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Midterm Exam 1

Midterm Exam 1. Monday 9-9:50a Chapters 4 and 5 multiple short answer problems testing basic concepts closed book, closed notes bring pen, pencil, calculator, ID study lectures, book, suggested problems look for seating chart on lecture room door don‘t cram! . Review Sessions:

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Midterm Exam 1

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  1. Midterm Exam 1 • Monday 9-9:50a • Chapters 4 and 5 • multiple short answer problems testing basic concepts • closed book, closed notes • bring pen, pencil, calculator, ID • study lectures, book, suggested problems • look for seating chart on lecture room door • don‘t cram! Review Sessions: Thurs 6-7:50p, ICS 174 (Sam) Fri 4:30-6:20p, ICS 174 (Lindsay)

  2. Materials Provided • Character tables also provided (unless the point of the question is to build the table).

  3. MO Diagrams for More Complex Molecules • Chapter 5 • Wednesday, October 16, 2013

  4. Boron trifluoride • 1. Point group D3h • 3. Make reducible reps for outer atoms z • 2. x y x y z y z x • 4. Get group orbital symmetries by reducing each Γ Γ2s = A1’ + E’ Γ2px = A2’ + E’ Γ2py = A1’ + E’ Γ2pz = A2’’ + E’’

  5. Boron trifluoride Γ2s = A1’ + E’ Γ2px = A2’ + E’ Γ2py = A1’ + E’ Γ2pz = A2’’ + E’’ • What is the shape of the group orbitals? • ? • ? • 2s: • A1’ • E’(y) • E’(x) • Which combinations of the three AOs are correct? • The projection operator methodprovides a systematic way to find how the AOs should be combined to give the right group orbitals (SALCs).

  6. BF3 - Projection Operator Method • In the projection operator method, we pick one AO in each set of identical AOs and determine how it transforms under each symmetry operation of the point group. • Fa • Fb • Fc A1’ = Fa + Fb + Fc + Fa + Fc+ Fb+ Fa + Fb + Fc+ Fa + Fc + Fb A1’ = 4Fa + 4Fb + 4Fc • A1’ • The group orbital wavefunctions are determined by multiplying the projection table values by the characters of each irreducible representation and summing the results.

  7. BF3 - Projection Operator Method • In the projection operator method, we pick one AO in each set of identical AOs and determine how it transforms under each symmetry operation of the point group. • Fa • Fb • Fc A2’ = Fa + Fb + Fc – Fa – Fc – Fb + Fa + Fb + Fc– Fa – Fc– Fb A2’ = 0 • There is no A2’ group orbital! • The group orbital wavefunctions are determined by multiplying the projection table values by the characters of each irreducible representation and summing the results.

  8. BF3 - Projection Operator Method • In the projection operator method, we pick one AO in each set of identical AOs and determine how it transforms under each symmetry operation of the point group. • Fa • Fb • Fc E’ = 2Fa – Fb – Fc + 0 + 0 + 0 + 2Fa – Fb – Fc + 0 + 0 + 0 E’ = 4Fa– 2Fb– 2Fc • E’(y) • The group orbital wavefunctions are determined by multiplying the projection table values by the characters of each irreducible representation and summing the results.

  9. BF3 - Projection Operator Method Γ2s = A1’ + E’ Γ2px = A2’ + E’ Γ2py = A1’ + E’ Γ2pz = A2’’ + E’’ • What is the shape of the group orbitals? • ? • ? • 2s: • A1’ • E’(y) • E’(x) • We can get the third group orbital, E’(x), by using normalization. • Normalization condition

  10. BF3 - Projection Operator Method • Let’s normalize the A1’ group orbital: • A1’ wavefunction • Normalization condition for group orbitals • nine terms, but the six overlap (S) terms are zero. • So the normalized A1’ GO is:

  11. BF3 - Projection Operator Method • Now let’s normalize the E’(y) group orbital: • E’(y) wavefunction • nine terms, but the six overlap (S) terms are zero. • So the normalized E’(y) GO is:

  12. BF3 - Projection Operator Method • the probability of finding an electron in in a group orbital, so for a normalized group orbital. • So the normalized E’(x) GO is:

  13. BF3 - Projection Operator Method Γ2s = A1’ + E’ Γ2px = A2’ + E’ Γ2py = A1’ + E’ Γ2pz = A2’’ + E’’ • What is the shape of the group orbitals? • notice the GOs are orthogonal (S = 0). • ? • 2s: • A1’ • E’(y) • E’(x) • Now we have the symmetries and wavefunctions of the 2s GOs. • We could do the same analysis to get the GOs for the px, py, and pz orbitals (see next slide).

