Chapter10

Chapter10 - Chapter 10: The Shapes of Molecules Chem 6A,...

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Unformatted text preview: Chapter 10: The Shapes of Molecules Chem 6A, Section D Nov 10, 2011 1 Chem 6A Michael J. Sailor, UC San Diego 1 Announcements: • Quiz next Thursday (Nov 17) is on Chapter 9. That quiz will also review concepts from chapters 7-8 • Practice Final is posted on the web: http://sailorgroup.ucsd.edu/Chem6A_sailor/Final_Exam_MASTER.pdf • Tues Nov 22 (Thanksgiving week) is review day • No office hours Weds Nov 23 Chem 6A Michael J. Sailor, UC San Diego 2 PROBLEM: Drawing Lewis structures Draw the Lewis structure of NH3 1. Count total valence electrons in the molecule 2. Make bonds to satisfy octet rule (a) Electrons left over? Make lone pairs (b) Too few electrons? Make double (or triple) bonds H Lone pair H N H 3 Chem 6A Michael J. Sailor, UC San Diego PROBLEM: Lewis Structures (see problem 10.5, 10.7, 10.11) Which of the following is the most plausible Lewis structure for the nitrate ion: (a) (b) (c) Chem 6A Michael J. Sailor, UC San Diego (d) (e) 4 Resonance in Lewis Structures (see problem 10.15) Resonance occurs when more than valid Lewis dot structure exists. Draw all the resonance structures for the nitrate ion: Chem 6A Michael J. Sailor, UC San Diego 5 Draw the Lewis Structure for XeF2 Draw the Lewis Structure for BF3 Chem 6A Michael J. Sailor, UC San Diego 6 Violations of the Octet Rule EXCEPTIONS TO THE OCTET RULE Too few electrons/atom: BF3 number of valence electrons: 3(7e-) + 3e- = 24 Too many electrons/atom: XeF2 number of valence electrons: 2(7e-) + 1(8e-) = 22 Chem 6A Michael J. Sailor, UC San Diego 7 Violations of the Octet Rule How can you exceed the octet? Use d-orbitals Some Guidelines for Octet Rule Violations: • Never violate: C, N, O, F • Often have too few electrons: B, Be • Never exceed octet: all 2nd row elements (Li, Be, B, C, N, O, F), because there are no 2d orbitals, and the 3d orbitals are too high in energy). Chem 6A Michael J. Sailor, UC San Diego 8 Pauling Electronegativity Scale Fig. 9.19 9 Chem 6A Michael J. Sailor, UC San Diego Using Electronegativity to Classify Bonds Fig. 9.21 Things that lead to increased covalent character: For anions: Polarizable large, highly negative For cations: Large polarizing power, small, highly positive Examples: Ba-Cl (Δ electronegativity) = 3.0 – 0.9 = 2.1 ionic Bi-I (Δ electronegativity) = 2.5 – 1.9 = 0.6 polar covalent Si-H (Δ electronegativity) = 2.1 – 1.8 = 0.3 covalent O-H (Δ electronegativity) = 3.5 – 2.1 = 1.4 polar covalent Chem 6A Michael J. Sailor, UC San Diego 10 Formal Charges (see problem 10.15) •Count number of electrons around each atom and compare it to the number of valence electrons for the atom. If the numbers are different, you must assign formal charges •Molecules generally try to minimize the number and magnitude of formal charges. The most electronegative atoms will hold negative formal charges. 11 Chem 6A Michael J. Sailor, UC San Diego Formal Charges (see problem 10.15) Example: Which is the best Lewis structure for BF3? F B F F B F F F F is too electronegative to hold a positive formal charge. Chem 6A Michael J. Sailor, UC San Diego 12 Formal Charges (see problem 10.15) Example: Which is the best Lewis structure for sulfate? 