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Chem 344 - Homework 9 – Due Friday, Apr Chem 344 - Homework 9 – due Friday, Apr. 11, 2014, 2 PM 1. In these examples, the valence (highest filled) electrons and orbitals are used. So CS is like C≡O except you use 3s,3p orbitals for S. the example thus ignores 1s on C and 1s,2s,2p on S 2. Similar point, Cl uses 3s,3p and Br uses 4s,4p. Cl (right) more electronegative, so orbitals lower in energy than Br (left) 3. orbitals are from interaction of sp hybrids, make linear C-C bonds, however the diagram should have included the Hs which also make two -bonds (C-H bonds) with the overlap of the sp hybrid on each C and the 1s on each H, which are linear extension of the C≡C bond, completed by orbitals from the left over px,py orbitals overlapping. The bonding interaction is much bigger than the s-p split, so b below * 4. This can be a lot of work, but the determinant simplifies by dividing through by , and letting x=-E/. this can be expanded by minors and a simpler polynomial results, th but since 5 order, you need help Note (if you had Ch.314) unlike benzene, naphthalene has no degenerate orbitals, this is due to its lower symmetry, D2h vs. D6h Degeneracy requires a Cn axis, n≥3 5. You should be able ot do the maximum analytically by taking the derivative, setting it to zero and solving: dSsp/dR = 0 The graph would then confirm your answer 6. This works for the low lying filled orbitals, but it misses on the higher ones because is seems to overcount the number of orbitals and violates our basic rule: # MOs = # AOs. For benzene this suggests the E=-2 orbital is degenerate, k=±3, but it is non-degenerate. 7. You would not necessarily know the6 signs in these LCAO-MO unless you worked them out (i.e. the coefficients could be +/-) but if you worked out propylene (C3H6 with Huckel theory, you would probably find this pattern. Since 5 is only on N what does that mean? Lone pair? 8. P16.5) The bond dissociation energies of the species NO, CF–, and CF+ follow the trend CF+ > NO > CF–. Explain this trend using MO theory. The configurations for the molecules are 2 2 1 NO: 12 2 *** 3 2 4 1 4 5 2 2 10 5 Bond Order = 2.5 2 2 2 2 CF: 12 2 *** 3 2 4 1 4 5 2 2 10 6 Bond Order = 2 2 22 CF: 12 2 ** 3 2 4 1 4 5 2 10 4 Bond Order = 3 2 We see that the calculated bond orders account for the relative bond strengths. 9. P16.9) Calculate the bond order in each of the following species. Which of the species in parts (a) through (d) do you expect to have the shorter bond length? a.Li2 or Li b. C2 or C 2 2 c. O2 or O2 d. F2 or F2 22* 2 Li2 : 1g 1 u 2 g 42 Bond Order = 1 2 21* 2 Li2 : 1g 1 u 2 g 32 Bond Order = 0.5 2 22**224 C2 : 1g 1 u 2 g 2 u 1 u 84 Bond Order = 2 2 22**223 C2 : 1g 1 u 2 g 2 u 1 u 74 Bond Order = 1.5 2 2**22 2 2 2 2 **11 O2 : 1g 1 u 2 g 2 u 3 g 1 u 1u 1 g 1 g 10 6 Bond Order = 2 2 2***2 2 2 2 22 1 O:2 1g 1 u 2 g 2 u 3 g 1 u 1 u 1 g 10 5 Bond Order = 2.5 2 2****2 2 2 2 22 2 2 F:2 1g 1 u 2 g 2 u 3 g 1 u 1 u 1 u 1 g 10 8 Bond Order = 1 2 2***2 2 2 2 22 3 * 2 F:2 1g 1 u 2 g 2 u 3 g 1 u 1 u 1 g 1 g 10 9 Bond Order = 0.5 2 Because the bond length is shorter for a greater bond order, the answers are Li2, C2, O2 , F2. 10. P16.18) Images of molecular orbitals for LiH calculated using the minimal basis set are shown here. In these images, the smaller atom is H. The H1s AO has a lower energy than the Li2s AO. The energies of the MOs are (left to right) –63.9, –7.92, and +2.14 eV. Make a molecular orbital diagram for this molecule, associate the MOs with the images, and designate the MOs in the images shown here as filled or empty. Which MO is the HOMO? Which MO is the LUMO? Do you expect the dipole moment in this molecule to have the negative end on H or Li? The left image corresponds to the 1 MO, because the MO is localized on the Li atom. From the MO diagram, the 2 MO should have a larger coefficient for the H1s AO than for the Li2s AO, following the rules outlined in Section 13.2. Therefore, the middle image is the 2 MO. Following the same reasoning, the 3* MO should have a larger coefficient for the Li2s AO, and more importantly a nodal plane separating H and Li. Therefore, the right image is the 3* MO. The 1 and 2 MOs are filled, and the 3* is empty. Because the coefficient for the H1s AO is larger than that for the Li2s AO in the 2s MO, the bonding electrons have a higher probability of being on the H than on the Li. Therefore, the negative end of the dipole is on the H atom. 11. P16.21) Sketch the molecular orbital energy diagram for the radical OH based on what you know about the corresponding diagram for HF. How will the diagrams differ? Characterize the HOMO and LUMO as antibonding, bonding, or nonbonding. The diagrams differ in the relative energies of the AOs involved and in that there are nonbonding valence electrons on the OH radical, whereas there are none on HF. The HOMO is the 1π, which is a nonbonding MO. The LUMO is the 5σ*, which is an antibonding MO. Extra Problems 1. 2. 3. 4. 5. 6. 7. 8. P16.6) What is the electron configuration corresponding to O , O , and ? What do you 2 2 expect the relative order of bond strength to be for these species? Which, if any, have unpaired electrons? 2****22 2 2 22 O:2 1g 1 u 2 g 2 u 3 g 1 u 1 u 11 g g 10 6 Bond Order = 2 2 22 2 2 2 22 2 1 **** O:2 1g 1 u 2 g 2 u 3 g 1 u 1 u 1 g 1 g 10 7 Bond Order = 1.5 2 2***2 2 2 2 22 1 O:2 1g 1 u 2 g 2 u 3 g 1 u 1 u 1 g 10 5 Bond Order = 2.5 2 Because the bond strengths increase with increasing bond order, the relative order of bond strength is and O2 > O2 > O2 . O2 has two unpaired electrons, and the positively and negatively charged species each have one. 9. P16.17) Sketch a molecular orbital energy diagram for CO and place the electrons in the levels appropriate for the ground state. The AO ionization energies are O2s: 32.3 eV; O2p: 15.8eV; C2s: 19.4 eV; and C2p: 10.9 eV. The MO energies follow the sequence (from lowest to highest) 1, 2, 3, 4, 1, 5, 2, 6. Assume that the 1s orbital need not be considered and define the 1 orbital as originating primarily from the 2s AOs on C and O. Connect each MO level with the level of the major contributing AO on each atom. O2 .
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