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Chapter 6-1 Chemistry 481, Spring 2014, LA Tech 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 Chemistry 481(01) Spring 2014

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Chapter 6-2 Chemistry 481, Spring 2014, LA Tech 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

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Chapter 6-3 Chemistry 481, Spring 2014, LA Tech Symmetry M.C. Escher

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Chapter 6-4 Chemistry 481, Spring 2014, LA Tech Symmetry Butterflies

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Chapter 6-5 Chemistry 481, Spring 2014, LA Tech Fish and Boats Symmetry Fish and Boats Symmetry

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Chapter 6-6 Chemistry 481, Spring 2014, 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 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) Identity (E) Proper rotation axis of order n (C n ) Proper rotation axis of order n (C n ) Plane of symmetry ( ) Plane of symmetry ( ) Improper axis (rotation + reflection) of order n (S n ), an inversion center is S 2 Improper axis (rotation + reflection) of order n (S n ), an inversion center is S 2

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Chapter 6-7 Chemistry 481, Spring 2014, LA Tech 2) What is a symmetry operation?

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Chapter 6-8 Chemistry 481, Spring 2014, 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. C 1 is the most common element leading to E, but other combination of symmetry operation are also possible for E.

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Chapter 6-9 Chemistry 481, Spring 2014, LA Tech C n - a proper rotation axis of order n The symmetry operation C n corresponds to rotation about an axis by (360/n) o. The symmetry operation C n corresponds to rotation about an axis by (360/n) o. H 2 O possesses a C 2 since rotation by 360/2 o = 180 o about an axis bisecting the two bonds sends the molecule into an H 2 O possesses a C 2 since rotation by 360/2 o = 180 o about an axis bisecting the two bonds sends the molecule into an indistinguishable form: indistinguishable form:

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Chapter 6-10 Chemistry 481, Spring 2014, LA Tech - a plane of reflection The symmetry operation corresponds to reflection in a plane. H 2 O possesses two reflection planes. Labels: h, d and v. Labels: h, d and v.

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Chapter 6-11 Chemistry 481, Spring 2014, 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):

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Chapter 6-12 Chemistry 481, Spring 2014, LA Tech S n - 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 = .

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Chapter 6-13 Chemistry 481, Spring 2014, LA Tech 2) What are four basic symmetry elements and operations?

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Chapter 6-14 Chemistry 481, Spring 2014, LA Tech 3) Draw and identify the symmetry elements in: a) NH 3 : b) H 2 O: c) CO 2 : d) CH 4 : e) BF 3 :

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Chapter 6-15 Chemistry 481, Spring 2014, LA Tech 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

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Chapter 6-16 Chemistry 481, Spring 2014, LA Tech 4) Draw, identify symmetry elements and assign the point group of following molecules: a) NH 2 Cl: b) SF 4 : c) PCl 5 : d) SF 6 : e) Chloroform f) 1,3,5-trichlorobenzene

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Chapter 6-17 Chemistry 481, Spring 2014, 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 T d Octahedral molecules have a point group O h Icosahedral molecules have a point group I h

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Chapter 6-18 Chemistry 481, Spring 2014, 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 D 4h point group irrespective of their chemical formula.

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Chapter 6-19 Chemistry 481, Spring 2014, LA Tech 5) Determine the point group to which each of the following belongs: a) CCl 4 b) Benzene c) Pyridine d) Fe(CO) 5 e) Staggered and eclipsed ferrocene, (η 5 -C 5 H 5 ) 2 Fe f) Octahedral W(CO) 6 g) fac- and mer-Ru(H 2 O) 3 Cl 3

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Chapter 6-20 Chemistry 481, Spring 2014, 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. C 2v E C 2 v v ' A 1 1 1 1 1 z x 2, y 2, z 2 A 2 1 1 -1 -1 R z xy B 1 1 -1 1 -1 x, R y xz B 2 1 -1 -1 -1 y, R x yz

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Chapter 6-21 Chemistry 481, Spring 2014, LA Tech Character Table T d

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Chapter 6-22 Chemistry 481, Spring 2014, LA Tech Information on Character Table The order of the group is the total number of symmetry elements and is given the symbol h. For C 2v h = 4. The order of the group is the total number of symmetry elements and is given the symbol h. For C 2v h = 4. First Column has labels for the irreducible representations. A 1 A 2 B 1 B 2 First Column has labels for the irreducible representations. A 1 A 2 B 1 B 2 The rows of numbers are the characters (1,-1)of the irreducible representations. The rows of numbers are the characters (1,-1)of the irreducible representations. p x, p y and p z orbitals are given by the symbols x, y and z respectively p x, p y and p z orbitals are given by the symbols x, y and z respectively d z2, d x2-y2, d xy, d xz and d yz orbitals are given by the symbols z 2, x 2 -y 2, xy, xz and yz respectively. d z2, d x2-y2, d xy, d xz and d yz orbitals are given by the symbols z 2, x 2 -y 2, xy, xz and yz respectively.

