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Reducible representation: How the basis objects transform. By adding and subtracting the basis objects the reducible representation can be reduced to a combination irreducible representations.
# of irreducible representations = # of classes of symmetry operations
Basis of a Representation: A set of objects capable of demonstrating the effects of the all the symmetry operations in a Point Group. A set of functions, atomic orbitals on a central atom or ligands, or common objects.
Representation: How the operations affect the basis objects by transforming them into themselves or each other by the operations. The representation is an array showing how the basis objects transform. Ususally we do not need the full matrices but only the trace of the matrices. These values are called the characters. The collected set of character represntations is called the character table.
Irreducible representations: A representation using a minimal set of basis objects. A single atomic orbital (2s, 2pz), a linear combination of atomic orbitals or hybrids (h1 + h2), etc. An irreducible representation may be minimally based on
• A single object, one dimensional, (A or B)• Two objects, two dimensional, (E)• Three objects, three dimensional, (T)
Effect of the 4 operations in the point group C2v
on a translation in the x direction. The translation is simplymultiplied by 1 or -1. It forms a 1 dimensional basis to show
what the operators do to an object.
Operation E C2 v v’
Transformation 1 -1 1 -1
Naming of Irreducible representations • One dimensional (non degenerate) representations are designated A or B. (A basis object is
only changed into itself or the negative of itself by the symmetry operations)
• Two-dimensional (doubly degenerate) are designated E. (Two basis object are required to repesent the effect of the operations for an E representation. In planar PtCl4
2- the px and py
orbitals of the Pt, an E representation, are transformed into each other by the C4 rotation, for instance.)
• Three-dimensional (triply degenerate) are designated T. (Three objects are interconverted by the symmetry operations for the T representations. In tetrahedral methane, Td, all three p orbitals are symmetry equivalent and interchanged by symmetry operations)
• Any 1-D representation symmetric with respect to Cn is designated A; antisymmétric ones are designated B
• Subscripts 1 or 2 (applied to A or B refer) to symmetric and antisymmetric representations with respect to C2 Cn or (if no C2) to v respectively
• Superscripts ‘ and ‘’ indicate symmetric and antisymmetric behavior respectively with respect to h.
• In groups having a center of inversion, subscripts g (gerade) and u (ungerade) indicate symmetric and antisymmetric representations with respect to i
But note that while this rationalizes the naming, the behavior with respect to each operation is provided in the character table.
Character Tables
• You have been exposed to symmetry considerations for diatomic molecules: or bonding.
• Characters indicate the behavior of an orbital or group of orbitals under the corresponding operations (+1 = orbital does not change; -1 = orbital changes sign; anything else = more complex change)
• Characters in the E column indicate the dimension of the irreducible representation (of degenerate orbitals having same energy)
• Irrecible representations are represented by CAPITAL LETTERS (A, B, E, T,...) whereas orbitals of that symmetry behavior are represented in lowercase (a, b, e, t,...)
• The identity of orbitals which a row represents is found at the extreme right of the row
• Pairs in brackets refer to groups of degenerate orbitals and, in those cases, the characters refer to the properties of the set
Definition of a Group
• A group is a set, G, together with a binary operation, *, such that the “product” of any two members of the group is a member of the group, usually denoted by a*b, such that the following properties are satisfied :
– (Associativity) (a*b)*c = a*(b*c) for all a, b, c belonging to G. – (Identity) There exists e belonging to G, such that e*g = g = g*e
for all g belonging to G. – (Inverse) For each g belonging to G, there exists the inverse of
g, g-1, such that g-1*g = g*g-1 = e.
• If commutativity ( a*b = b*a) for all a, b belonging to G, then G is called an Abelian group.
The symmetry operations of a Point Group comprise a “group”.
Example
Is there an identity element, e? a*e = a
Yes, 0
Consider the set of all integers and the operation of addition (“*” = +)
Is this set of objects (all integers) associative under the operation? (a*b)*c = a*(b*c)
Yes, (3 + 4) + 5 = 3 + (4+5)
For each element is there an inverse element, a-1? a-1 * a = e
Yes, 4 + (-4) = 0
Abelian? Is commutativity satisfied for each element? a * b = b * a
We have a group.
Yes. 3 + (-5) = (-5) + 3
As applied to our symmetry operators.
