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Table of Contents
IV.7 The Linear Modes of X-Pt-Pt-X of [Pt 2 (pop) 4 X 2 ] 4 -
- I.1 Symmetry operations and symmetry elements Chapter I. Symmetry and Group Theory
- I.2 Groups
- I.3 Similarity Transformations
- I.4 Point Groups
- I.5 Matrix Representations of Groups
- I.6 Point Group Representations
- I.7 Decomposing Reducible Representations
- I.8 Direct Products
- I.9 Symmetry Adapted Linear Combinations
- II.1 Quantum Theory – a brief tour Chapter II. Molecular Orbital Theory
- II.2 Wavefunctions as Bases for Irreducible Representations
- II.3 Quantum Mechanical Approach to Molecular Orbitals
- II.4 Homonuclear Diatomic Molecules
- II.5 Orbital Mixing in the Nondegenerate Case
- II.6 Orbital Energies
- II.7 Polyatomic Molecules
- II.8 Crystal Field Theory
- II.9 Molecular Orbitals in Inorganic Complexes
- II.9a Sigma Bonding in Octahedral ML
- II.9b Sigma Bonding in Tetrahedral ML
- II.9c Pi Bonding in Octahedral ML
- II.9d Pi Bonding in Tetrahedral ML
- II.9e Ferrocene
- II.9f Electron Deficient Bonds
- II.9g Linear Chains
- III.1 Electromagnetic Radiations Chapter III. General Spectroscopic Considerations
- III.2 Instrumentation
- III.3 Time Dependent States
- III.4 Experimental Quantities
- IV.1 The Harmonic Oscillator – a Classical view Chapter IV. Vibrational Spectroscopy – Part I. Theory
- IV.2 Quantum Mechanical Description of the Harmonic Oscillator
- IV.3 Selection Rules for the Harmonic Oscillator
- IV.4 The Anharmonic Oscillator
- IV.5 The Wilson FG Matrix Forumulation of Molecular Vibrations
- IV.6 Symmetry Coordinates
- IV.8 The Potential Energy Distribution
- IV.9 Overtones and Combinations
- IV.10 Raman Scattering
- IV.11 Raman Selection Rules
- IV.12 Infrared and Raman Intensities
- IV.13 A Complete Vibrational Assignment of SbCl
- IV.14 Pseudorotation in a Trigonal Bipyramid
- IV.15 Multi-minima Potential Functions
- IV.16 Use of IR and Raman to Deduce Isomeric Forms
- IV.17 Vibrational Spectra of Solids
Chapter V. Vibrational Spectroscopy Part II. Examples
V.I The Effect of Mass 85
V.2 The Effect of the Oxidation State of the Metal 86
V.3 Amines 86
V.4 Nitro and Nitrito Complexes 87
V.5 Carbonyl Complexes 87
V.6 Cyano Complexes 88
V.7 Nitrosyl Complexes 89
V.8 Olefin Complexes 90
V.9 Metal-Ligand Multiple Bonds 92
V.10 Metal-Metal Bonds 93
Chapter VI. Electronic Spectroscopy. Part I. Theory
VI.1 Electron Spin 94
VI.2 States 95
VI.3 Selection Rules 100
VI.4 Electronic Transitions 101
VI.5 Vibronic Coupling 104
VI.6 Jahn-Teller Distortions 105
VI.7 Spin Orbit Coupling 107
VI.8 Resonance Raman 109
VI.9 Relaxation Pathways 111
VI.10 Bandshapes 113
VI.11 Emission vs. Absorption 114
VI.12 Ligand Field Spectra 115
VI.13 Charge Transfer Spectra 121
Chapter VII. Electronic Spectroscopy. Part II. Examples
VII.1 Metal-ligand Multiple Bonding in trans - Dioxorhenium(V) 124
VII.2 Metal-Metal Bonding in Pt 2 (μ-P 2 O 5 H 2 )
4+ 129
VII.3 Mixed Valence Species 132
VII.4 Spatially Isolated Orbitals in Ru(bpy) 3
2+ 136
VII.5 Metalloporphyrins 139
Appendices
A. Character Tables of Selected Point Groups
B. Direct Products for Selected Point Groups
C. Fundamental Constants
D. Calculated Valence Orbital Energies, Hii
three axes are referred to as C 4 , C 4
2 = C 2 and C 4
3 , respectively.
