the Cn axis

xz plane is usually the plane of the moleculeExamine what happens after the molecule undergoes each symmetry

operation in the point group (E, C2, 2s)

T f ti M t i t i i th ff t f t


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3) Transformation Matrix = matrix expressing the effect of a symmetry

operation on the x,y,z axes

4) E Transformation Matrix

a. x,y,z x,y,zb. What matrix times x,y,z doesnt change anything?

z

y

x

???

???

???

z'

y'

x'

transformationmatrix

z

y

x

z

y

x

100

010

001

E Transformation Matrix


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5) C2 Transformation Matrix

a. x,y,z -x, -y, z

b. Correct matrix is:

6) sv(xz) Transformation Matrix

a. x,y,z x,-y,z


7) sv(yz) Transformation Matrix

a. x,y,z -x,y,z


z

y-

x-

z

y

x

100

010

001

z

y-x

z

yx

100

010001

z

y

x-

z

y

x

100

010

001

8) Th 4 m tri r th Matri Representation f th C p int r p


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8) These 4 matrices are the Matrix Representation of the C2vpoint group

a. All point group properties transfer to the matrices as well

b. Example: Esv(xz) = sv(xz)

B. Reducible and Irreducible Representations

Character = sum of diagonal from upper left to lower right (only defined

for square matrices)

The set of characters = a reducible representation (G) or shorthandversion of the matrix representation

For C2vPoint Group:

100

010

001

100

010

001

100

010

001

E C2 sv(xz) sv(yz)

3 -1 1 1

2) Reducible and Irreducible Representations


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2) Reducible and Irreducible Representations

a. Each matrix in the C2vmatrix representation can be block diagonalized

b. To block diagonalize, make each nonzero element into a 1x1 matrix

c. When you do this, the x,y, and z axes can be treated independently

Positions 1,1 always describe x-axis

Positions 2,2 always describe y-axis

Positions 3,3 always describe z-axis

d. Generate a partial character table from this treatment

100

010001

100

010001

100

010001

100

010001

E C2 sv(xz) sv(yz)

Axis used E C2 sv(xz) sv(yz)

x 1 -1 1 -1

y 1 -1 -1 1

z 1 1 1 1

G 3 -1 1 1

IrreducibleRepresentations

Reducible Repr.

Ch t T bl


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III.Character Tables The C2vCharacter Table

We have found three of the irreducible representations of the character tablethrough matrix math

One more (A2) irreducible representation is derived from the first three due tothe properties of character tables (below)

Rx, Ry, Rz stand for rotation about the x, y, z axes respectively

xs are p and d orbitals

4) Other symbols we need to know

R= any symmetry operation

c = character (#) i,j = different representations (A1, B2, etc)

h = order of the group (4 total operations in the C2vcase)

C2v E C2 sv(xz) sv(yz)

A1 1 1 1 1 z x2, y2, z2

A2 1 1 -1 -1 Rz xy

B1 1 -1 1 -1 x, Ry xz

B2 1 -1 -1 1 y, Rx yz


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B Molecular Vibrations


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B. Molecular Vibrations

1. To use symmetry, we must assign axes

to each atom of the molecule

a. The z-axis is usually the Cn axis

b. The x-axis is in the molecular planec. The y-axis is perpendicular to the molecular plane

2. Degrees of Freedom = possible atomic movements in the molecule

a. 3N degrees of freedom for a molecule of N atoms

b. Nonlinear molecules

3 translations (along x, y, z)

3 rotations (around x, y, z)

3N6 vibrations

c. Linear molecules

Only 2 rotations change the molecule

3N5 vibrations

3. We will use group theory to determine the symmetry of all nine motions andthen assign them to translation, rotation, and vibration

Look at the C character table


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Look at the C2vcharacter table

