Introduction to Molecular Orbital Theory

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8 hybridization Mixing 2s and 2p: requires degeneracy to maintain eigenfunctions of AOs. Otherwise, the hybrid orbital is an average value for the atom, not an exact solution. This makes sense when ligands impose directionality: guess of the mixing occurring in OMs.

Molecular Orbital (MO) Theory Diagram of molecular energy levels Magnetic and spectral properties Paramagnetic vs. Diamagnetic Electronic transitions Solid State - Conductance Predicts existence of molecules Bond Order

Molecular Orbital (MO) Theory Two atomic orbitals combine to form a bonding molecular orbital an anti-bonding molecular orbital e - in bonding MO’s = stability e - in anti-bonding MO’s = instability # atomic orbitals combined equals # of molecular orbitals formed

Central Themes Quantum mechanical level Molecule viewed as a collection of nuclei surrounded by delocalized molecular orbitals Atomic wave functions are summed to obtain molecular wave functions. If wave functions reinforce each other, a bonding MO is formed (region of high electron density exists between the nuclei). If wave functions cancel each other, an antibonding MO is formed (a node of zero electron density occurs between the nuclei).

An Analogy Amplitudes of wave functions added Amplitudes of wave functions subtracted.

MO: Molecular Hydrogen The bonding MO is lower in energy than an AO The anti- bonding MO is higher in energy than an AO

Considerations… Bond Order =1/2( # bonding e - – # antibonding e - ) Higher bond order = stronger bond Molecular electron configurations Highest Occupied Molecular Orbital = HOMO Lowest Unoccupied Molecular Orbital = LUMO An Example: H 2 (  1s ) 2

MO: Molecular Hydrogen

Predicting Stability: H 2 + & H 2 -  1s AO of H 1s MO of H 2 + bond order = 1/2(1-0) = 1/2 H 2 + does exist bond order = 1/2(2-1) = 1/2 H 2 - does exist 1s   MO of H 2 - 1s AO of H AO of H - configuration is (  1s ) 1 configuration is (  1s ) 2 (  1s ) 1

Helium: He 2 + vs He 2 Energy MO of He +  * 1s  1s AO of He + 1s MO of He 2 AO of He 1s AO of He 1s  * 1s  1s Energy He 2 + bond order = 1/2 He 2 bond order = 0 AO of He 1s

Bond Length vs. Bond Order

Next Row: 2s & 2p orbitals *2s*2s  2s 2s 1s *1s*1s  1s *1s*1s 1s 2s *2s*2s  2s Li 2 B.O. = 1Be 2 B.O. = 0 Bonding in s-block homonuclear diatomic molecules. Energy Li 2 Be 2

Combinations for p-orbitals Axial symmetry means  bond Non-axial symmetry means  bond

P orbital Complications Results in one  & one  * MO One pair from 2p z Results in two  & two  * MO’s One pair for 2p x and one pair for 2p y > half filled p orbitals (O, F, Ne) energy  2p < 2  2p < 1  * 2p < 2  * 2p  half filled p orbitals (B, C, N) energy 2  2p <  2p < 2  * 2p < 1  * 2p

MO – Now with S & P X 2

P orbital Complications Results in one  & one  * MO One pair from 2p z Results in two  & two  * MO’s One pair for 2p x and one pair for 2p y > half filled p orbitals (O, F, Ne) energy  2p < 2  2p < 1  * 2p < 2  * 2p  half filled p orbitals (B, C, N) energy 2  2p <  2p < 2  * 2p < 1  * 2p

S-P Energy Separation

E(2p)-E(2s): N 12.4 eV O 16.5 eV F 31.6 eV For N: “small” 2p/2s separation “big”  * (2s) /  (2pz) repulsion

S - P orbital mixing

Relative Energy Levels for 2s & 2p MO energy levels for O 2, F 2, and Ne 2 MO energy levels for B 2, C 2, and N 2 WITHOUT big 2s-2p repulsion WITH big 2s-2p repulsion

Triumph for MO Theory?

