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1 A7. Looking at the image and table provided below, it is apparent that the monomer and dimer are structurally almost identical. Although angular and dihedral data were not included, these data are also very similar. Comparing the hydrogen bond distance to the O-H distance, it is apparent that the proton has not moved to a position halfway between the two oxygen atoms, but has remained with the oxygen it was originally bound to. The dimerization energy, which is the energy of the dimer minus twice the energy of the monomer, corresponds approximately to the energy of the two hydrogen bonds formed. Acetic acid and trifluoroacetic acid have and kcal/mol dimerization energies, respectively. Dividing by two yields the approximate energy of each hydrogen bond. This is an order of magnitude smaller than the energy of a typical O-H bond, consistent with the observation that hydrogen bonding interactions are significantly weaker than covalent bonds. The magnitudes of the geometric changes upon dimerization are about the same for both compounds; however, the changes are ever so slightly larger for acetic acid. The acid with the larger dimerization energy also exhibits the larger structural changes. A8. The O-H stretching frequency is the highest frequency in both spectra, which can be validated by observing the stretching motion in Spartan. In the dimer, the O-H stretch is reduced in frequency and increased in intensity relative to the monomer. The reduced frequency implies that the force constant for vibration is reduced. This can be rationalized as a consequence of the attraction of the other (noncovalent-bonding) oxygen to the hydrogen. Additionally, the intensity of the peak is much larger in the dimer than in the monomer. Peak intensity is related to the change in the dipole during vibration. The O-H stretch in the dimer has a larger stretch magnitude and thus a larger dipole change.

2 A9. The distance between the two boron atoms is decreased upon removal of an electron (1.609 vs angstroms). This is contrary to the expectation based on the behavior of ethylene, where removal of an electron lengthens the distance between the two carbon atoms. Closer examination reveals that the HOMO of ethylene is a π bonding orbital formed the two p orbitals of C α and C β, while in diborane, the HOMO is a π* antibonding orbital formed from the two p orbitals of B α and B β. Removal of an

3 electron from a bonding MO increases the distance between to atoms, while the removal of an electron from an antibonding MO reduces the distance between two atoms. Note in addition that removal of the electron in diborane disturbs the symmetry of the system: the central hydrogen atoms fall out of the plane of symmetry. The proximity of the boron atoms may have resulted in angle bending strain (H-B-H), causing the central hydrogen atoms move out of plane to reduce this strain. A10. Both structures are energy minima, as stable geometries were obtained for both upon geometry optimization. Both structures maintained the initial symmetry during the optimization, although bond distances and angles were adjusted. However, the see-saw conformation is favored by kj/mol, and so at room temperature the population of the trigonal bipyramidal conformation is 6.16x10-26 %,

4 and thus would not be detected. The Boltzmann equation was used to determine the relative population of a species: where R is the Boltzmann constant, J/mol K, T is temperature in Kelvin, assumed to be room temperature (298 K), E is the difference between the energies of the two species, and N 1 and N 2 are the populations of the two species. A11. CaF 2 is stable for both a bent geometry and a linear geometry. However, the bent geometry is more stable by 2.38 kcal/mol. This is a failure of VESPR theory. A12. The geometry changes from trigonal planar to trigonal pyramidal as electrons are added to the cation. The geometry of the tertiary butyl cation is trigonal planar due to the sp 2 hybridized nature of the central carbon. The addition of electrons to create the tertiary butyl anion causes the geometry to become trigonal pyramidal. This is because the hybridization of the central carbon changes to sp 3 as the added electrons begin to occupy the cation s LUMO. The localization of the added electrons on the central carbon can be observed in both the neutral and anion molecules. VSEPR theory suggests that the most stable form is trigonal pyramidal, as the lone pair repels the electrons in the C-C bonds, and this is borne out by the calculations on the anion.

