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Unformatted text preview: Chemistry 1B Spring 2011 Unit 2, Problem Set 4 Answer Key Chapter 6 34. Suppose we remove three electrons from CO2 to create the ion CO23+. Is the dissociation energy of the resulting ion larger or smaller than that of CO2? How will the C‐O bonds change? Explain your answers. Electron configuration of CO2 (see figure 6.28): (2sA)2(2sB)2(σs)2(σp)2(πx,y)4(πnbx,y)4 Removing three electrons in this configuration would lead to the nonbonding orbital only having one electron. Since the nonbonding orbital does not participate in the bonding of the molecule, the C‐O bonds will not change and there will be little effect on the dissociation energy. 52. Draw a Lewis electron dot diagram for each of the following molecules and ions. Formulate the hybridization for the central atom in each case and give the molecular geometry sp2 Trigonal planar sp3 Tetrahedral sp3 Trigonal pyramidal sp Linear sp2 Trigonal planar 54. Describe the hybrid orbitals on the chlorine atom in the ClO4‐ and ClO3‐ molecular ions. Sketch the expected geometries of these ions. ClO4‐ sp hybridized Tetrahedral geometry (steric number =4) 3 ClO3‐ sp hybridized Trigonal pyramidal geometry (steric number =4) 3 62. Discuss the nature of the bonding in the nitrate ion (NO3‐). Draw the possible Lewis resonance diagrams for this ion. Use the VSEPR theory to determine the steric number, the hybridization of the central N atom, and the geometry of the ion. Show how the use of resonance structures can be avoided by introducing a de‐localized π MO. What bond order is predicted by the MO model for the N‐O bonds in the nitrate ion? steric number = 3 N atom undergoes sp2 hybridization (one p orbital is left over for π bonding, i.e. formation of a double bond) Geometry: trigonal planar Similar to what is done with benzene’s resonance structure (see figure 7.xx), four molecular orbitals can be formed from the unhybridized, “left over” N pz orbital and the three O pz orbitals: For the bond order of the nitrate anion, there are three “connections” formed (three N‐O linkages) using four bonds. Thus, the bond order is 4/3. 64. Calcium carbide (CaC2) is an intermediate in the manufacturing of acetylene (C2H2). It is the calcium salt of the carbide (also called acetylide) ion (C22‐). What is the electron configuration of this molecular ion? What is its bond order? Carbide resembles the molecular orbital diagram for C2 but with two extra electrons in the molecular orbitals (10 valence electrons exist in this ion). The molecular orbitals formed from the interaction of each carbon atom’s 2s orbitals are filled, so only the molecular orbitals formed from each atom’s set of 2 p orbitals is shown below. Therefore, the electron configuration is (1σg)2 (2σu*)2 (1πu)4 (3σg)2 and its bond order is 3. Chapter 7 12. Name the following hydrocarbons: a. 2,3‐dimethyl‐1,3‐butadiene b. 2,4‐hexadiene c. 2,2‐dimethylbutane d. methylpropene 14. State the hybridization of each of the carbon atoms in the hydrocarbon structures in problem 12. a. All sp2 except for two sp3 methyl (CH3) groups b. All are sp2, except for the two sp3 methyl (CH3) groups c. All sp3 d. The methyl carbons (CH3) are sp3, all others are sp2 15. To satisfy the octet rule, fullerenes must have double bonds. How many? Give a simple rule for one way of placing them in the structure shown in Figure 7.20a. Fullerenes are characterized by the hexagonal and pentagonal structure formed by the carbon atoms that make up the molecule. See figure 7.20 in your textbook for the structure. Double bonds exist on the bonds shared by two hexagonal faces (30 double bonds total). 16. It has been suggested that a compound of formula C12B24N24 might exist and have a structure similar to that of C60 (Buckminster fullerene). a. Explain the logic of this suggestion by comparing the number of valence electrons in C60 and C12B24N24. There are (60 x 4) = 240 valence electrons in C60 (each carbon atom has four valence electrons). There are (12 x 4) + (24 x 3) + (24 x 5) = 240 valence electrons in C12B24N24 (each nitrogen atom has five valence electrons, each boron atom has three valence electrons). Since the structures have equal numbers of valence electrons, the carbon/boron/nitrogen structure could exist. b. Propose the most symmetric pattern of carbon, boron, and nitrogen atoms in C12B24N24 to occupy the 60 atom sites in the buckminsterfullerene structure. Where could the double bonds be placed in such a structure? From the molecular composition, we know that there is one carbon atom for every two boron atoms and every two nitrogen atoms. We also know that each boron atom has three valence electrons, each carbon atom has four valence electrons, and each nitrogen atom has five valence electrons. Starting from the fullerene structure given in problem 15, we can arrange the atoms such that the structure is maintained – each atom makes itself available for three sigma bonds – and some extra electrons can form double bonds. A structure of the molecule is located below (ignore the alpha and beta notation). Half of the B‐N bonds are double bonds. Structure obtained from K. Kobayashi and N. Kurita, Bonding and Electronic Properties of a Multicomponent Fullerene C12B24N24 by a Non‐Local‐Density‐Functional Calculation, Physical Review Letters 70, 3542 (1993). 46. For each of the following molecules, construct the π MOs from the 2pz atomic orbitals perpendicular to the plane of the carbon atoms. Indicate which, if any, of these orbitals have identical energies from symmetry considerations. Show the number of electrons occupying each π MO in the ground state, and indicate whether either or both of the molecules are paramagnetic. Hint: refer to figures 7.17 and 7.18. a. Cyclobutadiene The pz atomic orbitals on each carbon atom can add constructively or destructively to make four molecular orbitals (since there are a total of four atomic pz orbitals being used to make molecular orbitals). The lowest energy molecular orbital will be the one where all of the orbitals add constructively – the orbital will only have one node. (Note: higher energy molecular orbitals will have an increasing number of nodes.) In this diagram, the atomic nuclei are represented as grey dots. Atomic orbitals constructively interfere
to produce molecular orbitals Applying the same formalism to obtain four molecular orbitals from the four atomic orbitals, we obtain the following MO diagram: The compound has two unpaired electrons so it is paramagnetic. b. Allyl radical The compound has one unpaired electron so it is paramagnetic. ...
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This note was uploaded on 09/10/2011 for the course CHEM 1B taught by Professor Pines during the Spring '08 term at Berkeley.
- Spring '08