Chemistry 481(01) Spring 2016 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 , 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 Chemistry 481, Spring 2016, LA Tech Chapter 6-1 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 Chemistry 481, Spring 2016, LA Tech Chapter 6-2 Symmetry

M.C. Escher Chemistry 481, Spring 2016, LA Tech Chapter 6-3 Symmetry Butterflies Chemistry 481, Spring 2016, LA Tech Chapter 6-4 Fish and Boats Symmetry Chemistry 481, Spring 2016, LA Tech Chapter 6-5 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 Chemistry 481, Spring 2016, LA Tech Chapter 6-6 2) What is a symmetry operation? Chemistry 481, Spring 2016, LA Tech Chapter 6-7

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. Chemistry 481, Spring 2016, LA Tech Chapter 6-8 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: Chemistry 481, Spring 2016, LA Tech Chapter 6-9 s - a plane of reflection The symmetry operation corresponds to reflection in a plane. H2O possesses two reflection planes. Labels: sh, sd and sv. Chemistry 481, Spring 2016, LA Tech 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): Chemistry 481, Spring 2016, LA Tech Chapter 6-11 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. Chemistry 481, Spring 2016, LA Tech Chapter 6-12 2) What are four basic symmetry elements and

operations? Chemistry 481, Spring 2016, LA Tech Chapter 6-13 3) Draw and identify the symmetry elements in: a) NH3: b) H2O: c) CO2: d) CH4: Chemistry 481, Spring 2016, LA Tech Chapter 6-14 Point Group

Assignment 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 Chemistry 481, Spring 2016, LA Tech Chapter 6-15 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 Chemistry 481, Spring 2016, LA Tech 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 Dh If there is no plane of symmetry, the point group is Cv Tetrahedral molecules have a point group Td Octahedral molecules have a point group Oh Icosahedral molecules have a point group Ih

Chemistry 481, Spring 2016, LA Tech Chapter 6-17 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. Chemistry 481, Spring 2016, LA Tech Chapter 6-18 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, (5-C5H5)2Fe f) Octahedral W(CO)6 g) fac- and mer-Ru(H2O)3Cl3 Chemistry 481, Spring 2016, LA Tech Chapter 6-19 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 Chemistry 481, Spring 2016, LA Tech Chapter 6-20

Character Table Td Chemistry 481, Spring 2016, LA Tech Chapter 6-21 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. Chemistry 481, Spring 2016, LA Tech

Chapter 6-22 H2O molecule belongs to C2v point group Chemistry 481, Spring 2016, LA Tech Chapter 6-23 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

Chemistry 481, Spring 2016, LA Tech Chapter 6-24 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 Chemistry 481, Spring 2016, LA Tech Chapter 6-25 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. Chemistry 481, Spring 2016, LA Tech Chapter 6-26 Meso-Tartaric Acid Chemistry 481, Spring 2016, LA Tech Chapter 6-27

Optical Activity Chemistry 481, Spring 2016, LA Tech 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.

Chemistry 481, Spring 2016, LA Tech Chapter 6-29 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. Chemistry 481, Spring 2016, LA Tech Chapter 6-30 Building a MO diagram for H2O O

H H z H O H a1 b2 y x b2 b1

b1 a1 b2 a1 a1 a1 b2 a1 Chemistry 481, Spring 2016, LA Tech Chapter 6-31 a1 orbital of H2O C2 sv2

sv1 E(1a1) C2(1a1) sv1 (1a1) sv2 (1a1) Chemistry 481, Spring 2016, LA Tech (+1)(1a1) (+1)(1a1) (+1)(1a1) (+1)(1a1) Chapter 6-32 b1 orbital of H2O sv2 E(1b1) sv2(1b1)

Chemistry 481, Spring 2016, LA Tech (+1)(1b1) (+1)(1b1) Chapter 6-33 b1 orbital of H2O, cont. C2 sv1 C2(1b1) s v1(1b1) Chemistry 481, Spring 2016, LA Tech (-1)(1b1) (-1)(1b1)

Chapter 6-34 b2 orbital of H2O sv1 E(1b2) sv1(1b2) Chemistry 481, Spring 2016, LA Tech (+1)(1b2) (+1)(1b2) Chapter 6-35 b2 orbital of H2O, cont. C2

s v2 C2(1b2) s v2(1b2) Chemistry 481, Spring 2016, LA Tech (-1)(1b2) (-1)(1b2) Chapter 6-36 [Fe(CN)6]4- Chemistry 481, Spring 2016, LA Tech Chapter 6-37 Reducing the Representation Use reduction formula