Add up how many vectors stay the same after an operation

If the atom moves, none of its vectors stay the same

If the atom stays and the vector is unchanged = +1

If the atom stays and the vector is reversed = -1

Reduce the reducible representation to its irreducible components


A1 1 1 1 1 z x2, y2, z2

A2 1 1 -1 -1 Rz xyB1 1 -1 1 -1 x, Ry xz

B2 1 -1 -1 1 y, Rx yz


G 9 -1 3 1

tionrepresentaeirreducibl

theofcharacter

tionrepresentareducible

theofcharacter

classin the

operationsofnumber1

given typeaof

tionsrepresentaeirreducibl

ofnumberThe

order

nA1 = [(1x9x1)+(1x-1x1)+(1x3x1)+(1x1x1)] = 3 A1

[(1 9 1) (1 1 1) (1 3 1) (1 1 1)] 1 A


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nA2 = [(1x9x1)+(1x-1x1)+(1x3x-1)+(1x1x-1)] = 1 A2

nA1 = [(1x9x1)+(1x-1x-1)+(1x3x1)+(1x-1x1)] = 3 B1

nA1 = [(1x9x1)+(1x-1x-1)+(1x3x-1)+(1x1x1)] = 2 B2

d) All motions of water match 3A1 + A2 + 3B1 + 2B2

e) Use the character table to remove translations

x, y, z = A1 + B1 + B2

f) Use the character table to remove rotations

Rx, Ry, Rz = A2 + B1 + B2

g) The motions remaining are the vibrations = 2A1 + B1

A1 = totally symmetric

B1 = antisymmetric to C2 and to reflection in yz plane


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Character Table (C2v)


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The functions to the right are called basis functions.

They represent mathematical functions such as orbitals,rotations, etc.


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Character Table Representations

1. Characters of +1 indicate that the basisfunction is unchanged by the symmetryoperation.

2. Characters of -1 indicate that the basis functionis reversed by the symmetry operation.

3. Characters of 0 indicate that the basis function

undergoes a more complicated change.


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Character Table Representations

1. AnA representation indicates that thefunctions are symmetric with respect to rotationabout the principal axis of rotation.

2. B representations are asymmetric with respect

to rotation about the principal axis.3. E representations are doubly degenerate.

4. Trepresentations are triply degenerate.

5. Subscrips uandgindicate asymmetric (ungerade)or symmetric (gerade) with respect to a center ofinversion.


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Applications of Group Theory

1. Predicting polarity of molecules. A moleculecannot have a permanent dipole moment if it

a) has a center of inversion

b) belongs to any of the D point groups

c) belongs to the cubic groups T or O


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2. Predicting chirality of molecules. Chiralmolecules lack an improper axis of rotation (Sn),a center of symmetry (i) or a mirror plane ().


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3. Predicting the orbitals used in bonds. Grouptheory can be used to predict which orbitals on acentral atom can be mixed to create hybrid

orbitals.


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4. Predicting the orbitals used in molecular orbitals.Molecular orbitals result from the combining oroverlap of atomic orbitals, and they encompass

the entire molecule.


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5. Determining the symmetry properties of allmolecular motion (rotations, translations andvibrations). Group theory can be used to

predict which molecular vibrations will be seenin the infrared or Raman spectra.


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Molecular Vibrations

Molecular motion includes translations,rotations and vibrations. The total number ofdegrees of freedom (types of molecular motion)

is equal to 3N, where N is the number of atomsin the molecule.


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Of the 3N types of motion, three representmolecular translations in the x, y or z directions.Linear molecules have two rotational degrees of

freedom, and non-linear molecules have threerotational degrees of freedom.


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To obtain red for all molecular motion, wemust consider the symmetry properties of thethree cartesian coordinates on all atoms of the

molecule.


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The molecule lies in the xz plane. The x axisis drawn in blue, and the y axis is drawn in black.

