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Consider Dinitrogen

20. 19. 18. 17. 16. 15. Ionization Energy (eV). Consider Dinitrogen. First ion state (X) = 2  g +. 2  u. Second ion state (A) = 2  u. 1  g. Third ion state (B) = 2  u +. 2p. 2p. 2  g +. 1  u. 2s. 2s. 1  u +. 1  g +. :N ≡N:. Ground state (X) = 1  g +. 2. +. S.

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Consider Dinitrogen

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  1. 20 19 18 17 16 15 Ionization Energy (eV) Consider Dinitrogen First ion state (X) = 2g+ 2u Second ion state (A) = 2u 1g Third ion state (B) = 2u+ 2p 2p 2g+ 1u 2s 2s 1u+ 1g+ :N≡N: Ground state (X) = 1g+

  2. 2 + S u 2 + P u 2 + S g Potential Well Description 2u 1g 2p 2p 2g+ 1u 2s 2s 1u+ 1g+ 1 S N 2 :N≡N: g Ground state (X) = 1g+

  3. Models to describe molecular electronic structure MO Theory compared to Valence Bond Theory

  4. CH CH 2p 4 4 sp3 2s 24 24 22 22 20 20 18 18 16 16 14 14 12 12 Ionization Energy (eV) Ionization Energy (eV) Consider methane. VSEPR gives 4 sp3 hybrid orbitals. Photoelectron Spectroscopy So why are there two valence ionizations separated by almost 10 eV?

  5. Use of reducible representations in M.O. theory Consider transformation properties of vectors aligned with the 4 C-H bonds.

  6. Apply Reduction Formula: C-H = A1 + T2 This is what the bonding MOs orbitals should be. http://www.mpip-mainz.mpg.de/~gelessus/group.html

  7. CH CH 4 4 24 24 22 22 20 20 18 18 16 16 14 14 12 12 Ionization Energy (eV) Ionization Energy (eV) LCAO Description of Methane, Td 2p (t2) t2 (1, 2, 3) a1 (1) 2s (a1) C CH4 H4 Obtained using the carbon AOs as basis objects. Obtained using the hydrogen AOs as basis objects. Orbitals of same symmetry combine and bond.

  8. Now back to our ordering of the MOs of diatomics. Orbital Energy Orbital Energy N2 O2 sg (2p) p*u (2p) pu (2p) sg (2p) pu (2p) Very involved in bonding (vibrational fine structure) s*u (2s) s*u (2s) (Energy required to remove electron, lower energy for higher orbitals)

  9. Very Simple Molecular Orbital Theory A molecular orbital, f, is expressed as a linear combination of atomic orbitals, holding two electrons. The multi-electron wavefunction and the multi-electron Hamiltonian are Where hi is the energy operator for electron i and involves only electron i

  10. Seek F such that Divide by F(1,2,3…) recognizing that hi works only on electron i. Since each term in the summation depends on the coordinates of a different electron then each term must equal a constant.

  11. Now our attention is on the equation that only involves one electron. We are going to express the molecular orbital in terms of the atomic orbitals. Multiply equation at top by one of the atomic orbitals, uk, k = 1,2,…#AOs, and integrate. Now use the expansion of the MO in the AOs. or These integrals are fixed numerical values. Define The al, l = 1, 2, 3…#AOs are unknowns. These values are known numerical quantities. There are #AO such equations. Secular equations.

  12. For k = 1 to AO These are the secular equations. The number of such equations is equal to the number of atomic orbitals, AO. The number of equations is equal to the number of AOs equations. The number of unknowns, the al, is also equal to the number of AOs. For there to be a nontrivial solution (all al not equal to zero) to the set of secular equations then the determinant below must equal zero

  13. Drastic assumptions can now be made. We will use the simple Huckel approximations. hi,i = a, if orbital i is on a carbon atom. Si,i = 1, normalized atomic orbitals, hi,j = b, if atom i bonded to atom j, zero otherwise Sij = 0 if I is not equal to j • Expand the secular determinant into a polynomial of degree AO in e, the energy. • Obtain the allowed values of e by finding the roots of the polynomial. • Choose one particular value of e, substitute into the secular equations. • Obtain the coefficients of the atomic orbitals within the molecular orbital.

  14. Example The allyl pi system. The secular equations: (a-e)a1 + b a2 + 0 a3 = 0 ba1 + (a-e) a2 + b a3 = 0 0 a1 +b a2 + (a-e) a3 = 0 Simplify by dividing every element by b and setting (a-e)/b = x

  15. For x = -sqrt(2) e = a + sqrt(2) b For x = 0; e = a Substitute into the secular equations normalized

  16. Verify that h f1 = e1 f1

  17. 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.

  18. 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:

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

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

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

  22. Mixes in anti-bonding Mixes in bonding Mixes in anti-bonding Mixes in bonding

  23. 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. Starting with the 1sA create a function of A1 sym ¼(E1sA + C21sA + sv1sA + sv’1sA) = ¼ (1sA + 1sB + 1sA + 1sB) = ( ½(1sA + 1sB)

  24. 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 GD 3 0 1

  25. 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

  26. Now get the SALC of E symmetry. Note that E is doubly degenerate. 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

  27. 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.

  28. Molecular orbitals of heteronuclear diatomic molecules

  29. 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

  30. 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

  31. 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

  32. 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)

  33. 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)

  34. Group theory is used in building molecular orbitals

  35. 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

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

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

  38. 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

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

  40. 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?

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

  42. 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

  43. 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 !

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

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

  46. Primary B1u interaction Primary Ag interaction

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

  48. LUMO The frontier orbitals of CO2 HOMO

  49. 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

  50. 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

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