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Chapter 6-1Chemistry 481, Spring 2017, LA Tech
Instructor: Dr. Upali Siriwardane
e-mail: [email protected]: CTH 311 Phone 257-4941
Office Hours:
M,W 8:00-9:00 & 11:00-12:00 am;
Tu,Th, F 9:30 - 11:30 a.m.
April 4 , 2017: Test 1 (Chapters 1, 2, 3, 4)
April 27, 2017: Test 2 (Chapters (6 & 7)
May 16, 2016: Test 3 (Chapters. 19 & 20)
May 17, Make Up: Comprehensive covering all Chapters
Chemistry 481(01) Spring 2017
Chapter 6-2Chemistry 481, Spring 2017, LA Tech
Chapter 6. Molecular symmetryAn introduction to symmetry analysis
6.1 Symmetry operations, elements and point groups 179
6.2 Character tables 183
Applications of symmetry
6.3 Polar molecules 186
6.4 Chiral molecules 187
6.5 Molecular vibrations 188
The symmetries of molecular orbitals
6.6 Symmetry-adapted linear combinations 191
6.7 The construction of molecular orbitals 192
6.8 The vibrational analogy 194
Representations
6.9 The reduction of a representation 194
6.10 Projection operators 196
Chapter 6-3Chemistry 481, Spring 2017, LA Tech
SymmetryM.C. Escher
Chapter 6-4Chemistry 481, Spring 2017, LA Tech
Symmetry Butterflies
Chapter 6-5Chemistry 481, Spring 2017, LA Tech
Fish and Boats Symmetry
Chapter 6-6Chemistry 481, Spring 2017, LA Tech
Symmetry elements and operations
• A symmetry operation is the process of
doing something to a shape or an object so
that the result is indistinguishable from the
initial state
• Identity (E)
• Proper rotation axis of order n (Cn)
• Plane of symmetry (s)
• Improper axis (rotation + reflection) of order
n (Sn), an inversion center is S2
2
Chapter 6-7Chemistry 481, Spring 2017, LA Tech
2) What is a symmetry operation?
Chapter 6-8Chemistry 481, Spring 2017, LA Tech
E - the identity element
The symmetry operation corresponds to
doing nothing to the molecule. The E
element is possessed by all molecules,
regardless of their shape.
C1 is the most common element leading to E,
but other combination of symmetry
operation are also possible for E.
Chapter 6-9Chemistry 481, Spring 2017, LA Tech
Cn - a proper rotation axis of order n
• The symmetry operation Cn corresponds to
rotation about an axis by (360/n)o.
• H2O possesses a C2 since rotation by 360/2o = 180o
about an axis bisecting the two bonds sends the
molecule into an
• indistinguishable form:
Chapter 6-10Chemistry 481, Spring 2017, LA Tech
s - a plane of reflectionThe symmetry operation corresponds to reflection in
a plane. H2O possesses two reflection planes.
Labels: sh, sd and sv.
Chapter 6-11Chemistry 481, Spring 2017, LA Tech
i - an inversion center
The symmetry operation corresponds to inversion
through the center. The coordinates (x,y,z) of every
atom are changed into (-x,-y,-z):
Chapter 6-12Chemistry 481, Spring 2017, LA Tech
Sn - an improper axis of order nThe symmetry operation is rotation by (360/n)o and
then a reflection in a plane perpendicular to the
rotation axis.
operation is
the same as
an inversion
is S2 = i
a reflection
so S1 = s.
3
Chapter 6-13Chemistry 481, Spring 2017, LA Tech
2) What are four basic symmetry elements and
operations?
Chapter 6-14Chemistry 481, Spring 2017, LA Tech
3) Draw and identify the symmetry elements in:
a) NH3:
b) H2O:
c) CO2:
d) CH4:
e) BF3:
Chapter 6-15Chemistry 481, Spring 2017, LA Tech
Point Group
AssignmentThere is a systematic
way of naming most
point groups
C, S or D for
the principal
symmetry axis
A number for the
order of the
principal axis
(subscript) n.
