Transcript C - Wits Structural Chemistry

Chem IV - Symmetry and Group Theory
Chapter 7
Part 1 - Introduction to symmetry elements,
symmetry operations and point groups
Symmetry in Nature
The Platonic Solids
Christian
Art
Islamic Art
Art Deco
Architecture
Photography
Ndebele
Zulu
Symmetry analysis
• A symmetry operation is an action that leaves a
molecule apparently unchanged
• Each symmetry operation is associated with a
symmetry element
H2 O
Point, line, or
plane
Symmetry element:
axis of rotation
Symmetry operation:
rotation
A symmetry operation leaves at least one point in the
molecule unmoved – they are operations of point group
symmetry
The identity operation, E – do nothing.
All molecules have at least E, and some have only the
symmetry element E
An n-fold rotation is a symmetry operation that leaves a molecule
apparently unchanged after rotation by 360o/n.
The symmetry element is an n-fold axis of rotation, Cn
BH3
Note that C33 = E
BrF5
Note that C42 = C2
Associated with the symmetry element C4 we have the
symmetry operations
C4
C42 ( = C2)
C43
C44 ( = E)
XeF4
The axis with the highest
order, here the C4 axis, is
called the principal axis
The principal axis defines
the z axis
When assigning axis of
the same order, we give
preference to those that
go through atoms, C2’,
followed by those
bisecting the bond angle,
C2’’.
Identify the axes of rotation of this snowflake.
List the symmetry operations associated with each of these
symmetry elements
Symmetry operation reflection through the
symmetry element mirror plane, σ
z
The mirror planes contain the principal C2
axis. They are therefore vertical mirror
planes (subscript “v”).
This mirror plane is
perpendicular to the
principal C4 axis. It is
therefore a horizontal
mirror (subscript “h”)
This mirror plane is
perpendicular to the
principal C4 axis. It is
therefore a vertical
mirror (subscript “v”).
Vertical mirror planes
are those that usually
go through atoms, and
again, you give those
that go through atoms
as higher priority, i.e.
sv. Those mirror
planes that do not go
through atoms are
sometimes more
accurately called
dihedral mirror planes.
This mirror plane is
parallel to (contains)
the principal C4 axis
and bisects the two C2'
axes. It is a dihedral
mirror (subscript “d”).
Mirror planes in molecules
σh(horizontal): plane perpendicular to principal axis
σd(dihedral), σv(vertical): plane colinear with principal axis
σd: σ parallel to Cn and bisecting two C2'
axes or two σv
σv: σ parallel to Cn and are often
coincident with lower rotation axes of high
priority.
Examples of difference between vertical and dihedral planes.
The vertical planes lie
parallel with the C2’
axes. The dihedral
planes are bisecting the
C2’ axes and in this case
also contain the C2’’
axes.
Examples of dihedral planes in the absence of C2 axes.
Compare staggered ethane with the
complex ML4ClBr. In the former
molecule, there are only the three
"horizontal" C2 axes and the planes
bisect the angle between them. In
the Newman projection the reason
for calling these planes "dihedral" is
clear. Finally we consider the
complex in figure 1.15c where there
are no "horizontal" C2 axes but
there are two sets of planes
containing the principal axis. Now
the planes on the atoms take priority
and are labelled v while the planes
between the ligands bisect the angle
between the vertical planes and
become dihedral. Again which way
round you do this is arbitrary but
having the atoms as the highest
priority pleases most chemists
Identify the mirror planes (σh, σd,σv) in the following examples
acetylene
Identify the mirror planes (σh, σd,σv) in the following examples
(Don’t forget the double bonds are delocalised)
Identify the mirror planes (σh, σd,σv) in the following examples
Os(cp)2
The symmetry operation inversion, i,
involves projecting each atom through a
point, the centre of inversion i, that is
located at the centre of the molecule