The red arrows indicate the z axis.

xyz


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The molecule lies in the xz plane. The x axis

is drawn in blue, and the y axis is drawn in black.The red arrows indicate the z axis.

yx


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If a symmetry operation changes theposition of an atom, all three cartesian

coordinates contribute a value of 0.

xyz


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For operations that leave an atom in place,the character is +1 for an axis that remains in

position, -1 for an axis that is reversed, and 0 foran axis that has been moved.

xyz


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Page 75

3

3

3

2

2

2

1

1

1

3

3

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2

2

2

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100000000

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z

y

x

z

y

x

zy

x

z

y

x

z

y

x

zy

x

E

For E c = + 9


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000000001 xx


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3

3

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z

y

x

z

y

x

z

y

x

z

y

x

z

y

x

z

y

x

yzs

Only require the characters: The sum of diagonal elements

For s(xz) c = + 1


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M l l Vib i


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E C2 v(xz) v(yz)

Identity leaves all 3atoms in position, sothe character will be

9.


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M l l Vib i


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E C2 v(xz) v(yz)

9

The C2 axis goesthrough the oxygenatom, and exchanges

the hydrogen atoms.

M l l Vib i


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E C2 v(xz) v(yz)

9

The z axis (red) onoxygen stays inposition. This axis

contributes +1towards thecharacter for C2.

M l l Vib i


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E C2 v(xz) v(yz)

9

The y axis (black) onoxygen is rotated by180o. This reverses

the axis, andcontributes -1 to thecharacter for C2.

M l l Vib i


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E C2 v(xz) v(yz)

9

The x axis (blue) onoxygen is also rotatedby 180o. This reverses

the axis, andcontributes -1 to thecharacter for C2.

M l l Vib i


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E C2 v(xz) v(yz)

9

The character for the C2 operation will be+1 (z axis on oxygen) -1 (y axis onoxygen) -1 (x axis on oxygen) = -1

M l l Vib i


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E C2 v(xz) v(yz)

9 -1

The character for the C2 operation will be+1 (z axis on oxygen) -1 (y axis onoxygen) -1 (x axis on oxygen) = -1


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M l l Vib i


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E C2 v(xz) v(yz)

9 -1

The z axis and the xaxis both lie within thexz plane, and remain

unchanged.

x

M l l Vib ti


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E C2 v(xz) v(yz)

9 -1

Each unchanged axiscontributes +1 to thecharacter for the

symmetry operation.

x

M l l Vib ti


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E C2 v(xz) v(yz)

9 -1

For 3 atoms, thecontribution to thecharacter will be:

3(1+1) =6

x

M l l Vib ti


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E C2 v(xz) v(yz)

9 -1

The y axis will bereversed by the mirrorplane, contributing a

value of -1 for each ofthe three atoms on theplane.

xy

M l l r Vibr ti n


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E C2 v(xz) v(yz)

9 -1 3

The character for thexz mirror plane will be:

6-3 = 3

xy

M l c l r Vibr ti n


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E C2 v(xz) v(yz)

9 -1 3

The yz mirror planebisects the molecule.Only the oxygen atom

lies in the plane.

xy


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E C2 v(xz) v(yz)

9 -1 3

The x axis on oxygen isreversed by thereflection, and

contributes a -1towards the character.

xy



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E C2 v(xz) v(yz)

9 -1 3 1

The character forreflection in the yzplane is:

1+1-1=1

xy

Character Table (C )


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E C2 v(xz) v(yz)

9 -1 3 13N


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This formula was derived from the Great orthorgonality theorem.

nA1 = [(1x9x1)+(1x1x1)+(1x3x1)+(1x1x1)] = 3 A1 (1 9 1) (1 1 1) (1 3 1) (1 1 1) 1 A


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nA2 = [(1x9x1)+(1x-1x1)+(1x3x-1)+(1x1x-1)] = 1 A2

nB1 = [(1x9x1)+(1x-1x-1)+(1x3x1)+(1x-1x1)] = 3 B1

nB2 = [(1x9x1)+(1x-1x-1)+(1x3x-1)+(1x1x1)] = 2 B2

All motions of water match

3A1 + A2 + 3B1 + 2B2



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E C2 v(xz) v(yz)

9 -1 3 1

The above reducible representation issometimes called 3N, because it reduces to all(3N) modes of molecular motion.

