1. III.Homonuclear Diatomics A.Why do we need MO’s 1)The N 2 Lewis structure explains bonding and reactivity quite well 2)Problem: The O 2 Lewis structure doesn’t explain why it is paramagnetic 3)Problem: The B 2 Lewis structure doesn’t have an octet 4)MO descriptions are better B.MO’s of the First Row Elements 1)Correct ordering only if identical orbitals interact 2)Relative energy depend on element 3)Electrons fill MO’s from bottom up 4)Bond Order a)BO = ½[bonding-antibonding] b)O 2 : BO = 2 5)MO symmetry labels a)g = gerade = symmetric to i b)u = ungerade = antisymmetric to i

2. C.Orbital Mixing 1)Orbitals of similar energies can interact if symmetry allows 2)This leads to adjustments in the energy ordering of diatomic MO’s a)Example:  g (2s) and  g (2p) can mix lowering  g (2s) and raising  g (2p) b)Example:  u (2s) and  u (2p) can mix lowering  u (2s) and raising  u (2p) 3)Effects of mixing a)Energy ordering changes as we move across the first row elements b)Bonding/Antibonding nature of a given MO may change

4. D.MO Theory and Properties of Homonuclear Diatomics 1)Magnetism a)Paramagnetic = strongly attracted by magnetic field due to upaired e- b)Diamagnetic = weakly repelled by magnetic field by all paired e- 2)H 2 [  g 2 (1s)] a)BO = 1 and BL = 74 pm b)H 2 + has BO = ½ and is less stable with BL = 106 pm 3)He 2 [  g 2 (1s)  u * 2 (1s)] BO = 0, Noble gas that does not exist as a diatomic 4)Li 2 [  g 2 (2s)] BO = 1 5)Be 2 [  g 2 (2s)  u * 2 (2s)] BO = 0, not stable 6)B 2 [  u 1  u 1 (2p)] a)Paramagnetism of B 2 is not explained by Lewis structure b)BO = 1 c)Without mixing, all e- paired = diamagnetic d)Mixing makes degenerate  u (2p) lower than  g (2p) e)Filling degenerate orbitals equally (Hund’s Rule) makes it paramagnetic

5. 7)C 2 [  u 2  u 2 (2p)] a.BO = 2, mixing suggest that both are  -bonds with no  -bond? b.C 2 is rare, but C 2 2- (with a  -bond as well BO = 3) is more stable 8)N 2 [  g 2  u 2  u 2 (2p)] a.BO = 3 agrees with short strong bond (BL = 109.8 pm, D = 945 kJ/mol) b.After nitrogen the “normal” energy ordering initially predicted returns i.Nuclear charge gets higher as we go across first row ii.Mixing becomes less important because the energy difference between 2s and 2p orbitals increases (s closer to nucleus, p more diffuse) iii.Large energy difference means mixing is less important 9)O 2 [  g 2  u 2  u 2  g * 1  g * 1 (2p)] a.BO = 2, paramagnetic (not explained by Lewis structure) b.O 2 ions correlate BO and BL very well

6. I.MO’s from d-orbitals Transition metals and other heavy elements use d-orbitals in their bonding interactions 1)d-orbitals may form , , or  bonds a)A  example is the d z2 /d z2 interaction b)A  example is the d yz /d yz interaction c)A d example is the d xy/ d xy interaction d)  bonds change signs upon C 4 rotation around the internuclear axis 10) F 2 [  g 2  u 2  u 2  g * 2  g * 2 (2p)] BO = 1, diamagnetic 11) Ne 2 BO = 0 Noble gas, diatomic not stable

7. 2)Examples II.Heteronuclear Diatomics A.Polar Bonds 1)MO pattern is same as homonuclear 2)One set of AO’s will be at a lower energy than the other 3)Valence Orbital Potential Energy a)Negative energies of attraction of e- to the nucleus b)Averaged for all e- on the same level (3p) c)As Z increases left to right, VOPE becomes larger

9. 4)The LCAO for heteronuclear diatomics uses different coefficients because the energies of the 2 atoms are no longer identical a.  = c a  a + c b  b (c a ≠ c b ) b.The AO closest in energy to the MO contributes most to it i.In CO the 2  MO is mostly O ii.The 2  * MO is mostly C c.The shape and energy of the MO is similar to the major contributing AO d.If  E > 12 eV, there is no interaction e.For CO, BO = 3 f.Mixing is still important 5)Molecular reactivity occurs at Frontier Orbitals a)HOMO = Highest Occupied Molecular Orbital b)LUMO = Lowest Unoccupied Molecular Orbital