i
Staggered form of ethane
Do these molecules have a centre of inversion?
The symmetry operations i and C2 should
not be confused
The symmetry operation improper rotation occurs about the
symmetry element Improper Axis, Sn
This is a compound operation
combining a rotation (Cn) with a
reflection through a plane
perpendicular to the Cn axis σh:
Read: Do Cn
followed by σh
Cn followed by σh or
σhCn = Sn
Neither the 90o rotation nor the reflection is
itself a symmetry operation for a tetrahedral
molecule, but their combined effect is the
symmetry operation S4
Do you see that S1 = σ and that S2 = i?
Use a CH4 molecule to verify that 2S4 = C2
Identify a S3 symmetry operation in BF3. What
is it equal to in this molecule?
Identify all the symmetry elements in the
(i)
eclipsed
(ii)
staggered
conformation of ethane
E, C3, C2, σh, σv, S3
E, C3, C2, σd, i, S6
Sketch the S4 axis of NH4+. Is there a C4 axis? How many S4
axes are there in the ion?
The Point Groups of Molecules
The set of symmetry elements of a molecule constitute a
group.
Since all symmetry operations leave at least one point in
the molecule unchanged, the group is called a point group
The point group is identified by it Schoenflies symbol
The process used to assign a molecule to a point group is
straightforward with a few exceptions. Use this schematic to
guide you.
CO2
OCS
SiIBrClF
Linear – No
2 of Cn, n>2 – No
Cn – No
σh – No
i - No
So…
Determine the point groups of the following molecules
CH2CBrCl
C2H2Cl2Br2
H2O2
H2O
S8
1,3,5,7-tetrafluorocyclooctatetrane
[Ni(en)3]2+
naphthalene
XeF4
Convention for orientation of a molecule in a coordinate system
1.
Centre of mass (or centre of symmetry) at origin.
2.
z axis = highest order axis and principal axis.
If several axes of highest order, the z axis is the one that passes through most
atoms.
3.
If a non-planar molecule contains a plane which contains more atoms than
any other plane, this is treated as it if where the molecular plane.
4.
Then assign x axis. If molecule is planar and z axis in plane, then the x axis is
 plane.
5.
If molecule planar and z axis  plane, then x axis will be in the plane and
chosen to pass through greatest number of atoms.
6.
The y axis is  to other two axes. Use right-hand rule, where your thumb = x,
index = y and middle = z.
Place the following in a cartesian coordinate system:
NH3 H2O cis-[Pt(NH3)2Cl2] trans-C2H2Cl2
Determine the point groups of the following molecules –
Homework
CCl4
Chloroform
Dichloromethane
SF6
BF5 (sq pyramidal)
1,2-dichloronaphthalene
1,5-dichloronapthalene
Ethane (staggered)
Ethane (eclipsed)
Os(cp)2 (staggered)
Os(cp)2 (eclipsed)
PPh3
Part 2 - Group Theory and Character Tables
Groups and Group Theory
Group: a collection of elements that obey the following
rules:
• Any combination of two elements must be equivalent
to another element in the group (the group shows
closure)
Show that σv(xz)C2 = σv’(yz) in the C2v point
group using H2O as a representation of that point
group
• There must exist an element that commutes with all
other elements and leaves them unchanged
For molecules this is the identity E since
EX = XE = X
• The associative law must hold
X(YZ) = (XY)Z
Use the symmetry elements C2, σv and σv’ of the
water molecule to show that the associative law
holds true
• Each X must have a reciprocal X-1 such that
XX-1 = X-1X = E
Find the reciprocals of C2 and σv’ of the water
molecule
Note the absence of the general requirement that the
commutative law holds (XY = YX). If it holds, the group is
an Abelian group
The set of symmetry operations constitute a group,
called a point group because at least one point is
unchanged.
Once we identify the point group → can look up all the
symmetry elements in character tables (see below).
The number of elements in a group: order of group, h.
Elements A, B, C, ..., M belong to the same class if, and
only if
X-1AX
X-1BX
X-1CX
▪
▪
▪
X1-MX







