Chapter 1 - Instructor Supplemental Solutions to Problems...

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Instructor Supplemental Solutions to Problems Marc Loudon Joseph G. Stowell to accompany Organic Chemistry 5th Edition This manual provides the solutions to the problems that are not provided in the Study Guide and Solutions Manual. These answers are provided as electronic files in Portable Document Format (PDF). Each chapter is provided as a separate file. For the conventions used in these solutions, see the Preface of the Study Guide and Solutions Manual. Permission is given to adopting instructors to post these answers only in electronic form for the benefit of their students. Distribution in print form or resale is a violation of copyright. Copyright © 2010 ROBERTS & COMPANY PUBLISHERS Greenwood Village, Colorado
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© 2010 Roberts and Company Publishers Chapter 1 Chemical Bonding and Chemical Structure Solutions to In-Text Problems 1.1 (b) The neutral calcium atom has a number of valence electrons equal to its group number, that is, 2. (d) Neutral Br, being in Group 7A, has 7 valence electrons; therefore, Br + has 6. 1.2 (b) The positive ion isoelectronic with neon must have 10 electrons and 11 protons, and therefore must have an atomic number = 11. This is the sodium ion, Na + . (d) Because Ne has atomic number = 10 and F has atomic number = 9, the neon species that has 9 electrons is Ne + . 1.3 (b) (d) 1.5 The structure of acetonitrile: 1.6 (b) The overall charge is –2. 1.8 (b) Formal charge does not give an accurate picture, because O is more electronegative than H; most of the positive charge is actually on the hydrogens. (d) An analysis of relative electronegativities would suggest that, because C is slightly more electronegative than H, a significant amount of the positive charge resides on the hydrogens. However, carbon does not have its full complement of valence electrons—that is, it is short of the octet by 2 electrons. In fact, both C and H share the positive charge about equally. 1.9 The bond dipole for dimethylmagnesium should indicate that C is at the negative end of the C—Mg bond, because carbon is more electronegative than magnesium. 1.10 (a) Water has bent geometry; that is, the H—O—H bond angle is approximately tetrahedral. Repulsion between the lone pairs and the bonds reduces this bond angle somewhat. (The actual bond angle is 104.5 ° .)
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INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 1 2 (c) The formaldehyde molecule has trigonal planar geometry. Thus, both the H—C—H bond angle and the H— C A O bond angle are about 120 ° . 1.11 (a) Bond angles: aa, ab, bc, bd, cd, de, df, and ef are all about 120 ° , because all are centered on atoms with trigonal planar geometry; fg is predicted to have the tetrahedral value of 109.5. The bond lengths increase in the order a ± g < e < b < d ± f < c (In Chapter 4, you’ll learn that C—H bonds attached to carbons of double bonds are shorter than C—H bonds attached to carbons of single bonds. For this reason, a < g .) 1.13 (a) Because the carbon has trigonal-planar geometry, the H—C
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This note was uploaded on 02/13/2010 for the course EGR 012 taught by Professor Witfield during the Spring '10 term at Aarhus Universitet.

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Chapter 1 - Instructor Supplemental Solutions to Problems...

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