Presentation on theme: "Molecular Symmetry Symmetry Elements Group Theory Photoelectron Spectra Molecular Orbital (MO) Diagrams for Polyatomic Molecules."— Presentation transcript:
Molecular Symmetry Symmetry Elements Group Theory Photoelectron Spectra Molecular Orbital (MO) Diagrams for Polyatomic Molecules
CHEM 3722 Symmetry 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
CHEM 3722 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 ( ) Proper rotation (C n ) Improper rotation (S n ) Inversion (i)
CHEM 3722 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 Plane of reflection (xz) (x,y,z) = (x,-y,z) (xy) (x,y,z) = (x,y,-z) (yz) (x,y,z) = (-x,y,z) (xz) (xy) x y z E
CHEM 3722 Symmetry 5 Symmetry Elements III Proper Rotation C n where n represents angle of rotation out of 360 degrees C 2 = 180°, C 3 = 120°, C 4 = 90° … C 2 (z) (x,y,z) = (-x,-y,z) C6C6 C3C3 C2C2 C6C6 C6C6 How many C 2 operations are there for benzene?
CHEM 3722 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) 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. S 4 (z) (x,y,z) = (y,-x,-z) Same as ( )(C 4 ) (x,y,z) Note, symmetry operations are just quantum operators: work from right to left 1 2 3 4 6 5 6 1 2 3 5 4 1 2 3 4 6 5 4 5 6 1 3 2 h C 6 = S 6
CHEM 3722 Symmetry 7 A Few To Try Determine which symmetry elements are applicable for each of the following molecules
CHEM 3722 Symmetry 8 Point Groups We can systematically classify molecules by their symmetry properties Call these point groups Use the flow diagram to the right Start C n axis Special groups: a) C v,D h (linear groups) b) T, T h, T d, O, O h, I, I h (1) (2) No proper or improper axes Only S n (n = even) axis: S 4, S 6 … (3) (4)(5) No C 2 ’s to C n n C 2 ’s to C n hh n v ’sNo ’s hh n d ’sNo ’s C nh C nv CnCn D nh D nv DnDn
CHEM 3722 Symmetry 9 Some Common Groups D 3h Point Group C 3, C 3 2, 3C 2, S 3, S 3 5, 3 v, h Trigonal planar C 3v Point Group C 3, 3 v Trigonal pyramid D 3h Point Group C 3, C 3 2, 3C 2, S 3, S 3 5, 3 v, h Trigonal bi-pyramid C 4h Point Group C 2, 2C 2 ’, 2C 2 ’’, C 4, S 4, S 4 2, 2 v, 2 d, h, i Square planar C 4v Point Group C 2, C 4, C 4 2, 2 v, 2 d Square pyramid AB 3 AB 5 AB 4 AB 5
CHEM 3722 Symmetry 10 Character Tables Character tables hold the combined symmetry and effects of operations For example, consider water (C 2v ) Point Group Symmetry operations available Symmetry of states A, B = singly degenerate E = doubly degenerate T = triply degenerate 1 = symmetric to C 2 rotation* 2 = antisymmetric to C 2 rotation* Coordinates and rotations of this symmetry Effect of this operation on an orbital of this symmetry.
CHEM 3722 Symmetry 11 The Oxygen’s p x 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 The p x orbital has B 1 symmetry
CHEM 3722 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 p x orbital for water’s oxygen atom It’s functional form is But it’s easier to use than this appears
CHEM 3722 Symmetry 13 Ammonia Let’s apply this mess to ammonia First, draw the structure and determine the number of and 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 C 3v point group (AB 3 )
CHEM 3722 Symmetry 14 Ammonia II The Character Table for C 3v is to the right (3) (0) (1) Take the bonds through the operations & see how many stay put We now have a representation ( ) of this group of orbitals that has the symmetry
CHEM 3722 Symmetry 15 Ammonia III This “reducible representation” of the hydrogen’s -bonds must be a sum of the symmetries available Only one possible sum will yield this reducible representation By inspection, we see that = A 1 + E From the character table, we now can get the symmetry of the orbitals in N N(2s) = A 1 -- x 2 + y 2 is same as an s-orbital N(2p z ) = A 1 N(2p x ) = N(2p y ) = E So, we can now set up the MO diagram and let the correct symmetries overlap
CHEM 3722 Symmetry 16 Ammonia, The MO Diagram ** A1A1 E x, E y A 1, E x, E y 2p’s 2s 2E x 2E y 3A 1 2A 1 1A 1
CHEM 3722 Symmetry 17 Methane The usual view of methane is one where four equivalent sp 3 orbitals are necessary for the tetrahedral geometry 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 sp 3 -s overlap for a MO Photoelectron spectrum (crude drawing) adapted from Roy. Soc. Chem., Potts, et al.
CHEM 3722 Symmetry 18 Methane: SALC Approach Methane is a tetrahedral, so use T d 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, SALC. The character table immediately gives us the symmetry of the s and p orbitals of the carbon: C(2s) = A 1 C(2p x, 2p y, 2p z ) = T 2 SALC = A 1 + T 2
CHEM 3722 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. 1s A 1 + T 2 2s 2p x 2p y 2p z 1s T2T2 A1A1 A1A1 22 11 44 55 33 6*6* 7*7* 8*8* 9*9* A1A1 A1A1 T2T2 T2T2 A1A1 C CH 4 4H’s
CHEM 3722 Symmetry 20 An Example: BF 3 BF 3 affords our first look at a molecule where -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 BF 3 is D 3h, with the following character table: We’ll begin by defining our basis sets of orbitals that do a certain type of bonding
CHEM 3722 Symmetry 21 F 3 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 reducing s
CHEM 3722 Symmetry 22 The Other Basis Sets
CHEM 3722 Symmetry 23 The MO Diagram for BF 3 No s-orbital interaction from F’s is included in this MO diagram!
CHEM 3722 Symmetry 24 The MO Diagram for BF 3 s-orbital interaction from F’s allowed
CHEM 3722 Symmetry 25 Using Hybrid Orbitals for BF 3 If we use sp 2 hybrids and the remaining p-orbital (p z ) of the boron, we see how hybridization yields the same exact picture. Build our sp 2 hybrids and take them through the operations Find the irreducible representations using the worksheet method and the reducible representation p z is found in the character table to be A 2 ”. 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.
CHEM 3722 Symmetry 26 -Bonding in Aromatic Compounds The -bonding in C 3 H 3 +1 (aromatic) 0 nodes 1 node The -bonding in C 4 H 4 +2 (aromatic) 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 1 node 2 nodes
CHEM 3722 Symmetry 27 Cyclopentadiene As the other examples showed, the actual geometric structure of the aromatic yields the general shape of the -MO region 0 nodes 1 node 2 nodes The -bonding in C 5 H 5 -1 (aromatic)
CHEM 3722 Symmetry 28 Benzene The -bonding in C 6 H 6 (aromatic) 0 nodes 1 node 2 nodes 3 nodes