Chemistry 481(01) Spring 2017 Instructor: Dr. Upali Siriwardane e-mail: upali@latech.edu Office: CT 311 Phone 257-4941 Office ours: 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 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 Chapter 6-1 Chapter 6-2 Symmetry Symmetry Butterflies M.C. Escher Chapter 6-3 Chapter 6-4 Fish and Boats Symmetry 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 (C n ) Plane of symmetry (s) Improper axis (rotation + reflection) of order n (S n ), an inversion center is S 2 Chapter 6-5 Chapter 6-6 1
2) What is a symmetry operation? 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. Chapter 6-7 Chapter 6-8 C n - a proper rotation axis of order n The symmetry operation C n corresponds to rotation about an axis by (360/n) o. 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: s - a plane of reflection The symmetry operation corresponds to reflection in a plane. 2 O possesses two reflection planes. Labels: s h, s d and s v. Chapter 6-9 Chapter 6-10 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): 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 S 2 = i a reflection so S 1 = s. Chapter 6-11 Chapter 6-12 2
2) What are four basic symmetry elements and operations? 3) Draw and identify the symmetry elements in: a) N 3 : b) 2 O: c) CO 2 : d) C 4 : e) BF 3 : Chapter 6-13 Chapter 6-14 Point Group Assignment There is a systematic way of naming most point groups C, S or D for the principal symmetry axis 4) Draw, identify symmetry elements and assign the point group of following molecules: a) N 2 Cl: b) SF 4 : A number for the order of the principal axis (subscript) n. A subscript h, d, or v for symmetry planes c) PCl 5 : d) SF 6 : e) Chloroform f) 1,3,5-trichlorobenzene Chapter 6-15 Chapter 6-16 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 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. Chapter 6-17 Chapter 6-18 3
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 5 ) 2 Fe 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 s v s 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 f) Octahedral W(CO) 6 g) fac- and mer-ru( 2 O) 3 Cl 3 Chapter 6-19 Chapter 6-20 Character Table T d 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. 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. 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. Chapter 6-21 Chapter 6-22 2 O molecule belongs to C 2v point group Symmetry Classes The symmetry classes for each point group and are labeled in the character table LabelSymmetry Class A B E T Singly-degenerate class, symmetric with respect to the principal axis Singly-degenerate class, asymmetric with respect to the principal axis Doubly-degenerate class Triply-degenerate class Chapter 6-23 Chapter 6-24 4
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 for molecules belonging to the point groups C n and C nv. Any of D groups, T, O and I groups will not be polar Chirality Only molecules lacking a S n axis can be chiral. This includes mirror planes and a center of inversion as S 2 =s, S 1 =i and D n groups. Not Chiral: D nh, D nd,t d and O h. Chapter 6-25 Chapter 6-26 Meso-Tartaric Acid Optical Activity Chapter 6-27 Chapter 6-28 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. Example: Valence MOs of Water 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, s v1, and s v2. Chapter 6-29 Chapter 6-30 5
Building a MO diagram for 2 O a 1 orbital of 2 O z y x Chapter 6-31 Chapter 6-32 b 1 orbital of 2 O b 1 orbital of 2 O, cont. Chapter 6-33 Chapter 6-34 b 2 orbital of 2 O b 2 orbital of 2 O, cont. Chapter 6-35 Chapter 6-36 6
[Fe(CN) 6 ] 4- Reducing the Representation Use reduction formula Chapter 6-37 Chapter 6-38 MO forml 6 diagram Molecules Group Theory and Vibrational Spectroscopy Chapter 6-39 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 6-40 Reducible Representations(3N) for 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. 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. Chapter 6-41 Chapter 6-42 7