=
A, B, C, ..., or M
Show that all the symmetry elements of NH3
belong to three classes. Hint: Identify ALL
symmetry elements first.
As will be seen, need only consider properties associated with
one representative from each class rather than all operations.
So, for NH3, where we have operations
E, C3, C32, σv, σv’, σv”
we can write as E, 2C3, 3σv
The symmetry of orbitals
Central importance for chemists: effect of symmetry on
properties of orbitals
Example: The effect of the symmetry elements of the C2v
point group (E, C2, σv, σv’) on the valence orbitals of O (2s,
2p)
z
E
C2
σv (xz)
σv’ (yz)
x
y
2s
2px
2py
2pz
Operators either
• change sign of wavefunction
• leave sign of wavefunction unchanged
(≡ ×(-1))
(≡ ×(+1))
For example:
(C2)(2px) = (-1)×(2px)
Here a 1 × 1
matrix
The operation C2 on
the wavefunction
2px
may be represented by
multiplying the
wavefunction by a
matrix
Such a matrix is
called a
transformation
matrix
So
Ψ´ = R(Ψ) = [M]×Ψ
Final
wavefunction
Initial
wavefunction
Symmetry
operator
Initial
wavefunction
Transformation
matrix
An aside
Multiplying two matrices
P = MN
 1 2  5 6   1 5  2  7 1 6  2  8  19 22


