III. The Molecular Orbital Model A.Problems with the L.E. Model 1.Assumes all e- are localized; this is not what is observed 2.Unpaired e- are not described very well at all 3.Gives us no clue to bond energies B.Molecular Orbital Model 1.Similar to atomic theory, but develops orbitals for a whole molecule 2.Gives molecular orbitals that electrons go into 3.Properties of the molecular orbitals (MO’s) a.Each MO can hold 2 e- of opposite spin b.Shape of the MO represents probability of where e- is C.H 2 Molecular Orbital Description 1.We can construct our MO’s from the AO’s we already know a.MO 1 = 1s a + 1s b b.MO 2 = 1s a - 1s b

2.Size, shape, and energy of MO’s 3.MO 1 =  -MO a.Large probability between H nuclei =  interaction b.Lower energy than H 1s orbitals = bonding MO c.The 2 total valence e- pair up, giving a stabilization = bond 4.MO 2 =  *-MO a.Large probability bracketing H nuclei =  interaction b.Higher energy than H 1s orbitals = antibonding MO c.No electrons in this MO  1s  1s E

5.H 2 - has three electrons a.Less stable than H 2 b.One e- in antibonding MO 6.Bond Order = Difference between the number of bonding electrons and number of antibonding electrons divided by two. 7.Example of B. O. a.H 2 = (2 – 0)/2 = 1 (single bond strength) b.H 2 - = (2 – 1)/2 = 0.5 (less than single bond strength, but stable) c.He 2 = (2 – 2)/2 = 0 (not stable, not a diatomic molecule)

I. Bonding in Homonuclear Diatomics A.Li, Be, B 1.Li 2 MO description a.Li 1s 2 2s 1, but we can only use valence e- (2s) in bonding b.1s orbital is core and can’t be reached for bonding c.MO’s for Li 2 d.Bond Order = (2 – 0)/2 = 1, so it is stable e.MO electron configuration would be  2s 2 f.Li 2 is stable, but only occurs as a gas

2.Be 2 uses the same MO’s as Li 2 a.but is  2s 2  2s 2 b.Bond Order = (2-2)/2 = 0 c.Be 2 is not a stable diatomic 3.B 2 must begin to use MO’s generated from 2p AO’s a.B = 1s 2 2s 2 2p 1 b.p-orbital MO combinations

c.Head-on overlap gives  2p and  * 2p interactions b.Side-on overlap gives  2p and  * 2p interactions. There are 2 such interactions each, because of the 2 Perpendicular p-orbitals not involved in the  -bond. c.2s--2s overlap is more favorable, at lower energy d.Bond Order = (4 – 2)/2 = 1, stable diatomic IV.Magnetism A.Magnetic Balance

1.Diamagnetic = repelled by a magnetic field a.Paired e- are diamagnetic b.H 2 is diamagnetic because all of its e- are paired c.Most compounds are diamagnetic because paired e- are favorable d.Repulsion is very weak 2.Paramagnetic = attracted by a magnetic field a.unpaired electrons are paramagnetic b.much stronger than diamagnetism c.Relatively few compounds have unpaired e- B.Magnetism and MO Theory 1.We predicted from MO theory that B 2 would be diamagnetic 2.Experiment shows us that it is paramagnetic. 3.We must adjust our model to explain this

4.We must allow our model to take into account 2s/2p interaction a.2s/2p mixing changes the ordering of some of our MO’s b.Spacing is no longer symmetric c.Predicts paramagnetism of B 2 d. Bond Order is still 1

5.Other homonuclear diatomics a.C 2 and N 2 use same ordering and MO set as B 2 b.O 2 and F 2 have different ordering because s/p mixing changes

6.Trends in 2 nd row diatomics a.As Bond Order increase, Bond Energy increases, Bond Length decreases b.Bond Order doesn’t completely predict Bond Energy i.B 2 > F 2 even though both have BO = 1 ii.F 2 has much electron repulsion, weakening the bond c.N 2 bond order = (8-2)/2 = 3. This triple bond is very strong and explains why N 2 is very unreactive. d.O 2 is paramagnetic. V.Heteronuclear Diatomics: Different elements have different A.O. energies A.Adjacent elements are the most similar (NO, CN, etc…) 1.We can use same MO’s as for homonuclear diatomics for adjacents 2.NO Bond Order = (8-3)/2 = 2.5

3.MO Theory better handles unpaired e- 4.Paramagnetic 5.NO + and CN - have the same configuration a.BO = (8-2)/2 = 3 (triple bonds) b.We can use the same figure c.Diamagnetic

B.Elements far apart have very different energies 1.We must devise new MO ordering for each case 2.HF (use H1s and F2p) 3.F much lower energy (more tightly bound e-) 4.F2p < H1s so the electrons will be closer to F (Predicts Electronegativity!!!)

VI. Combining LE and MO Models A.Resonance Structures 1.L.E. model is not very accurate in some cases where more than one Lewis Structure can be drawn 2.Resonance structures = equivalent, but not exact representations of the bonding in a molecule a.O 3 b.NO 3 - c.Benzene

B.  -bonds can be localized without problems; Use LE Model C.  -bonds need to Delocalize in some molecules for stability 1.Still use LE Model to describe the  -bonding 2.Use MO model to describe  -bonding 3.Simple as possible, but must be accurate Benzene

4. NO 3 -