1. Chemistry 481(01) Spring 2016 Instructor: Dr. Upali Siriwardane
Office: CTH 311 Phone
Office Hours:
M,W 8:00-9:00 & 11:00-12:00 am;
Tu,Th, F 9: :30 a.m.
April 7 , 2016: Test 1 (Chapters 1, 2, 3, 4)
April 28, 2016: Test 2 (Chapters  (6 & 7)
May 17, 2016: Test 3 (Chapters. 19 & 20)
May 18, Make Up: Comprehensive covering all Chapters

2. Chapter 6. Molecular symmetry
An 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

3. Symmetry
M.C. Escher

4. Symmetry Butterflies

5. Fish and Boats Symmetry

6. 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

7. 2) What is a symmetry operation?

8. 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.

9. 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:

10. s - a plane of reflection
The symmetry operation corresponds to reflection in a plane. H2O possesses two reflection planes.
Labels: sh, sd and sv.

11. 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):

12. Sn - an improper axis of order n
The 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.

13. 2) What are four basic symmetry elements and operations?

14. 3) Draw and identify the symmetry elements in:
a) NH3:
b) H2O:
c) CO2:
d) CH4:
e) BF3:

15. Point Group Assignment
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

16. 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

17. 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

18. 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.

19. 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

20. 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

21. Character Table Td

22. Information on Character Table
The 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.

23. H2O molecule belongs to C2v point group

24. Symmetry Classes
The symmetry classes for each point group and are labeled in the character table
Label Symmetry 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

25. Molecular Polarity and Chirality
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

26. Chirality Only 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.

27. Meso-Tartaric Acid

28. Optical Activity

29. 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.

30. 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, v1, and v2.

31. Building a MO diagram for H2Oz y x

32. a1 orbital of H2O

33. b1 orbital of H2O

34. b1 orbital of H2O, cont.

35. b2 orbital of H2O

36. b2 orbital of H2O, cont.

37. [Fe(CN)6]4-

38. Reducing the Representation
Use reduction formula

39. MO forML6 diagram Molecules

40. 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)

42. 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.

43. Summary of Operations by a set of four 9 x 9 transformation matrices.

44. Use Reduction Formula

45. Example H2O, C2v

46. Use Reduction Formula:
to show that here we have:
G3N = 3A1 + A B B2
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,

47. i.e. of the 3 (= 3N-6) vibrations for a molecule
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

48. 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.

49. Deduce G3N for our triatomic molecule, H2O in three lines:
E C sxz syz
unshifted atoms
c/unshifted atom s
\ G3N
For more complicated molecules this shortened
method is essential!!
Having obtained G3N, we now must reduce it to find
which irreducible representations are present.

50. Example H2O, C2v

51. Use Reduction Formula:
to show that here we have:
G3N = 3A1 + A B B2
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,

52. i.e. of the 3 (= 3N-6) vibrations for a molecule
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

53. Further examples of the determination of Gvib, via G3N:
NH3 (C3v)
C3v E C sv
\ G3N
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)

54. Reduction formula ® G3N = A1 + E + T1 + 3T2
CH4 (Td)
Td E C C S sd
\ G3N
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)

55. G3N = A1g + A2g + B1g + B2g + Eg + 2A2u + B2u + 3Eu
XeF4 (D4h)
D4h E 2C4 C2 2C2' 2C2" i 2S4 sh 2sv 2sd
\G3N
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.

56. 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.

57. Let us return to the C2v molecule:
Use as bases for stretches:
Dr1, Dr2.
Use as basis for bend: Dq
C2v E C sxz syz
Gstretch
Gbend
N.B. Transformation matrices for Gstretch :
E, syz:
C2, sxz :
i.e. only count UNSHIFTED VECTORS (each of these ® +1 to c).

58. 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.

59. Bases for stretches: Dr1, Dr2, Dr3.
Other examples:
NH3
Bases for stretches: Dr1, Dr2, Dr3.
Bases for bends: Dq1, Dq2, Dq3.
C3v E C s
Gstretch
Gbend
Reduction formula ® Gstretch = A1 + E
Gbend = A1 + E
(Remember Gvib (above) = 2A1 + 2E)

60. Bases for stretches: Dr1, Dr2, Dr3, Dr4.
6 angles q1,.....q6, where q1
lies between r1 and r2 etc.
Bases for stretches: Dr1, Dr2, Dr3, Dr4.
Bases for bends: Dq1, Dq2, Dq3, Dq4, Dq5, Dq6.
Td E C C S sd
Gstretch
Gbend
Reduction formula ® Gstretch = A1 + T2
Gbend = A1 + E + T2
But G3N (above) = A1 + E + 2T2

61. Deals with polarizability
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

62. Raman Spectroscopy Each Raman band has wavenumber:
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)

63. 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.

64. 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.