1. Lecture 6

2. Molecular orbitals of heteronuclear diatomic molecules

3. 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

4. Combinations of two s orbitals in a homonuclear molecule (e.g. H 2 ) Antibonding Bonding In this case, the energies of the A.O.’s are identical

5. More generally:  c a  (1s a )  c b  (1s b )] n A.O.’sn M.O.’s The same principle is applied to heteronuclear diatomic molecules But the atomic energy levels are lower for the heavier atom

6. 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)

7. The molecular orbitals of carbon monoxide  c c  (C)  c o  (O)] 2s2p C-19.43-10.66 O-32.38-15.85 E(eV) Each MO receives unequal contributions from C and O (c c ≠ c o )

8. Group theory is used in building molecular orbitals

9. A1A1 A1A1 A1A1 A1A1 B1 B2B1 B2 B1 B2B1 B2 “C-like MO” “O-like MO” Frontier orbitals “C-like MO’s” “O-like MO’s” mixing Larger homo lobe on C Bond order 3

10. A related example: HF s (A 1 )2p(A 1, B 1, B 2 ) H-13.61 (1s) F-40.17 (2s)-18.65 No s-s int. (  E > 13 eV) Non-bonding (no E match) Non-bonding (no symmetry match)

11. Extreme cases: ionic compounds (LiF) Li transfers e - to F Forming Li + and F - A1A1 A1A1

12. Molecular orbitals for larger molecules 1. Determine point group of molecule (if linear, use D 2h and C 2v 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 p x, p y, p z. (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

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

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

15. -13.6 eV -40.2 eV -13.6 eV Good E match Strong interaction Poor E match weak interaction

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

17. CO 2 D ∞h, use D 2h (O O) group orbitals the same as for F F But C has more AO’s to be considered than H !

18. CO 2 D ∞h, use D 2h No match Carbon orbitals

19. A g -A g interactions B 1u -B 1u interactions All four are symmetry allowed

20. Group orbitals 1(A g ) and 2 (B 1u ) Group orbitals 3(A g ) -4 (B 1u ) 2s(C) (A g ) and Groups 1 & 3 (A g ) 2s(C) Group 1(s) < 2s(C) Group 3(p z ) Primary A g interaction 2p z (C) (B 1u ) and Groups 2 & 4 (B 1u ) 2p z (C) Group 2(s) << 2p z (C) Group 4(p z ) Primary B 1u interaction

21. Primary A g interaction Primary B 1u interaction

22. Bonding  Bonding  Non-bonding  Non-bonding  4 bonds All occupied MO’s are 3c-2e

23. LUMO HOMO The frontier orbitals of CO 2

24. Molecular orbitals for larger molecules: H 2 O 1. Determine point group of molecule: C 2v 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 p x, p y, p z. (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

25. For H H group orbitals  v ’ two orbitals interchanged  E two orbitals unchanged C 2 two orbitals interchanged 2 200  v two orbitals unchanged    

26. No match

27. pzpz bonding slightly bonding antibonding pxpx bonding antibonding pypy non-bonding

28.  3 10 Find reducible representation for 3H’s Irreducible representations:  Molecular orbitals for NH 3

30. pzpz bonding Slightly bonding anti-bonding bonding anti-bonding LUMO HOMO

31. 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. l j is the dimension of the irreducible rep. h the order order of the group. Starting with the 1s A create a function of A 1 sym ¼(E1s A + C 2 1s A +  v 1s A +  v ’1s A ) = ¼ (1s A + 1s B + 1s B + 1s A )

32. Consider the bonding in NF 3  A 3 0 -1 A B C D  B 3 0 1  C 3 0 1  D 3 0 1  A = A 2 + E  B =  C =  D = A 1 + E 1 23

33. Now construct SALC  A = A 2 + E P A2 (p 1 ) = 1/6 (p 1 + p 2 + p 3 + (-1)(-p) 1 + (-1)(-1p 3 ) + (-1)(-p 2 ) No AO on N is A 2

34. E: P A2 (p 1 ) = 2/6 (2p 1 - p 2 - p 3 ) = E 1 Apply projection operator to p 1 But since it is two dimensional, E, there should be another SALC P A2 (p 2 ) = 2/6 (2p 2 - p 3 - p 1 ) = E’ But E 1 and E’ should be orthogonal want sum of products of coefficients to be zero. E 2 = E’ + k E 1. = (-1 +k*2) p 1 + (2 + k(-1)) p 2 + (-1 + k(-1)) = 0 Have to choose k such that they are orthogonal. 0 = (2/6) 2 (2(-1 + k*2) -1 (2 + k(-1)) -1 (-1 + k(-1)) k = ½ E 2 = 2/6 (3/2 p 2 - 3/2 p 3 )

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

37. Hybrid orbitals s + p = 2 sp hybrids (linear) s + 2p = 3 sp 2 hybrids trigonal planar s + 3p = 4 sp 3 hybrids tetrahedral s + 3p + d = 5 dsp 3 hybrids trigonal bipyramidal s + 3p + 2d = 6 d 2 sp 3 hybrids octahedral