5 Geometric data for A12 Bond distances are provided in angstroms

6 A21. For HF/6-311+G**, the error in the frequencies ranges from 1.8% to 32.8%, with an average value of 15.64%. Note that the HF value is larger than experiment for all the molecules. B3LYP/6-311+G* is far more accurate for calculating the frequencies, as the average unsigned error is 3.63%. Three of the calculated frequencies are larger than the experimental result, while one is smaller. Interestingly, the percent error increases in the order of the molecules given, which corresponds to the degree of charge separation: LiF possesses an ionic bond, CO is polar covalent, and N 2 and F 2 are purely covalent. Can you postulate a reason for the correlation? Frequencies in cm -1 experimental HF/6-311+G** HF % err B3LYP/6-311+G** B3LYP % err LiF % % CO % % N % % F % % Avg. % err % 3.63% Highlight shows the only example where the computational method provided a frequency less than the experimental value. 1 The average is of the unsigned error (i.e. the average of the absolute values).

7 C2. There are 5 bonding orbitals and 2 antibonding orbitals. Bonding order is thus 5-2=3; 3 bonds. Reconsider. Orbitals below -30 are probably not valence and should not be considered in bonding. From theory, each Mo atom has a configuration Mo:5s 2 4d 4. The homodiatomic MOs formed from overlap of the s, d xy, d yz, d zx, d z2, and d x2-y2 orbitals are 1, 1 *, 2, 2 *, 1 yz, 1 zx, 1 yz *, 1 zx *, 1 xy, 1 x2- y2, 1 xy *, 1 x2-y2 *. I suggest (1) reevaluating the next-highest -bonding, (2) visualizing some of the LUMOs, (3) eliminating the MOs below -30. Then let s talk.

8 C4. The molecular orbitals of diborane and ethylene are very similar. However, there are some obvious differences. First, the order of the similar orbitals is different, as illustrated in the accompanying image. For example, the HOMO for ethylene is similar to one of the degenerate E 2 molecular orbitals of diborane (the orbitals are labeled with subscripts beginning with 0 for the lowest-energy orbital depicted). The shapes of the similar MO s are also somewhat different. For example, the lowest-energy MO for each structure is composed of s atomic orbitals, interfering constructively, but since diborane has 2 central hydrogens that are out of plane, the shape is someone different in order to encompass these 2 hydrogens. This same kind of consideration leads to a small shape difference between the HOMO of ethylene and the associated E 2 orbital of diborane. This structural difference leads to small differences between the members of most pairs of corresponding orbitals. Another example of this is the E 3 MO of ethylene and the E 3 MO of diborane. The E 3 MO of ethylene has σ bonding between the two carbon atoms; however, in E 3 of diborane, the corresponding σ interaction contains contributions from the central hydrogens, and thus has a slightly different shape. In regard to the corresponding π orbitals, the two degenerate E 2 orbitals of diborane are similar to the E 2 and the HOMO of ethylene. As can be seen in the illustration below and as mentioned above, the E 2 of ethylene is nearly identical to the corresponding E 2 of of diborane, while the HOMO of ethylene differs from the corresponding E 2 of diborane due to the central protons in diborane, which affect the shape of the orbital. The associated π* orbitals are the ethylene E 4 orbital and the diborane HOMO. These are nearly identical, as there is no contribution to the MO from the AOs of diborane s central hydrogen atoms.

