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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) Materials Provided Character tables also provided (unless the point of the question is to build the table). MO Diagrams for More Complex Molecules Chapter 5 Wednesday, October 16, 2013 Boron trifluoride 1. Point group D3h 2. x y z 3. Make reducible reps for outer atoms z x y y z x 4. Get group orbital symmetries by reducing each Γ Γ2s 3 0 1 3 0 1 Γ2pz 3 0 -1 -3 0 1 Γ2px 3 0 -1 3 0 -1 Γ2py 3 0 1 3 0 1 Γ2s = A1’ + E’ Γ2pz = A2’’ + E’’ Γ2px = A2’ + E’ Γ2py = A1’ + E’ Boron trifluoride Γ2s = A1’ + E’ Γ2pz = A2’’ + E’’ Γ2px = A2’ + E’ Γ2py = A1’ + 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 method provides a systematic way to find how the AOs should be combined to give the right group orbitals (SALCs). 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 σh S3 S32 Fb Fa Fb Fc Fa Fc Fb 1 1 1 1 1 1 1 C32 C2(a) C2(b) C2(c) AO E C3 Fa Fa Fb Fc Fa Fc A1’ 1 1 1 1 1 σv(a) σv(b) σv(c) A 1 ’ = Fa + F b + F c + F a + Fc + Fb + Fa + F b + F c + Fa + F c + F b 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. 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 σh S3 S32 Fb Fa Fb Fc Fa Fc Fb -1 1 1 1 -1 -1 -1 C32 C2(a) C2(b) C2(c) AO E C3 Fa Fa Fb Fc Fa Fc A2’ 1 1 1 -1 -1 σv(a) σv(b) σv(c) A 2 ’ = Fa + F b + F c – Fa – Fc – Fb + Fa + F b + F c – 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. 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 σh S3 S32 Fb Fa Fb Fc Fa Fc Fb 0 2 -1 -1 0 0 0 C32 C2(a) C2(b) C2(c) AO E C3 Fa Fa Fb Fc Fa Fc E’ 2 -1 -1 0 0 σv(a) σv(b) σv(c) 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. BF3 - Projection Operator Method Γ2s = A1’ + E’ Γ2pz = A2’’ + E’’ Γ2px = A2’ + E’ Γ2py = A1’ + 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. 𝜓 2 𝑑𝜏 = 1 Normalization condition BF3 - Projection Operator Method Let’s normalize the A1’ group orbital: 𝜓𝐴′1 = 𝑐a [𝜙(2𝑠Fa ) + 𝜙(2𝑠Fb ) + 𝜙(2𝑠Fc )] 𝜓 2 𝑑𝜏 = 1 𝑐a2 𝑐a2 𝑐a2 A1’ wavefunction Normalization condition for group orbitals [𝜙(2𝑠Fa ) + 𝜙(2𝑠Fb ) + 𝜙(2𝑠Fc )]2 𝑑𝜏 = 1 𝜙 2 (2𝑠Fa )𝑑𝜏 + 1+1+1 =1 𝜙 2 (2𝑠Fb )𝑑𝜏 + 𝑐a = So the normalized A1’ GO is: 𝜓𝐴′1 = nine terms, but the six overlap (S) terms are zero. 𝜙 2 (2𝑠Fc )𝑑𝜏 = 1 1 3 1 3 [𝜙(2𝑠Fa ) + 𝜙(2𝑠Fb ) + 𝜙(2𝑠Fc )] BF3 - Projection Operator Method Now let’s normalize the E’(y) group orbital: 𝜓𝐸′ (𝑦) = 𝑐a [2𝜙(2𝑠Fa ) − 𝜙(2𝑠Fb ) − 𝜙(2𝑠Fc )] 𝑐a2 [2𝜙(2𝑠Fa ) − 𝜙(2𝑠Fb ) − 𝜙(2𝑠Fc )]2 𝑑𝜏 = 1 𝑐a2 4 𝑐a2 𝜙 2 (2𝑠Fa )𝑑𝜏 + 4+1+1 =1 𝜙 2 (2𝑠Fb )𝑑𝜏 + 𝑐a = E’(y) wavefunction nine terms, but the six overlap (S) terms are zero. 