Reduction of a Reducible Representation. Given a Reducible Rep how do we find what Irreducible reps it contains?

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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, i 1,2.
Irred. Reps
Reducible Rep S (ai IrreducibleRepi)
i
The Irreducible reps are orthogonal. Hence for
the reducible rep and a particular irreducible
rep j S(character of Reducible Rep)(character of
Irreducible Repi) aj h Or aj S(character
of Reducible Rep)(character of Irreducible Repi)
/ h
Sym ops
Sym ops
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Reducible Representations in Cs E and sh
Use the two sp hybrids as the basis of a
representation
h1
h2
sh operation.
E operation.
h1 becomes h1 h2 becomes h2.
h1 becomes h2 h2 becomes h1.


The reflection operation interchanges the two
hybrids.
The hybrids are unaffected by the E operation.
Obtainthe trace of the matrix representation.
0 0 0
1 1 2
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Lets 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


The reflection operation interchanges the two
hybrids.
The hybrids are unaffected by the E operation.
Proceed using the trace of the matrix
representation.
0 0 0
1 1 2
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The Irreducible Representations for Cs.
Cs E sh
A A 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.
G 2 0 (h1, h2)
Note that G A A
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The Irreducible Representations for Cs.
Cs E sh
A A 1 1 -1 x, y,Rz z, Rx,Ry x2,y2,z2,xy yz, xz
Lets verify some things. Order of the group
sym operations 2 A and A are orthogonal
11 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)
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The Irreducible Representations for Cs.
Cs E sh
A A 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.
G 2 0 (h1, h2)
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Now lets 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 x, y,Rz z, Rx,Ry x2,y2,z2,xy yz, xz
G 2 0 (h1, h2)
aA (1 2 1 0)/2 1
aA (1 2 1 0)/2 1
Or again G 1A 1A. Note that this holds
for any reducible rep G as above and not limited
to the hybrids in any way.
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These are block-diagonalized matrices (x, y, z
coordinates are independent of each other)
Reducible Rep
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Another Example of Reduction
Sulfur Difluoride belongs to the C2v point group.
What this means is that the molecular
structure behaves at A1 for C2v This means that
the molecular structure is transformed into
itself by all the sym operations.
We can apply these symmetry operations to any
kind of object and see how the object behaves.
Additional objects may be needed to show what
the symmetry operations do.
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We know that each irreducible rep is 1
dimensional but lets assume that we do not know
the dimensions of the irreducible reps.