For the C3v point group
What is the inverse of each operator? A * A-1 = E
E C3(120) C3(240) v (1) v (2) v (3)
E C3(240) C3(120) v (1) v (2) v (3)
Examine the matrix representation of the elements of the C2v point group
1 0 0
0 1 0 0 0 1
x y z
=
x y z
-1 0 0
0 -1 0 0 0 1
x y z
=
-x -y
z
C2
1 0 0
0 -1 0 0 0 1
x y z
=
x -y
z
v(xz)
E
- x y z
’v(yz)
Multiplying two matrices (a reminder)
a11 a12 a21 a22 a31 a32
b11 b12 b13 b21 b22 b23
=
c11 c12 c13 c21 c22 c23 c31 c32 c33
c11 = a11b11 + a12b21c12 = a11b12 + a12b22c13 = a11b13 + a12b23
c21 = a21b11 + a22b21c22 = a21b12 + a22b22c23 = a31b13 + a32b23
c31 = a31b11 + a32b21c32 = a31b12 + a32b22c33 = a31b13 + a32b23
±1 0 0
0 ±1 0 0 0 ±1
Most of the transformationmatrices we use have the form
-1 0 0
0 -1 0 0 0 1
x y z
=
1 0 0
0 1 0 0 0 1
x y z
=
1 0 0
0 -1 0 0 0 1
x y z
=
C2E v(xz) ’v(yz)
What is the inverse of C2? C2
What is the inverse of v? v
-1 0 0
0 -1 0 0 0 1
x y z
=
-1 0 0
0 -1 0 0 0 1
x y z
==
1 0 0
0 1 0 0 0 1
x y z
=
1 0 0
0 -1 0 0 0 1
x y z
=
1 0 0
0 -1 0 0 0 1
x y z
==
1 0 0
0 1 0 0 0 1
x y z
=
-1 0 0
0 -1 0 0 0 1
x y z
=
1 0 0
0 1 0 0 0 1
x y z
=
1 0 0
0 -1 0 0 0 1
x y z
=
C2E v(xz) ’v(yz)
What of the products of operations?
E * C2 = ? C2
1 0 0
0 1 0 0 0 1
x y z
=
-1 0 0
0 -1 0 0 0 1
x y z
==
-1 0 0
0 -1 0 0 0 1
x y z
=
v * C2 = ? ’v
1 0 0
0 -1 0 0 0 1
x y z
=
-1 0 0
0 -1 0 0 0 1
x y z
==
Classes
Two members, c1 and c2, of a group belong to the same class if there is a member, g, of the group such that
g*c1*g-1 = c2
Consider PtCl4
C2 C2
C2(x) C2(y)
So these operations belong to the same class?
C2C2
C4
C2(x) =
100
010
001
C2(y) =
100
010
001
Since C4 moves C2(x) on top of C2(y) it is an obvious choice for g
100
001
010
100
001
010
C4 =
C43 =
C2(x)
100
010
001
C2(y)
100
010
001
100
001
010
100
001
010
C4 C43
=
Belong to same class!
How about the other two C2 elements?
Properties of Characters of Irreducible Representations in Point Groups
• Total number of symmetry operations in the group is called the order of the group (h). For C3v, for example, it is 6.
1 + 2 + 3 = 6
• Symmetry operations are arranged in classes. Operations in a class are grouped together as they have identical characters. Elements in a class are related.
This column represents three symmetry operations having identical characters.
Properties of Characters of Irreducible Representations in Point Groups - 2
The number of irreducible reps equals the number of
classes. The character table is square.
3 by 3
The sum of the squares of the dimensions of the each irreducible rep equals the order of the group, h.
1 + 2 + 3 = 6 = h
1
1
22
6 = h
Properties of Characters of Irreducible Representations in Point Groups - 3
For any irreducible rep the squares of the characters summed over the symmetry operations equals the order of the group, h.
A1: 12 + 2 (12) + 3 (12) = 6 = # of sym operations = 1+2+3
A2: 12 + 2 (12) + 3((-1)2) = 6
E: 22 + 2 (-1)2 + 3 (0)2 = 6
Properties of Characters of Irreducible Representations in Point Groups - 4
Irreducible reps are orthogonal. The sum over the symmetry operations of the products of the characters for two different irreducible reps is zero.
For A1 and E:
1 * 2 + 2 (1 *(-1)) + 3 (1 * 0) = 0
Note that for any single irreducible rep the sum is h, the order of the group.
Properties of Characters of Irreducible Representations in Point Groups - 5
Each group has a totally symmetric irreducible rep having all characters equal to 1
Reduction of a Reducible Representation. Given a Reducible Rep how do we find what Irreducible reps it contains?
Irreducible reps may be regarded as orthogonal vectors. The magnitude of the vector is h-1/2
Any representation may be regarded as a vector which is a linear combination of the irreducible representations.