A subscript h, d, or v
for symmetry planesChapter 6-16Chemistry 481, Spring 2017, LA Tech
4) Draw, identify symmetry elements and assign the
point group of following molecules:
a) NH2Cl:
b) SF4:
c) PCl5:
d) SF6:
e) Chloroform
f) 1,3,5-trichlorobenzene
Chapter 6-17Chemistry 481, Spring 2017, LA Tech
Special Point Groups
Linear molecules have a C∞ axis - there are an
infinite number of rotations that will leave a linear
molecule unchanged
If there is also a plane of symmetry perpendicular to
the C∞ axis, the point group is D∞h
If there is no plane of symmetry, the point group is
C∞v
Tetrahedral molecules have a point group Td
Octahedral molecules have a point group Oh
Icosahedral molecules have a point group Ih
Chapter 6-18Chemistry 481, Spring 2017, LA Tech
Point groups
It is convenient to classify molecules
with the same set of symmetry elements
by a label. This label summarizes the
symmetry properties. Molecules with the
same label belong to the same point
group.
For example, all square molecules
belong to the D4h point group
irrespective of their chemical formula.
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Chapter 6-19Chemistry 481, Spring 2017, LA Tech
5) Determine the point group to which each of
the following belongs:
a) CCl4
b) Benzene
c) Pyridine
d) Fe(CO)5
e) Staggered and eclipsed ferrocene, (η5-C5H5)2Fe
f) Octahedral W(CO)6
g) fac- and mer-Ru(H2O)3Cl3Chapter 6-20Chemistry 481, Spring 2017, LA Tech
Character tablesSummarize a considerable amount of information
and contain almost all the data that is needed to
begin chemical applications of molecule.
C2v E C2 sv sv'
A1 1 1 1 1 z x2, y2, z2
A2 1 1 -1 -1 Rz xy
B1 1 -1 1 -1 x, Ry xz
B2 1 -1 -1 -1 y, Rx yz
Chapter 6-21Chemistry 481, Spring 2017, LA Tech
Character Table Td
Chapter 6-22Chemistry 481, Spring 2017, LA Tech
Information on Character TableThe order of the group is the total number of
symmetry elements and is given the symbol h. For
C2v h = 4.
First Column has labels for the irreducible
representations. A1 A2 B1 B2
The rows of numbers are the characters (1,-1)of the
irreducible representations.
px, py and pz orbitals are given by the symbols x, y
and z respectively
dz2, dx2-y2, dxy, dxz and dyz orbitals are given by the
symbols z2, x2-y2, xy, xz and yz respectively.
Chapter 6-23Chemistry 481, Spring 2017, LA Tech
H2O molecule belongs to C2v point
group
Chapter 6-24Chemistry 481, Spring 2017, LA Tech
Symmetry Classes
The symmetry classes for each point group and are
labeled in the character table
LabelSymmetry Class
A Singly-degenerate class, symmetric with respect to the principal axis
B Singly-degenerate class, asymmetric with respect to the principal axis
E Doubly-degenerate class
T Triply-degenerate class
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Chapter 6-25Chemistry 481, Spring 2017, LA Tech
Molecular Polarity and Chirality
Polarity:
Only molecules belonging to the point groups Cn,
Cnv and Cs are polar. The dipole moment lies along
the symmetry axis for molecules belonging to the
point groups Cn and Cnv.
• Any of D groups, T, O and I groups will not be
polar
Chapter 6-26Chemistry 481, Spring 2017, LA Tech
ChiralityOnly molecules
lacking a Sn axis
can be chiral.
This includes
mirror planes
and a center of
inversion as
S2=s , S1=i and Dn
groups.
Not Chiral: Dnh,
Dnd,Td and Oh.
Chapter 6-27Chemistry 481, Spring 2017, LA Tech
Meso-Tartaric Acid
Chapter 6-28Chemistry 481, Spring 2017, LA Tech
Optical Activity
Chapter 6-29Chemistry 481, Spring 2017, LA Tech
Symmetry allowed combinations
• Find symmetry species spanned by a set of
orbitals
• Next find combinations of the atomic orbitals on
central atom which have these symmetries.
