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# Perturbation Theory

Download Presentation ## Perturbation Theory

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1. Perturbation Theory H0 is the Hamiltonian of for a known system for which we have the solutions: the energies, e0, and the wavefunctions, f0. H0f0 = e0f0 We now change the system slightly (change a C into a N, create a bond between two atoms). The Hamiltonian will be changed slightly. For the changed system H = H0 + H1 H1 is the change to the Hamiltonian. We want to find out what happens to the molecular orbital energies and to the MOs.

2. How are energies and wave functions affected by a change? Energy to ei0: Zero order (no change, no correction): ei0 First Order correction: Wave functions corrections to f0i Zero order (no correction): f0i First order correction:

3. Example: Creating allyl system out of ethylene plus methyl radical Pi system only: “Unperturbed” system: ethylene + methyl radical “Perturbed” system: allyl system

4. Working out of the predicted values for the “perturbed” system. MO 1 only. Energy e1 Orbital 1 Note the stabilizing interaction. Bonding!

5. Mixes in anti-bonding Mixes in bonding Mixes in anti-bonding Mixes in bonding

6. Projection Operator Algorithm to create an object forming a basis of an irreducible rep from an arbitrary function. Where the projection operator results from using the symmetry operators, R, multiplied by characters of the irreducible reps. j indicates the desired symmetry. lj is the dimension of the irreducible rep (number in the E column). Starting with the 1sA create a function of A1 sym ¼(E1sA + C21sA + sv1sA + sv’1sA) = ¼ (1sA + 1sB + 1sA + 1sB) = ( ½(1sA + 1sB)

7. Consider the bonding in NF3,C3v. We can see what Irreducible Reps can be obtained from the atomic orbitals. Then get the Symm Adapted Linear Combinations by using Projection Op. These p orbitals are set-up as sigma and pi with respect to the internuclear axis. 3 2 1 A B C D GA 3 0 -1 GA = A2 + E } GB 3 0 1 GB = GC = GD = A1 + E GC 3 0 1 Now we know which irreducible reps we can get out of each kind of atomic orbital. Have to get the correct linear combination of the AOs. The E irred. rep is the bad one. GD 3 0 1

8. Now construct SALCs first A2 then E. See what the different symmetry operations do to the one of the types of p orbitals. GA = A2 + E Now construct an SALC of A2 sym PA2(p1) = 1/6 (p1 + p2 + p3 + (-1)(-p)1 + (-1)(-1p3) + (-1)(-p2) No AO on N is A2

9. Now get the SALC of E symmetry. Note that E is doubly degenerate. Apply projection operator to p1 PE(p1) = (2p1 - p2 - p3) = E1 But since it is two dimensional, E, there should be another SALC. Apply PE to p2. PE(p2) = (2p2 - p3 - p1) = E’ But E1 and E’ should be orthogonal. We want sum of products of coefficients to be zero. Create a linear combination of E2 = E’ + k E1.= (-1 +k*2) p1 + (2 + k(-1)) p2 + (-1 + k(-1))p3 Have to choose k such that they are orthogonal. 0 = (2(-1 + k*2) -1 (2 + k(-1)) -1 (-1 + k(-1)) k = ½ E2 = (3/2 p2 - 3/2 p3) = p2 – p3

10. For now we show the interaction of the N p orbitals (px and py) of E symmetry with the SALC of the F p orbitals of E symmetry. We will return to C3v molecules later on….

11. Molecular orbitals of heteronuclear diatomic molecules

12. The general principle of molecular orbital theory Interactions of orbitals (or groups of orbitals) occur when the interacting orbitals overlap. the energy of the orbitals must be similar the interatomic distance must be short enough but not too short A bonding interaction takes place when: regions of the same sign overlap An antibonding interaction takes place when: regions of opposite sign overlap

13. Antibonding Bonding Combinations of two s orbitals in a homonuclear molecule (e.g. H2) In this case, the energies of the A.O.’s are identical

14. More generally: Y = N[caY(1sa) ± cbY (1sb)] n A.O.’s n M.O.’s The same principle is applied to heteronuclear diatomic molecules But the atomic energy levels are lower for the heavier atom

15. Orbital potential energies (see also Table 5-1 in p. 134 of textbook) Average energies for all electrons in the same level, e.g., 3p (use to estimate which orbitals may interact)

16. The molecular orbitals of carbon monoxide Y = N[ccY(C) ± coY (O)] E(eV) Each MO receives unequal contributions from C and O (cc ≠ co)

17. Group theory is used in building molecular orbitals

18. B1B2 “C-like MO’s” A1 Frontier orbitals Larger homo lobe on C B1B2 A1 mixing “O-like MO’s” “C-like MO” A1 “O-like MO” A1 Bond order 3

