Immediate Consequences of SymmetryI. Optical Activity (Chirality)

• If a mirror image of a molecule cannot be superimposed on the original the molecule is chiral.

• Chiral molecules are optically active in the sense that they rotate the plane of polarized light.

• For complex molecules this is difficult to visualize and symmetry elements may assist in the determination of optical activity.

Cannot Have Sn

• A molecule which contains an n-fold improper axis is always superimposable on its mirror image. Recall that Sn consists of a rotation followed by a reflection.

• Since a reflection creates the mirror image, Sn is equivalent to rotating in space the mirror image. By definition, a molecule containing an Sn axis is brought into coincidence with itslef by the Sn operation and hence its mirror image, after rotation, is superimposable.

Watch out!

• It follows from symmetry that a molecule may only be chiral if it does not possess an improper rotation axis, Sn .

• This element may be implied by other elements present. Recall that S2 is equivalent to i . Also, S1 is equivalent to .

• Therefore, molecules with any one of , i, or Sn cannot be optically active.

2. Dipole moment

• A polar molecule is one with a permanent electric dipole moment.

• This can be the case only if the center of negative and positive charges do not coincide.

• Since a symmetry operation leaves a molecule in a configuration physically indistinguishable from before, the direction of the dipole moment vector must also remain unchanged.

Must coincide with symmetry elements

• The direction of the dipole moment must be coincident with each of the symmetry elements. It cannot be perpendicular to any mirror plane or axis of rotation.

• If a molecule has a Cn axis, the dipole must lie along this axis. If there is a , it must lie in this plane; if several planes, it must lie at their intersection. E.g., for NH3, dipole moment is along C3 axis, which is also intersection of 3 v.

Limits point groups with dipoles

• It follows that molecules with an i (operation reverses direction of vectors) or with two or more non-coincident Cn axes (dipole moment cannot lie on two axes at the same time) cannot have a dipole moment (this rules out the D and higher symmetry families).

• Only molecules belonging to Cn , Cnv , or Cs may have a dipole moment. For Cn and Cnv , the dipole moment must lie along the symmetry axis, and n may be any value from 1 to .

• Thus, we must include C1 and Cv in these categories.

In-class activity

• Which of the following are:a) polar b) optically active

• CCl4 CHCl3 CO CO2

• H2O2 (skew) H2O2 (staggered)

Point Groups and Multiplication Tables

• A point group is a set of symmetry operations that form a complete multiplication table containing all products and reciprocals of its elements.

• Multiplication of two symmetry operations is defined as a sequence of symmetry operations. – The product C4 means “do the operation and

then do the C4 on the figure that results from doing the operation.”

– Note that the order of execution of the symmetry operations is from right to left.

• Every multiplication table must contain the identity.

• Every symmetry operation must have a reciprocal which is defined as the symmetry operation done in reverse.

• The product of a symmetry operation and its reciprocal is the identity operation.

Definition of a group

• A group is any collection of elements which together with some well-defined combining operation (ordinary algebra, matrix algebra, one operation followed by another, etc.) obey a certain set of rules.

1. The ‘product’ or combination of any two elements in a group must produce an element which is also in the group.

2. The group must contain the identity element, which combines with any element in the group, leaving that element unchanged.

3. The associative law must hold for all elements of the group.

4. Every element must have an inverse (combined, they yield identity) which is also a member of the group.

1. PQ = R R in the group

2. RE = ER = R R in the group

3. P(QR) = (PQ)R For all elements

4. RR1 = R1R = E R1 in the group

• The number of symmetry operations in a group is called the order of the group, h.

• A group always consists of at least two subgroups, each of which satisfies the requirements for a group.– The identity is always a subgroup of order 1.– The order, g, of a subgroup, is always an

integral divisor of the order, h, of the group to which it belongs.

• A group always consists of two or more classes, each of which contains symmetry operations which can be transformed into one another by a symmetry operation of the group.