1 Rotation about the z-axis will
not change the phase of the pz orbitals. The X, Y, a and b axes defined in figure I-1 are also C 2
rotational axes. It will be shown later that the C 2 rotations in this group can be grouped into
three classes which are differentiated with the use of ' and " {C 2 (Z)},{C 2 '(X) & C 2 '(Y)}, {C 2 "(a)
& C 2 "(b)}. Since the C 2 ' and C 2 " axes are perpendicular to the z-axis, rotation about any one of
them will invert the pz orbitals as shown in figure I-2.
Figure I-2. The effect of some of the Cn operations on the atoms and the
chlorine pz orbitals in PtCl 4
2 - .
Thus the ion contains C 4 , C 4
3 , C 4
2 = C 2 , C 2 '(x) ,C 2 '(y), C 2 "(a) and C 2 "(b) or 7 rotational axes.
- Reflections can be made through three different types of planes: vertical planes (σ v )
contain the principle axis , horizontal planes (σ h ) are perpendiuclar to the principle axis and
dihedral planes (σ d ) contain the principle axis and bisect two C 2 axes. The distinction between
vertical and dihedral is often unclear. Where appropriate, planes bisecting bond angles will be
designated as dihedral while those containing bonds will be designated as vertical. See figure I-
1 Since the clockwise C 4
3 operation is equivalent to a counterclockwise C 4 rotation, the C 4 and C 4
3 operations are
also referred to as the C 4
and C 4
- operations, respectively.
Figure I-3. The effect of reflection through the symmetry planes on the atoms
and the chlorine pz orbitals in PtCl 4
2 - .
In PtCl 4
2 - , the planes containing the z-axis (σv and σd) will not change the phase of the pz-
orbitals while reflection through the plane perpendicular to the z-axis (σh) does invert them
(figure I-3). Thus, PtCl 4
2 - contains: σv(XZ )and σv(YZ ), σh(XY) and σd (a) and σd (b) or five
planes of symmetry. Note that the a and b planes are defined as those planes perpendicular to the
plane of the ion and containing the a and b rotational axes.
- An improper rotation or a rotary reflection ( Sn ) is a Cn followed by a σh. Since
PtCl 4
2 - is a planar ion, the Z-axis is an element for both proper and improper rotations. See
figure I-4. Note that an S 4 results in the same numbering as a C 4 , but the phases of the pz orbitals
are changed.
- A Center of Inversion ( i ) takes all (x,y,z) → (-x,-y,-z). This operation can be
performed by a C 2 (z) which takes (x,y,z) → (-x,-y,+z) followed by a σ(xy) which inverts z, i.e., i
= C 2 σ h = S 2. Since i and S 2 are equivalent, S 2 is not usually used.
Figure I-4. The effect of improper rotations on the atoms and the
chlorine pz orbitals in PtCl 4
2 - .
PtCl 4
2 - contains E, C 4 , C 4
3 , C 4
2 = C 2 , C 2 '(x) ,C 2 '(y), C 2 "(a), C 2 "(b), i, S 4 , S 4
3 ,σv(XZ ),
σv(YZ ), σh, σd (a) and σd (b). These sixteen symmetry elements specify the symmetry of the
ion.
Show that trigonal bipyramidal MX 5 contains the following symmetry elements:
E,C 3 , C 3
2 , C 2 , C 2 ', C 2 ", σv, σv', σv", S 3 , and S 3
2
Successive Operations. In some of the following discussion, the result of applying more
than one operation will be important. The result of performing a C 4 rotation followed by a
reflection through the XZ plane (σvC 4 ) is the same as a single σd(a) operation.
However, if the order of the operations is reversed, i.e ., C 4 σv, the result is equivalent to a σd(b).
C 2 C 2 C 4
3 E C 4
C 4
3 C 4
3 E C 4 C 2
These operations satisfy all of the requirements of a group of order 4 (h=4). Indeed, they
comprise the C 4 point group. Since all of the members of the C 4 point group are also found in
the D4h point group (h=16), C 4 is said to be a subgroup of D4h. Note that the order of a subgroup
must be an integral divisor of the order of the group.