We do know that the d orbitals on an isolated
atom are equivalent. Lets guess that we need
all five for a representation. It will turn out
to be reducible but again assume that we do not
know that.
Remember what these matrices mean! Transformed d
orbitals matrix original d orbitals
We have to find out how each d orbital transforms
under the C2v operations. Formulate the matrices
of the representation based on the d orbitals in
this order dz2, dxz, dyz, dxy, dx2-y2.
Operation E C2 sv(xz) sv(yz)
Representation, G
Trace 5 1 1 1
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Now do Reduction.
We have four irreducible reps A1, A2, B1, B2.
The reducible rep, G, should be expressible as a
linear combination of the irreducible reps.
G aA1 A1 aA2 A2 aB1 B1 aB2 B2
Or using vector notation.
(5,1,-1,-1) aA1 (1,1,1,1) aA2 (1,1,-1,-1)
aB1 (1,-1, 1,-1) aB2 (1,-1, -1,1)
Find one of the a coefficents, say, aB1, by
multiplying everything by the elements of the B1
rep and summing over sym elements.
(1,-1, 1,-1) x
(1,-1, 1,-1) x
(1,-1, 1,-1) x
(1,-1, 1,-1) x
(1,-1, 1,-1) x
(5,1,1,1) aA1 (1,1,1,1) aA2 (1,1,-1,-1)
aB1 (1,-1, 1,-1) aB2 (1,-1, -1,1)
aB1 4
0
5 -1 1 -1 4
0
0
a B1 1 and similarly aA1 2 aA2 1 aB2
1
Or the d orbitals transform as G 2A1 A2
B1 B2
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The d orbitals transform as G 2 A1 A2 B1
B2
Note that dz2 and dx2-y2 transform as A1 dxy as
A2 dxz as B1 dyz as B2
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Vibrational Modes of water, C2v
Symmetry operations
Point group
Characters 1 symmetric behavior -1 antisymmetric
Mülliken symbols
Each row is an irreducible representation
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Lets 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
Linear Molecules Linear Molecules Linear Molecules Linear Molecules Linear Molecules
N 3 x N 3 2 3N-5
Example 3 (HCN) 9 3 2 4
Non linear Molecules Non linear Molecules Non linear Molecules Non linear Molecules Non linear Molecules
N 3 x N 3 3 3N-6
Example 3 (H2O) 9 3 3 3
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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
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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.
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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
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Finding the reducible representation
E 3 s 1 C2 -1 i -3 C3 0
3x3 9
1x-1 -1
3x1 3
1x1 1
( atoms not moving x char. contrib.)
G
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Now separate the reducible representation into
irreducible ones to see how many there are of
each type
S
A1 1/4 (1x9x1 1x(-1)x1 1x3x1 1x1x1) 3
A2
1/4 (1x9x1 1x(-1)x1 1x3x(-1) 1x1x(-1)) 1
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Symmetry of molecular movements of water
Vibrational modes
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Which of these vibrations having A1 and B1
symmetry are IR or Raman active?
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Often you analyze selected vibrational modes
Example C-O stretch in C2v complex.
n(CO)
2 x 1 2
0 x 1 0
2 x 1 2
0 x 1 0
G
Find vectors remaining unchanged after
operation.
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Both A1 and B1 are IR and Raman active
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
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What about the trans isomer?
Only one IR active band and no Raman active bands
Remember cis isomer had two IR active bands and
one Raman active
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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
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Molecular Orbitals
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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
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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
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Combinations of two s orbitals (e.g. H2)
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Both s (and s) notation means symmetric with
respect to rotation about internuclear axis
s
s
s
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p orbitals can combine in two ways, s and p.
s (and s) notation means no change of sign upon
rotation
p (and p) notation means change of sign upon C2
rotation
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Combinations of two p orbitals
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Combinations of two sets of p orbitals
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Combination of s and p orbitals
Combination of two misaligned p orbitals in p
mode.
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Combinations of d orbitals
No interaction different symmetry
d means change of sign upon C4
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Is there a net interaction?
NO
NO
YES
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Relative energies of interacting orbitals must be
similar
Weak interaction
Strong interaction
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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)
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O2 (2 x 8e)
1/2 (10 - 6) 2 A double bond
Or counting only valence electrons 1/2 (8 - 4)
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Note subscripts g and u symmetric/antisymmetric up
on i
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Place labels g or u in this diagram
su
pg
pu
sg
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su
sg
g or u?
pg
pu
du
dg
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Orbital mixing
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s orbital mixing
Notice the change in ordering here caused by the
mixing.
When two MOs of the same symmetry mix the one
with higher energy moves higher and the one with
lower energy moves lower
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Molecular orbitals for diatomic molecules From H2
to Ne2
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In Li2 through N2 sg above pu. O2 through Ne2 pu
above sg. -The s orbitals are more affected by
effective nuclear charge than are the p. -Less
mixing for O, F and Ne due to smaller orbitals
and greater bond distance.
Neutral O2 pu2 pu2 pg1 pg1 (double bond,
paramagnetic) Dianion O22- pu2 pu2 pg2 pg2
(single bond, diamagnetic)
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Bond lengths in diatomic molecules
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Photoelectron Spectroscopy
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O2
N2
sg (2p)
pu (2p)
pu (2p)
sg (2p)
pu (2p)
Very involved in bonding (vibrational fine
structure)
su (2s)
su (2s)
(Energy required to remove electron, lower energy
for higher orbitals)