Order and Classes

B

F

F

FC2, v

C2, v

C2, v

C3, S3

Molecular plane is h.

The three C2 axes are equivalent so the C2 operations are in the same class.

The three v planes are equivalent so the v operations are in the same class.

Note that the separate existence of a C3 and a h requires the existence of an S3 axis.

The C3 and C32 operations

are in the same class.

The S3 and S35 operations

are in the same class.

Rules For Symmetry Operations and Classes

• The Inversion:

– There is only one inversion and it is always in a class by itself.

• Reflections:

– Reflection in a plane perpendicular to the axis of highest symmetry is designated h and forms a separate class.

– Reflection in a plane that contains the axis of highest symmetry and passes through several atoms is designated v. All are in the same class.

– Reflection in a plane that contains the axis of highest symmetry and passes between sets of atoms is designated d. All are in the same class.

Proper Rotations:

– In the Cn point groups, Cn, Cn2, Cn

3, etc are in separate classes.

– In groups of higher symmetry, Cnm and

Cnn-m are in the same class.

Systematic Generation of Point Groups

• We will examine the generation of some of the simpler point groups by constructing multiplication tables for symmetry operations.

• Each point group is generated by starting with the identity and at least one other symmetry operation.

• The additional symmetry operations needed to complete the point group will arise automatically since all products must be members of the group.

• Multiplication is not necessarily commutative (AB is not necessarily the same as BA).

• Multiplication is always associative (A[BC] = [AB]C).• No two rows and no two columns in a multiplication

table can be the same.

Groups of Low Symmetry

No symmetry: Only the identity is a symmetry operation.

C1 E

EE

The designation of this as the C1 point group means that the identity operation is considered as a C1 rotation axis. Since C1 is rotation by 2/1, this corresponds to doing nothing to the object.

Schönflies symbol for the point group.

Cs

E

E

E h

h

h h

ENote that, like the number 1, the identity multiplied by anything else leaves the thing it is multiplied by unchanged.

Multiplication by the identity is always commutative.

h is reflection in the molecular plane. Since there is only one reflection plane, it is automatically the molecular plane.

C3

E

E

E E

C3

C3

C3 C32

C32 C3

2

C3

C3

C32

C32

E

C32 C3

2 = four C3 operations in sequence = C3

4 = C33C3 (associative law)

Since C33 = E, C3

2C32 = EC3 = C3

(multiplication by the identity doesn’t change anything)

Note that the product C3C3 has generated the new product C3

2 which must also be a member of the group.

Vectors, Matrices and Point Groups

• In the quantum description of a chemical bond, the wavefunctions for the electrons in the bond have the same directional properties as the bond.

• Since wavefunctions for the electrons in a bond have directional properties, they can be treated as vectors.

• Group Theory provides a systematic classification of vectors in terms of their geometric properties.

• Any vector can be written as the product of a column matrix of unit vectors and a square matrix describing the geometric properties of the vector.

• The properties of this square geometric matrix provide the quantitative link between Group Theory and chemical bonding.

Representations of Point Groups

• Each symmetry operation has geometric properties that can be expressed in a geometric transformation matrix that is specific for that operation.

• A complete set of geometric transformation matrices for the operations of a point group as applied to a particular object is called the representation of the group.

• In many cases, the full geometric transformation matrix consists of smaller square matrices arranged along the diagonal of the larger matrix. Corresponding sets of these smaller matrices are also representations of the group.

• Representations consisting of matrices that do not factor into smaller ones are called irreducible representations.

• The dimension of a representation is the number of rows or columns in its geometric transformation matrices.

a b c

d e f

g h i

j k

l m

n o p q

r s t u

v w x y

z

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0

0 0 0 0

0 0

0 0

0 0

0 0

0 0 0

0 0 0

0 0 0

0 0 0

0 0 0

0 0 0

The Characters of Matrices

• Using the full geometric matrices is very cumbersome – Each atom in a molecule requires three unit vectors (three

coordinates) to locate it.