Reducible Rep = (ai * IrreducibleRepi)
The Irreducible reps are orthogonal. Hence for the reducible rep and a particular irreducible rep i
(character of Reducible Rep)(character of Irreducible Repi) = ai * h
Or
ai = (character of Reducible Rep)(character of Irreducible Repi) / h
Sym ops
Sym ops
Reducible Representations in Cs = E and h
Use the two sp hybrids as the basis of a representation
h1 h2
10
01
01
10
E operation. h operation.
2
1
h
h=
2
1
h
h
2
1
h
h=
1
2
h
h
The hybrids are unaffected by the E operation.
The reflection operation interchanges the two hybrids.
Proceed using the trace of the matrix representation.
1 + 1 = 2 0 + 0 = 0
h1 becomes h1; h2 becomes h2. h1 becomes h2; h2 becomes h1.
10
01
01
10
2
1
h
h=
2
1
h
h
2
1
h
h=
1
2
h
h
The hybrids are unaffected by the E operation.
The reflection operation interchanges the two hybrids.
Proceed using the trace of the matrix representation.
1 + 1 = 2 0 + 0 = 0
h1 , h1; become themselves
h1 h2; do not become themselves, interchange
Let’s observe one helpful thing here.
Only the objects (hybrids) that remain themselves, appear on the diagonal of the transformation of the symmetry operation, contribute to the trace. They commonly contribute +1 or -1 to the trace depending whether or not they are multiplied by -1.
The Irreducible Representations for Cs.
Cs E h
A’
A”
1 1
1 -1
x, y,Rz
z, Rx,Ry
x2,y2,z2,xy
yz, xz
The reducible representation derived from the two hybrids can be attached to the table.
2 0 (h1, h2)
Note that = A’ + A”
h1h2
, ,h1 - h2 h1 + h2
The Irreducible Representations for Cs.
Cs E h
A’
A”
1 1
1 -1
x, y,Rz
z, Rx,Ry
x2,y2,z2,xy
yz, xz
The reducible representation derived from the two hybrids can be attached to the table.
2 0 (h1, h2)
Let’s verify some things.
Order of the group = # sym operations = 2
A’ and A’’ are orthogonal: 1*1 + 1+(-1) = 0
Sum of the squares over sym operations = order of group = h
The magnitude of the A’ and A’’ vectors are each (2) 1/2: magnitude2 = ( 12 + (+/- 1)2)
Now let’s do the reduction.
We assume that the reducible rep G can be expressed as a linear combination of A’ and A’’
= aA’ A’ + aA’’ A’’; our task is to find out the coefficients aA’ and aA’’
Cs E h
A’
A”
1 1
1 -1
x, y,Rz
z, Rx,Ry
x2,y2,z2,xy
yz, xz
2 0 (h1, h2)
aA’ = (1 * 2 + 1 *0)/2 = 1
aA’’ = (1 * 2 + 1 *0)/2 = 1
Or again G = 1*A’ + 1*A’’. Note that this holds for any reducible rep G as above and not limited to the hybrids in any way.
Point group Symmetry operations
Characters+1 symmetric behavior
-1 antisymmetricMülliken symbols
Each row is an irreducible representation
Water is C2v. Let’s use the Character Table
# of atoms degrees of freedom
Translational modes
Rotational modes
Vibrational modes
N (linear) 3 x N 3 2 3N-5
Example
3 (HCN)
9 3 2 4
N (non- linear)
3N 3 3 3N-6
Example
3 (H2O)
9 3 3 3
Let’s determine how many independent vibrations a molecule can have. It depends on how many atoms, N, and whether
the molecule is linear or non-linear.
Symmetry and molecular vibrations
A molecular vibration is IR activeonly if it results in a change in the dipole moment of the molecule
A molecular vibration is Raman activeonly if it results in a change in the polarizability of the molecule
In group theory terms:
A vibrational mode is IR active if it corresponds to an irreducible representationwith the same symmetry of a x, y, z coordinate (or function)
and it is Raman active if the symmetry is the same asA quadratic function x2, y2, z2, xy, xz, yz, x2-y2
If the molecule has a center of inversion, no vibration can be both IR & Raman active
How many vibrational modes belong to each irreducible representation?
You need the molecular geometry (point group) and the character table
Use the translation vectors of the atoms as the basis of a reducible representation.
Since you only need the trace recognize that only the vectors that are either unchanged or have become the negatives of themselves by a symmetry operation contribute to the character.
Apply each symmetry operation in that point group to the moleculeand determine how many atoms are not moved by the symmetry operation.