Problem I.1. Water belongs to the C2v point group, {C2vE, C 2 , σ(XZ), σ(YZ)}. Define the
molecular plane as the XZ plane and generate the multiplication table for the C2v point group.
In the following section, extensive use of the multiplication table will be made, but since
the point group is so large, its multiplication table is cumbersome (16x16). We will, therefore,
consider the ammonia molecule which has lower symmetry. NH 3 belongs to the C3v point group
of order 6, {C3v| E, C 3 , C 3
2 , σv, σv',σv"}. The effect of each of the symmetry operations of the
C3v point group on the ammonia molecule is shown in figure I.
Figure I.5. The effect of each of the symmetry operations of the C3v point group
on the ammonia molecule as viewed down the C 3 axis.
The student should verify that the multiplication table for the C3v point group is,
first →
second ↓
E C 3 C 3
2 σv σv' σv"
E E C 3 C 3
2 σv σv' σv"
C 3 C 3 C 3
2 E σv" σv σv'
C 3
2 C 3
2 E C 3 σv' σv" σv
σv σv σv' σv" E C 3 C 3
2
σv' σv' σv" σv C 3
2 E C 3
σv" σv" σv σv' C 3 C 3
2 E
I.3 Similarity Transformations
The operations, X and Y are said to be conjugate if they are related by a similarity
transfomation , i.e ., if Z
- 1 XZ = Y , where Z is at least one operation of the group. A class is a
complete set of operations which are conjugate to one another. The operations of a class have a
"similar" effect and are therefore treated together. To determine which operations of the group
are in the same class as C 3 , one must determine which operations are conjugate to C 3. The
results of the similarity transformation of C 3 with every other member of the group are
determined from the multiplication table above to be,
E
- 1 C 3 E = C 3 C 3 - 1 C 3 C 3 = C 3
2 C 3 C 3 = C 3
2 C 3
2 = C 3
(C 3
2 )
2 = C 3 C 3 C 3
2 = C 3
2 C 3
2 = C 3 σv
- 1 C 3 σv = σvC 3 σv = σvσv" = C 3
2
(σv')
- 1 C 3 σv' = σv'C 3 σv' = σv'σv = C 3
2 (σv")
- 1 C 3 σv" = σv"C 3 σv" = σv"σv' = C 3
2
Thus, C 3 and C 3
2 are conjugate and are members of the same class. In a similar manner, it
can be shown that σv, σv', and σv" are also conjugate and these three operations form another
class of the C3v point group. The C3v point group is then written as {C 3 vE, 2C 3 , 3σv}. The
order of a class must be an integral divisor of the order of the group. Similar considerations
allow us to write the D4h point group as {D4hE, 2C 4 , C 2 , 2C 2 ', 2C 2 ", i, 2S 4 , σh, 2σv, 2σd}.
Problem I.2 Identify the identity operator and the inverse of each function, and determine the
classes for a group with the following multiplication table.
M N P Q R S
M P S Q M N R N R Q S N M P
P Q R M P S N Q M N P Q R S
R S P N R Q M
S N M R S P Q
I.4 Point Groups
The following groups all contain the identity operation and only the minimum operations
required to define the group are given. In many cases, these minimum operations lead to other
operations. In the following, "k" is an integer ≥ 2.
C 1 : No symmetry Ck: Only a Ck rotational axis:
Cs: Only a plane of symmetry. Ci: Only an inversion center.
Ckh : A Ck rotational axis and a σh. Ckv : A Ck rotational axis and a σv.
Dk : One Ck and kC 2 rotational axes. The kC 2 axes are perpendicular to the Ck and at equal
angles to one another.
Dkh : The Dk operations plus a σh but this combination also results in kσv's.
Dkd : The Dk operations plus kσd's containing the Ck and bisecting the angles between
adjacent C 2 's.
Sk : Only the improper rotation, Sk. Note k must be an even number since an odd number
would require a σh.
Td : The tetrahedral point group contains three mutally perpendicular C 2 axes, four C 3 axes
and a σd through each pair of C 3 's.
Oh : The octahedral point group has three mutally perpendicular C 4 axes, four C 3 axes and a
center of symmetry.
