chemistry 481(01) spring 2014
DESCRIPTION
Chemistry 481(01) Spring 2014. Instructor: Dr. Upali Siriwardane e-mail: [email protected] Office: CTH 311 Phone 257-4941 Office Hours: M,W 8:00-9:00 & 11:00-12:00 am; Tu,Th , F 10:00 - 12:00 a.m . April 10 , 2014: Test 1 (Chapters 1, 2, 3,) - PowerPoint PPT PresentationTRANSCRIPT
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Chapter 6-1Chemistry 481, Spring 2015, LA Tech
Instructor: Dr. Upali Siriwardane
e-mail: [email protected]
Office: 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 7 , 2015: Test 1 (Chapters 1, 2, 3)
April 30, 2015: Test 2 (Chapters 5, 6 & 7)
May 19, 2015: Test 3 (Chapters. 19 & 20)
May 19, Make Up: Comprehensive covering all Chapters
Chemistry 481(01) Spring 2015
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Chapter 6-2Chemistry 481, Spring 2015, LA Tech
Chapter 6. Molecular symmetry An introduction to symmetry analysis 6.1 Symmetry operations, elements and point groups 1796.2 Character tables 183Applications of symmetry 6.3 Polar molecules 1866.4 Chiral molecules 1876.5 Molecular vibrations 188The symmetries of molecular orbitals 6.6 Symmetry-adapted linear combinations 1916.7 The construction of molecular orbitals 1926.8 The vibrational analogy 194Representations 6.9 The reduction of a representation 1946.10 Projection operators 196
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Chapter 6-3Chemistry 481, Spring 2015, LA Tech
SymmetryM.C. Escher
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Chapter 6-4Chemistry 481, Spring 2015, LA Tech
Symmetry Butterflies
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Chapter 6-5Chemistry 481, Spring 2015, LA Tech
Fish and Boats Symmetry
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Chapter 6-6Chemistry 481, Spring 2015, 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
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Chapter 6-7Chemistry 481, Spring 2015, LA Tech
2) What is a symmetry operation?
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Chapter 6-8Chemistry 481, Spring 2015, 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.
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Chapter 6-9Chemistry 481, Spring 2015, 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:
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Chapter 6-10Chemistry 481, Spring 2015, 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.
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Chapter 6-11Chemistry 481, Spring 2015, LA Tech
i - an inversion centerThe symmetry operation corresponds to inversion through the center. The coordinates (x,y,z) of every atom are changed into (-x,-y,-z):
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Chapter 6-12Chemistry 481, Spring 2015, 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.
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Chapter 6-13Chemistry 481, Spring 2015, LA Tech
2) What are four basic symmetry elements and operations?
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Chapter 6-14Chemistry 481, Spring 2015, LA Tech
3) Draw and identify the symmetry elements in:
a) NH3:
b) H2O:
c) CO2:
d) CH4:
e) BF3:
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Chapter 6-15Chemistry 481, Spring 2015, LA Tech
Point GroupAssignment
There 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
planes
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Chapter 6-16Chemistry 481, Spring 2015, 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
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Chapter 6-17Chemistry 481, Spring 2015, LA Tech
Special Point GroupsLinear molecules have a C∞ axis - there are an infinite number of rotations that will leave a linear molecule unchangedIf 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
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Chapter 6-18Chemistry 481, Spring 2015, 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 2015, 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)3Cl3
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Chapter 6-20Chemistry 481, Spring 2015, LA Tech
Character tables Summarize 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
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Chapter 6-21Chemistry 481, Spring 2015, LA Tech
Character Table Td
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Chapter 6-22Chemistry 481, Spring 2015, 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 respectivelydz2, dx2-y2, dxy, dxz and dyz orbitals are given by the symbols z2, x2-y2, xy, xz and yz respectively.
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Chapter 6-23Chemistry 481, Spring 2015, LA Tech
H2O molecule belongs to C2v point group
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Chapter 6-24Chemistry 481, Spring 2015, LA Tech
Symmetry ClassesThe symmetry classes for each point group and are labeled in the character tableLabelSymmetry Class
A Singly-degenerate class, symmetric with respect to the principal axisB Singly-degenerate class, asymmetric with respect to the principal axisE Doubly-degenerate classT Triply-degenerate class
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Chapter 6-25Chemistry 481, Spring 2015, 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
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Chapter 6-26Chemistry 481, Spring 2015, LA Tech
ChiralityOnly molecules lacking a Sn axis can be chiral.This includes mirror planesand a center of inversion as S2=s , S1=i and Dn groups.Not Chiral: Dnh, Dnd,Td and Oh.