– For a molecule containing n atoms the geometric matrix would be 3n x 3n for each symmetry operation in the group.

• It is much more convenient to describe the molecule in terms of the characters of the geometric transformation matrices.

• The character, , of a square matrix is the sum of its diagonal elements.

• The representations of molecules are nearly always written as the characters of the symmetry operations, not as the actual geometric transformation matrices.

Notes

• The principal rotation axis is chosen as the z-axis.

• There is always one completely symmetric irreducible representation (each element is 1).

Reflection in a Plane:Initial Vector

Reflected Vector

x, y plane

The x and y unit vectors are in the reflection plane and are unchanged.

The z unit vector is perpendicular to the plane and reverses.

x2 = x1 + 0 y1 + 0 z1

y2 = 0 x1 + y1 + 0 z1

z2 = 0 x1 + 0 y1 – z1

x2

y2

z2

1 0 0

0 1 0

0 0 -1

x1

y1

z1

=

The Geometric Transformation Matrices

Transformation matrix

In general, for a reflection transform, all off-diagonal elements are 0, on-diagonal are 1 for coordinates in plane, 1 for the other.

Inversion:

x2 = -x1 + 0 y1 + 0 z1

y2 = 0 x1 - y1 + 0 z1

z2 = 0 x1 + 0 y1 - z1

x2

y2

z2

-1 0 0

0 -1 0

0 0 -1

x1

y1

z1

=

0 if off-diagonal, 1 on-diagonal (all coordinates are inverted).

Proper Rotation by an angle :

x2

y2

z2

=

cos -sin 0

sin cos 0

0 0 1

x1

y1

z1

We now have non-zero off-diagonal elements, which sometimes complicates matters, which we’ll try to avoid in this course.

Improper Rotation:

x2 = x1 cos - y1 sin + 0 z1

y2 = x1 sin + y1 cos + 0 z1

z2 = 0 x1 + 0 y1 – z1

x2

y2

z2

=

cos -sin 0

sin cos 0

0 0 -1

x1

y1

z1

The only difference from the proper rotation is the effect on the z vector of the reflection perpendicular to the rotation axis.

Rules for Manipulating Representations

1. The sum of the squares of the dimensions (li2)

of the irreducible representations of a group is equal to the order (h) of the group.

hllli 22

21

tionsrepresenta all

2

2. The sum of the squares of the characters ([i(R)]2) in any one irreducible representation of a group equals the order (h) of the group. (R is any symmetry operation, is the character, i and j for different representations)

hRi R operations all

2

3. The vectors whose components are the characters [i(R)] of two different irreducible representations (i and j) are orthogonal.

0R operations all

RR ji

4. In any given representation, either reducible or irreducible, the characters of all matrices belonging to the same class are identical.

5. The number of irreducible representations in a group is equal to the number of classes in the group.

Application to the C2v Point Group

• This point group has four symmetry operations, E, C2, v(x,z) and v(y,z). Each is in a separate class.

Number of irreducible representations = ?? (Rule 5) 4

Possible dimensions of representations are ?? (Rule 1) 1

• Since the order of the group is 4, the sum of the li2 = 4.

• There are four irreducible representations and the smallest possible matrix is 1x1 so the smallest possible value of l is 1.

• Only l = 1 will sum to 4 when squared and added four times.

x2

y2

z2

1 0 0

0 1 0

0 0 1

x1

y1

z1

=E

x2

y2

z2

=

cos -sin 0

sin cos 0

0 0 1

x1

y1

z1

x2

y2

z2

-1 0 0

0 -1 0

0 0 1

x1

y1

z1

=

C2

For C2 : θ = 180°Therefore sin 180° = 0and cos 180° = -1

σv(xz)

x2

y2

z2

1 0 0

0 -1 0

0 0 1

x1

y1

z1

=

x2

y2

z2

-1 0 0

0 1 0

0 0 1

x1

y1

z1

=

σv’(yz)

E C2

σv(xz) σv’(yz)

We now “block diagonalize” each transformation matrix- break it down into smaller matrices including the non-zero elements along the diagonal.