2- O O S O 2- O O S O O O O 2- O S O O Molecules try to minimize the number and size of formal charges. The most electronegative atoms will hold negative formal charges. 13 Chem 6A Michael J. Sailor, UC San Diego PROBLEM: Formal Charges (see problem 10.15) Which Lewis structure of chlorate shown below has the most stable distribution of formal charges? (a) (b) (c) (d) (e) For all Lewis structures, you must consider: (1) Number of electrons (2) Octet rule (3) Distribution of formal charges Chem 6A Michael J. Sailor, UC San Diego 14 Valence Shell Electron Pair Repulsion (VSEPR) Valence Shell Electron Pair Repulsion Model Used to predict structures of main group compounds. The main concept is that electrons in bonds and lone electron pairs repel each other and will arrange themselves around the molecule to maximize their distance apart. Chem 6A Michael J. Sailor, UC San Diego 15 Valence Shell Electron Pair Repulsion (VSEPR) Example: What is the shape of BF3? 1.Draw Lewis dot structure, figure out how many lone pairs + bonding pairs you have (the coordination number). 2.Use idealized structures for number of atoms + lone pairs 3.Give lone pairs more room Chem 6A Michael J. Sailor, UC San Diego 16 18 17 Chem 6A Michael J. Sailor, UC San Diego PROBLEM: Drawing VSEPR structures from Lewis structures Recall the Lewis structure of NF3: F Lone pairs F Total electrons: 26 Total pairs of electrons: 13 Total bonding pairs: 3 Total lone pairs: 10 N F Chem 6A Michael J. Sailor, UC San Diego 18 PROBLEM: Drawing VSEPR structures from Lewis structures the Lewis structure of NH3: H Lone pair H N H What is the structure? 19 Chem 6A Michael J. Sailor, UC San Diego Valence Shell Electron Pair Repulsion (VSEPR) Structure of NH3: N H H H N H 109.5° IDEAL H H 107.3° ACTUAL Chem 6A Michael J. Sailor, UC San Diego 20 Valence Shell Electron Pair Repulsion (VSEPR) Why give lone pairs more room? Lone pairs require more space than bonding pairs. Example: H2O Draw the VSEPR structure of the H2O molecule. Is the H-O-H angle greater than or less than 109.5°, and why? H O H Chem 6A Michael J. Sailor, UC San Diego 21 Valence Shell Electron Pair Repulsion (VSEPR) ANSWER: H 104.5° O H H-O-H angle is < 109.5 because the lone pair electrons require more space than bonding pairs Chem 6A Michael J. Sailor, UC San Diego 22 30 23 Chem 6A Michael J. Sailor, UC San Diego PROBLEM: Molecular Shape and Polarity (VSEPR) Is ClF3 polar? Draw the VSEPR diagram and show the direction of the molecular dipole. Atom Cl 3xF Total: Valence electrons F 7 3x7=21 Cl 28 F F T-shaped Chem 6A Michael J. Sailor, UC San Diego 24 PROBLEM: Drawing VSEPR structures Draw the VSEPR structures for XeF3+, SbBr3, and GaCl3. Indicate the dipole moment of each molecule (if it has one) with a dipolar arrow. 25 Chem 6A Michael J. Sailor, UC San Diego PROBLEM: Drawing VSEPR structures Draw the VSEPR structures for XeF3+, SbBr3, and GaCl3. Indicate the dipole moment of each molecule (if it has one) with a dipolar arrow. Atom Xe 3xF Valence electrons + F 8 3x7=21 charge Total: Xe Less 1 electron 28 F F T-shaped Chem 6A Michael J. Sailor, UC San Diego 26 PROBLEM: Drawing VSEPR structures Draw the VSEPR structures for XeF3+, SbBr3, and GaCl3. Indicate the dipole moment of each molecule (if it has one) with a dipolar arrow. Atom Sb 3 x Br Total: Valence electrons 5 3x7=21 Sb Br 26 Br Br Trigonal pyramidal 27 Chem 6A Michael J. Sailor, UC San Diego PROBLEM: Drawing VSEPR