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Chapter 6-27Chemistry 481, Spring 2015, LA Tech
Meso-Tartaric Acid
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Chapter 6-28Chemistry 481, Spring 2015, LA Tech
Optical Activity
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Chapter 6-29Chemistry 481, Spring 2015, 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.
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Chapter 6-30Chemistry 481, Spring 2015, 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 2015, LA Tech
Building a MO diagram for H2OO
H HO
H H
a1
a1
a1
a1
a1
b1b1
b2
b2
b2
b2
a1
x
z
y
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Chapter 6-32Chemistry 481, Spring 2015, LA Tech
a1 orbital of H2O
E(1a1) (+1)(1a1)C2(1a1) (+1)(1a1)sv1(1a1) (+1)(1a1)sv2(1a1) (+1)(1a1)
C2
sv1
sv2
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Chapter 6-33Chemistry 481, Spring 2015, LA Tech
b1 orbital of H2O
sv2
sv2(1b1) (+1)(1b1)E(1b1) (+1)(1b1)
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Chapter 6-34Chemistry 481, Spring 2015, LA Tech
b1 orbital of H2O, cont.
C2
sv1
C2(1b1) (-1)(1b1)sv1(1b1) (-1)(1b1)
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Chapter 6-35Chemistry 481, Spring 2015, LA Tech
b2 orbital of H2O
sv1
sv1(1b2) (+1)(1b2)E(1b2) (+1)(1b2)
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Chapter 6-36Chemistry 481, Spring 2015, LA Tech
b2 orbital of H2O, cont.
C2
C2(1b2) (-1)(1b2)sv2(1b2) (-1)(1b2)
sv2
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Chapter 6-37Chemistry 481, Spring 2015, LA Tech
[Fe(CN)6]4-
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Chapter 6-38Chemistry 481, Spring 2015, LA Tech
Reducing the RepresentationUse reduction formula
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Chapter 6-39Chemistry 481, Spring 2015, LA Tech
MO forML6 diagram Molecules
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Chapter 6-40Chemistry 481, Spring 2015, 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)
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Chapter 6-41Chemistry 481, Spring 2015, LA Tech
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Chapter 6-42Chemistry 481, Spring 2015, 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 2015, LA Tech
Summary of Operations by a set of four 9 x 9 transformation matrices.
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Chapter 6-44Chemistry 481, Spring 2015, LA Tech
Use Reduction Formula
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Chapter 6-45Chemistry 481, Spring 2015, LA Tech
Example H2O, C2v
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Chapter 6-46Chemistry 481, Spring 2015, LA Tech
Use Reduction Formula:
R
pp )R()R(g1a
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,
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Chapter 6-47Chemistry 481, Spring 2015, 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
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Chapter 6-48Chemistry 481, Spring 2015, 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 2015, 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.
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Chapter 6-50Chemistry 481, Spring 2015, LA Tech
Example H2O, C2v
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Chapter 6-51Chemistry 481, Spring 2015, LA Tech
Use Reduction Formula:
R
pp )R()R(g1a
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,
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Chapter 6-52Chemistry 481, Spring 2015, 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
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Chapter 6-53Chemistry 481, Spring 2015, 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)
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Chapter 6-54Chemistry 481, Spring 2015, 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 2015, 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.
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Chapter 6-56Chemistry 481, Spring 2015, 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 6-57Chemistry 481, Spring 2015, 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 00 1
C2, sxz :
0 11 0
i.e. only count UNSHIFTED VECTORS (each of these ® +1 to ).
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Chapter 6-58Chemistry 481, Spring 2015, 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.
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Chapter 6-59Chemistry 481, Spring 2015, 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 3s
Gstretch 3 0 1
Gbend 3 0 1
Reduction formula ® Gstretch = A1 + E
Gbend = A1 + E
(Remember Gvib (above) = 2A1 + 2E)
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Chapter 6-60Chemistry 481, Spring 2015, LA Tech
CH4H
C
HH
H
r1
r2r3
r4 6 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
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Chapter 6-61Chemistry 481, Spring 2015, LA Tech
IR AbsorptionsInfra-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
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Chapter 6-62Chemistry 481, Spring 2015, 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)
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Chapter 6-63Chemistry 481, Spring 2015, 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.
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Chapter 6-64Chemistry 481, Spring 2015, 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.
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Chapter 6-65Chemistry 481, Spring 2015, LA Tech