Determining the point group to which a molecule belongs will be the first step in a
treatment of the molecular orbitals or spectra of a compound. It is therefore important that this
be done somewhat systematically. The flow chart in figure I-6 is offered as an aid, and a few
examples should clarify the process. We will first determine the point groups for the following
Pt(II) ions,
A contains three C 2 axes, i.e ., [Ck?] is yes with k=2. It contains a plane of symmetry so [σ?] is
yes. The three C 2 axes are perpendicular, i.e , there is a C 2 axis and two perpendicular C 2 's which
means that [⊥C 2 ?] is yes. There is a plane of symmetry perpendicular to the C 2 so [σh?] is yes
and we arrive at the D2h point group. B contains only one C 2 axis, no ⊥C 2 's, no σh, but it does
Problem I.3 Determine the point group to which each of the following belongs.
I.5 Matrix Representations of Groups
There are an infinite number of ways of choosing matrices to represent the symmetry
operations. The choice of representation is determined by its basis , i.e ., by the labels or
functions attached to the objects. The number of basis functions or labels is called the
dimension of the representation.
A convenient basis to use when dealing with the motions of molecules is the set of
Cartesian displacement vectors. Each atom has three degrees of motional freedom so a molecule
with N atoms will generate a basis of dimension 3N. For the water molecule, a 9-dimensional
basis results and thus each operation will be represented by a 9x9 matrix. These 9-basis vectors
are shown below along with the results of the C 2 (z) rotation.
The result of this operation is: (xi→ - xj), (yi→ - yj), and (zi→ +zj) where i = j for the oxygen atom
coordinates since the oxygen lies on the C 2 axis and therefore does not change its position, but i
≠ j for the hydrogen atoms since they are do not lie on the C 2 axis and are therefore rotated into
one another, e.g ., x 2 → - x 3. We can represent this transformation in matrix notation where each
atom will have a 3x3 matrix,
x i
y i
z i
x j
y j
z j
which must be placed into the 9x9 matrix representation of the C 2 operation. The oxygen atom
is not moved by the rotation (i = j = 1), so its 3x3 matrix remains in its original position (1,1) on
the diagonal while the hydrogen atoms are exchanged by the rotation so their 3x3 matrices are
rotated off of the diagonal to the (2,3) and (3,2) positions.
The matrix representation of this C 2 rotation is:
For reflection through the plane of the molecule (σv = YZ), only the x-coordinate is changed and
no atoms are moved so the matrix representation is
Thus, 9x9 matrices like those above could serve as the representations of the operations for the
water molecule in this basis. Fortunately, only the trace of this matrix needs to be specified, i.e .,
the sum of the diagonal elements. The resulting number is called the character , and the
character of an operation R is given the symbol χ( R ). In this example, χ(C 2 ) = - 1 and χ(σv) = 3.
Two important points about the character should be apparent from the above:
- Only those atoms which remain in the same position can contribute to the trace since
otherwise their 3x3 matrices will be rotated off of the diagonal.
- Each operation contributes the same amount to the trace for each atom since all atoms
have the same 3x3 matrix.
For a reflection through the plane bisecting the H-O-H bond angle, χ(σv') = +1 since only the O
is unshifted and a plane contributes +1 for each unshifted atom. The character for the identity
element will always be the dimension of the basis since all labels are unchanged. For water
then, χ(E) = 9.
The representation (Γ) for water in this basis is:
E C 2 σv = YZ σv' = X
Z
The s-orbitals can also serve as a basis,
In this basis, a C 2 rotation and a reflection through the plane perpendicular to the molecular plane
do not change A or B, and change only the sign of C while reflection through the molecular plane
leaves all three unchanged.
E = σ =
while C 2 = σ' =
. In this basis, no basis vector is changed into
another by a symmetry operation, i.e ., this basis is symmetry adapted. Now, our 3x
representation consists of three 1x1 matrices and we have converted our reducible representation Γ
into three irreducible representations, Γ 1 ; Γ 2 and Γ 3. i.e ., Γ= Γ 1 + Γ 2 + Γ 3
E C 2 σ^ σ'
The term "irreducible representation" is used so frequently that it is often abbreviated as "irrep".
We will also use this abbreviation throughout this book,
irrep ≡ irreducible representation
Decomposing a reducible representation into irreps is a very important process, and a procedure
to accomplish the decomposition will be described later in this chapter.