3N for water reduces to:3A1 + A2 + 3B1 + 2B2



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3N for water = 3A1 + A2 + 3B1 + 2B2

Note that there are 9 modes of motion.

These include vibrations, rotations andtranslations.



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3N for water = 3A1 + A2 + 3B1 + 2B2

Translations have the same symmetry

properties as x, y and z.



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3N for water = 3A1 + A2 + 3B1 + 2B2





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3N for water = 3A1 + A2 + 3B1 + 2B2



2



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3N for water = 3A1 + A2 + 3B1 + 2B2



2 2



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3N for water = 3A1 + A2 + 3B1 + 2B2



2 2



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3N for water = 3A1 + A2 + 3B1 + 2B2



2 2 1



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rot & vib = 2A1 + A2 + 2B1 + 1B2



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rot & vib = 2A1 + A2 + 2B1 + 1B2

Rotations have the same symmetry as

Rx, Ryand Rz.



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rot & vib = 2A1 + A2 + 2B1 + 1B2


Rx, Ryand Rz.



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rot & vib = 2A1 + 2B1 + 1B2


Rx, Ryand Rz.

1



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rot & vib = 2A1 + 1B1 + 1B2


Rx, Ryand Rz.

Rotations and Translations


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Rotations and Translations

Rz

Rx

Ry

Transz

Transy

Transx



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vib = 2A1 + B1The three vibrational modes remain.

Two have A1 symmetry, and one has B1

symmetry.



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vib = 2A1 + B1Two vibrations are symmetric with

respect to all symmetry operations of the

group.



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vib = 2A1 + B1One vibration is asymmetric with

respect to rotation and reflection

perpendicular to the molecular plane.



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vib

= 2A1

+ B1

A1 symmetric stretch

A1 bend

B1 asymmetric stretch

TINJAUAN SIMETRI PADA SiH2Cl2


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Example - SiH2Cl2 Point group C2v

Character table

C2v E C2 sv(xz) sv(yz) h= 4A1 +1 +1 +1 +1 z x

2, y2, z

A2 +1 +1 -1 -1 Rz xy

B1 +1 -1 +1 -1 x, Ry xz

B2 +1 -1 -1 +1 y, Rx yzSi

Cl2

H1

Cl1

H2

z

x

y

Draw x, yand zvectors on all atoms

Count +1, -1, 0 if vector transforms to itself, minus itself, or moves

Perform symmetry operations

Character table

C2v E C2 sv(xz) sv(yz) h= 4H1 H2z

x

y


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A1 +1 +1 +1 +1 z x2, y2, z2

A2 +1 +1 -1 -1 Rz xyB1 +1 -1 +1 -1 x, Ry xz

B2 +1 -1 -1 +1 y, Rx yz

Si

Cl2

1

Cl1

2 x

Operation E

Si atom xtransforms into Si x count +1

ytransforms into Si y count +1

ztransforms into Si z count +1

total +3

Same for other 4 atoms grand total +15

Character table


x

y


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A1 +1 +1 +1 +1 z x2, y2, z2

A2 +1 +1 -1 -1 Rz xyB1 +1 -1 +1 -1 x, Ry xz

B2 +1 -1 -1 +1 y, Rx yz

Si

Cl2

1

Cl1

2 x

Operation C2 Si atom xtransforms into Si -x count -1

ytransforms into Si -y count -1


total -1

H1 and H2 move - swap places count 0

Cl1 and Cl2 swap places count 0

grand total -1

Character table


x

y


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A1 +1 +1 +1 +1 z x2, y2, z2

A2 +1 +1 -1 -1 Rz xyB1 +1 -1 +1 -1 x, Ry xz

B2 +1 -1 -1 +1 y, Rx yz

Si

Cl2

1

Cl1

2 x

Operation sv(xz) Si atom xtransforms into Si x count +1ytransforms into Si -y count -1