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Chapter 6-23 Chemistry 481, Spring 2014, LA Tech H 2 O molecule belongs to C 2v point group

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Chapter 6-24 Chemistry 481, Spring 2014, LA Tech Symmetry Classes The symmetry classes for each point group and are labeled in the character table LabelSymmetry Class ASingly-degenerate class, symmetric with respect to the principal axis BSingly-degenerate class, asymmetric with respect to the principal axis EDoubly-degenerate class TTriply-degenerate class

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Chapter 6-25 Chemistry 481, Spring 2014, LA Tech Molecular Polarity and Chirality Polarity: Only molecules belonging to the point groups C n, C nv and C s are polar. The dipole moment lies along the symmetry axis formolecules belonging to the point groups C n and C nv. Any of D groups, T, O and I groups will not be polar Any of D groups, T, O and I groups will not be polar

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Chapter 6-26 Chemistry 481, Spring 2014, LA Tech Chirality Only molecules lacking a S n axis can be chiral. This includes mirror planes and a center of inversion as S 2 = , S 1 =I and D n groups. Not Chiral: D nh, D nd,T d and O h.

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Chapter 6-27 Chemistry 481, Spring 2014, LA Tech Meso-Tartaric Acid

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Chapter 6-28 Chemistry 481, Spring 2014, LA Tech Optical Activity

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Chapter 6-29 Chemistry 481, Spring 2014, LA Tech Symmetry allowed combinations Find symmetry species spanned by a set of orbitals Find symmetry species spanned by a set of orbitals Next find combinations of the atomic orbitals on central atom which have these symmetries. 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). 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. 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 6-30 Chemistry 481, Spring 2014, LA Tech Example: Valence MOs of Water H 2 O has C 2v symmetry. The symmetry operators of the C 2v group all commute with each other (each is in its own class). Molecualr orbitals should have symmetry operators E, C 2, v1, and v2.

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Chapter 6-31 Chemistry 481, Spring 2014, LA Tech Building a MO diagram for H 2 O x z y

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Chapter 6-32 Chemistry 481, Spring 2014, LA Tech a 1 orbital of H 2 O

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Chapter 6-33 Chemistry 481, Spring 2014, LA Tech b 1 orbital of H 2 O

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Chapter 6-34 Chemistry 481, Spring 2014, LA Tech b 1 orbital of H 2 O, cont.

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Chapter 6-35 Chemistry 481, Spring 2014, LA Tech b 2 orbital of H 2 O

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Chapter 6-36 Chemistry 481, Spring 2014, LA Tech b 2 orbital of H 2 O, cont.

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Chapter 6-37 Chemistry 481, Spring 2014, LA Tech [Fe(CN) 6 ] 4-

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Chapter 6-38 Chemistry 481, Spring 2014, LA Tech Reducing the Representation Use reduction formula 1

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Chapter 6-39 Chemistry 481, Spring 2014, LA Tech MO forML 6 diagram Molecules

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Chapter 6-40 Chemistry 481, Spring 2014, 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). 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. 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 Linear molecules: 3N - 5 vibrations Non-linear molecules: 3N - 6 vibrations (N is the number of atoms) Non-linear molecules: 3N - 6 vibrations (N is the number of atoms)

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Chapter 6-41 Chemistry 481, Spring 2014, LA Tech

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Chapter 6-42 Chemistry 481, Spring 2014, LA Tech Hence we can deduce 3N for our triatomic molecule in three lines: E C2 xz yz unshifted atoms 3 1 1 3 /unshifted atom 3 -1 1 1 3N 9 -1 1 3 For more complicated molecules this shortened method is essential!! Having obtained 3N, we now must reduce it to find which irreducible representations are present.

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Chapter 6-43 Chemistry 481, Spring 2014, LA Tech Reducible Representations(3N) for H 2 O: Normal Coordinate Method If we carry out the symmetry operations of C 2v on this set, we will obtain a transformation matrix for each operation. If we carry out the symmetry operations of C 2v on this set, we will obtain a transformation matrix for each operation. E.g. C 2 effects the following transformations: E.g. C 2 effects the following transformations: x 1 -> -x 2, y 1 -> -y 2, z 1 -> z 2, x 2 -> -x 1, y 2 -> -y 1, z 2 -> z 1, x 3 -> -x 3, y 3 -> -y 3, z 3 -> z 3. x 1 -> -x 2, y 1 -> -y 2, z 1 -> z 2, x 2 -> -x 1, y 2 -> -y 1, z 2 -> z 1, x 3 -> -x 3, y 3 -> -y 3, z 3 -> z 3.

