710 lecture 04
TRANSCRIPT
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Lecture 4
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Point group Symmetry operations
Characters
+1 symmetric behavior
-1 antisymmetricMlliken symbols
Each row is an irreducible representation
Character Table
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x, y, zSymmetry of translations (p orbitals)
Rx, Ry, Rz: rotations
Classes of operations
dxy, dxz, dyz, as xy, xz, yz
dx2- y2 behaves as x2y2
dz2 behaves as 2z2 - (x2 + y2)
px,py,pz behave as x, y, z
s behaves as x2 + y2 + z2
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Symmetry of Atomic Orbitals
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Effect of the 4 operations in the point group C2von a translation in the x direction
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Naming of Irreducible representations
One dimensional (non degenerate) representations are designated A or B.
Two-dimensional (doubly degenerate) are designated E.
Three-dimensional (triply degenerate) are designated T.
Any 1-D representation symmetric with respect to Cn is designatedA;antisymmtric ones are designated B
Subscripts 1 or 2 (applied to A or B refer) to symmetric and antisymmetricrepresentations with respect to C2Cn or (if no C2) to svrespectively
Superscripts and indicate symmetric and antisymmetric operations withrespect to sh, respectively
In groups having a center of inversion, subscripts g(gerade) and u(ungerade) indicate symmetric and antisymmetric representations withrespect to i
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Character Tables
Irreducible representations are the generalized analogues of s or p symmetry indiatomic molecules.
Characters in rows designated A, B,..., and in columns other than E indicate thebehavior of an orbital or group of orbitals under the corresponding operations (+1 =orbital does not change; -1 = orbital changes sign; anything else = more complexchange)
Characters in the column of operation Eindicate the degeneracy of orbitals
Symmetry classes are represented by CAPITAL LETTERS (A, B, E, T,...) whereasorbitals 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
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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 isa member of the group, usually denoted by a*b, suchthat the following properties are satisfied :
(Associativity) (a*b)*c= a*(b*c) for all a, b, cbelonging to G.
(Identity) There exists e belonging to G, such that e*g= g= g*efor all gbelonging to G.
(Inverse) For each gbelonging to G, there exists the inverse of
g,g
-1
, such that g
-1
*g= g*g
-1
= e. Ifcommutativityis satisfied, i.e. a*b = b*a for all a, b
belonging to G, then G is called an abelian group.
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Examples
The set of integers Z, is an abelian group
under addition.
What is the element e, identity, such that
a*e = a?
What is the inverse of the a element?
0
-a
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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)
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Examine the matrix represetation of the C2v point group
1 0 0
0 1 00 0 1
xyz
=
xyz
-1 0 0
0 -1 00 0 1
xyz
=
-x-y
z
C2
1 0 0
0 -1 0
0 0 1
xy
z
=
x-y
z
sv(xz)
E
- xy
z
sv(yz)
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Multiplying two matrices (a reminder)
a11 a12a21 a22a31 a32
b11 b12 b13b21 b22 b23
=
c11 c12 c13c21 c22 c23c31 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 00 0 1
Most of the transformation
matrices we use have the
form
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-1 0 0
0 -1 00 0 1
1 0 0
0 1 00 0 1
1 0 0
0 -1 00 0 1
C2E sv(xz) sv(yz)
What is the inverse of C2? C2
What is the inverse ofsv? sv
-1 0 0
0 -1 00 0 1
-1 0 0
0 -1 00 0 1
=
1 0 0
0 1 00 0 1
1 0 0
0 -1 00 0 1
1 0 0
0 -1 00 0 1
=
1 0 0
0 1 00 0 1
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-1 0 0
0 -1 00 0 1
1 0 0
0 1 00 0 1
1 0 0
0 -1 00 0 1
C2E sv(xz) sv(yz)
What of the products of operations?
E * C2 = ?C2
1 0 0
0 1 00 0 1
-1 0 0
0 -1 00 0 1
=
-1 0 0
0 -1 00 0 1
sv * C2 = ? sv
1 0 0
0 -1 00 0 1
-1 0 0
0 -1 00 0 1
=
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ClassesTwo members, c
1and c
2, of a group belong to the same class if there is a member,
g, of the group such that
g*c1*g-1 = c2
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Properties of Characters of Irreducible Representations in
Point Groups
Total number ofsymmetry 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.
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Properties of Characters of Irreducible
Representations in Point Groups - 2
The number of irreducible reps equals the number of
classes. The character table issquare.
3 by 3
The sum of the squares of the dimensions of the each irreducible rep equals theorder of the group, h.
1 + 2 + 3 = 6
1
1
22
6
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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 + (12 + 12 ) + = 6
A2: 12 + (12 + 12 ) + ((-1)2 + (-1)2 + (-1)2 ) = 6
E: 22 + (-1)2 + (-1)2 = 6
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Properties of Characters of Irreducible
Representations in Point Groups - 4
Irreducible reps are orthogonal. The sum of the products of the characters for
each symmetry operation is zero.