• Combining these are known as symmetry adapted
linear combinations (or SALCs).
• The characters show their behavior of the
combination under each of the symmetry
operations. several methods for finding the
combinations.
Chapter 6-30Chemistry 481, Spring 2017, LA Tech
Example: Valence MOs of Water
• H2O has C2v symmetry.
• The symmetry operators of the C2v group all
commute with each other (each is in its own
class).
• Molecualr orbitals should have symmetry
operators E, C2, sv1, and sv2.
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Chapter 6-31Chemistry 481, Spring 2017, LA Tech
Building a MO diagram for H2O
x
z
y
Chapter 6-32Chemistry 481, Spring 2017, LA Tech
a1 orbital of H2O
Chapter 6-33Chemistry 481, Spring 2017, LA Tech
b1 orbital of H2O
Chapter 6-34Chemistry 481, Spring 2017, LA Tech
b1 orbital of H2O, cont.
Chapter 6-35Chemistry 481, Spring 2017, LA Tech
b2 orbital of H2O
Chapter 6-36Chemistry 481, Spring 2017, LA Tech
b2 orbital of H2O, cont.
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Chapter 6-37Chemistry 481, Spring 2017, LA Tech
[Fe(CN)6]4-
Chapter 6-38Chemistry 481, Spring 2017, LA Tech
Reducing the Representation
Use reduction formula
Chapter 6-39Chemistry 481, Spring 2017, LA Tech
MO forML6 diagram Molecules
Chapter 6-40Chemistry 481, Spring 2017, LA Tech
Group Theory and Vibrational
Spectroscopy
• When a molecule vibrates, the symmetry of the
molecule is either preserved (symmetric
vibrations) or broken (asymmetric vibrations).
• The manner in which the vibrations preserve or
break symmetry can be matched to one of the
symmetry classes of the point group of the
molecule.
• Linear molecules: 3N - 5 vibrations
• Non-linear molecules: 3N - 6 vibrations (N is the
number of atoms)
Chapter 6-41Chemistry 481, Spring 2017, LA Tech Chapter 6-42Chemistry 481, Spring 2017, LA Tech
Reducible Representations(3N) for
H2O: Normal Coordinate Method
• If we carry out the symmetry operations of C2v on this
set, we will obtain a transformation matrix for each
operation.
• E.g. C2 effects the following transformations:
• x1 -> -x2, y1 -> -y2, z1 -> z2 , x2 -> -x1, y2 -> -y1, z2 ->
z1, x3 -> -x3 , y3 -> -y3, z3 -> z3.
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Chapter 6-43Chemistry 481, Spring 2017, LA Tech
Summary of Operations by a set of four
9 x 9 transformation matrices.
Chapter 6-44Chemistry 481, Spring 2017, LA Tech
Use Reduction Formula
Chapter 6-45Chemistry 481, Spring 2017, LA Tech
Example H2O, C2v
Chapter 6-46Chemistry 481, Spring 2017, LA Tech
Use Reduction Formula:
R
pp )R()R(g
1a
to show that here we have:
G3N = 3A1 + A2 + 2B1 + 3B2
This was obtained using 3N cartesian coordinate vectors.
Using 3N (translation + rotation + vibration) vectors would
have given the same answer.
But we are only interested in the 3N-6 vibrations.
The irreducible representations for the rotation and
translation vectors are listed in the character tables,
e.g. for C2v,
Chapter 6-47Chemistry 481, Spring 2017, LA Tech
GT = A1 + B1 + B2
GR = A2 + B1 + B2
i.e. GT+R = A1 + A2 + 2B1 + 2B2
But Gvib = G3N - GT+R
Therefore Gvib = 2A1 + B2
i.e. of the 3 (= 3N-6) vibrations for a molecule
like H2O, two have A1 and one has B2 symmetry
Chapter 6-48Chemistry 481, Spring 2017, LA Tech
INTERNAL COORDINATE METHOD
We used one example of this earlier - when we used
the "bond vectors" to obtain a representation
corresponding to bond stretches.