total +1H1 and H2 also lie in xzplane, and behave as Si count +1 each

Cl1 and Cl2 swap places count 0

grand total +3

Character table


x

y


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A1 +1 +1 +1 +1 z x2, y2, z2

A2 +1 +1 -1 -1 Rz xyB1 +1 -1 +1 -1 x, Ry xz

B2 +1 -1 -1 +1 y, Rx yz

Si

Cl2Cl1

Operation sv(yz) Si atom xtransforms into Si -x count -1ytransforms into Si y count +1


total +1

H1 and H2 swap places count 0

Cl1 and Cl2 also lie in yzplane, and behave as Si count +1 each

grand total +3

No. of modes of each symmetry species


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Example - SiH2Cl2 Point group C2v

Overall we have:

E C2 sv(xz) sv(yz)+15 -1 +3 +3

This is the reducible representationof the setof 3N (=15) atomic displacement vectors

We reduce it to the irreducible representations,using a formula

1

Reduce the reducible representation


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Character table

C2v 1E 1C2 1sv(xz) 1sv(yz) h= 4A1 +1 +1 +1 +1 z x

2, y2, z2

A2 +1 +1 -1 -1 Rz xy

B1 +1 -1 +1 -1 x, Ry xz

B2 +1 -1 -1 +1 y, Rx yz

Formula is )(.)(.1

RRgh

aR

Ri cc

Reduciblerepresentation15 -1 3 3

No. of A1 motions = 1/4 [1.15.1 + 1.(-1).1 + 1.3.1 + 1.3.1] = 5

Formula is )(.)(.1 RRgh

aR

Ri cc


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Character table


2, y2, z2

A2

+1 +1 -1 -1 Rz

xy

B1 +1 -1 +1 -1 x, Ry xz

B2 +1 -1 -1 +1 y, Rx yz


No. of A1

motions = 1/4 [1.15.1 + 1.(-1).1 + 1.3.1 + 1.3.1] = 5

No. of A2 motions = 1/4 [1.15.1 + 1.(-1).1 + 1.3.(-1) + 1.3.(-1)] = 2

Formula is )(.)(.1 RRgh

aR

Ri cc


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Character table


2, y2, z2

A2 +1 +1 -1 -1 Rz xyB1 +1 -1 +1 -1 x, Ry xz

B2 +1 -1 -1 +1 y, Rx yz


No. of A1

motions = 1/4 [1.15.1 + 1.(-1).1 + 1.3.1 + 1.3.1] = 5

No. of A2 motions = 1/4 [1.15.1 + 1.(-1).1 + 1.3.(-1) + 1.3.(-1)] = 2

No. of B1 motions = 1/4 [1.15.1 + 1.(-1).(-1) + 1.3.1 + 1.3.(-1)] = 4

No. of B2 motions = 1/4 [1.15.1 + 1.(-1).(-1) + 1.3.(-1) + 1.3.1] = 4

Translations, rotations, vibrations


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Symmetry speciesof all motions are:-

5A1 + 2A2 + 4B1 + 4B2 - the irreducible representation

3 of these are translationsof the whole molecule

3 are rotations

Symmetry species of translations are given by

vectors (x, y, z) in the character table

Symmetry species of rotations are given by Rx,Ryand Rz in the character table

ymmetry species of all motions are:-

Translations, rotations, vibrations


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ymmetry speciesof all motions are:-

5A1 + 2A2 + 4B1 + 4B2

Character table

C2v 1E 1C2 1sv(xz) 1sv(yz) h= 4A

1

+1 +1 +1 +1 z x2, y2, z2

A2 +1 +1 -1 -1 Rz xy

B1 +1 -1 +1 -1 x, Ry xz

B2 +1 -1 -1 +1 y, Rx yz

Translationsare:- A1 + B1 + B2

otationsare:- A2 + B1 + B2

so vibrationsare:- 4A1 + A2 + 2B1 + 2B2

Vibrational modes of SiH2Cl2


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Symmetry speciesof vibrations

are:- 4A1 + A2 + 2B1 + 2B2

What does each of these modes look like?

