Transcript Document 7317530

Lecture 7
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)
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
Effect of the 4 operations in the point group C2v
on a translation in the x direction. The translation is simply
multiplied by 1 or -1. It forms a 1 dimensional basis to show
what the operators do to an object.
Operation
Transformation
E
1
C2
-1
sv
1
sv ’
-1
Symmetry of Atomic Orbitals
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 PtCl42- 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  sv respectively
•
Superscripts ‘ and ‘’ indicate symmetric and antisymmetric behavior respectively with respect to
sh.
•
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: s or p 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
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)
Is there an identity element, e? a*e = a
Yes, 0
For each element is there an inverse element, a-1? a-1 * a = e
Yes, 4 + (-4) = 0
We have a group.
Abelian? Is commutativity satisfied for each element? a * b = b * a
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)
sv (1)
sv (2)
sv (3)
E
C3(240)
C3(120)
sv (1)
sv (2)
sv (3)
Examine the matrix representation of the elements of the C2v point group
-1
0 0
x
0 -1
0
y
0
1
z
0
=
0 0
0
y
0
1
z
sv(xz)
x
-y
0 1
0
y
z
0 0
1
z
x
=
y
z
-x
x
x
0 -1
0
1 0 0
E
C2
1
-x
=
y
-y
z
z
s’v(yz)
Multiplying two matrices (a reminder)
a11
a12
a21
a22
a31
a32
c11 = a11b11 + a12b21
c12 = a11b12 + a12b22
c13 = a11b13 + a12b23
b11 b12 b13
b21 b22 b23
c11 c12 c13
= c21 c22
c23
c31 c32
c33
c21 = a21b11 + a22b21
c22 = a21b12 + a22b22
c23 = a31b13 + a32b23
Most of the transformation
matrices we use have the
form
c31 = a31b11 + a32b21
c32 = a31b12 + a32b22
c33 = a31b13 + a32b23
±1 0 0
0 ±1
0 0
0
±1
1 0 0
x
0 1
0
y
0 0
1
z
E
-1
=
0 0
0 -1
0
y
0
1
z
0
1
x
=
0 0
0 -1
0
y
0
1
z
0
-1
What is the inverse of sv?
=
sv(xz)
C2
What is the inverse of C2?
x
C2
sv
s’v(yz)
0 0
x-1
0 0
x
1 0 0
0 -1
0
y0 = -1
0
=
y
=0
1
0
0
0
1
z0
1
z
0 0
1
1
0 0
0
1x 0 0
x
0 -1
0 0y -1
=
0
=
y
0
1 0z
1
z
0
0
1 0 0
=0
x
1
0
y
0 0
1
z
What of the products of operations?
1 0 0
x
0 1
0
y
0 0
1
z
-1
=
0
y
0
1
z
0
C2
sv * C2 = ?
1
x
0 -1
E
E * C2 = ?
0 0
=
0 0
x
0 -1
0
y
0
1
z
0
=
sv(xz)
C2
s’v(yz)
1 0 0
-1
x
0 0
0 1
0
y0 =-1
0
y=
0 0
1
z0
1
z
0 0
x
0
-1
x
=
s’v
1
0 0
x-1
0 -1
0
y0 = -1
0
y= =
0
1
z0
1
z
0
0
0 0
x
0 -1
0
y
0
1
z
0
=
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(x)
C2
C2(y)
So these operations belong to the same class?
C2
C2
1 0 0 

C2(x) = 
0 1 0 
 0 0  1


 1 0 0 


C2(y) =  0 1 0 
 0 0  1


C4
C4
=
Since C4 moves C2(x)
on top of C2(y) it is an
obvious choice for g
C43
 0 1 0


1 0 0
0 0 1


 0 1 0

=
 1 0 0
 0 0 1


C4
C43
C2(y)
 0 1 0  1 0 0 

 