We will give more examples of these, and also the other
main type of vibration - bending modes.
For stretches we use as internal coordinates changes
in bond length, for bends we use changes in bond angle.
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Chapter 6-49Chemistry 481, Spring 2017, LA Tech
Deduce G3N for our triatomic molecule, H2O
in three lines:
E C2 sxz syz
unshifted atoms 3 1 1 3
/unshifted atom s 3 -1 1 3
\ G3N 9 -1 1 3
For more complicated molecules this shortened
method is essential!!
Having obtained G3N, we now must reduce it to find
which irreducible representations are present.
Chapter 6-50Chemistry 481, Spring 2017, LA Tech
Example H2O, C2v
Chapter 6-51Chemistry 481, Spring 2017, LA Tech
Use Reduction Formula:
R
pp )R()R(g
1a
to show that here we have:
G3N = 3A1 + A2 + 2B1 + 3B2
This was obtained using 3N cartesian coordinate vectors.
Using 3N (translation + rotation + vibration) vectors would
have given the same answer.
But we are only interested in the 3N-6 vibrations.
The irreducible representations for the rotation and
translation vectors are listed in the character tables,
e.g. for C2v,
Chapter 6-52Chemistry 481, Spring 2017, LA Tech
GT = A1 + B1 + B2
GR = A2 + B1 + B2
i.e. GT+R = A1 + A2 + 2B1 + 2B2
But Gvib = G3N - GT+R
Therefore Gvib = 2A1 + B2
i.e. of the 3 (= 3N-6) vibrations for a molecule
like H2O, two have A1 and one has B2 symmetry
Chapter 6-53Chemistry 481, Spring 2017, LA Tech
Further examples of the determination of Gvib, via G3N:
NH3 (C3v) N
HH
H
C3v E 2C3 3sv
12 0 2
\ G3N 12 0 2
Reduction formula G3N = 3A1 + A2 + 4E
GT+R (from character table) = A1 + A2 + 2E,
\ Gvib = 2A1 + 2E
(each E "mode" is in fact two vibrations (doubly degenerate)
Chapter 6-54Chemistry 481, Spring 2017, LA Tech
CH4 (Td)
H
C
HH
H
Td E 8C3 3C2 6S4 6sd
15 0 -1 -1 3
\ G3N 15 0 -1 -1 3
Reduction formula G3N = A1 + E + T1 + 3T2
GT+R (from character table) = T1 + T2,\ Gvib = A1 + E + 2T2
(each E "mode" is in fact two vibrations (doubly degenerate),
and each T2 three vibrations (triply degenerate)
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Chapter 6-55Chemistry 481, Spring 2017, LA Tech
XeF4 (D4h)Xe
F
F F
F
D4h E 2C4 C2 2C2' 2C2" i 2S4 sh 2sv 2sd
15 1 -1 -3 -1 -1 -1 5 3 1
\G3N 15 1 -1 -3 -1 -1 -1 5 3 1
Reduction formula G3N = A1g + A2g + B1g + B2g + Eg + 2A2u + B2u + 3Eu
GT+R (from character table) = A2g + Eg + A2u + Eu,
\ Gvib = A1g + B1g + B2g + A2u + B2u + 2Eu
For any molecule, we can always deduce the overall symmetry
of all the vibrational modes, from the G3N representation.
To be more specific we need now to use the
INTERNAL COORDINATE method.
Chapter 6-56Chemistry 481, Spring 2017, LA Tech
INTERNAL COORDINATE METHOD
We used one example of this earlier - when we used
the "bond vectors" to obtain a representation
corresponding to bond stretches.
We will give more examples of these, and also the other
main type of vibration - bending modes.
For stretches we use as internal coordinates changes
in bond length, for bends we use changes in bond angle.
Chapter 6-57Chemistry 481, Spring 2017, LA Tech
Let us return to the C2v molecule:
H
O
H
r1 r2
Use as bases for stretches:
Dr1, Dr2.