2 rules

(i) there is 1 stretching vibration per bond

(ii) must treat symmetry-related atoms together



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2 rules

(i) there is 1 stretching vibration per bond

(ii) we must treat symmetry-related atoms together

We therefore have:-

two stretching modes of the SiCl2 group

two of the SiH2 group

The remaining five modes must be deformations(angle bending vibrations)



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We therefore have:-


We can stretch the two Si-Cl bonds

together in phase

or together out of phase

Is vibration symmetrical with respectto each symmetry operation?


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- if yes +1, if no -1

From the character table, this belongsto the symmetry species A1

We call the mode of vibrationsymSiCl2

E C2 sxz syz+1 +1 +1 +1

x

z

y

Is vibration symmetrical withrespect to each symmetryoperation?


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operation?

- if yes +1, if no -1

E C2 sxz syz

From the character table, thisbelongs to the symmetry speciesB2

We call the mode of vibrationasymSiCl2

+1 -1 -1 +1

x

z

y



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We therefore have:-


We can stretch the two Si-H bonds

together in phase

or together out of phase

and two stretching modes of the SiH2 group


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x

z

y

From the character table, this belongsto the symmetry species A1

We call the mode of vibrationsym SiH2

E C2 sxz syz+1 +1 +1 +1


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From the character table, this belongs tothe symmetry species B1

We call the mode of vibrationasym SiH2

E C2 sxz syz+1 -1 +1 -1

x

z

y



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We now have:-


two of the SiH2 group

The remaining five modes must be deformations(angle bending vibrations)

As with stretches, we must treat symmetry-

related atoms together


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From the character table, this belongs tothe symmetry species A1

We call the mode of vibrationsym SiCl2(or SiCl2 scissors)

E C2 sxz syz+1 +1 +1 +1

x

z

y


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We call the mode of vibrationsym SiH2(or SiH2 scissors)

+1 +1 +1 +1

E C2 sxz syz

x

z

y


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We call the mode of vibration SiH2 (orSiH2 wag)

E C2 sxz syz+1 -1 +1 -1

x

z

y


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We call the mode of vibration SiH2 (orSiH2 rock)

+1 -1 -1 +1

E C2 sxz syz

x

z

y


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y

x


We call the mode of vibration SiH2 (orSiH2 twist)

E C2 sxz syz+1 +1 -1 -1


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For a molecular vibration to be seen in theinfrared spectrum (IR active), it must change thedipole moment of the molecule. The dipole

moment vectors have the same symmetryproperties as the cartesian coordinates x, y and z.



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Raman spectroscopy measures thewavelengths of light (in the IR range) scatted bya molecule. Certain molecular vibrations will

cause the frequency of the scattered radiation tobe less than the frequency of the incidentradiation.

http://www.uncp.edu/home/mcclurem/ptable/transelm.jpg


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For a molecular vibration to be seen in theRaman spectrum (Raman active), it must changethe polarizability of the molecule. The

polarizability has the same symmetry propertiesas the quadratic functions:

xy, yz, xz, x2, y2 and z2

Molecular Vibrations of Water

2A + B


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vib = 2A1 + B1

The two vibrations with A1 symmetryhave z as a basis function, so they will be

seen in the infrared spectrum of water.This will result in two peaks (at differentfrequencies) in the IR spectrum of water.


2A + B


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vib = 2A1 + B1

The two vibrations with A1 symmetryalso have quadratic basis functions, so they

will be seen in the Raman spectrum ofwater as well.


2A + B


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vib = 2A1 + B1

The two vibrations with A1 symmetrywill appear as two peaks in both the IR and

Raman spectra. The two frequenciesobserved in the IR and Raman for thesevibrations will be the same in both spectra.