[1] 0 0

0 [1] 0

0 0 [1]

[-1] 0 0

0 [-1] 0

0 0 [1]

[-1] 0 0

0 [1] 0

0 0 [1]

[1] 0 0

0 [-1] 0

0 0 [1]

In this case the x, y, z axes are also block diagonalized, and can be treated independently.

E C2 v(xz) v(yz)

1 -1 1 -1

3 -1 1 1

1 1 1 1

1 -1 -1 1

coordinate

x

y

z

Γ

Here we have composed representations by simply transcribing the element (the character of a 1x1 matrix) in the 1,1 position of each transformation matrix, which describes the result of the symmetry operation on the x axis; and similarly for 2,2 (y) and 3,3 (z). By adding the values in each column above, we can get a reducible representation.

4th representation

•We obtain this by following all the rules:

The character for the identity must be 1

The sum of squares of the characters in each representation must equal the order (4)

No two representations may be the same

The representations must all be

orthogonal to each other.

C2v E C2 v(xz) v(yz)

1

2

1 1 1 1

1 -1 1 -1

3

4 1 -1 -1 1

1 1 -1 -1

The Structure of Character Tables

• The left-hand column gives the Schönflies symbol for the point group with the Mulliken symbols for the irreducible representations below it.

• The main part of the table has the symmetry operations at the top, arranged in classes, with the identity at the left followed by the other symmetry operations starting with the highest order rotation axis and followed in order of decreasing symmetry. Below this are listed the characters for the symmetry operations in each irreducible representation.

• At the far right are two columns showing the transformation properties of coordinates and their squares and cross-products.

Rules for Assigning Mulliken Symbols

1. All one-dimensional representations are either A or B. Two-dimensional representations are E and three-dimensional ones are T or F.

2. One-dimensional representations that are symmetric (positive character) with respect to rotation about the principal axis are A while those that are antisymmetric (negative character) are B.

3. The subscript 1 designates an A or B representation that is symmetric with respect to rotation about a C2 Cn or, in its absence, is symmetric with respect to a v. The subscript 2 indicates antisymmetric behaviour.

4. A ‘ designates symmetry with respect to a h. A “ indicates antisymmetric behaviour.

5. The subscript g (for gerade meaning even in German) indicates symmetry with respect to inversion. The subscript u indicates antisymmetric behaviour.

The column containing x, y, z, Rx, Ry and Rz gives the irreducible representations for translation along the three Cartesian axes (x, y, z) and for rotation about the three Cartesian axes (Rx, Ry, Rz).

The column at the far right containing the squares and cross-products gives the irreducible representations for properties (such as d-orbital wavefunctions and molecular polarizability) that depend on these squares and cross-products.

C2v E C2 v(xz)

A1

A2

B1

1 1 1 1

1 1 -1 -11 -1 1 -11 -1 -1 1

z

Rz

x, Ry

x2, y2, z2

xz

v(yz)

B2 y, Rx yz

xy

Groups with Classes Larger than 1

• Recall that the characters of all the matrices in a class are identical. This means that– Character tables can be compressed by lumping all the

operations in a class in a single column.

– The sums over symmetry operations required in analyzing representations can be taken over classes provided we multiply the product for a given class by the order, g, of the class.

hgR kki k classes, all

2Rule 2:

0k classes, all

kkji gRR Rule 3:

Remember that the class order for each symmetry operation can be different.

Consider the C3v Point GroupC3v E 2C3 3v

1

3

2

1 1 1

1 1 -1

2 -1 0

Rule 1: There are 6 operations so l2 = 6 l =?? 1, 1, 2

Rule 5: There are 3 classes so the number of irreducible representations = ? 3

hgR kki k classes, all

2For the one dimensional representations, can be??

For the two dimensional representation, (E) has to be??