  
  

 3 4  7 8   3  5  4  7 3  6  4  8   43 50
In linear algebra, the trace of an n-by-n square matrix is defined
to be the sum of the elements on the main diagonal (the
diagonal from the upper left to the lower right)
1
0
0
x
0
1
0
y
0
0
1
z
=
Let a suitable set of functions exist that are the basis set
of a point group.
There exists a collection of matrices that represents the
operation of the symmetry operators.
If
AB(Ψ) = C(Ψ)
 (there must exist)
[MA], [MB] and [MC]
such that
[MA][MB](Ψ) = [MC](Ψ)
A collection of matrices that obeys the multiplication laws of
the symmetry operators of a point group is a
representation of that group, symbol Γ (Gk., Gamma).
Let’s return to this:
Then Γ = (1, -1, 1, -1) is a
representation of the C2v point group
Take the 2px wavefunction as
a basis set for the C2v point
group
z
y
E
x
C2
2px
What if our basis set is the set of valence orbitals on O (2s, 2p)?
Then we need a set of 4 × 4 matrices.
 s 
1 0 0
 

 px 
 0 1 0
C2    
p
0 0 1
 y

0 0 0
p 

 z
0  s 
 
0  p x 
0  p y 
 
1  p z 
 s 
1 0 0
 

 px 
0 1 0
s v ( xz)   
py
0 0 1
 

0 0 0
p 

 z
0  s 
 
0  p x 
0  p y 
 
1  p z 
Write down the matrices corresponding to the symmetry
operators E and σv’(yz)
Use the two H 1s orbitals in H2O as basis set and show
that the transformation matrices are
 1 0

E
0 1
0 1

C2 
1 0
0 1

s v ( xz)
 1 0
 1 0

s v ' ( yz)
0 1
and determine the trace of each one.
In general, for n basis functions of a point group we will
need a set of n × n matrices to represent the point group
IMPORTANT Beware of how many elements your basis
set consists off.
For a point group with a symmetry operator greater than
two-fold a complication arises – the rotation will “mix”
elements of the basis set.
x´ = x cos θ – y sin θ
y´ = x sin θ + y cos θ
 x   cos
Cn    
 y   sin 
 sin   x 
 
cos  y 
Example. Take the valence orbitals of N in ammonia as the
basis set for the C3v point group. Then the transformation
matrix for the C3 operation is
 s  1
 
 p x  0
 p 
 y  0
 p 
 z  0
0
1
3
0
2
2
0
 3
1
0
2
2
0

0

0
1 
Write down the transformation matrices for all other
symmetry operators of the C3v point group
If X, A and B  of a group G:
... then X-1  G
(why?)
Suppose X-1AX = B
We say
- A is converted to B by a similarity transformation
- B is the similarity transform of A by X
- A and B are conjugate
Reducing matrices
For symmetry operators A, B and C  square transformation
matrices [A], [B] and [C].
Suppose AB = C; then [A][B] = [C]
Let [A]´ and [B]´ be similarity transforms of [A] and [B]
So
[A]´ = [X]-1[A][X]
[B]´ = [X]´[B][X]
Now
[A]´[B]´ = [X]-1[A][X][X]-1[B][X]
= [X]-1[A][B][X]
= [X]-1[C][X]
= [C]´
So [A]´, [B]´ and [C]´ are true representations of G
Suppose as the result of suitable transformations we end up
with matrices of the form
 | a1 | | 0 | | 0 | 