structures Draw the VSEPR structures for XeF3+, SbBr3, and GaCl3. Indicate the dipole moment of each molecule (if it has one) with a dipolar arrow. Atom Ga 3 x Cl Total: Valence electrons Cl 3 3x7=21 Not polar 24 Ga Cl Cl Trigonal planar Chem 6A Michael J. Sailor, UC San Diego 28 The Energy Spectrum Chem 6A Michael J. Sailor, UC San Diego 29 The Solar Spectrum Chem 6A Michael J. Sailor, UC San Diego 30 Environmental Chemistry 1995 NOBEL PRIZE in CHEMISTRY: "for their work in atmospheric chemistry, particularly concerning the formation and decomposition of UCSD ozone". http://nobelprize.org/nobel_prizes/chemistry/laureates/1995/ 31 Chem 6A Michael J. Sailor, UC San Diego Upper atmosphere ozone depletion Crutzen, Molina, and Rowland won the Nobel prize for identifying the important man-made sources of chemicals that can deplete ozone, and for realizing that those chemicals are removing ozone faster than Nature can supply it. Cl O F O ozone O O C N O F F Cl Chem 6A Michael J. Sailor, UC San Diego N O 32 Final Exam Final Exam: Fri Dec 9, 7 pm Bring: (1) Calculator (no iPhones, PDAs, anything that can transmit or receive RF) (2) #2 Pencils (3) Binder or other hard surface to write on (4) 5x7 note card (your handwritten notes can be on both sides) (5) Your UCSD ID card 33 Chem 6A Michael J. Sailor, UC San Diego The Ozone Layer Ozone (O3) found between 10 and 30 km above earth's surface. (a 747 cruises at an altitude of ~10 km) Ozone is depleted by reaction with light: O3 + UV light (λ< 320 nm) → O2 + O Ozone is also regenerated by reaction with light: O2 + UV light (λ< 240 nm) → 2 O O2 + O + M → O3 + M (M is some other hem 6A moleculeUC San2Diego O2) air Michael J. Sailor, (N or C Image: www.nasa.gov 44 34 Ozone absorbs UV-C light O3 + UV light (λ< 320 nm) → O2 + O Molina’s UV-C absorption spectrum of ozone (from “Chemistry of the Environment,” R.A. Bailey) Chem 6A Michael J. Sailor, UC San Diego 35 Upper atmosphere ozone depletion by chlorofluorocarbons 1974, Mario Molina and Sherwood Rowland identified the threat to the ozone layer from chlorofluorocarbon (CFC) gases - "freons" - from spray bottles, refrigerators, air conditioners, and plastic foam manufacture. Based on 2 prior observations: •James Lovelock (England) showed that man-made, chemically inert, CFC gases had spread globally throughout the atmosphere. •Richard Stolarski and Ralph Cicerone (USA): free chlorine atoms in air decompose ozone catalytically. Chem 6A Michael J. Sailor, UC San Diego 36 How do chlorofluorocarbons deplete ozone? Initiation: Cl + UV light C F F F + C F Cl F F Freon-13 Propagation: Cl. + O3 → O2 + OCl. OCl. + O → O2 + Cl. Net: O3 + O → 2O2 Catalytic reaction: Cl. is not used up After initiation, no more UV light is needed Chem 6A Michael J. Sailor, UC San Diego 37 Nitrogen oxides also deplete ozone by a catalytic reaction Paul Crutzen (1970) showed that the nitrogen oxides NO and NO2 react catalytically with ozone: Propagation: NO + O3 → NO2 + O2 NO2 + O → NO + O2 Net: O3 + O → 2O2 Catalytic reaction: NO is not used up No initiation step Chem 6A Michael J. Sailor, UC San Diego 38 Upper atmosphere ozone depletion • 1957 Team of British scientists led by Joseph Farman began measuring Ozone at Halley Bay, Antarctica. • They noticed a drop in Ozone level beginning in 1977, correlated with the appearance of CFCl3 and CF2Cl2 • NASA, using NIMBUS satellites with more sensitive instruments, did not report a similar drop. 39 Chem 6A Michael J. Sailor, UC San Diego Farman’s Halley Bay, Antarctica data • Drop in ozone level correlates with increase in appearance of CFCl3 and CF2Cl2 Chem 6A Michael J. Sailor, UC San Diego Increasing CFCl3 and CF2Cl2 • Ozone at Halley Bay, Antarctica over 26 years. 