Name Section Rec TA Ch 1a, Problem Set Seven Due Friday, Nov. 14, 2014 at 4 PM in the Ch1a Box Side 1 of 4 Problems 3 and 6 are designated P as no collaboration problems. 1. (50 points) Consider the neutral, anion (–1), and cation (+1) species for each of the following homonuclear diatomics. Using the general MO diagram given for homonuclear diatomics, list the orbital configuration (e.g., 1sσb2 1sσ*2, …, 2px,yπb4 2pσ 2 b …) and bond order (e.g., 0.5, 1.5, 2, etc.) for each of the following molecules and ions. Also label which species are paramagnetic and which are diamagnetic. (For Li and higher, you may omit MOs formed from the 1s orbitals) Use a table for your responses (species, config, bond order, P/D). The orbitals in your configuration must be in the proper filling order. a. Hydrogen (i.e. H+2 , H2 , H−2 ) e. Nitrogen (…) b. Helium (i.e. He+2 , He2 , He−2 ) f. Oxygen (…) g. Fluorine (…) c. Beryllium (…) € d. Carbon (…) € For oxygen and fluorine, remember to put the 2pσb below the 2pxyπb MO. 2p 2pσ* 2pxyπ* 2p 2pσb 2pxyπb 2s 2sσ * 2s 2sσ b 1s 1sσ * 1sσ b 1s Name Section Rec TA 2. Ch 1a, Problem Set Seven Due Friday, Nov. 14, 2014 at 4 PM in the Ch1a Box Side 2 of 4 (40 points) Consider the series of homonuclear dihalogens: F2, Cl2, Br2, I2. Molecule Bond Energy1 eV Bond Length1 pm F2 1.610 141.2 Cl2 Br2 I2 2.488 1.972 1.540 198.8 228.1 266.6 1 2 Ionization Potential2 eV data not provided 11.6 10.56 9.4 Electron Affinity2 eV data not provided 2.4 2.6 2.6 Color (phase) nearly colorless (g) (absorbs UV) green/yellow (g) red/brown (ℓ) violet (s) A. Haaland, Molecules and Models, Oxford University Press (2008), p. 58. G. Wulfsberg, Inorganic Chemistry, University Science Books (2000), p. 521. a) (7 pts) Explain the trends in the data for bond length of the series of dihalogens. Support your explanation with quantitative evidence – use atomic radii to predict bond lengths. Atomic radii values can be found in Gray. b) (10 pts) Explain the trends in the data for bond energy of the series of dihalogens. Also explain why the bond energy of F2 breaks the general trend. Hint: Something was slightly off for F2 in part a, and this may help you explain what’s going on. c) (15 pts) Identify and sketch the shape of the Lowest Unoccupied Molecular Orbital (LUMO). Label the node(s). To a first approximation, homonuclear dihalogens all have the same shape LUMO. Why? d) (8 pts) When a molecule absorbs energy with the correct wavelength, an electron will move from the HOMO to the LUMO. This HOMO-LUMO gap determines the color of homonuclear dihalogens. Use the color data in the table to qualitatively describe the trend for the HOMO-LUMO gap (energy difference and what wavelength that difference corresponds to) going down the column. Remember, the perceived color is the complement of the color absorbed. Page 836 of OGC has a depiction of the color wheel. P 3. (20 points) Consider the molecules NO and N2. a) (14 points) Draw populated molecular orbital diagrams for each molecule using proper relative orbital energies. Use the table of valence orbital energies on the next page. The electron configuration of NO is available on page 100 of Gray. (Note that πb,2px,2py are lower in energy than σb,2pz.) b) (6 points) Which molecule should have the higher ionization energy? Guidelines for your diagrams: • Label all AOs and MOs with the proper designation (1s, 1σb, etc.) • Show which AOs mix to form each MO. • Fill in the electrons on both the AOs and the MOs • Relative orbital energies are important! • Basically, it should look similar in style to the diagram on the first page Name Section Rec TA Ch 1a, Problem Set Seven Due Friday, Nov. 14, 2014 at 4 PM in the Ch1a Box Side 3 of 4 4. (40 points) Carbon monoxide forms numerous complexes as a ligand to transition metals. Now that you have learned about molecular orbital (M.O.) theory, we can look at CO from that perspective and compare it to the Lewis structure. a. (8 pts) Construct an M.O. diagram