  14. BF3 - Projection Operator Method • boron orbitals • 2s: • A1’ • E’(y) • E’(x) • A1’ • 2py: • E’ • A1’ • E’(y) • E’(x) • E’ • 2px: • A2’ • E’(y) • E’(x) • A2’’ • 2pz: • A2’’ • E’’(y) • E’’(x)

  15. BF3 - Projection Operator Method • boron orbitals • 2s: • A1’ • E’(y) • E’(x) • A1’ • 2py: • E’ • A1’ • E’(y) • E’(x) • little overlap • E’ • 2px: • A2’ • E’(y) • E’(x) • A2’’ • 2pz: • A2’’ • E’’(y) • E’’(x)

  16. –8.3 eV –14.0 eV –18.6 eV –40.2 eV Boron trifluoride • F 2s is very deep in energy and won’t interact with boron. Al Si B Ca P Li C S Mg Cl Be N H 3p Al Ar O 3s B 2p F Si 1s P C Ne 2s He S N Cl Ar O F Ne

  17. E′ A2′ +E′ –18.6 eV A2″ + E″ Energy A1′ + E′ –8.3 eV A2″ A1′ A2″ E′ A2″ –14.0 eV –40.2 eV A1′ + E′ A1′ Boron Trifluoride σ* σ* π* nb π σ σ nb

  18. d orbitals • l= 2, so there are 2l+ 1 = 5 d-orbitals per shell, enough room • for 10 electrons. • This is why there are 10 elements in each row of the d-block.

  19. σ-MOs for Octahedral Complexes 1. Point group Oh The six ligands can interact with the metal in a sigma or pi fashion. Let’s consider only sigma interactions for now. • 2. pi sigma

  20. σ-MOs for Octahedral Complexes • 2. • 3. Make reducible reps for sigma bond vectors Γσ = A1g + Eg + T1u 4. This reduces to: six GOs in total

  21. σ-MOs for Octahedral Complexes Γσ = A1g + Eg + T1u 5. Find symmetry matches with central atom. • Reading off the character table, we see that the group orbitals match the metal s orbital (A1g), the metal p orbitals (T1u), and the dz2and dx2-y2 metal d orbitals (Eg). We expect bonding/antibonding combinations. • The remaining three metal d orbitals are T2g andσ-nonbonding.

  22. σ-MOs for Octahedral Complexes • We can use the projection operator method to deduce the shape of the ligand group orbitals, but let’s skip to the results: L6 SALC symmetry label σ1 + σ2 + σ3 + σ4 + σ5 + σ6A1g(non-degenerate) σ1 - σ3 , σ2 - σ4 , σ5 - σ6T1u(triply degenerate) σ1 - σ2 + σ3 - σ4 , 2σ6 + 2σ5 - σ1 - σ2 - σ3 - σ4Eg(doubly degenerate)

  23. σ-MOs for Octahedral Complexes • There is no combination of ligand σ orbitals with the symmetry of the metal T2gorbitals, so these do not participate in σ bonding. L + T2g orbitals cannot form sigma bonds with the L6 set. S = 0. non-bonding

  24. σ-MOs for Octahedral Complexes • 6. Here is the general MO diagram for σ bonding in Oh complexes:

  25. Summary • MO Theory • MO diagrams can be built from group orbitals and central atom orbitals by considering orbital symmetries and energies. • The symmetry of group orbitals is determined by reducing a reducible representation of the orbitals in question. This approach is used only when the group orbitals are not obvious by inspection. • The wavefunctions of properly-formed group orbitals can be deduced using the projection operator method. • We showed the following examples: homonuclear diatomics, HF, CO, H3+, FHF-, CO2, H2O, BF3, and σ-ML6

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