Molecular Vibrations of Water2A B


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vib = 2A1 + B1

The vibration with B1 symmetry has xand xz as basis functions. This vibration will

be both IR active and Raman active. Thisvibration will appear as a peak (at the samefrequency) in both spectra.

Molecular Vibrations of Water 2A B


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vib = 2A1 + B1

Both the IR and Raman spectrashould show three different peaks.

Summary


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1. Obtain the point group of the molecule.2. Obtain 3N by considering the three cartesian

coordinates on all atoms that arent moved by

the symmetry operation.3. Reduce 3N .

4. Eliminate translations and rotations.

5. Determine if remaining vibrations are IRand/or Raman active.

Application: Carbonyl Stretches


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Can IR and Raman spectroscopy determine thedifference between two square planarcomplexes: cis-ML2(CO)2 and trans-ML2(CO)2?

cisand transML2(CO)2


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cisisomerC2v transisomerD2h

cis- ML2(CO)2


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C2v: E C2 xz yz

CO: 2 0 2 0

cis- ML2(CO)2


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CO reduces to A1 + B1.A1 is a symmetricstretch, and B1 is an

asymmetric stretch.

cis- ML2(CO)2


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CO reduces to A1 + B1.The symmetricstretch (A1) is IR and

Raman active.

cis- ML2(CO)2


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CO reduces to A1 + B1.The asymmetricstretch (B1) is both IR

and Raman active.

transML2(CO)2


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transisomerD2h

The transisomer lies inthe xy plane. The pointgroup D2h has thefollowing symmetry

elements:

D2h E C2(z) C2(y) C2(x) i xy xz yz

y

x

transML2(CO)2


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transisomerD2h

The transisomer lies inthe xy plane. CO isobtained by lookingonly at the two C-O

bonds.

D2h E C2(z) C2(y) C2(x) i xy xz yz

CO 2 0 0 2 0 2 2 0

y

x

transML2(CO)2


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transisomerD2h

CO

reduces to Ag

(asymmetric stretch) andB3u(an asymmetricstretch).

y

x

transML2(CO)2


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transisomerD2h

CO

reduces to Ag

(asymmetric stretch) andB3u(an asymmetricstretch).

Aghas x2, y2 and

z2 as basis functions, sothis vibration is Ramanactive.

y

x

transML2(CO)2


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transisomerD2h

Ag

has x2, y2 andz2 as basis functions, sothis vibration is Ramanactive.

B3u has x as abasis function, so thisvibration is IR active.

y

x

transML2(CO)2


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Aghas x2, y2 and z2 as basis functions, sothis vibration is Raman active.

B3u has x as a basis function, so this

vibration is IR active.

The IR and Raman spectra will each showone absorption at different frequencies.

Exclusion Rule


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If a molecule has a center of symmetry, none of its modes

of vibration can be both infrared and Raman active.


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Chemical Applications of GroupTheory


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We have learnt the point group theory of molecularsymmetry. We shall learn how to use this theory inour chemical research.

1. Representation of groups1.1 Matrix representation and reducible representation


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1.2 Reducing of representations

Suppose that we have a set of n-dimensional matrices, A, B,


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C, , which form a representation of a group. These n-D

matrices themselves constitute a matrix group.

If we make the same similarity transformation on each matrix,

we obtain a new set of matrices,

This new set of matrices is also a representation of the group.

If A is a blocked-factored matrix, then it is easy to prove that

B,C are also blocked-factored matrices.

...CC;BBAA 111 '';' GGGGGG

,.....','

4

3

2

1

4

3

2

1

B

B

B

B

B

A

A

A

A

A

A1 A2 A3 are n1 n2 n3 D submatrices with n n1 + n2 + n3 +

Furthermore, it is also provable that the various sets of

submatrices


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{A1,B1,C1}, {A2,B2,C2}, {A3,B3,C3}, {A4,B4,C4},are in themselves representations of the group.

We then call the set of matrices A,B,C, a reducible

representation of the group.

If it is not possible to find a similarity transformation to reduce

a representation in the above manner, the representation is

said to be irreducible.