Summary of Operations by a set of four 9 x 9 transformation matrices. Use Reduction Formula Chapter 6-43 Chapter 6-44 Example 2 O, C 2v Use Reduction Formula: 1 ap (R) p(r) g R to show that here we have: G 3N = 3A 1 + A 2 + 2B 1 + 3B 2 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 C 2 v, Chapter 6-45 Chapter 6-46 G T = A 1 + B 1 + B 2 G R = A 2 + B 1 + B 2 i.e. G T+R = A 1 + A 2 + 2B 1 + 2B 2 But G vib = G 3N - G T+R Therefore G vib = 2A 1 + B 2 i.e. of the 3 (= 3N-6) vibrations for a molecule like 2 O, two have A 1 and one has B 2 symmetry INTERNAL COORDINATE METOD 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 6-47 Chapter 6-48 8
Deduce G 3N for our triatomic molecule, 2 O in three lines: Example 2 O, C 2v 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!! aving obtained G 3N, we now must reduce it to find which irreducible representations are present. Chapter 6-49 Chapter 6-50 Use Reduction Formula: 1 ap (R) p(r) g to show that here we have: G 3N = 3A 1 + A 2 + 2B 1 + 3B 2 R G T = A 1 + B 1 + B 2 G R = A 2 + B 1 + B 2 i.e. G T+R = A 1 + A 2 + 2B 1 + 2B 2 But G vib = G 3N - G T+R 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. Therefore G vib = 2A 1 + B 2 i.e. of the 3 (= 3N-6) vibrations for a molecule like 2 O, two have A 1 and one has B 2 symmetry The irreducible representations for the rotation and translation vectors are listed in the character tables, e.g. for C 2 v, Chapter 6-51 Chapter 6-52 Further examples of the determination of G vib, via G 3N : N3 (C3v) N C3v E 2C3 3sv 12 0 2 C 4 (T d ) C Td E 8C3 3C2 6S4 6sd 15 0-1 -1 3 \ G 3N 12 0 2 Reduction formula G 3N = 3A 1 + A 2 + 4E G T+R (from character table) = A 1 + A 2 + 2E, \ G vib = 2A 1 + 2E (each E "mode" is in fact two vibrations (doubly degenerate) \ 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) Chapter 6-53 Chapter 6-54 9
XeF4 (D4h) F Xe F F F INTERNAL COORDINATE METOD 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 We used one example of this earlier - when we used the "bond vectors" to obtain a representation corresponding to bond stretches. Reduction formula G3N = A1g + A2g + B1g + B2g + Eg + 2A2u + B2u + 3Eu GT+R (from character table) = A2g + Eg + A2u + Eu, We will give more examples of these, and also the other main type of vibration - bending modes. \ 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. For stretches we use as internal coordinates changes in bond length, for bends we use changes in bond angle. To be more specific we need now to use the INTERNAL COORDINATE method. Chapter 6-55 Chapter 6-56 Let us return to the C2v molecule: r 1 O r 2 Use as bases for stretches: Dr1, Dr2. Gbend is clearly irreducible, i.e. A1. Use as basis for bend: D Gstretch reduces to A1 + B2 C2v E C2 sxz syz Gstretch 2 0 0 2 Gbend 1 1 1 1 We can therefore see that the three vibrational modes of 2O divide into two stretches (A1 + B2) and one bend (A1). N.B. Transformation matrices for Gstretch : E, syz: 1 0 C2, sxz : 0 1 0 1 1 0 We will see later how this information helps in the vibrational assignment. i.e. only count UNSIFTED VECTORS (each of these +1 to ). Chapter 6-57 Chapter 6-58 Other examples: N3 r 1 N r 2 1 opposite to r 1 r 3 2 opposite to r 2 3 opposite to r 3 C4 r 1 6 angles 1,... 6, where 1 r C r 4 lies between r1 and r2 etc. 2 r 3 Bases for stretches: Dr1, Dr2, Dr3. Bases for stretches: Dr1, Dr2, Dr3, Dr4. Bases for bends: D 1, D 2, D 3. Bases for bends: D 1, D 2, D 3, D 4, D 5, D 6. C3v E 2C3 3s Gstretch 3 0 1 Gbend 3 0 1 Td E 8C3 3C2 6S4 6sd Gstretch 4 1 0 0 2 Gbend 6 0 2 0 2 Reduction formula Gstretch = A1 + E Gbend = A1 + E Reduction formula Gstretch = A1 + T2 Gbend = A1 + E + T2 (Remember Gvib (above) = 2A1 + 2E) But G3N (above) = A1 + E + 2T2 Chapter 6-59 Chapter 6-60 10
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 6-61 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 n 0 = wavenumber of incident radiation n vib = a vibrational wavenumber of the molecule (in general several of these) Chapter 6-62 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. 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. Chapter 6-63 Chapter 6-64 Chapter 6-65 11