19. Non-bonding (no symmetry match) Non-bonding (no E match) A related example: HF No s-s int. (DE > 13 eV)

20. A1 A1 Li transfers e- to F Forming Li+ and F- Extreme cases: ionic compounds (LiF)

21. Molecular orbitals for larger molecules 1. Determine point group of molecule (if linear, use D2h and C2v instead of D∞h or C∞v) 2. Assign x, y, z coordinates (z axis is higher rotation axis; if non-linear y axis in outer atoms point to central atom) 3. Find the characters of the representation for the combination of 2s orbitals on the outer atoms, then for px, py, pz. (as for vibrations, orbitals that change position = 0, orbitals that do not change =1; and orbitals that remain in the same position but change sign = -1) 4. Find the reducible representations (they correspond to the symmetry of group orbitals, also called Symmetry Adapted Linear Combinations SALC’s of the orbitals). 5. Find AO’s in central atom with the same symmetry 6. Combine AO’s from central atom with those group orbitals of same symmetry and similar E

22. F-H-F- D∞h, use D2h 1st consider combinations of 2s and 2p orbitals from F atoms 8 GROUP ORBITALS DEFINED

23. Group orbitals can now be treated as atomic orbitals and combined with appropriate AO’s from H 1s(H) is Ag so it matches two group orbitals 1 and 3 Both interactions are symmetry allowed, how about energies?

24. Good E match Strong interaction Poor E match weak interaction -13.6 eV -13.6 eV -40.2 eV

25. Lewis structure F-H-F- implies 4 e around H ! MO analysis defines 3c-2e bond (2e delocalized over 3 atoms) Bonding e Non-bonding e

26. CO2 D∞h, use D2h (O O) group orbitals the same as for F F But C has more AO’s to be considered than H !

27. No match CO2 D∞h, use D2h Carbon orbitals

28. B1u-B1uinteractions Ag-Aginteractions All four are symmetry allowed

29. Primary B1u interaction Primary Ag interaction

30. Non-bonding p Bonding p 4 bonds All occupied MO’s are 3c-2e Bonding s Non-bonding s

31. LUMO The frontier orbitals of CO2 HOMO

32. 1. Determine point group of molecule: C2v 2. Assign x, y, z coordinates (z axis is higher rotation axis; if non-linear y axis in outer atoms point to central atom - not necessary for H since s orbitals are non-directional) 3. Find the characters of the representation for the combination of 2s orbitals on the outer atoms, then for px, py, pz. (as for vibrations, orbitals that change position = 0, orbitals that do not change =1; and orbitals that remain in the same position but change sign = -1) 4. Find the irreducible representations (they correspond to the symmetry of group orbitals, also called Symmetry Adapted Linear Combinations SALC’s of the orbitals). 5. Find AO’s in central atom with the same symmetry 6. Combine AO’s from central atom with those group orbitals of same symmetry and similar E Molecular orbitals for larger molecules: H2O

33. G 0 2 0 2 For H H group orbitals G = A1+ B1 E two orbitals unchanged C2 two orbitals interchanged sv two orbitals unchanged sv’ two orbitals interchanged

34. No match

35. antibonding antibonding px non-bonding pz py slightly bonding bonding bonding

36. Molecular orbitals for NH3 Find reducible representation for 3H’s G 0 1 3 Irreducible representations: G = A1 + E

37. anti-bonding anti-bonding LUMO pz Slightly bonding HOMO bonding bonding

38. Projection Operator Algorithm of creating an object forming a basis for an irreducible rep from an arbitrary function. Where the projection operator sums the results of using the symmetry operations multiplied by characters of the irreducible rep. j indicates the desired symmetry. lj is the dimension of the irreducible rep. h the order order of the group. Starting with the 1sA create a function of A1 sym ¼(E1sA + C21sA + sv1sA + sv’1sA) = ¼ (1sA + 1sB+ 1sB + 1sA)

39. Consider the bonding in NF3 GA 3 0 -1 GA = A2 + E } GB 3 0 1 GB = GC = GD = A1 + E GC 3 0 1 GD 3 0 1 3 2 1 A B C D

40. Now construct SALC See what the different symmetry operations do to the one of the types of p orbitals. GA = A2 + E Now construct an SALC of A2 sym PA2(p1) = 1/6 (p1 + p2 + p3 + (-1)(-p)1 + (-1)(-1p3) + (-1)(-p2) No AO on N is A2

41. E: Apply projection operator to p1 PA2(p1) = (2p1 - p2 - p3) = E1 But since it is two dimensional, E, there should be another SALC PA2(p2) = (2p2 - p3 - p1) = E’ But E1 and E’ should be orthogonal. We want sum of products of coefficients to be zero. Create a linear combination of E2 = E’ + k E1.= (-1 +k*2) p1 + (2 + k(-1)) p2 + (-1 + k(-1))p3 Have to choose k such that they are orthogonal. 0 = (2(-1 + k*2) -1 (2 + k(-1)) -1 (-1 + k(-1)) k = ½ E2 = (3/2 p2 - 3/2 p3) = p2 – p3

42. The geometries of electron domains Molecular shapes: When we discussed VSEPR theory Can this be described in terms of MO’s?

43. s + p = 2 sp hybrids (linear) s + 3p = 4 sp3hybrids tetrahedral s + 2p = 3 sp2hybrids trigonal planar s + 3p + d = 5 dsp3hybrids trigonal bipyramidal s + 3p + 2d = 6 d2sp3hybrids octahedral Hybrid orbitals