[ A]   | 0 | | a 2 | | 0 | 
 | 0 | | 0 | | a |
3 

 | b1 | | 0 | | 0 | 


[ B]   | 0 | | b2 | | 0 | 
 | 0 | | 0 | | b |
3 

where |0| is a region of the matrix where all elements are 0.
These are said to be reduced matrices.
Submatrices |an| and |bn| have the same order.
It can be shown that if [A]´[B]´ = [C]´ then |an||bn| = |cn|
The submatrices are also true representations of G
Thus smaller representations can be found within larger ones.
If this is impossible, then the original matrix is irreducible.
The character of a matrix
Character χ = sum diagonal elements
χ is unaltered by similarity transforms
Larger matrices may be reduced to irreducible components
by similarity transforms alone. So
n
 R   ai  i ( R )
i 1
- χR is the character of a reducible matrix for the Rth
operation of the point group
- χi(R) is the character of the ith irreducible
representation of the Rth operation
- ai is the number of times χi(R) is contained in χR.
Character Tables
This is a typical character table
Character table: tabulation of
χi(R)  i irreducible
representations of G.
Symmetry elements of the same class
related by a similarity transform 
matrices have the same character.
This is a 1-dimensional representation (1×1 matrix)
This is a 2-dimensional representation (2×2 matrix)
Mulliken symbols give information about symmetry
elements of the irreducible representations
One dimensional representations symmetric w.r.t. rotation about the
principal axis (+1 under Cn) are designated A; those that are
antisymmetric (-1 under Cn) are designated B.
If the one dimensional representation is symmetric w.r.t. C2  Cn
subscript “1” is added to A or B. If C2 is absent then the symmetry
w.r.t. a vertical plane is used instead.
Two dimensional representations are designated E and three
dimensional ones are designated T
Primes and double primes indicate representations that are
respectively symmetric and antisymmeric w.r.t. σh
If i  G then subscript “g” (Ger., gerade, even) is added if the
representation is symmetric w.r.t. inversion; otherwise “u” (Ger.,
ungerade, odd) is used
Direct products
As we shall see, it is often necessary to multiply together
representations of two sets of functions. This is a direct
product of the two representations.
Example: B1 × B2 in C2v
B1 × B2
= (1×1
= (1
= A2
-1×-1
1
1×-1
-1
-1×1)
-1)
Reduction of matrices using character tables
It can be shown that the number of times, aj, which is the
irreducible representation j, occurs in a reducible
representation is given by:
1
a j   n R  ( R)  j ( R)
h g
• h = order of group (= number of elements in group = sum of squares of
characters entered under E in the character table;
• g = number of classes;
• nR = number of elements in a class (number in front of class symbol)
• χ(R) = character of reducible representation;
• χj(R) = character of the irreducible representation.
Usually not necessary to write out a full transformation matrix for each
symmetry operator – merely the character.
Diagonal elements describes to what extent a function remains in its
original position after a symmetry operation: 0 – moves completely away; 1
– remains unchanged; -1 – changes sign; and so on.
Example: The p orbitals of C in the carbonate anion. How to they
transform in the D3h point group? (So we use the 2p orbitals as a basis set.)
Operator
Effect
E
Nothing
C3
z→z
x → x cos 120o – y sin 120o
y → x in 120o + y cos 120o
C2’
Contribution to
χ
χ
3
1
-1/2
-1/2
0
z → -z
x→x
y→ -y
-1
1
-1
-1
σh
z → -z
x,y unchanged
-1
2
1
S3
z → -z
x → x cos 120o – y sin 120o
y → x in 120o + y cos 120o
-1
-1/2
-1/2
-2
σv
z→z
x→x
y→ -y
1
1
-1
1
on diagonal
off diagonal
Hence Γ2p = (3, 0, -1, 1, -2, 1) is a reducible representation of
the C3v point group.
Let’s reduce this representation and express this in terms of the
irreducible representations of the point group.
aj 
1
 n R  ( R)  j ( R)
h g
Γ2p = (3 0 -1 1 -2 1)
a(a1’)
= 1/12[1.3.1 + 2.0.1 + 3.-1.1 + 1.1.1 + 2.-2.1 + 3.1.1]
= 1/12[3
+ 0 + -3 + 1 + -4 + 3]
=0
a(a2’)
= 1/12[1.3.1 + 2.0.1 + 3.-1.-1 + 1.1.1 + 2.-2.1 + 3.1.-1]
= 1/12[ 3 + 0 + 3
+ 1 - 4
- 3 ]
=0
a(e’)
= 1/12[1.3.2 + 2.0.-1 + 3.-1.0 + 1.1.2 + 2.-2.-1 + 3.1.0]
= 1/12[ 6 + 0
+ 0
+ 2 + 4
+ 0 ]
= 1/12[12]
=1
a(a1”)
= 1/12[1.3.1 + 2.0.1 + 3.-1.1 + 1.1.-1 + 2.-2.-1 + 3.1.-1]
= 1/12[ 3 + 0 - 3 - 1
+ 4
- 3 ]
=0
This means that a1’ does
not appear at all in Γ2p
aj 
1
 n R  ( R)  j ( R)
h g
Γ2p = (3 0 -1 1 -2 1)
a(a2”)
= 1/12[1.3.1 + 2.0.1 + 3.-1.-1 + 1.1.-1 + 2.-2.-1 + 3.1.1]
= 1/12[ 3 + 0 + 3
- 1 + 4
+ 3 ]
= 1/12[12]
=1
a(e”)
= 1/12[1.3.2 + 2.0.-1 + 3.-1.0 + 1.1.-2 + 2.-2.1 + 3.1.0]