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10 C5. In the diagram below, the similarities between S 2 O and H 2 O are very apparent. Both molecules HOMO is an atomic p orbital on the heavy atom. This is a nonbonding interaction and represents the occupancy of the lone pairs. The second-highest occupied MO for both molecules is a σ-bonding orbital, with contributions from a p and an s orbital of the heavy atom and the s orbitals of the hydrogen atoms. It is also represents some lone pair occupancy. Although the 2 highest MOs for both these molecules are very similar in structure, energetically the S 2 O orbitals are higher in energy. This is because these orbitals are utilizing the sulfur 3p and 3s electrons, as opposed to the 2p and 2s electrons from oxygen in the two highest MOs in H 2 O. C6. Looking at the two highest MOs of LiOH and NaOH shown in the Figure below, and comparing them to the two highest MOs of water in the Figure above (C5), it is evident that the shape of these MOs are about the same. What is different is the position of these MOs, and their bonding. In water, the HOMO has the character of an atomic p orbital on the oxygen representing the lone pair occupancy. This lone pair occupancy repels the bonding hydrogen atoms and leads to the bent shape. In NaOH and LiOH, this orbital has the same shape and is centered about the oxygen, but is being shared by the oxygen and the other atoms. In LiOH, this is more evident, as lithium only has 3 electrons and thus a smaller atomic radius. Lithium and oxygen are in close enough proximity that there is a clear π-bond between the p orbitals of both these atoms. In NaOH, the significantly larger size of Na doesn t allow for this clear interaction, however, the image of the HOMO of NaOH does show some small occupancy on the particular p atomic orbital of Na capable of π bonding with the p occupancy on the oxygen, however, this is insignificant. The π bonding interaction of the HOMO in both LiOH and NaOH is responsible for the linear shape. The second highest MO of water is also similar in shape to the second highest MO of Li and OH. The bonding, however, is different due to the bent vs linear atomic geometry. All of the 2 nd highest MOs possess p character on the central oxygen. In water, this p character is along the axis of symmetry and interacts with the s orbitals of the hydrogens to provide σ bonding character. In LiOH and NaOH, the p

11 character is parallel to the plane, and thus doesn t provide any bonding interaction in NaOH nonbonding, but provides π-bonding in LiOH due to the close proximity of lithium and oxygen. C7. a. For ethylene, since the HOMO is a π-bonding orbital between the p atomic orbitals of the 2 carbon atoms, the C=C bond should lengthen since the strength of the bond has been reduced, and the planarity should remain as one of the electrons of the π-bond still remains. The radical should be evenly delocalized. b. For formaldimine, the HOMO has occupancy representative of the lone pair of the N, as well as a π-antibonding interaction between the nitrogen and carbon moieties. It is then expected that the H-N-C bond angle will increase due to the reduced electron density where the lone pair was previously situated. Additionally, the N=C double bond should lengthen slightly due to the double bond contributing to stabilizing the radical, resulting in delocalization of the radical. c. For formaldehyde, the HOMO is an antibonding orbital; it also has σ bonding character for the C-H bonds, as well as occupancy for the lone pairs. Prediction of geometric changes for the cation is difficult because of competing effects. Reduction of the bond order from loss of an electron in an antibonding orbital should reduce the C=O bond distance. On the other hand, delocalization of the electron density to stabilize the radicalas there is lone pair occupancy in the HOMO, and a radical is more stable if it is delocalized through conjugation, it is likely that loss of an electron from this MO would be compensated by electrons from the other MOs of the conjugated π-orbitals, thereby delocalizing the radical and stabilizing the overall structure. We see that there is a net bond lengthening upon loss of the electron, but to a lesser extent than for ethylene or foraldimine.

12 Ethylene Formaldimine Formaldehyde D5. The model without the B-B bond provides the better description. The electron density in between the two boron atoms is concave. I have included two representations of the electron density surface to provide the same message. It does not matter how the structure is drawn, as bonds that are drawn by the user are not detected by the software, and the same density is provided. The software is only aware of the relative position of the atoms and will fill the electrons in the most efficient manner (lowest energy). The software also leaves the bonds as they were entered by the user, which is why the image below and to the left shows a bond where there is clearly no bond. D6. The unpaired electron appears to be the most localized to the most delocalized in the following order: Ethynyl, phenyl, vinyl, and then ethyl. The order of the bond dissociation energies parallels the order of spin delocalized in the corresponding radicals as expected. Ethyl Radical Spin Density (from Ethane) Vinyl Radical Spin Density (from Ethylene)

13 Ethynyl Radical Spin Density (from acetylene) Phenyl Radical Spin Density (from benzene) hartrees deprotonated H atom bonded CH Bond Energy radical Bonded - (radical + H atom) C2H C2H C2H C6H D The resulting structure does has some delocalized structure as is evident by the neighboring C-C bonds of Å, however it is no longers delocalized across all the carbon atoms. The beta LUMO of the benzene radical cation is located on the carbon atoms where the C-C bond is Å. Additionally, the second highest alpha HOMO has the same MO orientation/position. This tells us that this is where the radical is delocalized and why the the remaining delocalized structure still exists.