40 Upper atmosphere ozone depletion Nature 315, 207 - 210 (16 May 1985) Large losses of total ozone in Antarctica reveal seasonal ClOx/ NOx interaction J. C. FARMAN, B. G. GARDINER & J. D. SHANKLIN “…We suggest that the very low temperatures which prevail from midwinter until several weeks after the spring equinox make the Antarctic stratosphere uniquely sensitive to growth of inorganic chlorine, ClX, primarily by the effect of this growth on the NO2/NO ratio. This, with the height distribution of UV irradiation peculiar to the polar stratosphere, could account for the O3 losses observed.” 41 Chem 6A Michael J. Sailor, UC San Diego http://toms.gsfc.nasa.gov The hole in the ozone layer Chem 6A Michael J. Sailor, UC San Diego 42 Chlorofluorocarbons (CFCs) release chlorine atoms that deplete ozone CFCs have been used as refrigerants 75% of atmospheric ClO comes from CFCs Montreal Protocol: Phaseout of CFC production and use: December 31, 1995 marked the end of the production of CFCs in the industrial world. 43 Chem 6A Michael J. Sailor, UC San Diego Chlorine in stratosphere Montreal Protocol of 1987 Predicted change in chlorine content in the stratosphere for three different scenarios: a) Without restrictions on release b) Limitations according to the original Montreal Protocol of 1987 c) Release limitations now internationally agreed. Chlorine content is a measure of the magnitude of ozone depletion. Chem 6A Michael J. Sailor, UC San Diego 44 Results of the ban on CFCs “CFCs are no longer accumulating in the atmosphere at an accelerating rate” “Currently, we are experiencing depletion of approximately 3 percent at Northern Hemisphere mid-latitudes and 6 percent at Southern Hemisphere mid-latitudes…if no action had been taken to limit CFCs, ozone depletion at midlatitudes would eventually have reached 20 percent or more…If international agreements are adhered to, the ozone layer is expected to recover around 2050.” –US EPA, http://www.epa.gov/ozone/geninfo/benefits.html 45 Chem 6A Michael J. Sailor, UC San Diego What to use in place of chlorofluorocarbons (CFCs)? hydrochlorofluorocarbons (HCFCs) hydrofluorocarbons (HFCs) HCFCs and HFCs are more chemically reactive than CFCs and decompose in the troposphere before reaching the stratosphere. H C H H H Cl C F C H F HCFC-142b Chem 6A Michael J. Sailor, UC San Diego F C F F F HFC-134a 46 Why are hydrofluorocarbons more reactive than chlorofluorocarbons? Bond Energy kJ/mol C-C C-H 347 413 C-F 453 C-Cl 339 O-Cl O-H 203 467 O-F 190 HCFC-142b: 3(C-H)-3(OH) 3 x 413 – 3 x 467 = -162 kJ/mol Freon-113: 2(C-Cl) + 1(C-F) – 2(O-Cl) – 1(O-F) 2 x 339 + 453 – 2(203) – 190 = +535 H Cl C H H C Cl F F HCFC-142b Chem 6A Michael J. Sailor, UC San Diego Cl C Cl C F F F Freon-113 47 Extras Chem 6A Michael J. Sailor, UC San Diego 48 ...
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This note was uploaded on 11/16/2011 for the course CHEM 6A taught by Professor Pomeroy during the Spring '08 term at UCSD.

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