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MO Diagrams for More Complex Molecules Chapter 5 Wednesday, October 22, 2014.

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Presentation on theme: "MO Diagrams for More Complex Molecules Chapter 5 Wednesday, October 22, 2014."— Presentation transcript:

1 MO Diagrams for More Complex Molecules Chapter 5 Wednesday, October 22, 2014

2 Boron trifluoride 1. Point group D 3h Make reducible reps for outer atoms Γ 2s Γ 2pz Γ 2px Γ 2py Get group orbital symmetries by reducing each Γ Γ 2s = A 1 ’ + E ’ Γ 2pz = A 2 ’’ + E ’’ Γ 2px = A 2 ’ + E ’ Γ 2py = A 1 ’ + E ’ x y z x y z x y z

3 Boron trifluoride Γ 2s = A 1 ’ + E ’ Γ 2pz = A 2 ’’ + E ’’ Γ 2px = A 2 ’ + E ’ Γ 2py = A 1 ’ + E ’ What is the shape of the group orbitals? 2s: A1’A1’ E’(y) ?? E’(x) The projection operator method provides a systematic way to find how the AOs should be combined to give the right group orbitals (SALCs). Which combinations of the three AOs are correct?

4 BF 3 - Projection Operator Method A 1 ’ = F a + F b + F c + F a + F c + F b + F a + F b + F c + F a + F c + F b FaFa 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. AOEC3C3 C32C32 C 2(a) C 2(b) C 2(c) σhσh S3S3 S32S32 σ v(a) σ v(b) σ v(c) FaFa FaFa FbFb FcFc FaFa FcFc FbFb FaFa FbFb FcFc FaFa FcFc FbFb FbFb FcFc The group orbital wavefunctions are determined by multiplying the projection table values by the characters of each irreducible representation and summing the results. A1’A1’ A 1 ’ = 4F a + 4F b + 4F c A1’A1’

5 BF 3 - Projection Operator Method A 2 ’ = F a + F b + F c – F a – F c – F b + F a + F b + F c – F a – F c – F b FaFa 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. AOEC3C3 C32C32 C 2(a) C 2(b) C 2(c) σhσh S3S3 S32S32 σ v(a) σ v(b) σ v(c) FaFa FaFa FbFb FcFc FaFa FcFc FbFb FaFa FbFb FcFc FaFa FcFc FbFb FbFb FcFc The group orbital wavefunctions are determined by multiplying the projection table values by the characters of each irreducible representation and summing the results. A2’A2’ A 2 ’ = 0 There is no A 2 ’ group orbital!

6 BF 3 - Projection Operator Method E’ = 2F a – F b – F c F a – F b – F c FaFa 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. AOEC3C3 C32C32 C 2(a) C 2(b) C 2(c) σhσh S3S3 S32S32 σ v(a) σ v(b) σ v(c) FaFa FaFa FbFb FcFc FaFa FcFc FbFb FaFa FbFb FcFc FaFa FcFc FbFb FbFb FcFc The group orbital wavefunctions are determined by multiplying the projection table values by the characters of each irreducible representation and summing the results. E’E’ E’ = 4F a – 2F b – 2F c E’(y)

7 Γ 2s = A 1 ’ + E ’ Γ 2pz = A 2 ’’ + E ’’ Γ 2px = A 2 ’ + E ’ Γ 2py = A 1 ’ + E ’ What is the shape of the group orbitals? 2s: A1’A1’ E’(y) ?? E’(x) We can get the third group orbital, E’(x), by using normalization. BF 3 - Projection Operator Method Normalization condition

8 BF 3 - Projection Operator Method nine terms, but the six overlap (S) terms are zero. A 1 ’ wavefunction Normalization condition for group orbitals So the normalized A 1 ’ GO is: Let’s normalize the A 1 ’ group orbital:

9 BF 3 - Projection Operator Method nine terms, but the six overlap (S) terms are zero. E’(y) wavefunction So the normalized E’(y) GO is: Now let’s normalize the E’(y) group orbital:

10 BF 3 - Projection Operator Method So the normalized E’(x) GO is:

11 Γ 2s = A 1 ’ + E ’ Γ 2pz = A 2 ’’ + E ’’ Γ 2px = A 2 ’ + E ’ Γ 2py = A 1 ’ + E ’ What is the shape of the group orbitals? 2s: A1’A1’ E’(y) ? E’(x) Now we have the symmetries and wavefunctions of the 2s GOs. BF 3 - Projection Operator Method We could do the same analysis to get the GOs for the p x, p y, and p z orbitals (see next slide). notice the GOs are orthogonal (S = 0).

12 2s: A1’A1’E’(y)E’(x) BF 3 - Projection Operator Method 2p y : A1’A1’E’(y)E’(x) 2p x : A2’A2’ E’(y)E’(x) 2p z : A 2 ’’ E’’(y)E’’(x) boron orbitals A1’A1’ A 2 ’’ E’

13 2s: A1’A1’E’(y)E’(x) BF 3 - Projection Operator Method 2p y : A1’A1’E’(y)E’(x) 2p x : A2’A2’ E’(y)E’(x) 2p z : A 2 ’’ E’’(y)E’’(x) boron orbitals A1’A1’ A 2 ’’ E’ little overlap

14 F 2s is very deep in energy and won’t interact with boron. H He Li Be B C N O F Ne B C N O F Na Mg Al Si P S Cl Ar Al Si P S Cl Ar 1s1s 2s2s 2p2p 3s3s 3p3p –18.6 eV –40.2 eV –14.0 eV –8.3 eV Boron trifluoride

15 π* Boron Trifluoride π nb σ σ σ* A1′A1′ E′E′ A2″A2″ A2″A2″ A1′A1′ A2″A2″ E′E′ Energy –8.3 eV –14.0 eV A 1 ′ + E′ A 2 ″ + E″ A 2 ′ + E′ –18.6 eV –40.2 eV

16 d orbitals l = 2, so there are 2 l + 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.

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

18 2. 3. Make reducible reps for sigma bond vectors σ- MOs for Octahedral Complexes 4. This reduces to: Γ σ = A 1g + E g + T 1u six GOs in total

19 σ- MOs for Octahedral Complexes Γ σ = A 1g + E g + T 1u Reading off the character table, we see that the group orbitals match the metal s orbital (A 1g ), the metal p orbitals (T 1u ), and the d z2 and d x2-y2 metal d orbitals (E g ). We expect bonding/antibonding combinations. The remaining three metal d orbitals are T 2g and σ-nonbonding. 5. Find symmetry matches with central atom.

20 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: L 6 SALC symmetry label σ 1 + σ 2 + σ 3 + σ 4 + σ 5 + σ 6 A 1g (non-degenerate) σ 1 - σ 3, σ 2 - σ 4, σ 5 - σ 6 T 1u (triply degenerate) σ 1 - σ 2 + σ 3 - σ 4, 2σ 6 + 2σ 5 - σ 1 - σ 2 - σ 3 - σ 4 E g (doubly degenerate)

21 σ- MOs for Octahedral Complexes There is no combination of ligand σ orbitals with the symmetry of the metal T 2g orbitals, so these do not participate in σ bonding. L + T 2g orbitals cannot form sigma bonds with the L 6 set. S = 0. T 2g are non-bonding

22 σ- MOs for Octahedral Complexes 6. Here is the general MO diagram for σ bonding in O h complexes:

23 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, H 3 +, FHF -, CO 2, H 2 O, BF 3, and σ-ML 6


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