of CO and use it to ascertain the bond order. When combining the p orbitals, the πb molecular orbital has a lower energy than the σb molecular orbital. b. (8 pts) Draw the Lewis dot structure of CO and assign formal charges based on the bond order from part a. In terms of electronegativity, why is this unusual? c. (8 pts) Based on your answer to part b, would positive metal ions bind to the lone pairs of carbon (Mn+-CO) or the lone pairs of oxygen (Mn+-OC)? Explain. Although CO has “lone pairs” in the Lewis sense, the more rigorous molecular orbital theory assigns it only bonding and antibonding orbitals. These Lewis “lone pairs” are in fact molecular orbitals formed from the 2s atomic orbitals from carbon and oxygen. Due to their different energies, these 2s atomic orbitals combine poorly. The lowest σb retains most of the character of one of the 2s atomic orbitals, while the lowest σ* retains most of the character of the other 2s atomic orbital. In this sense, these are the molecular orbitals that correspond to the Lewis “lone pairs”. d. (8 pts) Using the table of valence orbital energies, predict which atom donates the majority of its character to the σb orbital and which atom contributes to the σ*. Be sure to explain your prediction! e. (8 pts) CO forms a bond to the metal by donating “lone pair” electrons from the MOs arising from the 2s orbitals of oxygen and carbon into empty d orbitals of the metal. Energetically, would you expect this bonding arrangement to be metalCO or metal-OC? Explain. Compare this to the bonding arrangement (Mn+-CO, or Mn+-OC) acquired with the Lewis model from part c. (Remember, the energy levels are given in the table below!) This is a table of valence orbital energies (given in eV). Use this for relative energies in your MO diagrams, but only as a qualitative guide to the relative energies: you do not need to draw your molecular orbital diagrams to scale. Only parts 4d and 4e require you to quantitatively compare energies. The electron configuration of CO is given in Gray on page 100. Atom H He Li Be B C N O F Ne metals 1s -13.6 (eV) -24.6 2s 2p -5.4 -9.3 -14 -19.4 -25.6 -32.3 -40.2 -48.5 -8.3 -10.6 -13.2 -15.8 -18.6 -21.6 3d about -15 Name Section Rec TA Side 4 of 4 Ch 1a, Problem Set Seven Due Friday, Nov. 14, 2014 at 4 PM in the Ch1a Box 5. (30 points, 5 pts each) The hydroxyl radical, OH•, is important not only in atmospheric chemistry but also in cell biology, as it participates in biochemical reactions as a free radical. (For an atmospheric chemistry example, check out Professor Mitchio Okumura’s work on OH• + NO2•) a. Formulate the electronic structure of OH• in terms of M.O. theory. Use only the oxygen 2p and hydrogen 1s atomic orbitals. b. Which type of M.O. (σb, σ*, πb, π*, nonbonding) contains the unpaired electron? c. Why don’t the other oxygen and hydrogen atomic orbitals play a role in the M.O. diagram for OH•? d. Is the orbital you specified in part (b.) associated with both constituent atoms or is it localized on a single atom (if so, which one)? e. Based on your M.O. diagram what is the bond order of OH•? f. Compare your answers in (d.) and (e.) to the Lewis dot structure of OH•. P 6. (20 points) Problem Five teaches us that when atomic orbitals combine poorly the new molecular orbitals retain much of their original atomic orbital character. That is to say, these molecular orbitals closely resemble Lewis lone pairs. a. (10 pts) Consider fluorine as it forms bonds in a Lewis dot structure. Usually fluorine is written as having one bond and three lone pairs. Would you expect any of the molecular orbitals in F2 to behave as lone pairs as written in a Lewis dot structure? Explain why. b. (10 pts) Repeat this treatment for HF. Refer to the table of valence energies on the previous page (and maybe even look in one of the textbooks for an MO diagram for HF). Explain the possibility that any of the molecular orbitals in HF behave like fluorine lone pairs in the Lewis dot structure.