14 E1. Charges are anticipated by atomic electronegativities, as represented by Figures 1 and 2 below. As electronegativity increases, charge increases at a diminishing rate. The first row hydrides have larger charges than their parallel second row hydrides. By observing the halogen hydrides in increasing order of electronegativity (HBr, HCl, then HF), this relationship is made more clear. It shows the changes in charge when the columns of the periodic table are observed in addition to the rows. This is represented in Figure 3 below. Thus, since electronegativity and charge increases as you go up and to the right on the periodic table, atomic electronegativity can be used to anticipate charge. Figure 1. Row 1 Hydrides H charge vs Electronegativity

15 H charge H charge H charge Row 1 Hydrides NH3 H2O HF CH4 H2 BH LiH Electronegativity Figure 2. Row 2 Hydrides: H charge vs Electronegativity Row 2 Hydrides AlH NaH SiH4 PH3 Electronegativity H2S HCl HBR Figure 3. Halogen Hydrides H charge vs Electronegativity HBr HCl HF y = x Electronegativity E2. Without looking at the electrostatic potential map, one may expect the dipole moment of ammonia to be very similar in magnitude, but opposite in direction to trifluoroamine. The reason the magnitudes are expected to be similar in magnitude because the difference in electronegativity between each of the three N-H in NH 3 and three N-F bonds in NF 3 are 0.9 and 1.0 respectively.

16 Additionally, the geometry in both structures is trigonal pyramidal. The direction of the dipole moment of ammonia is expected to be opposite to that of trifluoroamine since N is more electronegative than H, but less electronegative than F. When predicting the electrostatic map of ammonia, one could expect the majority of the electron density to be on the nitrogen as described by drawing the dipole moments as follows: And as expected, we obtain the following result: As mentioned above, we expect the opposite for trifluoroamine since: The result, however, is not as expected.

17 It is apparent from these results that the nitrogen lone pairs are significantly contributing to the electron density of trifluoroamine. A better representation of the dipoles is thus: Where the dipole of the non-bonded electron pair is included in addition to the dipoles from the bonded interactions. Since the dipole of the non-bonded electron pair has opposite direction to that of the N-F dipoles, the result is an evenly distributed electron density.

18 E3. Considering the Lewis definition of an acid, acidity is determined by the ability of a molecule to accept a pair of non-bonded electrons. As shown in the figure below, where red is negative electrostatic potential and blue is positive electrostatic potential, hydrogen fluoride has the most potential to accept an electron pair at the hydrogen, and is thus the most acidic of the three, followed by water, and then ammonia. Electrostatic potential maps of HF, H2O, and NH3

19 E4. The electrostatic potential maps of ethanol, acetic acid, sulfuric acid, and nitric acid (shown below) reveal the known ordering of acid strength in these compounds. The region of strong blue color shows positive electrostatic potential, and thus these regions are good electron pair acceptors. Looking at the electrostatic maps of each compound, it can be seen that the acidic hydrogen increases in positive electrostatic potential with each given compound in the aforementioned order, and thus the compounds are ordered in increasing acidity. Additionally, below you will find an electrostatic potential map of trifluoromethanesulfonic acid, which is a stronger acid than both nitric acid and sulfuric acid. The electrostatic map of trifluoromethanesulfonic acid shows a region of stronger positive electrostatic potential on the acidic hydrogen than both sulfuric and nitric acid.

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