Molecular Symmetry

1. Molecular Symmetry Symmetry Elements Group Theory Photoelectron Spectra Molecular Orbital (MO) Diagrams for Polyatomic Molecules

2. The Symmetry of Molecules • The shape of a molecule influences its physical properties, reactivity, and its spectroscopic behavior • Determining the symmetry of a molecule is fundamental to gaining insight into these characteristics of molecules • The chemist’s view of symmetry is contained in the study of group theory • This branch of mathematics classifies the properties of a molecule into groups, defined by the symmetry of the molecule • Each group is made up of symmetry elements or operations, which are essentially quantum operators disguised as matrices • Our goal is to use group theory to build more complex molecular orbital diagrams Symmetry

3. Symmetry Elements • You encounter symmetry every day • A ball is spherically symmetric • Your body has a mirror image (the left and right side of your body) • Hermite polynomials are either even (symmetric on both sides of the axis) or odd (symmetric with a twist). • Symmetry operations are movements of a molecule or object such that after the movement the object is indistinguishable from its original form • Symmetry elements are geometric representations of a point, line, or plane to which the operation is performed • Identity element (E) • Plane of reflection (s) • Proper rotation (Cn) • Improper rotation (Sn) • Inversion (i) Symmetry

4. Symmetry Elements II • Identity • If an object (O) has coordinates (x,y,z), then the operation E(x,y,z) (x,y,z) • The object is unchanged E • Plane of reflection • s(xz) (x,y,z) = (x,-y,z) • s(xy) (x,y,z) = (x,y,-z) • s(yz) (x,y,z) = (-x,y,z) s(xz) z s(xy) y x Symmetry

5. Symmetry Elements III • Proper Rotation • Cn where n represents angle of rotation out of 360 degrees • C2 = 180°, C3 = 120°, C4 = 90° … • C2(z) (x,y,z) = (-x,-y,z) C6 C6 C3 How many C2 operations are there for benzene? C6 C2 Symmetry

6. Symmetry Elements IV • Inversion • Takes each point through the center in a straight line to the exact distance on the other side of center • i (x,y,z) = (-x,-y,-z) 1 4 6 2 3 5 5 3 2 6 4 1 • Improper Rotation • A two step operation that first does a proper rotation and then a reflection through a mirror plane perpendicular to the rotational axis. • S4(z) (x,y,z) = (y,-x,-z) • Same as (s)(C4) (x,y,z) • Note, symmetry operations are just quantum operators: work from right to left 1 6 6 2 5 1 sh C6 = S6 5 3 4 2 4 3 Symmetry

7. A Few To Try • Determine which symmetry elements are applicable for each of the following molecules Symmetry

8. Point Groups • We can systematically classify molecules by their symmetry properties • Call these point groups • Use the flow diagram to the right Special groups: a) Cv,Dh (linear groups) b) T, Th, Td, O, Oh, I, Ih (1) Start (2) No proper or improper axes (3) Only Sn (n = even) axis: S4, S6… Cn axis (4) (5) No C2’s  to Cn n C2’s  to Cn sh n sv’s No s’s sh n sd’s No s’s Cnh Cnv Cn Dnh Dnv Dn Symmetry

9. Some Common Groups AB3 • D3h Point Group • C3, C32, 3C2, S3, S35, 3sv, sh • Trigonal planar • C3v Point Group • C3, 3sv • Trigonal pyramid • D3h Point Group • C3, C32, 3C2, S3, S35, 3sv, sh • Trigonal bi-pyramid • C4h Point Group • C2, 2C2’, 2C2’’, C4, S4, S42, 2sv, 2sd, sh, i • Square planar • C4v Point Group • C2, C4, C42, 2sv, 2sd • Square pyramid AB3 AB5 AB4 AB5 Symmetry

10. Character Tables Point Group • Character tables hold the combined symmetry and effects of operations • For example, consider water (C2v) Symmetry operations available Effect of this operation on an orbital of this symmetry. Coordinates and rotations of this symmetry Symmetry of states A, B = singly degenerate E = doubly degenerate T = triply degenerate 1 = symmetric to C2 rotation* 2 = antisymmetric to C2 rotation* Symmetry

11. The Oxygen’s px orbital • For water, we can look at any orbital and see which symmetry it is by applying the operations and following the changes made to the orbital • If it stays the same, it gets a 1 • If it stays in place but gets flipped, -1 • If it moves somewhere else, 0 1 -1 1 -1 The px orbital has B1 symmetry Symmetry

12. The Projection Operator • Since we want to build MO diagrams, our symmetry needs are simple • Only orbitals of the same symmetry can overlap to form bonds • Each point group has many possible symmetries for an orbital, and thus we need a way to find which are actually present for the particular molecule of that point group • We’ll also look at collections of similar atoms and their collective orbitals as a group • The projection operator lets us find the symmetry of any orbital or collection of orbitals for use in MO diagrams • It will also be of use in determining the symmetry of vibrations, later • We just did this for the px orbital for water’s oxygen atom • It’s functional form is • But it’s easier to use than this appears Symmetry