The irreducible representations of a group is of fundamental

importance.

2. Character Tables of Point Groups


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These translation vectors constitute a set of bases of C2v group.


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.2 symmetry species: Mulliken symbols


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All 1-D irreducible reps. are labeled by either A or B, 2-D irreducible rep.by E, 3-D irreducible rep. by T and so on.

A: symmetric with respect to Cn rotation, i.e., c(Cn)=1.

B: asymmetric with respect to Cn rotation, i.e., c(Cn)=-1.

Subscriptions 1 or 2 designates those symmetric or asymmetric withrespect to a C2 or a sv .

Subscripts g or u for universal parity or disparity.

Superscripts or designates those symmetric or asymmetric withrespect to sh


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This formula was derived from the Great orthorgonality theorem.


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CO2 has 3 modes of vibration

Vibrational spectroscopy


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O=C=O O=C=O O=C=O

Infra-red inactive - nodipole change IR active IR active

H2O has 3 modes of vibration

IRactive

IR active IR active

HO

H HO

H HO

H

Numberof active modes tells us about symmetry

Molecular vibrations - number of modes

yz


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4 atoms - can move independently in x, y, z directions

x

xyz

xy

zxy

z

3N degrees of freedom for a N-atom molecule.

If atoms fixed, there are: 3 translational degrees

3 rotational degrees

and the rest (3N-6) are vibrational modes


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Good Luck In the Final Exam!

Final Exam


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Content: Chapters 5-9Time: June 13, 8:00-10:00

Venue: -102

Tools: : June 10-12,

316()

4) Symmetry and IRa. IR only sees a vibration if the vibration changes the molecules dipole

b. Motion along the x, y, z axes creates a changed dipole

a. Infrared Active vibrations match up with x, y, z on character table

I f d I ti ib ti d t


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b. Infrared Inactive vibrations dont

c. For water, all three vibrations are infrared active

5) Examples and Exercises pages 113-116

C. Molecular Vibrations of ML2(CO)2 complexes

The symmetry ofcis- ML2(CO)2 complexes is C2v

The C=O stretch has only one possible direction of motion

Instead of using xyz vectors at each atom, we can use a single vector


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G 2 0 2 0

2) The symmetry oftrans- ML2(CO)2 complexes is D2ha. Symmetry operations on the vectors generate a reducible representation

b. Reduction formula give 2 irreducible representations


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c. Only the B3u representation is IR Active

d. We can tell cisfrom transby the number of C=O IR bands


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2) Though more complex, the C3vCharacter Table can be generated similarly to thatof the C2vgroup


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C. Notes on Character Tables

Multiple operations in the same class are listed together

Different C2axes are listed separately with primes ()

Those through outer atoms are

Those not through outer atoms are

Symmetry of orbitals are listed except for s orbitals, which are always in the firstlisted A irreducible representation

Irreducible Representation Labels

Degeneracy (dimension) is determined by the character of E operation A if E = 1 and c of Cn = 1

B if E = 1 and c of Cn = -1

E if E = 2 (doubly degenerate)

T if E = 3 (triply degenerate)

b) Subscripts 1 if symmetric to perpendicular C2 axis (or sv)

2 if antisymmetric to perpendicular C2 axis (or sv)

g if symmetric to i

u if antisymmetric to i


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u if antisymmetric to i

Primes

if symmetric to sh

if antisymmetric to sh

IV.Applications of Symmetry Chiral Molecules

Molecules not superimposable with their mirror images are called chiral ordissymetric

They may still have some symmetry operations: E, Cn

Chiral molecules cannothave i, s, or Sn symmetry operations

IV.Applications of Symmetry Chiral Molecules

Molecules not superimposable with their mirror images are called chiral or


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p p g

dissymetric They may still have some symmetry operations: E, Cn

Chiral molecules cannothave i, s, or Sn symmetry operations

(1) Teori Grup Dan Vibrasi-revisi

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