For A1 and E:
1 * 2 + (1 *(-1) + 1 *(-1)) + (1 * 0 + 1 * 0 + 1 * 0) = 0
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Properties of Characters of Irreducible
Representations in Point Groups - 5
Each group has a totally symmetric irreducible rep having all characters equal
to 1
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Reduction of a Reducible
RepresentationIrreducible 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
S(character of Reducible Rep)(character of Irreducible Repi) = ai * h
Or
ai =S(character of Reducible Rep)(character of Irreducible Repi) / h
Sym ops
Sym ops
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These are block-diagonalized matrices
(x, y, z coordinates are independent of each other)
1 0 0
0 -1 00 0 1
1 0 0
0 1 00 0 1
-1 0 0
0 -1 00 0 1
C2 sv(xz)E
-1 0 0
0 1 00 0 1
sv'(yz)
C2 sv(xz) sv'(yz)
x
y
z
E
1
1
1
3
-1
-1
1
-1
1
-1
1
1
-1
1
1
1
Irreducible
representations
Reducible Rep
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Point group Symmetry operations
Characters+1 symmetric behavior
-1 antisymmetricMlliken symbols
Each row is an irreducible representation
C2vCharacter Table to be used for water
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Lets use character tables!
Symmetry and molecular vibrations
# of atoms degrees offreedom
Translationalmodes
Rotationalmodes
Vibrationalmodes
N (linear) 3 x 2 3 2 3N-5 = 1
Example
3 (HCN)
9 3 2 4
N (non-
linear)
3N 3 3 3N-6
Example
3 (H2O)
9 3 3 3
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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 active
only if it results in a change in the polarizability of the molecule
In group theory terms:
A vibrational motion is IR active if it corresponds to an irreducible representationwith the same symmetry as anx, y, zcoordinate (or function)
and it is Raman active if the symmetry is the same as
x2, y2, z2, or one of the rotational functions Rx, Ry, Rz
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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 areeither unchanged or have become the negatives of themselves by a
symmetry operation contribute to the character.
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Apply each symmetry operation in that point group to the molecule
and determine how many atomsare not movedby 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
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.
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3x3
9
1x-1
-1
3x1
3
1x1
1
Finding the reducible representation
(# atoms not moving x char. contrib.)
E= 3
s = 1
C2 = -1
i = -3
C3 = 0
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Now separate the reducible representation into irreducible ones
to 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
S
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Symmetry of molecular movements of water
Vibrational modes
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IR activeRaman active
Which of these vibrations having A1 and B1 symmetry are IR or Raman active?
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ML C
CLO
O
C2
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 sv(xz) sv(yz)
Often you analyze selected vibrational modes
n(CO)
Find: # vectors remaining unchanged after operation.
2 x 1
2
0 x 1
0
2 x 1
2
0 x 1
0
Example: C-O stretch in C2v complex.
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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
A1 is IR active
B1 is IRandRamanactive
= A1 + B1
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MC L
CLO
1
2O
What about the trans isomer?
C2(z) sv(xz) sv'(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
sv(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
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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 orbitalsfollowing the same rules as for atomic orbitals
In terms of approximate solutions to the Scrdinger equation
Molecular Orbitals are linear combinations of atomic orbitals (LCAO)
Y = caya + cbyb (for diatomic molecules)
Interactions depend on thesymmetry 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 orbitalsymmetry must be such that regions of the same sign overlap
the energy of the orbitals must besimilar
the interatomic distance must be short enough but not too short
If the total energy of the electrons in the molecularorbitals
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)
Antibonding
Bonding
More generally:Y = N[caY(1sa) cbY (1sb)]
nA.O.s nM.O.s
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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)
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Both s (and s*) notation means symmetric/antisymmetric with respect to rotation
s s* s*zC2 zC2 zC2
zC2Not s
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Combinations of twop orbitals (e.g. H2)
p (and p*) notation means
change of sign upon C2 rotation
s (and s*) notation means no
change of sign upon rotation
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Combinations of twop orbitals
zC2
zC2
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Combinations of two sets ofp orbitals
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Combinations ofs andp orbitals
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Combinations ofdorbitals
No interactiondifferent symmetry
d means change of sign upon C4
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NO NOYES
Is there a net interaction?
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Relative energies of interacting orbitals must be similar
Strong interaction Weak 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 highestMaximum spin multiplicity
Electrons have different quantum
numbers including spin (+ , -
)
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Bond order = # of electronsin bonding MO's# of electrons inantibonding MO's12
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O2 (2 x 8e)
1/2 (10 - 6) = 2A double bond
Or counting only
valence electrons:1/2 (8 - 4) = 2
Note subscripts
gand usymmetric/antisymmetric
upon i
Place labels or in this diagram
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Place labelsgoru in this diagram
sg
p*g
s*u
pu
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dg
d*u
p*g
pu
s*u
sg
goru?
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Orbital mixing
Same symmetry and similar energies !
shouldnt they interact?
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sorbital mixing
When two MOs of the same symmetry mix
the one with higher energy moves higher and the one with lower energy moves lower
M l l bi l
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H2sg2 (single bond)
He2
sg
2 s*u
2 (no bond)
Molecular orbitals
for diatomic molecules
From H2 to Ne2
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E (Z*)
DE s > DE pParamagneticdue to mixing
C2pu2 pu
2 (double bond)
C22- pu
2 pu2sg
2(triple bond)
O2pu2 pu
2 p*g1p*g
1 (double bond)
paramagnetic
O22- pu
2 pu2 p*g
2p*g2 (single bond)
diamagnetic
B d l th i di t i l l
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Bond lengths in diatomic molecules
Filling bonding orbitals
Filling antibonding orbitals
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Photoelectron Spectroscopy
hn
(UV o X rays) e-
Ionization
energy
hn
photons
kinetic energy of
expelled electron=-
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N2O2
s*u (2s)
pu (2p)
sg (2p)
s*u (2s)
sg (2p)pu (2p)
p*u (2p)
Very involved in bonding
(vibrational fine structure)