± 1

2 since each diagonal element has to be 1 (the unit vectors are unchanged). This means that (C3) =??-1 and (v) =?? 0.

x2

y2

z2

1 0 0

0 1 0

0 0 1

x1

y1

z1

=E

x2

y2

z2

=

cos -sin 0

sin cos 0

0 0 1

x1

y1

z1

2C3

For C3 : θ = 120° (2/3)Therefore sin 120° = √3/2

and cos 120° = -1/2

3σv(xz)

x2

y2

z2

1 0 0

0 -1 0

0 0 1

x1

y1

z1

=

x2

y2

z2

-1/2 -√3/2 0

√3/2 -1/2 0

0 0 1

x1

y1

z1

=

E 2C3

3σv(xz)

The transformation matrix for C3 cannot be block diagonalized into 1x1 because it has off-diagonal elements. It can be blocked into 2x2 and 1x1 matrices; the others must follow the same pattern for consistency.

1 0 0

0 1 0

0 0 [1]

1 0 0

0 -1 0

0 0 [1]

The C3 matrix must be blocked this way because the (x,y) combination is needed for the new x,y coordinates.

-1/2 -√3/2 0

√3/2 -1/2 0

0 0 [1]

Building the representationsC3v E 2C3 3v

1

3

2

1 1 1

1 1 -1

2 -1 0

Here we will compose the 2-D representation by taking the characters of the 2x2 matrices (summing along the diagonal), and one of the 1-D representations from the characters of the 1x1 matrices.

coordinate

x, y

z

The third representation can be found by using the defining properties of group tables.

Practice problem

• For the D2d point group (character table on next slide):

• A) determine the order of the group• B) test the E representation for orthogonality

with the other 4 representations (Rule 3)• C) Confirm that the sum of the squares of the

characters in each irreducible representation equals the order of the group (Rule 2)

D2d E 2S4 C2 2C2’ 2d

A1 1 1 1 1 1 A2 1 1 1 1 1 B1 1 1 1 1 1

B2 1 1 1 1 1 E 2 0 2 0 0

Reducing Reducible Representations

• Virtually every practical application of group theory starts with a reducible representation.

• All reducible representations contain some integer multiple of each of the irreducible representations of the point group.

• The application of group theory to problems begins by determining how many of each kind of irreducible representation are contained in the reducible representation.

• The irreducible representations can be thought of as being like components of a vector and determine the geometric features of the phenomenon being analyzed.

• We could construct a reducible representation for a point group by multiplying each irreducible representation by some integer, a, and adding the result for each symmetry operation, R, over all irreducible representations. The character of each symmetry operation in the reducible representation would then be

tionsrepresenta

eirreducibl

RaR ii

• We normally start with a knowledge of (R) in the reducible representation and need to determine the value of ai for each irreducible representation in the group.

• Since irreducible representations are orthogonal, if we multiply the characters in the reducible representation by those in a given irreducible representation, only the terms for the “test” irreducible representation will survive.

These observations can be summarized in the equation

RR

ii gRRh

a 1

in which (R) is the character for symmetry operation R in the reducible representation and i(R) is the character for the same symmetry operation in the ith irreducible representation. ai is the number of times the ith irreducible representation is repeated in the reducible one.

A C3v Example

C3v E 2C3 3v

1

3

2

1 1 1

1 1 -1

2 -1 0

reducible 7 1 -3

06

927

133112176

1

326

1131311

vvCCEEa For 1:

For 2:

36

18

133112176

12

a

For 3:

26

12

033112276

13

a

Application to Molecular Vibrations and Infrared and Raman

Spectroscopy

• For a molecule containing n atoms, 3n coordinates (component vectors) are needed to completely define the positions of all the atoms.

• Translational motion of the molecule through space accounts for three coordinates, representing the location of the center of mass.

• Rotational motion of the molecule accounts for– Two coordinates if the molecule is linear.– Three coordinates if the molecule is not linear.