13. Ammonia • Let’s apply this mess to ammonia • First, draw the structure and determine the number of s and p bonds • Then look at how each bond changes for the group of hydrogen atoms, building a set of “symmetry adapted linear combinations of atomic orbitals” (SALC) to represent the three hydrogen's by symmetry (not by their individual atomic orbitals) • Finally, we’ll compare these symmetries to those of the s and p atomic orbitals of the nitrogen to see which overlap, thus building our MO diagram from the SALC 3 sigma bonds and no pi bonds, thus we’ll build our SALC’s from the projection operator and these 3 MO’s Note, ammonia is in C3v point group (AB3) Symmetry

14. Ammonia II • The Character Table for C3v is to the right • Take the s bonds through the operations & see how many stay put • We now have a representation (G) of this group of orbitals that has the symmetry (3) (0) (1) Symmetry

15. Ammonia III • This “reducible representation” of the hydrogen’s s-bonds must be a sum of the symmetries available • Only one possible sum will yield this reducible representation • By inspection, we see that Gs = A1 + E • From the character table, we now can get the symmetry of the orbitals in N • N(2s) = A1 -- x2 + y2 is same as an s-orbital • N(2pz) = A1 • N(2px) = N(2py) = E • So, we can now set up the MO diagram and let the correct symmetries overlap Symmetry

16. Ammonia, The MO Diagram 3A1 2Ex 2Ey s* 2p’s A1, Ex, Ey 2A1 Ex, Ey A1 2s 1A1 Symmetry

17. Methane • The usual view of methane is one where four equivalent sp3 orbitals are necessary for the tetrahedral geometry sp3-s overlap for a s MO sp3-s overlap for a s MO Photoelectron spectrum (crude drawing) adapted from Roy. Soc. Chem., Potts, et al. • However, the photoelectron spectrum shows two different orbital energies with a 3:1 population ratio • Maybe the answer lies in symmetry • Let’s build the MO diagram using the SALC method we saw before Symmetry

18. Methane: SALC Approach • Methane is a tetrahedral, so use Td point group • The character table is given below • To find the SALC’s of the 4H’s, count those that do not change position for each symmetry operation and create the reducible representation, GSALC. GSALC = A1 + T2 • The character table immediately gives us the symmetry of the s and p orbitals of the carbon: • C(2s) = A1 • C(2px, 2py, 2pz) = T2 Symmetry

19. The MO Diagram of Methane • Using the symmetry of the SALC’s with those of the carbon orbitals, we can build the MO diagram by letting those with the same symmetry overlap. s7* s8* s9* T2 A1 s6* 2px 2py 2pz T2 1s A1 + T2 2s s3 s4 s5 A1 T2 s2 A1 s1 1s A1 A1 C CH4 4H’s Symmetry

20. An Example: BF3 • BF3 affords our first look at a molecule where p-bonding is possible • The “intro” view is that F can only have a single bond due to the remaining p-orbitals being filled • We’ll include all orbitals • The point group for BF3 is D3h, with the following character table: • We’ll begin by defining our basis sets of orbitals that do a certain type of bonding Symmetry

21. F3 Residue Basis Sets • Looking at the 3 F’s as a whole, we can set up the s-orbitals as a single basis set: Regular character table Worksheet for reducingGss Symmetry

22. The Other Basis Sets Symmetry

23. The MO Diagram for BF3 No s-orbital interaction from F’s is included in this MO diagram! Symmetry

24. The MO Diagram for BF3 s-orbital interaction from F’s allowed Symmetry

25. Using Hybrid Orbitals for BF3 • If we use sp2hybrids and the remaining p-orbital (pz) of the boron, we see how hybridization yields the same exact picture. • Build our sp2hybrids and take them through the operations • Find the irreducible representations using the worksheet method and the reducible representation • pz is found in the character table to be A2”. • Result: identical symmetries for boron’s orbital’s in both cases • This is how it should be, since hybridization is an equivalent set of orbitals that are simply oriented in space differently. Symmetry

26. p-Bonding in Aromatic Compounds The -bonding in C3H3+1 (aromatic) 1 node 0 nodes The -bonding in C4H4+2 (aromatic) 2 nodes 1 node Aromatic compounds must have a completely filled set of bonding -MO’s. This is the origin of the Hückel (4N+2) -electron definition of aromaticity. 0 nodes Symmetry

27. Cyclopentadiene • As the other examples showed, the actual geometric structure of the aromatic yields the general shape of the p-MO region The -bonding in C5H5-1 (aromatic) 2 nodes 1 node 0 nodes Symmetry

28. Benzene 3 nodes The -bonding in C6H6 (aromatic) 2 nodes 1 node 0 nodes Symmetry