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Chapter 6-44 Chemistry 481, Spring 2014, LA Tech Summary of Operations by a set of four 9 x 9 transformation matrices. Summary of Operations by a set of four 9 x 9 transformation matrices.

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Chapter 6-45 Chemistry 481, Spring 2014, LA Tech Use Reduction Formula

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Chapter 6-46 Chemistry 481, Spring 2014, LA Tech Example H 2 O, C 2v

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Chapter 6-47 Chemistry 481, Spring 2014, LA Tech Use Reduction Formula : to show that here we have: 3N = 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 6-48 Chemistry 481, Spring 2014, LA Tech T = A1 + B1 + B2 R = A2 + B1 + B2 i.e. T+R = A1 + A2 + 2B1 + 2B2 But vib = 3N - T+R Therefore vib = 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 6-49 Chemistry 481, Spring 2014, LA Tech Further examples of the determination of vib, via 3N: NH3(C3v) C3v E 2C3 3 v 12 0 2 3N 12 0 2 Reduction formula 3N = 3A1 + A2 + 4E T+R (from character table) = A1 + A2 + 2E, vib = 2A1 + 2E (each E "mode" is in fact two vibrations (doubly degenerate)

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Chapter 6-50 Chemistry 481, Spring 2014, LA Tech CH 4 (T d ) Td E 8C3 3C2 6S4 6 d 15 0 -1 -1 3 3N 15 0 -1 -1 3 Reduction formula 3N = A1 + E + T1 + 3T2 T+R (from character table) = T1 + T2, vib = 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 6-51 Chemistry 481, Spring 2014, LA Tech XeF 4 (D 4h ) D4h E 2C4 C2 2C2' 2C2" i 2S4 h 2 v 2 d 15 1 -1 -3 -1 -1 -1 5 3 1 3N 15 1 -1 -3 -1 -1 -1 5 3 1 Reduction formula 3N = A1g + A2g + B1g + B2g + Eg + 2A2u + B2u + 3Eu T+R (from character table) = A2g + Eg + A2u + Eu, vib = A1g + B1g + B2g + A2u + B2u + 2Eu For any molecule, we can always deduce the overall symmetry of all the vibrational modes, from the 3N representation. To be more specific we need now to use the INTERNAL COORDINATE method.

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Chapter 6-52 Chemistry 481, Spring 2014, 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 6-53 Chemistry 481, Spring 2014, LA Tech Let us return to the C2v molecule: Use as bases for stretches: r1, r2. Use as basis for bend: C2v E C2 xz yz stretch 2 0 0 2 bend 1 1 1 1 N.B. Transformation matrices for stretch : E, yz:C2, xz : i.e. only count UNSHIFTED VECTORS (each of these +1 to )

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Chapter 6-54 Chemistry 481, Spring 2014, LA Tech bend is clearly irreducible, i.e. A1. stretch 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 6-55 Chemistry 481, Spring 2014, LA Tech Other examples: Bases for stretches: r1, r2, r3. Bases for bends: 1, 2, 3. C3v E 2C3 3 stretch 3 0 1 bend 3 0 1 Reduction formula stretch = A1 + E bend = A1 + E (Remember vib (above) = 2A1 + 2E)

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Chapter 6-56 Chemistry 481, Spring 2014, LA Tech CH 4 6 angles 1,..... 6, where 1 lies between r1 and r2 etc. Bases for stretches: r1, r2, r3, r4. Bases for bends: 1, 2, 3, 4, 5, 6. Td E 8C3 3C2 6S4 6 d stretch 4 1 0 0 2 bend 6 0 2 0 2 Reduction formula stretch = A1 + T2 bend = A1 + E + T2 But 3N (above) = A1 + E + 2T2

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Chapter 6-57 Chemistry 481, Spring 2014, 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

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Chapter 6-58 Chemistry 481, Spring 2014, LA Tech Raman Spectroscopy Named after discoverer, Indian physicist C.V.Raman (1927). It is a light scattering process. 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, 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). 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 : Each Raman band has wavenumber : where = Raman scattered wavenumber 0 = wavenumber of incident radiation vib = a vibrational wavenumber of the molecule (in general several of these)

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Chapter 6-59 Chemistry 481, Spring 2014, 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. 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 6-60 Chemistry 481, Spring 2014, 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 x 2 or yz) then the fundamental transition for this normal mode will be Raman active.

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Chapter 6-61 Chemistry 481, Spring 2014, LA Tech

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