Multiply that number by the character contribution of that operation:
E = 3 = 1C2 = -1i = -3C3 = 0
That will give you the reducible representation
A shorter method can be devised. Recognize that a vector is unchanged or becomes the negative of itself if the atom does not move. A reflection will leave two vectors unchanged and multiply the other by -1 contributing +1.For a rotation leaving the position of an atom unchanged will invert the direction of two vectors, leaving the third unchanged.Etc.
3x39
1x-1-1
3x13
1x11
Finding the reducible representation
(# atoms not moving x char. contrib.)
E = 3 = 1C2 = -1i = -3C3 = 0
Now separate the reducible representation into irreducible onesto see how many there are of each type
A1 = 1/4 (1x9x1 + 1x(-1)x1 + 1x3x1 + 1x1x1) = 3
A2 = 1/4 (1x9x1 + 1x(-1)x1 + 1x3x(-1) + 1x1x(-1)) = 1
9 -1 3 1
ML C
CLO
OC2
E
1
2
ML C
CLO
O
1
2
ML C
CLO
O1
2
ML C
CLO
O
1
2
ML C
CLO
O1
2
C2 v(xz) v(yz)
Often you analyze selected vibrational modes
(CO)
Find: # vectors remaining unchanged after operation.
2 x 12
0 x 10
2 x 12
0 x 10
Example: C-O stretch in C2v complex.
A1 = 1/4 (1x2x1 + 1x0x1 + 1x2x1 + 1x0x1) = 1
A2 = 1/4 (1x2x1 + 1x0x1 + 1x2x-1 + 1x0x-1) = 0
2 0 2 0
B1 = 1/4 (1x2x1 + 1x0x1 + 1x2x1 + 1x0x1) = 1
B2 = 1/4 (1x2x1 + 1x0x1 + 1x2x-1 + 1x0x1) = 0
ML C
CLO
O Both A1 and B1 are
IR and Raman active
= A1 + B1
MC L
CLO
1
2O
What about the trans isomer?
C2(z) v(xz) v'(yz)
x
E
1
1
2
1
-1
0
1
-1
0
D2h
Ag
B3u
C2(y) C2(x)
1
1
2
i
1
-1
0
v(xy)
1
1
2
1
1
2
1
-1
0
Only one IR active band and no Raman active bands
Remember cis isomer had two IR active bands and one Raman active
Symmetry and NMR spectroscopy
The # of signals in the spectrumcorresponds to the # of types of nuclei not related by symmetry
The symmetry of a molecule may be determinedFrom the # of signals, or vice-versa
Atomic orbitals interact to form molecular orbitals
Electrons are placed in molecular orbitalsfollowing the same rules as for atomic orbitals
In terms of approximate solutions to the Scrödinger equation
Molecular Orbitals are linear combinations of atomic orbitals (LCAO)
caacbb (for diatomic molecules)
Interactions depend on the symmetry properties
and the relative energies of the atomic orbitals
As the distance between atoms decreases
Atomic orbitals overlap
Bonding takes place if:the orbital symmetry must be such that regions of the same sign overlapthe energy of the orbitals must be similarthe interatomic distance must be short enough but not too short
If the total energy of the electrons in the molecular orbitalsis less than in the atomic orbitals, the molecule is stable compared with the atoms
Combinations of two s orbitals (e.g. H2)
Antibonding
Bonding
More generally:ca(1sa)cb(1sb)]
n A.O.’s n M.O.’s
Electrons in bonding orbitals concentrate between the nuclei and hold the nuclei together(total energy is lowered)
Electrons in antibonding orbitals cause mutual repulsion between the atoms(total energy is raised)
Combinations of two p orbitals (e.g. H2)
and notation meanschange of sign upon C2 rotation
and notation means nochange of sign upon rotation
Molecular orbitalsfor diatomic molecules
From H2 to Ne2
Electrons are placed
in molecular orbitals
following the same rules
as for atomic orbitals:
Fill from lowest to highest
Maximum spin multiplicityElectrons have
different quantum numbers including spin
(+ ½, - ½)
O2 (2 x 8e)
1/2 (10 - 6) = 2A double bond
Or counting onlyvalence electrons:1/2 (8 - 4) = 2
Note subscriptsg and u
symmetric/antisymmetricupon i
orbital mixing
When two MO’s of the same symmetry mixthe one with higher energy moves higher and the one with lower energy moves lower
E (Z*)
E > E Paramagneticdue to mixing
C2 u2 u
2 (double bond)
C22- u
2 u2 g
2(triple bond)
O2 u2 u
2 g1 g1 (double bond)paramagneticO2
2- u2 u
2 g2 g2 (single bond)diamagnetic
Photoelectron Spectroscopy
h(UV o X rays) e-
Ionization energy
hphotons
kinetic energy of expelled electron= -