I.6 Point Group Representations
A point group representation is a basis set in which the irreducible representations are the
basis vectors. As i , j , and k form a complete, orthonormal basis for three-dimensional space, so
too do the irreps form the a complete orthonormal basis for an m-dimensional space, where m is
the number of irreducible representations and is equal to the number of classes in the group.
These considerations are summarized by the following rules.
- The number of basis vectors or irreps (m) equals the number of classes.
- The sum of the squares of the dimensions of the m irreps equals the order,
d h i
2
i
m
=
1
The character of the identity operation equals the dimension of the representation,
χ(E) = di which is referred to as the degeneracy of the irrep. The degeneracy of most
irreducible representations is 1(non-degenerate representations are 1x1 matrices) but
can sometimes be 2 or 3. The icosahedral point group to which C 60 (Buckminster
fulerene or "Buckyball") belongs, has irreducible representations with degeneracies
of 4 and 5. No character in an irreducible representation can exceed the dimension
of the representation. Thus, in non-degenenerate representations, all characters
must be ±1.
- The member irreps are orthonormal, i.e ., the sum of the squares of the characters in
any irrep is equal to the order (row normalization), while the sum of the product of
the characters over all operations in two different irreps is zero (orthogonality).
g(R)! i
(R)
R
"^!^ j (R)^ =^ h#ij
where the sum is over all of the classes of operations, g(R) is the number of
operations, R, in the class, χi(R) and χj(R) are the characters of the operation R in the
i
th and j
th irreps, h is the order of the group and δij is the Kroniker delta (0 when i≠j
and 1 when i=j).
- The sum of the squares of the characters of any operation over all of the irreps times
the number of operations in the class is equal to h, i.e ., columns of the representation
are also normalized.
g(R)! i
2
i = 1
m
" (R)^ =^ h
where m is the number of irreducible representations.
- The sum of the products of the characters of any two different operations over all of
the irreps is zero, i.e ., columns of the representation are also orthoganol.
! i
(R' )!
i i = 1
m
" (R)^ =^0
- The first representation is always the totally symmetric representation in which all
characters are +1.
- Any reducible representation in the point group can be expressed as a linear
combination of the irreducible representations - the completeness of the set.
We will now generate the C2v point group, {C2vE, C 2 , σ, σ'}. The order of the group is
four and the number of classes in the group is four, i.e ., h = m = 4. Each class has only one
operation, i.e ., g(R)=1 in all cases. Rule 2 states that d 1
2 + d 2
2
2
2 = 4 so that d 1 = d 2 = d 3
= d 4 = 1 - there are no degenerate representations in C2v. The character of the identity operation
is always the dimension of the representation, χi(E) = di. Therefore, all of the characters under
the E operation are known. From rule 6 we may write the characters of Γ 1 as +1's only. Thus,
we may write the following,
C2v E C 2 σv σv'
Γ 2 1 a b c
Γ 3 1 d e f
Γ 4 1 g h i
and since there are no degenerate representations, characters a through i must all be +1 or - 1. In
order to maintain orthogonality of rows and columns, only one of the remaining characters in
each row and in each column may be +1 while the other two in each column and row must each
be - 1, i.e ., there are only three remaining +1's and no two can be in the same column or row. We
will make a , e and i the +1's and then others must be - 1.
The C2v character table is then determined to be:
C2v E C 2 σv σv'
px +1 - 1 +1 - 1
py +1 - 1 - 1 +
Thus, Γp = A 1 + B 1 + B 2. The px orbital is said to
- form the basis for the B 1 represention,
- have B 1 symmetry, or
- transform as B 1.
Translations along the x, y and z directions (x, y, z) transform in the same way as px, py and pz.
To see this simply translate the water molecule slightly in the x-direction without moving any of
the symmetry elements. In this new position, C 2 and σ(yz) are destroyed and the characters are
those of px above.
Rotation of the water molecule slightly about the z-axis moves the H's out of the plane.
In this orientation, the C 2 axis is still preserved, but both planes of symmetry are destroyed, i.e .,
Rz transforms as A 2. Rotation about the x-axis preserves the yz plane but destroys both the C 2
rotation and the xz reflection while rotation about the y-axis preserved the xz-plane and destroys
the C 2 rotation and yz reflection.