= 1/12[ 6 + 0
+ 0
- 2 - 4
+ 0 ]
=0
Hence Γ2p = a2” + e’. Two of the p orbitals (2px and 2py are
degenerate) and the other (2pz) has a different energy.
NOTE: When describing the symmetry species of
orbitals, lower case letters are used.
We could have told this straight away by consulting the 2nd last column of
the character table, where we see the functions (x,y) are degenerate and
transform as e’ whilst the function (z) transforms as a2”.
Recall that
ψ2px = n1(x).f(r)
ψ2py = n2(y).f(r)
ψ2pz = n3(z).f(r)
Hence ψ2px has the same symmetry properties as (x), and so on.
Notice this typo in Atkins: should be (xz, yz)
How do the d orbitals transform under the D3h point
group? And the 4s orbital?
How do the metal d orbitals transform in trans-tetraammineaquachlorocobalt(III)?
Show that the valence orbitals of nitrogen in ammonia
transform as 2a1 + e
Summary of important character table concepts
Taken from
http://mutuslab.cs.uwindsor.ca/macdonald/Teaching/0359-250.htm
Part 3 - Some Applications of Group Theory
in Chemistry
Polarity of molecules
Polar molecule: permanent electric dipole moment
δ-
δ+
Cannot be polar if molecule posses a centre of inversion i
Dipole moment cannot be  to a mirror plane
Dipole moment cannot be  to a Cn axis
 A polar molecule cannot belong to a group that includes i,
any of the groups D and their derivatives, the cubic groups
(Oh,, Td, Ih) and their modifications
Are these molecules polar?
Chirality of molecules
 A molecule cannot be chiral if it possesses an improper
axis of rotation, Sn. These are groups such as Dnd, Dnh, Td,
Oh, etc
A mirror plane σ ≡ S1
A centre of inversion i ≡ S2
So molecules with σ and i are not chiral too.
Are these chiral molecules (assume locked in these
conformations)?
Symmetry adapted linear combination of atomic orbitals
Molecular orbitals are built up from the combination of atomic
orbitals of the correct symmetry. This is a symmetry adapted
linear combination (SALC) of atomic orbitals.
Example. The diagram shows a molecular orbital made up of
a linear combination of H 1s orbitals in ammonia. To what
symmetry species does φ belong?
φ1 = ψA1s + ψB1s + ψC1s
Example. Identify the symmetry label of the orbital φ = ψ0 – ψ0’
in the C2v molecule NO2 where ψ0 is an O2px orbital on one
oxygen atom and ψ0’ is an O2px orbital on the other O atom.
The construction of molecular orbitals
Now that we know something about group theory let’s return to a topic in
Chem III Inorganic: the MO description of the bonding in transition metal
complexes
Consider a metal from the first row of the d block in an Oh complex such
as high-spin [Fe(H2O)6]2+.
Available orbitals:
4s: a1g
4p: t1u
3d: eg, t2g
Γmetal = a1g + t1u +
eg + t2g
3d, 4s, 4p
Now the ligands. Assume for the moment σ bonding only (appropriate for
H2O as ligands).
Exact nature of ligand orbital use unimportant – only the symmetry of the
orbitals is important.
+
+
+
+
+
+
Six ligand-based
orbitals to be
combined – in this
case as it sigma
bonding only, can use
spheres.
+
+
+
+
+
+
Now apply the symmetry operations of the Oh point group to these 6
orbitals (Your basis set is 6 now).
E
8C3
6C2
6C4
3C2
i
6S4
8S6
3σh
6σd
6
0
0
2
2
0
0
0
4
2
and we find that Γσ(ligand)= (6 0 0 2 2 0 0 0 4 2) which is a reducible
representation of the Oh point group. Apply the methods we have learned
to express as the sum of irreducible representations and find
Γσ(ligand) = a1g + eg + t1u
Γmetal = a1g + t1u + eg + t2g
Γσ(ligand) = a1g + eg + t1u
and notice that there are no ligand orbitals with symmetry t2g.
Therefore the t2g orbitals (xy, xz, yz) will be non-bonding.
The MO diagram is therefore the familiar one we met in Chem III – but
hopefully you will now have a deeper understanding of its origins.
What if we include π bonding?
This will involve overlap of ligand orbitals which are  to the M–L σ bond.
σ
π
π
The p orbitals have the same symmetry
properties as a vector orthogonal to the main σ
bond framework.
E
8C3
6C2
6C4
3C2
i
6S4
8S6
3σh
6σd
12
0
0
0
-4
0
0
0
0
0
and reducing this into irreducible representations we find
Γπ(ligand) = t1g + t2g + t1u + t2u
Γπ(ligand) = t1g + t2g + t1u + t2u
t1g and t2u – non bonding because no metal orbitals with this symmetry
t1u – largely non-bonding because metal t1u orbitals (4p) already involved
in σ bonding
t2g orbitals can overlap with non-bonding metal t2g orbitals (xy, xz, yz)
to form M–L π bonds
If the ligands are π donors such as
Cl– the ligand π orbitals with be
lower in energy that the metal t2g
orbitals.
The ligand field splitting is
now between t2g* and eg*.