Figure 9.37 Diagram of the Kind of Apparatus Used to Measure the Paramagnetism of a Sample

Can MO Theory Explain Bonding? SOLUTION: PROBLEM:As the following data show, removing an electron from N 2 forms an ion with a weaker, longer bond than in the parent molecules, whereas the ion formed from O 2 has a stronger, shorter bond: PLAN:Find the number of valence electrons for each species, draw the MO diagrams, calculate bond orders, and then compare the results. Explain these facts with diagrams that show the sequence and occupancy of MOs. Bond energy (kJ/mol) Bond length (pm) N2N2N2N2 N2+N2+N2+N2+ O2O2O2O2 O2+O2+O2+O N 2 has 10 valence electrons, so N 2 + has 9. O 2 has 12 valence electrons, so O 2 + has 11.

Can MO Theory Explain Bonding?  2s  2s  2p  2p  2p  2p N2N2 N2+N2+ O2O2 O2+O2+ bond orders 1/2(8-2)=31/2(7-2)=2.51/2(8-4)=2  2s  2s  2p  2p  2p  2p bonding e - lost 1/2(8-3)=2.5 antibonding e - lost

Real World Applications Most molecules are heteroatomic What needs to be considered? Orbitals involved Electronegativity (Orbital energies) Hybridization (Group Theory) Mixing

Heteronuclear Diatomic Molecules Molecular orbital diagrams for heteronuclear molecules have skewed energies for the combining atomic orbitals to take into account the differing electronegativities. The more electronegative elements are lower in energy than those of the less electronegative element.

Energy Level Diagram for NO

Heteronuclear Diatomic Molecules The energy differences between bonding orbitals depend on the electronegativity differences between the two atoms The larger the difference the more polar the bond that is formed (ionic character increases) The difference reflects the amount of overlap between the bonding orbitals. If the difference is too great the orbitals cannot overlap effectively and nonbonding orbitals will be formed.

Formation of MO’s in HF The bond in HF involves the 1s electron of H and the 2p orbital of F A bonding and antibonding MO are produced  sp and MO’s The remaining 2p orbitals on F have no overlap with H orbitals. They are termed as ‘nonbonding’ orbitals. These orbitals retain the characteristics of the F 2p atomic orbitals. Lack of overlap to produce nonbonding orbitals is much more pronounced for side-on bonding

The Energy Level Diagram for HF

Let’s Start Slowly: HF Valence electrons H – 1s 1 F – 1s 2 2s 2 2p 5 Focus on the valence interactions Accommodate for differences in electronegativity Allow mixing between symmetry- allowed states

Let’s Start Slowly: HF Energy MO of HF AO of H 1s1s  2p x 2p y  AO of F 2p

Energy The MO diagram for NO MO of NO 2s AO of N 2p  * 2s  2s 2s AO of O 2p  2p  2p  * 2p  * 2s possible Lewis structures

Figure 9.39 The Molecular Orbtial Energy-Level Diagrams, Bond Orders, Bond Energies, and Bond Lengths for the Diatomic Molecules B 2 Through F 2

Figure 9.41 The Molecular Orbital Energy-Level Diagram for the NO Molecule

Figure 9.42 The Molecular Orbital Energy-Level Diagram for Both the NO + and CN - ions

Figure 9.43 A Partial Molecular Orbital Energy-Level Diagram for the HF Molecule

Figure 9.44 Electron Distribution

Figure 9.48 (a) The Molecular Orbital System in Benzene is Formed by Combining the six p Orbtials From the Six sp 2 Hybridized Carbon Atoms (b) The Electrons in the Resulting pi Molecular Orbitals are Delocalized Over the Entire Ring of Carbon Atoms

Figure 9.48

Figure 9.49 (a) The p Orbitals Used to Form the  Bonding System in the NO 3 -. (b) A Representation of the Delocalization of the Electrons in the Pie Molecular Orbital System of the No 3 - ion

Figure 9.49

Try this: Generate an MO Diagram for CO What is the bond order? What is the HOMO? highest occupied molecular orbital What is the LUMO? lowest unoccupied molecular orbital Draw a corresponding Lewis dot structure Bonus: Based on your answers above, what can you envision for the bonding interaction of CO with a transition metal (like Fe)?

HOMO is lone pair on C. CO always binds to metals via the C end

2 H. H:H Energy Diagram

BENZENE PI MOLECULAR ORBITALS 11 33 22 44 55 66 six p’s six mo’s top view

CYCLOPENTADIENYL ANION 11 33 22 44 55