C2v E C 2 (z) σv(xz) σv(yz)
Rz +1 +1 - 1 - 1
Rx +1 - 1 - 1 +
Ry +1 - 1 +1 - 1
Thus, the rotations in the C2v point group transform as A 2 + B 1 + B 2. In most character tables,
C2v has the following form:
C2v E C 2 σv σv'
Α 1 1 1 1 1 z x
2 , y
2 , z
2
Α 2 1 1 - 1 - 1 Rz xy
Β 1 1 - 1 1 - 1 x, Ry xz
Β 2 1 - 1 - 1 1 y, Rx yz
The final column gives the squares and binary products of the coordinates and represent the
transformation properties of the d-orbitals.
A set of character tables is given in the appendix A. This set is only a minimal set having
only those point groups encountered in this course. More complete sets are available in
other texts.
As a final example, the C3v point group will be generated. The operations of C3v are E,
C 3 , C 3
2 , σv, σv', and σv", which can now be writen as E, 2C 3 , and 3σv since C 3 and C 3
2 are
conjugate as are all three σv. This group has an order of six and contains three classes (h=6,
m=3) ⇒ d 1
2
2
2 = 6 ⇒ d 1 = d 2 = 1 and d 3 = 2. Since the dimensions of the irreps are the
χ(E) and every group contains the totally symmetric irrep,
C3v 1 E 2 C 3 3 σv
Γ 2 1 j k
Γ 3 2 m n
Orthogonality with Γ 1 requires that Σg(R)χ(R) = 0 for Γ 2 [( 1 )(1)(1) + ( 2 )(1)(j) + ( 3 )(1)(k) = 0]
and for Γ 3 [( 1 )(1)(2) + ( 2 )(1)(m) + ( 3 )(1)(n) = 0]. Γ 2 is non-degenerate (j and k must each be ±1)
so the orthogonality condition implies that j = +1 and k = - 1. Normalization of Γ 3 means (1)(2)
2
2 ) + 3(n
2
) = 6 so m = - 1 and n =0. Alternatively, we can use the fact that g(R)Σχ(R)
2 = h
down any column so 2(
2
2
2 ) = 6 or m
2 = 1 and 3(
2
2
2 ) = 6 so n
2 = 0.
C3v E 2C 3 3 σv
The ammonia molecule (C3v point group) and the coordinate system used in the following
discussion is given below (one N-H bond in the XZ plane).
The pz orbital is not changed by any of the operations of the group, i.e ., it is totally
symmetric and transforms as A 1. However, it should be apparent that px and py are neither
symmetric nor antisymmetric with respect to the C 3 or σv operations, but rather go into linear
combinations of one another and must therefore be considered together as components of a 2
dimensional representation. The matrices in this irreducible representation will be 2x2 and not
1x1. The character of the identity operation will then be 2 (the trace of a 2x2 matrix with 1's on
the diagonal), i.e ., χ(E)=2. A rotation through an angle 2π/n can be represented by the following
transformation:
x'
y'
& =^
cos(2 ' / n) sin(2' / n)
-sin(2 ' / n) cos(2 ' / n)
x
y
(^) & the trace of the Cn rotation matrix is
2cos(2π/n) which for n=3 is 2cos(2π/3) = 2(-0.5) = - 1, i.e ., χ(C 3 ) = - 1. The character for
reflection through a plane can be determined by the effect of reflection through any one of the
three planes since they are all in the same class. The easiest operation to use is the reflection
through the XZ plane which results in px → px and py → - py or
x'
y'
& =^
x
y
(^) & which has a
trace of 0, χ(σv)=0. The transformation properties of the px and py orbitals are represented as,
E 2C 3 3 σv
(x,y) 2 - 1 0
which is the E irreducible representation. The px and py orbitals are degenerate in C3v symmetry
and are taken together to form a basis for the two-dimensional irreducible representation, E.
Treating rotations and binary products as before, we can represent the C3v point group as
C3v E 2C 3 3 σv
Α 1 1 1 1 z x
2 +y
2 ; z
2
Γred 7 1 1
a(E) =
1
6
1
6
The reducible representation can be decomposed as follows: Γred = 2A 1 + A 2 + 2E. The results
can be verified by adding the characters of the irreps,
C3v E 2C 3 3 σv
Γred 7 1 1
Problem I.7 Decompose the following reducible representations of the C4v point
group.