Use as basis for bend:
D
C2v E C2 sxz syz
Gstretch 2 0 0 2
Gbend 1 1 1 1
N.B. Transformation matrices for Gstretch :
E, syz: 1 0
0 1
C2, sxz : 0 1
1 0
i.e. only count UNSHIFTED VECTORS (each of these +1 to ).
Chapter 6-58Chemistry 481, Spring 2017, LA Tech
Gbend is clearly irreducible, i.e. A1.
Gstretch reduces to A1 + B2
We can therefore see that the three vibrational
modes of H2O divide into two stretches (A1 + B2)
and one bend (A1).
We will see later how this information helps
in the vibrational assignment.
Chapter 6-59Chemistry 481, Spring 2017, LA Tech
Other examples:
NH3 N
HH
Hr1
r2
r31 opposite to r1
2 opposite to r2
3 opposite to r3
Bases for stretches: Dr1, Dr2, Dr3.
Bases for bends: D1, D2, D3.
C3v E 2C3 3sGstretch 3 0 1
Gbend 3 0 1
Reduction formula Gstretch = A1 + E
Gbend = A1 + E
(Remember Gvib (above) = 2A1 + 2E)
Chapter 6-60Chemistry 481, Spring 2017, LA Tech
CH4H
C
HH
H
r1
r2r3
r46 angles 1,.....6, where 1
lies between r1 and r2 etc.
Bases for stretches: Dr1, Dr2, Dr3, Dr4.
Bases for bends: D1, D2, D3, D4, D5, D6.
Td E 8C3 3C2 6S4 6sd
Gstretch 4 1 0 0 2
Gbend 6 0 2 0 2
Reduction formula Gstretch = A1 + T2
Gbend = A1 + E + T2
But G3N (above) = A1 + E + 2T2
11
Chapter 6-61Chemistry 481, Spring 2017, LA Tech
IR Absorptions
Infra-red absorption spectra arise when a molecular
vibration causes a change in the dipole moment of
the molecule. If the molecule has no permanent
dipole moment, the vibrational motion must create
one; if there is a permanent dipole moment, the
vibrational motion must change it.
Raman Absorptions
Deals with polarizability
Chapter 6-62Chemistry 481, Spring 2017, LA Tech
Raman Spectroscopy• Named after discoverer, Indian physicist C.V.Raman (1927).
It is a light scattering process.
• Irradiate sample with visible light - nearly all is transmitted; of the rest, most scattered at unchanged energy (frequency) (Rayleigh scattering), but a little is scattered at changed frequency (Raman scattering). The light has induced vibrational transitions in molecules (ground excited state) - hence some energy taken from light,
• scattered at lower energy, i.e. at lower wavenumber. Raman scattering is weak - therefore need very powerful light source - always use lasers (monochromatic, plane polarised, very intense).
• Each Raman band has wavenumber:where n = Raman scattered wavenumber
n0 = wavenumber of incident radiation
nvib = a vibrational wavenumber of the molecule
(in general several of these)
Chapter 6-63Chemistry 481, Spring 2017, LA Tech
Molecular Vibrations
• At room temperature almost all molecules are in
their lowest vibrational energy levels with
quantum number n = 0. For each normal mode, the
most probable vibrational transition is from this
level to the next highest level (n = 0 -> 1). The
strong IR or Raman bands resulting from these
transitiions are called fundamental bands. Other
transitions to higher excited states (n = 0 -> 2, for
instance) result in overtone bands. Overtone
bands are much weaker than fundamental bands.
Chapter 6-64Chemistry 481, Spring 2017, LA Tech
If the symmetry label of a normal mode corresponds
to x, y, or z, then the fundamental transition for this
normal mode will be IR active.
If the symmetry label of a normal mode corresponds
to products of x, y, or z (such as x2 or yz) then the
fundamental transition for this normal mode will be
Raman active.
Chapter 6-65Chemistry 481, Spring 2017, LA Tech