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Molecular Symmetry Symmetry Elements Group Theory Photoelectron Spectra Molecular Orbital (MO) Diagrams for Polyatomic Molecules 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 2 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) CHEM 3722 Symmetry 3 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 H H H B B H H E H H H H B B H H H 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 y s(xy) x CHEM 3722 Symmetry 4 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) 6 C6 5 1 4 2 3 C6 1 How many C2 operations are there for benzene? 5 6 2 5 3 C3 4 6 3 1 4 2 C6 4 C2 3 5 2 6 1 CHEM 3722 Symmetry 5 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 2 5 3 4 CHEM 3722 6 sh C6 = S6 5 1 4 2 3 Symmetry 6 A Few To Try Determine which symmetry elements are applicable for each of the following molecules O H H H H N H Cl Cl Cl Ru Cl CHEM 3722 Cl Cl O Cl Ru Cl Cl Cl Symmetry 7 Point Groups We can systematically classify molecules by their symmetry properties Special groups: a) Cv,Dh (linear groups) b) T, Th, Td, O, Oh, I, Ih (1) Start (2) No proper or improper axes (3) Call these point groups Use the flow diagram to the right Only Sn (n = even) axis: S4, S6… (4) Cn axis (5) No C2’s to Cn CHEM 3722 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 8 Some Common Groups AB3 D3h Point Group AB3 C3v Point Group AB5 C2, 2C2’, 2C2’’, C4, S4, S42, 2sv, 2sd, sh, i Square planar C4v Point Group CHEM 3722 C3, C32, 3C2, S3, S35, 3sv, sh Trigonal bi-pyramid C4h Point Group AB5 C3, 3sv Trigonal pyramid D3h Point Group AB4 C3, C32, 3C2, S3, S35, 3sv, sh Trigonal planar C2, C4, C42, 2sv, 2sd Square pyramid Symmetry 9 Character Tables Character tables hold the combined symmetry and effects of operations For example, consider water (C2v) Point Group Symmetry operations available C 2v E C2 s ( xz ) s ( yz ) A1 1 1 1 1 z x ,y ,z A2 1 1 1 1 Rz xy B1 1 1 1 1 x, R y xz B2 1 1 1 1 y,Rx yz 2 2 2 Effect of this operation on an orbital of this Coordinates and rotations symmetry. 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* CHEM 3722 Symmetry 10 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 z H O x H y E C2 s (x z ) s (y z ) CHEM 3722 1 -1 1 C 2v E C2 s ( xz ) s ( yz ) A1 1 1 1 1 z x ,y ,z A2 1 1 1 1 Rz xy B1 1 1 1 1 x,R y xz B2 1 1 1 1 y,R x yz 2 2 The px orbital has B1 symmetry -1 Symmetry 11 2 The Projection Operator Since we want to build MO diagrams, our symmetry needs are simple 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 Only orbitals of the same symmetry can overlap to form bonds 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 l It’s functional form is j j ˆ P ( R ) Rˆ h j But it’s easier to use than this appears CHEM 3722 Symmetry 12 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 z N H H x H y 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) CHEM 3722 Symmetry 13 Ammonia II The Character Table for C3v is to the right E 2C 3 3s v A1 1 1 1 z B1 1 1 1 Rz E 2 1 0 ( x , y ), ( R x , R y ) x y ,z 2 2 2 ( x y ), xy , xz , yz 2 2 Take the s bonds through the operations & see how many stay put s3 H H s3 H H s3 H H C 3v N s1 s2 H C3 H N s1 s2 H s2 H H sv N s1 s2 H H N s1 s2 s3 E H N s3 s1 s2 H N s1 H s3 H H (3) (0) (1) We now have a representation (G) of this group of orbitals that has the symmetry Gs CHEM 3722 E 2C 2 3s V 3 0 1 Symmetry 14 Ammonia III This “reducible representation” of the hydrogen’s s-bonds must be a sum of the symmetries available C 3v E 2C 3 3s v A1 1 1 1 z B1 1 1 1 Rz E E 2 2C12 30 sV ( x , y ), ( R x , R y ) Gs 3 0 1 x y ,z 2 2 2 ( x y ), xy , xz , yz 2 2 From the character table, we now can get the symmetry of the orbitals in N Only