• The remaining coordinates, (3n-5) or (3n-6), are required to describe vibrations of the atoms relative to each other.

Probing Structure With Infrared Spectroscopy

• When a molecule vibrates, its dipole moment may oscillate.

• This oscillating dipole moment interacts with the electric vector of electromagnetic radiation (light).

• Only those vibrations which have the same irreducible representations as translational motion along x, y, or z will interact with light. These will be vibrations which cause an oscillation in the (temporary) dipole moment.

• The reducible representation of the molecule contains the relation between molecular structure and this interaction.

• When a molecular vibration has the same frequency as the light shining on it, the vibration will absorb energy from the light beam.

• The frequency (energy) at which the light is absorbed depends on the strength of the vibrating bond and the masses of the atoms involved in the bond.• The energies of molecular vibrations are of the same magnitude as the energy of light in the infrared part of the spectrum.

• The number of peaks in the infrared spectrum of a molecule can be determined by using group theory and this provides a link between molecular structure and the infrared spectrum.

The Recipe

• Determine the reducible representation for all motions of the molecule.– Attach three unit vectors (x, y, z) to each atom.– For each symmetry operation, only those atoms

that do not change position contribute to the character for that symmetry operation in the reducible representation.

– A unit vector that does not change its orientation contributes +1 to the character in the reducible representation.

– A unit vector that reverses its direction contributes –1.

• Reduce the reducible representation to find the number of each kind of irreducible representation it contains.

• Subtract the irreducible representations for translation (x, y, z) and rotation (Rx, Ry, Rz) leaving those for the vibrations.

•For Cn operations, the sum of the diagonal elements of the geometric matrix is always 1+2cos per unshifted atom for rotation by the angle .•For Sn operations, the sum of the diagonal elements of the geometric matrix is always –1+2cos per unshifted atom for rotation by the angle .

The Ammonia MoleculeThe ammonia molecule belongs to the

C3v point group.

N

HH

H

C3

v

v

v

x

zy

x

zy

x

zyx

zy

There are 12 vectors all of which are unchanged by E so (E) = 12.E:-

C3:- Only N is unshifted by C3. The rotation angle is 2/3 so the character for C3 is 1+2cos(2/3) = 1 + 2x(-0.5) = 0.v:- For any reflection plane, the N and one H are un-shifted. Since the character per atom for v is 1 (z and x stay the same, y changes sign), the total character for the two atoms is 2.

The reducible representation is red. 12 0 2

This factors into the following irreducible representations.

36

18123102112

6

1)( 1 An

16

6123102112

6

12 An

46

24023102212

6

1En

red = 3A1 + 1A2 + 4E

C3v E 2C3 3v

A1

A2

E

1 1 1

1 1 -1

2 -1 0

C3v E 2C3 3v

A1

A2

E

1 1 1

1 1 -1

2 -1 0

z

Rz

(x,y) (Rx, Ry)

x2 + y2, z2

(x2 - y2, xy) (xz, yz)

red = 3A1 + 1A2 + 4E

Translation contributes A1 + E (remember that E is doubly degenerate so one E accommodates translation along x and y).Rotation contributes A2 + E (again E is doubly degenerate and so includes both Rx and Ry).

The irreducible representations for the vibrations of NH3 are therefore 2A1 + 2E.

The number of vibrational modes is 2 (from A1) plus 2x2 = 4 (from the doubly degenerate E) for a total of 6. Since the molecule is nonlinear, the number of vibrations is 3x4 – 6 = 6 which agrees with the group theory calculation as required.

Infrared active transitions have to transform like x, y, or z so we conclude that all these vibrations are infrared active.

The A1 vibrations are symmetric with respect to both C3 and v and so describe the symmetric N-H bend and the symmetric N-H stretch.

N

HH

H

Symmetric Stretch

N

HH

H

Symmetric Bend

The two doubly-degenerate E vibrations are antisymmetric with respect to the C3 axis. They describe asymmetric bends and stretches.

