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Chemistry 481(01) Spring 2014

Chemistry 481(01) Spring 2014. Instructor: Dr. Upali Siriwardane e-mail: upali@latech.edu 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,)

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Chemistry 481(01) Spring 2014

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  1. Chemistry 481(01) Spring 2014 • Instructor: Dr. Upali Siriwardane • e-mail: upali@latech.edu • 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,) • May 1, 2014: Test 2 (Chapters  6 & 7) • May 20, 2014: Test 3 (Chapters. 19 & 20) • May 22, 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 GroupAssignment • There is a systematic way of naming most point groupsC, 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 groupIh

  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 C2svsv' 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 Polarity: Only molecules belonging to the point groups Cn, Cnv and Cs are polar. The dipole moment lies along the symmetry axis formolecules 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 H2O z 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 1

  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)

  41. Hence we can deduce G3N for our triatomic molecule in three lines: E C2 sxzsyz unshifted atoms 3 1 1 3 c/unshifted atom 3 -1 1 1 \ 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.

  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 + 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,

  47. 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. Further examples of the determination of Gvib, via G3N: NH3 (C3v) 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)

  49. CH4 (Td) 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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