one possible sum will yield this reducible representation By inspection, we see that Gs = A1 + E 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 CHEM 3722 Symmetry 15 Ammonia, The MO Diagram 3A1 2Ex 2Ey s* 2p’s A 1 , E x, E y 2A1 Ex, Ey A1 H H H 2s 1A1 z N H H CHEM 3722 x H y Symmetry 16 Methane The usual view of methane is one where four equivalent sp3 orbitals are necessary for the tetrahedral geometry H sp3-s overlap for a s MO H sp3-s overlap for a s MO C H H 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 CHEM 3722 Symmetry 17 Methane: SALC Approach Methane is a tetrahedral, so use Td point group The character table is given below Td E 8C 3 3C 2 6S 4 6s d A1 1 1 1 1 1 A2 1 1 1 1 1 E 2 1 2 0 0 T1 3 0 1 1 1 (R x , R y , R z ) T2 3 0 1 1 1 x, y , z 2 2 2 2z x y , x y 2 2 2 2 2 xy , yz , xz 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. H H C H x y z H Td E 8C 3 3C 2 6S 4 6s d G SALC 4 1 0 0 2 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 CHEM 3722 Symmetry 18 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* A1 2px 2py T2 s9* T2 s6* 2pz 1s A 1 + T2 2s A1 1s A1 C CHEM 3722 s3 s4 s5 s2 A1 s1 A1 CH4 Symmetry T2 4H’s 19 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 F F B F The point group for BF3 is D3h, with the following character table: D3h E 2C 3 3C 2 sh 2S 3 3s v A1 1 1 1 1 1 1 A 2 1 1 1 1 1 1 Rz E 2 1 0 2 1 0 x, y A1 1 1 1 1 1 1 A 2 1 1 1 1 1 1 z E 2 1 0 2 1 0 R x ,R y x y ,z 2 xy , x y 2 2 2 2 xz , yz We’ll begin by defining our basis sets of orbitals that do a certain type of bonding CHEM 3722 Symmetry 20 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: s s o rb ita ls E 2C 3 3C 2 sh 2S 3 3s v G ss 3 0 1 3 0 1 G s s A1 E B Regular character table Worksheet for reducing Gss CHEM 3722 D3h D3h E 2C 3 3C 2 1s h 2S 3 3s v G ss 3 0 1 3 0 1 A1 1 1 1 1 1 1 A 2 1 1 1 1 1 1 E 2 1 0 2 1 0 A1 1 1 1 1 1 1 A 2 1 1 1 1 1 1 E 2 1 0 2 1 0 A1 3 0 3 3 0 3 12 1 A 2 3 0 3 3 0 3 0 0 E 6 0 0 6 0 0 12 1 A1 3 0 3 3 0 3 0 0 A 2 3 0 3 3 0 3 0 0 E 6 0 0 6 0 0 0 0 Symmetry 12 21 The Other Basis Sets p s o rb ita ls B B D3 h E 2C 3 3C 2 sh 2S 3 3s v G ps 3 0 1 3 0 1 G ps A1 E p n b o rb ita ls B p p o rb ita ls B B D3h E 2C 3 3C 2 sh 2S 3 3s v D3h E 2C 3 3C 2 sh 2S 3 3s v G pp 3 0 1 3 0 1 G pnb 3 0 1 3 0 1 G pp A 2 E CHEM 3722 G p n b A 2 E Symmetry 22 The MO Diagram for BF3 No s-orbital interaction from F’s is included in this MO diagram! B (2 p ) E A 2 p nb pp ps E E E A 2 B(s) A1 A 2 A1 ss E A1 CHEM 3722 Symmetry 23 The MO Diagram for BF3 s-orbital interaction from F’s allowed B (2 p ) E A 2 p nb pp ps E E E A 2 B(s) A1 A 2 A1 ss E A1 CHEM 3722 Symmetry 24 Using Hybrid Orbitals for BF3 If we use sp2 hybrids and the remaining p-orbital (pz) of the boron, we see how hybridization yields the same exact picture. Build our sp2 hybrids and take them through the operations Find the irreducible representations using the worksheet method and the reducible representation B D3h E 2C 3 3C 2 sh 2S 3 3s v G sp 2 3 0 1 3 0 1 G sp 2 A1 E 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. CHEM 3722 Symmetry 25 p-Bonding in Aromatic Compounds The p-bonding in C3H3+1 (aromatic) 1 node 0 nodes The p-bonding in C4H4+2 (aromatic) 2 nodes 1 node 0 nodes CHEM 3722 Symmetry Aromatic compounds must have a completely filled set of bonding p-MO’s. This is the origin of the Hückel (4N+2) p-electron definition of aromaticity. 26 Cyclopentadiene As the other examples showed, the actual geometric structure of the aromatic yields the general shape of the p-MO region The p-bonding in C5H5-1 (aromatic) 2 nodes 1 node 0 nodes CHEM 3722 Symmetry 27 Benzene 3 nodes The p-bonding in C6H6 (aromatic) 2 nodes 1 node 0 nodes CHEM 3722 Symmetry 28