Chemistry 481, Spring 2016, LA Tech Chapter 6-38 MO forML6 diagram Molecules Chemistry 481, Spring 2016, LA Tech Chapter 6-39 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) Chemistry 481, Spring 2016, LA Tech Chapter 6-40 Chemistry 481, Spring 2016, LA Tech Chapter 6-41 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 , Chemistry 481, Spring 2016, LA Tech y3 -> -y3, z3 -> z3. Chapter 6-42 Summary of Operations by a set of four 9 x 9 transformation matrices. Chemistry 481, Spring 2016, LA Tech Chapter 6-43 Use Reduction Formula

Chemistry 481, Spring 2016, LA Tech Chapter 6-44 Example H2O, C2v Chemistry 481, Spring 2016, LA Tech Chapter 6-45 Use Reduction Formula: 1 a p ( R ) p ( R ) g R 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, Chemistry 481, Spring 2016, LA Tech Chapter 6-46 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 Chemistry 481, Spring 2016, LA Tech Chapter 6-47 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. Chemistry 481, Spring 2016, LA Tech

Chapter 6-48 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. Chemistry 481, Spring 2016, LA Tech Chapter 6-49 Example H2O, C2v Chemistry 481, Spring 2016, LA Tech Chapter 6-50

Use Reduction Formula: 1 a p ( R ) p ( R ) g R 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,

Chemistry 481, Spring 2016, LA Tech Chapter 6-51 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 Chemistry 481, Spring 2016, LA Tech Chapter 6-52 Further examples of the determination of Gvib, via G3N: NH3

N (C3v) H C3v H H E \ G3N Reduction formula 2C3 3sv 12

0 2 12 0 2 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)

Chemistry 481, Spring 2016, LA Tech Chapter 6-53 H CH4 (Td) C H H Td E \ G3N

H 8C3 3C2 15 0 -1 15 0 -1 Reduction formula 6S4

6 sd -1 3 -1 3 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) Chemistry 481, Spring 2016, LA Tech

Chapter 6-54 F XeF4 F (D4h) Xe F D4h \G3N E 15

F 2C4 C2 2C2' 15 1 -1 1 -1 2C2" -3

-3 i -1 -1 -1 2S4 -1 -1 sh -1 2sv 2 sd

5 3 5 3 1 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 Chemistry 481, Spring 2016, LA Tech INTERNAL COORDINATE method. Chapter 6-55 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. Chemistry 481, Spring 2016, LA Tech

Chapter 6-56 Let us return to the C2v molecule: O r1 H r2 Use as bases for stretches: Dr1, Dr2. H q

Use as basis for bend: Dq C2v E C2 sxz syz Gstretch 2 0 0

2 Gbend 1 1 1 1 N.B. Transformation matrices for Gstretch E, syz: 1 0 0

1 C2, sxz : : 0 1 1 0 i.e. only count UNSHIFTED VECTORS (each of these +1 to ). Chemistry 481, Spring 2016, LA Tech Chapter 6-57

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. Chemistry 481, Spring 2016, LA Tech Chapter 6-58 Other examples: r1 NH3

r3 H N H r2 H q1 opposite to r1 q2 opposite to r2 q3 opposite to r3 Bases for stretches: Dr1, Dr2, Dr3. Bases for bends: Dq1, Dq2, Dq3. 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) Chemistry 481, Spring 2016, LA Tech Chapter 6-59 H r1 CH4 r2 r3

H 6 angles q1,.....q6, where q1 r4 C H lies between r1 and r2 etc. H Bases for stretches: Dr1, Dr2, Dr3, Dr4. Bases for bends: Dq1, Dq2, Dq3, Dq4, Dq5, Dq6. Td E

8C3 3C2 6S4 6 sd 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 Chemistry 481, Spring 2016, LA Tech

Chapter 6-60 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 Chemistry 481, Spring 2016, LA Tech 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 n0 = wavenumber of incident radiation nvib = a vibrational wavenumber of the molecule (in general several of these) Chemistry 481, Spring 2016, LA Tech 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. Chemistry 481, Spring 2016, LA Tech Chapter 6-63 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. Chemistry 481, Spring 2016, LA Tech Chapter 6-64 Chemistry 481, Spring 2016, LA Tech Chapter 6-65