N

HH

H

Asymmetric Stretch

N

HH

H

Asymmetric Bend

Practice Problem

• Using the x, y, and z coordinates for each atom in CH4, determine the reducible representation; reduce it; classify the irreducible representations into translational, rotational and vibrational modes; and decide which vibrational modes are IR active.

Analysis of Specific VibrationsSuppose we want to find the number of infrared active stretching vibrations in Ni(CO)4. The molecule is tetrahedral as shown below.

Ni

COCO

CO

COWe start by drawing vectors pointing in the direction of the stretch along each bond axis.

The vibrations will have the transformation properties of these vectors so we deduce the reducible representation for this set of vectors in the normal way and then decompose it to find what irreducible representations it contains.

Td Point Group

Td E 8C3 3C2 6S4 6d

A1 1 1 1 1 1 x2 + y2 + z2

A2 1 1 1 -1 -1

E 2 -1 2 0 0 (2z2-x2-y2, x2-y2)

T1 3 0 -1 1 -1 (Rx,Ry,Rz)

T2 3 0 -1 -1 1 (x, y, z) (xy, xz, yz)vib

Ni

COCO

CO

CO

C3

C2,S4,d

E leaves all four vectors in place and unchanged in direction so it contributes +4.

4

C3 leaves one vector in place and unchanged in direction so it contributes +1.

1

C2 and S4 move all the vectors and so do not contribute.

0 0

d leaves two vectors in place and unchanged in direction so it contributes +2.

2

Td E 8C3 3C2 6S4 6d

A1 1 1 1 1 1 x2 + y2 + z2

A2 1 1 1 -1 -1

E 2 -1 2 0 0 (2z2-x2-y2, x2-y2)

T1 3 0 -1 1 -1 (Rx,Ry,Rz)

T2 3 0 -1 -1 1 (x, y, z) (xy, xz, yz)vib 4 1 0 0 2

Next decompose the reducible representation.

124

241260011814

24

11

An

0)1(261181424

12

An

0026)1(182424

1En

0)1(260183424

11

Tn

11260183424

12

Tn

There are 1 A1 and 1 triply degenerate T2 (four vibrations).

Only T2 transforms like x, y, z so only it is infrared active.

Raman Spectroscopy

• Raman effect depends on change in polarisability .(See far-right column in character tables; explicitly given in Hollas spectroscopy text).

• General Rule: If a vibrational mode has the same symmetry as one or more of the binary combinations of x, y and z the transition from this mode will be Raman active.

• Infrared and Raman are based on two DIFFERENT phenomena and therefore there is no necessary relationship between the two activities.

• The higher the molecular symmetry the fewer “co-incidences” between Raman and infrared active modes.

C2v E C2 v(xz)

A1

A2

B1

1 1 1 1

1 1 -1 -11 -1 1 -11 -1 -1 1

z

Rz

x, Ry

x2, y2, z2

xz

v(yz)

B2 y, Rx yz

xy

For C2v x2, y2, z2, xy, xz and yz

Transform as: A1, A1, A1, A2, B1 and B2

Therefore, any vibrational mode in a C2v molecule is Raman active (while A1, B1, and B2 modes are also IR active).

C3vE 2C3 v

A1 1 1 1 Tz x2 + y2, z2

A2 1 1 -1 Rz

E 2 -1 0 (Tx,Ty) or (Rx,Ry)

(x2 - y2, xy)

(yz, xz)

For C3v, only A1 and E modes would be

Raman active (same for IR activity).

Td E 8C3 3C2 6S4 6d

A1 1 1 1 1 1 x2 + y2 + z2

A2 1 1 1 -1 -1

E 2 -1 2 0 0 (2z2-x2-y2, x2-y2)

T1 3 0 -1 1 -1 (Rx,Ry,Rz)

T2 3 0 -1 -1 1 (x, y, z) (xy, xz, yz)

Self-test: Which modes in a tetrahedral molecule are Raman active? Which are IR active?