C4v E 2C 4 C 2 2 σv 2 σd
The reducible representation of the Cartesian displacement vectors for water was
determined earlier (see page I.8 - I.10) and is given in the following table as Γcart
C2v E C 2 σv σ'v
A 1 1 1 1 1 z
A 2 1 1 - 1 - 1 Rz
B 1 1 - 1 1 - 1 x, Ry
B 2 1 - 1 - 1 1 y,Rx
Γcart 9 - 1 3 1
Decomposition of Γcart yields,
a(A 1 ) = 1/4 {(1)(1)(9) + (1)( 1)(-1) + (1)( 1)(3) + (1)( 1)(1)} = 1/4 {12} = 3
a(A 2 ) = 1/4 {(1)(1)(9) + (1)( 1)(-1) + (1)(-1)(3) + (1)(-1)(1)} = 1/4 { 4} = 1
a(B 1 ) = 1/4 {(1)(1)(9) + (1)(-1)(-1) + (1)( 1)(3) + (1)(-1)(1)} = 1/4 {12} = 3
a(B 2 ) = 1/4 {(1)(1)(9) + (1)(-1)(-1) + (1)(-1)(3) + (1)( 1)(1)} = 1/4 { 8} = 2
i.e ., Γcart = 3A 1 + A 2 + 3B 1 + 2B 2
Linear combinations of the 3N displacement vectors represent the degrees of motional
freedom of the molecule. Of these 3N degrees of freedom, three are translational, three are
rotational and the remaining 3N-6 are the vibrational degrees of freedom. Thus, to get the
symmetries of the vibrations, the irreducible representations of translation and rotation need only
be subtracted from Γ cart , but the irreps of rotation and translation are available from the character
table. For the water molecule, Γvib = Γcart - Γtrans - Γrot = {3A 1 + A 2 + 3B 1 + 2B 2 } - {A 1 + B 1 +
B 2 } - {A 2 + B 1 + B 2 } = 2A 1 + B 1. Construction of these "symmetry coordinates" will be
discussed in detail in the Vibrational Spectroscopy chapter, but we will draw them below and
note that S 1 (symmetric stretch) and S 2 (bending mode) both preserve the symmetry of the
molecule, i.e ., are totally symmetric, while S 3 (antisymmetric stretch) destroys the plane
perpendicular to the molecule and the C 2 axis but retains the plane of the molecule, i.e ., it has B 1
symmetry.
Problem I.8 Determine the symmetries of the vibrations of NH 3 , PtCl 4
2 - and SbF 5.
I.8 Direct Products
It is often necessary to determine the symmetry of the product of two irreps, i.e ., to
determine their direct product.
Direct Products: The representation of the product of two representations is given by the
product of the characters of the two representations.
Verify that under C2v symmetry A 2 ⊗ B 1 = B 2
C2v E C 2 σv σv'
A 2 1 1 - 1 - 1
B 1 1 - 1 1 - 1
A 2 ⊗ B 1 1 - 1 - 1 1
As can be seen above, the characters of A 2 ⊗ B 1 are those of the B 2 irrep.
Verify that A 2 ⊗ B 2 = B 1 , B 2 ⊗ B 1 = A 2. Also verify that
- the product of any non degenerate representation with itself is totally symmetric and
- the product of any representation with the totally symmetric representation yields the
original representation.
Note that,
- A x B = B; while A x A = B x B = A
- "1" x "2" = "2" while "1" x "1" = "2" x "2" = "1"
- g x u = u while g x g = u x u =g.
A table of Direct Products for the groups pertinent to this course is given in Appendix B.
The basis of selection rules (see Chapter III) is that the transition between two states a
and b is electric dipole allowed if the electric dipole moment matrix element is non-zero, i.e .,
a μ b =! a
"
μ! b d$ % 0
where μ = μx, + μy + μz is the electric dipole moment operator which transforms in the same
manner as the p-orbitals (x, y and z in the character table). A necessary condition for this
inequality is that the direct product of the integrand, ψa⊗μ⊗ψb= ψa⊗(μx, + μy + μz )⊗ψb, must
contain the totally symmetric representation.