This splitting has
decreased relative to the σonly case.
Ligand such as Cl– that act as π
donors are low down in the
spectrochemical series.
If the ligands are π acceptors such
as CO the ligand π orbitals with be
higher in energy that the metal t2g
orbitals.
The ligand field splitting between t2g
and eg*. has increased relative to
the σ-only case.
Ligand such as CO that act as π
acceptors are high up in the
spectrochemical series and
produce low-spin complexes.
Molecular vibrations – Infrared and Raman spectroscopy
IR absorption occurs when a vibration results in change in electric
dipole moment of a molecule.
Raman transition occurs when the polarisability of a molecule
changes during a vibration
Exclusion rule
If a molecule has a centre of inversion none of its normal vibrational
modes are both IR and Raman active
Example
H3N
Cl
Pt
H3N
Cl
C2v
H3N
Cl
Pt
Cl
NH3
D2h
Pd-Cl stretching occurs between 200 and 400 cm-1.
Trans isomer: modes cannot be both IR and Raman active
Molecular vibrations – Infrared and Raman spectroscopy
No. of
atoms
degrees of
freedom
Translational
modes
Rotational
modes
Vibrational
modes
N (linear)
3N
3
2
3N-5
Example
3 (HCN)
9
3
2
4
N (nonlinear)
3N
3
3
3N-6
Example
3 (H2O)
9
3
3
3
A molecular vibration is IR active only if it results in a change in the
dipole moment of the molecule
A molecular vibration is Raman active only if it results in a change in
the polarisability of the molecule
In group theory terms:
A vibrational motion is IR active if it corresponds to an irreducible
representation with the same symmetry as an x, y, z coordinate (or
function)
and it is Raman active if the symmetry is the same as x2, y2, z2, xy, etc
The vibration will have
the same symmetry
properties as these
vectors. The two
vectors together
constitute the
molecular vibration.
symmetric stretch
Γ = (1 1 1 1)
 A1
Both IR and
Raman active
Both IR and
Raman active
anti-symmetric stretch
Γ = (1 -1 -1 1)
 B2
y
z
x
symmetric stretch
Γ = (1 1 1 1 1 1 1 1)
 Ag
Raman active
IR active
anti-symmetric stretch
Γ = (1 -1 1 -1 -1 1 1 -1 )
 B2u
In H2O, how many vibrational modes belong to each irreducible
representation?
You need the point group and the character table
•
Centre of mass (or centre of symmetry) at origin.
•
z axis = highest order axis and principal axis. If several axes of highest
order, the z axis is the one that passes through most atoms.
•
Then assign x axis. If molecule is planar and z axis in plane, then the
x axis is  plane
In H2O, how many vibrational modes belong to each irreducible
representation?
You need the point group and the character table
degrees of
freedom
Use the translation vectors of the atoms as the basis of a reducible
representation.
Since you only need the trace recognise that only the vectors that are either
unchanged or have become the negatives of themselves by a symmetry
operation contribute to the character.
Note that a vector will either be unchanged or the negative of itself if the
atom does not move.
Now consider only the atoms that don’t move
A reflection can either
leave the vector unchanged (multiply by +1)
or
change direction (multiply by -1)
A rotation can either
invert the direction of the vector (multiply by -1)
or
leave it unchanged (multiply by +1)
Nothing x -1
Γ moves y -1
z1
9
-1
x1
y -1
z1
1
x -1,-1,-1
y 1,1,1,
z 1,1,1
3
Now express the reducible representation as the sum of
the irreducible representations of the point group.
You should find that
Γ = 3A1 + A2 + 2B1 + 3B2
Three are translations (t): A1, B1, B2.
Three are rotations (r): A2, B1, B2
The remaining three are vibrations
(v): 2A1 + B2
Three are translations (t): A1, B1, B2.
Three are rotations (r): A2, B1, B2
Which of these vibrations with A1
or B2 symmetry are IR and/or
Raman active?
Vibrational modes of water
Raman active
IR active
Suppose we were only interested in the stretching modes and not all the
vibrational modes. Then a reasonable basis set would be the following
two vectors:
O
H
H
It should be easy to see that
Γ = (2 0 0 2)
= A1 + B2
Don’t confuse with the following:
To what symmetry species* does the following vibration belong? Is it IRor Raman-active or both?
N
O
*or
O
“how does the following vibration transform”
“what us the symmetry of”
etc.
Note that this is one single vibration not two vectors representing the
motion of the molecule as in the previous example.
Review question
a) Draw a Lewis structure of XeF3+ and use VSEPR theory to
predict two plausible structures.
b) Use group theory to explain how vibrational spectroscopy
could be used to determine the actual structure