1
0
0

 0 1 0
 0 0 1   0 0  1


 
 0 1 0


 1 0 0
 0 0 1


C2(x)
1 0 0 
= 0 1 0 


 0 0  1


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.
1
+
2
+
3 = 6=h
3 by 3
1
1
22
6=h
The sum of the squares of the dimensions of the each irreducible rep equals the
order of the group, 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 = S (ai * IrreducibleRepi)
The Irreducible reps are orthogonal. Hence for the reducible rep and a
particular irreducible rep i
Sym ops
S(character of Reducible Rep)(character of Irreducible Repi)
= ai * h
Or
Sym ops
ai =
S(character of Reducible Rep)(character of Irreducible Repi) / h
Reducible Representations in Cs = E and sh
Use the two sp hybrids as the basis of a representation
h1
h2
E operation.
h1 becomes h1; h2 becomes h2.
 1 0   h1 

  
 0 1   h2 
sh operation.
h1 becomes h2; h2 becomes h1.
 0 1   h1 

  
 1 0   h2 
h
=  1 
h 
 2
The hybrids are unaffected
by the E operation.
h 
 1
The reflection operation
interchanges the two hybrids.
Proceed using the trace of the matrix representation.
1+1 = 2
h
=  2 
0 + 0 = 0
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.
h1 h2; do not become
themselves, interchange
h1 , h1; become themselves
 1 0   h1 