𝜙 2 (2𝑠Fc )𝑑𝜏 = 1 1 6 So the normalized E’(y) GO is: 𝜓𝐸′ (𝑦) = 1 6 [2𝜙(2𝑠Fa ) − 𝜙(2𝑠Fb ) − 𝜙(2𝑠Fc )] BF3 - Projection Operator Method 𝜓𝐴′1 = 1 3 𝜓𝐸′ (𝑦) = [𝜙(2𝑠Fa ) + 𝜙(2𝑠Fb ) + 𝜙(2𝑠Fc )] 1 6 [2𝜙(2𝑠Fa ) − 𝜙(2𝑠Fb ) − 𝜙(2𝑠Fc )] 𝒄𝟐𝒊 𝐢𝐬 the probability of finding an electron in 𝝓𝒊 in a group orbital, so 𝒄𝟐𝒊 = 𝟏 for a normalized group orbital. So the normalized E’(x) GO is: 𝜓𝐸′ (𝑥) = 1 2 [𝜙(2𝑠Fb ) − 𝜙(2𝑠Fc )] BF3 - Projection Operator Method Γ2s = A1’ + E’ Γ2pz = A2’’ + E’’ Γ2px = A2’ + E’ Γ2py = A1’ + 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). BF3 - Projection Operator Method boron orbitals 2s: A1’ E’(y) E’(x) 2py: A1’ E’(y) E’(x) A1’ E’ E’ 2px: A2’ E’(y) E’(x) A2’’ 2pz: A2’’ E’’(y) E’’(x) BF3 - Projection Operator Method boron orbitals 2s: A1’ E’(y) E’(x) A1’ 2py: A1’ E’(y) E’ E’(x) little overlap 2px: A2’ E’(y) E’ E’(x) A2’’ 2pz: A2’’ E’’(y) E’’(x) Boron trifluoride F 2s is very deep in energy and won’t interact with boron. B Li –8.3 eV Ca C –14.0 eV N B 2p Al O F 1s C Si P Mg Be H Al –18.6 eV Ne 2s He Si S 3p 3s Cl P S N Cl Ar O –40.2 eV F Ne Ar Boron Trifluoride σ* E′ σ* π* –8.3 eV A2″ Energy A2″ –14.0 eV A2′ + E′ A1′ nb A2″ + E″ –18.6 eV A1′ + E′ π A2″ σ σ A1′ E′ A1′ + E′ nb –40.2 eV 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. 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 σ-MOs for Octahedral Complexes 2. 3. Make reducible reps for sigma bond vectors 4. This reduces to: Γσ = A1g + Eg + T1u six GOs in total σ-MOs for Octahedral Complexes 5. Find symmetry matches with central atom. Γσ = A1g + Eg + T1u Reading off the character table, we see that the group orbitals match the metal s orbital (A1g), the metal p orbitals (T1u), and the dz2 and dx2-y2 metal d orbitals (Eg). We expect bonding/antibonding combinations. The remaining three metal d orbitals are T2g and σ-nonbonding. 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 + σ6 A1g (non-degenerate) σ1 - σ3 , σ2 - σ4 , σ5 - σ6 T1u (triply degenerate) σ1 - σ2 + σ3 - σ4 , 2σ6 + 2σ5 - σ1 - σ2 - σ3 - σ4 Eg (doubly degenerate) σ-MOs for Octahedral Complexes There is no combination of ligand σ orbitals with the symmetry of the metal T2g orbitals, so these do not participate in σ bonding. L + T2g orbitals cannot form sigma bonds with the L6 set. S = 0. non-bonding σ-MOs for Octahedral Complexes 6. Here is the general MO diagram for σ bonding in Oh complexes: 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