  
 0 1   h2 
 0 1   h1 

  
 1 0   h2 
h
=  1 
h 
 2
The hybrids are unaffected
by the E operation.
h 
 1
The reflection operation
interchanges the two hybrids.
Proceed using the trace of the matrix representation.
1+1 = 2
h
=  2 
0 + 0 = 0
The Irreducible Representations for Cs.
Cs
E
sh
A’
A”
1
1
1 x, y,Rz x2,y2,z2,xy
-1 z, Rx,Ry yz, xz
The reducible representation derived from the two hybrids can be attached to the table.
G
2
0 (h1, h2)
Note that G = A’ + A”
,
h1
,
h2
h1 - h2
h1 + h2
The Irreducible Representations for Cs.
Cs
E
sh
A’
A”
1
1
1 x, y,Rz x2,y2,z2,xy
-1 z, Rx,Ry yz, xz
The reducible representation derived from the two hybrids can be attached to the table.
G
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’’
G = aA’ A’ + aA’’ A’’; our task is to find out the coefficients aA’ and aA’’
Cs
E
sh
A’
A”
1
1
1
-1
x, y,Rz
z, Rx,Ry
G
2
0
(h1, h2)
x2,y2,z2,xy
yz, xz
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.
Water is C2v. Let’s use the Character Table
Symmetry operations
Point group
Mülliken symbols
Characters
+1 symmetric behavior
-1 antisymmetric
Each row is an irreducible representation
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.
# of atoms
degrees of
freedom
Translational
modes
Rotational
modes
Vibrational
modes
N (linear)
3xN
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
Symmetry and molecular vibrations
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 polarizability of the molecule
In group theory terms:
A vibrational mode is IR active if it corresponds to an irreducible representation
with the same symmetry of a x, y, z coordinate (or function)
and it is Raman active if the symmetry is the same as
A 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.
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.
Apply each symmetry operation in that point group to the molecule
and determine how many atoms are not moved by the symmetry operation.
Multiply that number by the character contribution of that operation:
E=3
s=1
C2 = -1
i = -3
C3 = 0
That will give you the reducible representation
Finding the reducible representation
E=3
s=1
C2 = -1
i = -3
C3 = 0
G
3x3
9
1x-1
-1
3x1
3
1x1 (# atoms not moving x char. contrib.)
1
Now separate the reducible representation into irreducible ones
to see how many there are of each type
S
G
9
-1
3
1
A1 = 1/4 (1x9x1 + 1x(-1)x1 + 1x3x1 + 1x1x1) = 3
A2 = 1/4 (1x9x1 + 1x(-1)x1 + 1x3x(-1) + 1x1x(-1)) = 1
Symmetry of molecular movements of water
Vibrational modes
Which of these vibrations having A1 and B1 symmetry are IR or Raman active?
Raman active
IR active
Often you analyze selected vibrational modes
Example: C-O stretch in C2v complex.
1
n(CO)
L
L
C2
M
C
C
2
1
L
L
M
C
C
2
2x1
2
O
O
2
L
L
M
C
C
1
0x1
0
O
sv(yz)
sv(xz)
C2
E
O
O
O
1
L
L
M
C
C
2
2x1
2
O
O
L
L
M
C
C
2 O
1
O
0x1
0
Find: # vectors remaining unchanged after operation.
G
L
L
M
C
C
O
Both A1 and B1 are
O
IR and Raman active
G
2
0
2
0
= A1 + B1
A1 = 1/4 (1x2x1 + 1x0x1 + 1x2x1 + 1x0x1) = 1
A2 = 1/4 (1x2x1 + 1x0x1 + 1x2x-1 + 1x0x-1) = 0
B1 = 1/4 (1x2x1 + 1x0x1 + 1x2x1 + 1x0x1) = 1
B2 = 1/4 (1x2x1 + 1x0x1 + 1x2x-1 + 1x0x1) = 0
What about the trans isomer?
1
L
C
O
O
C
L
M
2
D2h E C2(z) C2(y) C2(x) i sv(xy) sv(xz) sv'(yz)
Ag
1
1
1
1
1
1
1
1
B3u
1
-1
-1
1
-1
1
1
-1
G
2
0
0
2
0
2
2
0
x
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 spectrum
corresponds to the # of types of nuclei not related by symmetry
The symmetry of a molecule may be determined
From the # of signals, or vice-versa
Molecular Orbitals
Atomic orbitals interact to form molecular orbitals
Electrons are placed in molecular orbitals
following 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)
Y = caya + cbyb (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 overlap
the energy of the orbitals must be similar
the interatomic distance must be short enough but not too short
If the total energy of the electrons in the molecular orbitals
is 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:
Y = N[caY(1sa)  cbY (1sb)]
n A.O.’s
n M.O.’s
Electrons in antibonding orbitals cause mutual repulsion between the atoms
(total energy is raised)
Electrons in bonding orbitals concentrate between the nuclei and hold the nuclei together
(total energy is lowered)
Both s (and s*) notation means symmetric/antisymmetric with respect to rotation
z
C2
z
C2
s
C2
s*
C2
z
Not s
z
s*
Combinations of two p orbitals (e.g. H2)
s (and s*) notation means no
change of sign upon rotation
p (and p*) notation means
change of sign upon C2 rotation
Combinations of two p orbitals
C2
z
C2
z
Combinations of two sets of p orbitals
Combinations of s and p orbitals
Combinations of d orbitals
No interaction – different symmetry
d means change of sign upon C4
Is there a net interaction?
NO
YES
NO
Relative energies of interacting orbitals must be similar
Strong interaction
Weak interaction
Molecular orbitals
for 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 multiplicity
Electrons have different quantum
numbers including spin (+ ½, ½)
1
Bond order =
2
# of electrons
in bonding MO's
# of electrons in
- antibonding MO's
O2 (2 x 8e)
1/2 (10 - 6) = 2
A double bond
Or counting only
valence electrons:
1/2 (8 - 4) = 2
Note subscripts
g and u
symmetric/antisymmetric
upon i
Place labels g or u in this diagram
s*u
p*g
pu
sg
s*u
g or u?
sg
p*g
pu
d*u
dg
Orbital mixing
Same symmetry and similar energies !
shouldn’t they interact?
s orbital mixing
When two MO’s of the same symmetry mix
the one with higher energy moves higher and the one with lower energy moves lower
Molecular orbitals
for diatomic molecules
From H2 to Ne2
H2 sg2 (single bond)
He2 sg2 s*u2 (no bond)
E (Z*)
Paramagnetic
due to mixing
DE s > DE p
C2 pu2 pu2 (double bond)
C22- pu2 pu2 sg2(triple bond)
O2 pu2 pu2 p*g1 p*g1 (double bond)
paramagnetic
O22- pu2 pu2 p*g2 p*g2 (single bond)
diamagnetic
Bond lengths in diatomic molecules
Filling bonding orbitals
Filling antibonding orbitals
Photoelectron Spectroscopy
hn
(UV o X rays)
Ionization
energy
e-
hn
= photons -
kinetic energy of
expelled electron
O2
N2
sg (2p)
pu (2p)
s*u (2s)
p*u (2p)
sg (2p)
s*u (2s)
pu (2p)
Very involved in bonding
(vibrational fine structure)
(Energy required to remove electron, lower energy for higher orbitals)