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Unformatted text preview: P1: PBU/OVY JWCL234-01 P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Bind Rite JWCL234-Solomons-v1 1 December 8, 2009 THE BASICS: BONDING AND MOLECULAR STRUCTURE SOLUTIONS TO PROBLEMS Another Approach to Writing Lewis Structures When we write Lewis structures using this method, we assemble the molecule or ion from the constituent atoms showing only the valence electrons (i.e., the electrons of the outermost shell). By having the atoms share electrons, we try to give each atom the electronic structure of a noble gas. For example, we give hydrogen atoms two electrons because this gives them the structure of helium. We give carbon, nitrogen, oxygen, and fluorine atoms eight electrons because this gives them the electronic structure of neon. The number of valence electrons of an atom can be obtained from the periodic table because it is equal to the group number of the atom. Carbon, for example, is in Group IVA and has four valence electrons; fluorine, in Group VIIA, has seven; hydrogen, in Group 1A, has one. As an illustration, let us write the Lewis structure for CH3 F. In the example below, we will at first show a hydrogen’s electron as x, carbon’s electrons as o’s, and fluorine’s electrons as dots. Example A 3 H , C , and F are assembled as H H H C F or H C F H H If the structure is an ion, we add or subtract electrons to give it the proper charge. As an example, consider the chlorate ion, ClO3 − . Example B Cl , and O and an extra electron × are assembled as Ο Ο Cl Ο − or Ο Ο Cl Ο − 1 CONFIRMING PAGES 20:10 QC: PBU/OVY T1: PBU Printer: Bind Rite JWCL234-Solomons-v1 THE BASICS: BONDING AND MOLECULAR STRUCTURE 1.1 14 N, 7 protons and 7 neutrons; 15 N, 7 protons and 8 neutrons 1.2 (a) one (b) seven 1.3 (a) covalent (c) four (b) ionic (d) three (c) covalent (e) eight (f ) five (d) covalent •• O 1.4 − O P O O − •• •• 2 December 8, 2009 •• JWCL234-01 P2: PBU/OVY − P1: PBU/OVY O •• •• •• •• (d) H O •• N O (g) H • F •• •• • 1.5 (a) H •• O •• • • •• (b) •• F •• •• (e) H F •• •• O O •• (f ) F •• •• C H H H (h) H •• O •• O •• H H O C •• O •• H – H H (c) H S O •• •• • • O •• P B H H O H 1.6 (a) H C O − (c) − C C (e) H O (f ) H N C O − H O (b) H N H − (d) H C O C − − CONFIRMING PAGES 20:10 P1: PBU/OVY JWCL234-01 P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Bind Rite JWCL234-Solomons-v1 December 8, 2009 THE BASICS: BONDING AND MOLECULAR STRUCTURE H 1.7 (a) H C+ C (d) H C H H − O H H H O H H (c) H O (f ) H C O H N H + C O + (h) H H C + N N H H H H O C − C H O N − O 1.8 (a) H C H H C (e) H C H H + (g) H H H (b) H 3 − O (b) and (c). Since the two resonance structures are equivalent, each should make an equal contribution to the overall hybrid. The C—O bonds should therefore be of equal length (they should be of bond order 1.5), and each oxygen atom should bear a 0.5 negative charge. Ο O 1.9 (a) H C H H O (b) H − C C C + − H O H C H − C H H H H (c) H + C N H H H H H H − H N H H (d) C + C C C N C N − H CONFIRMING PAGES 20:10 P1: PBU/OVY JWCL234-01 P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Bind Rite JWCL234-Solomons-v1 4 December 8, 2009 THE BASICS: BONDING AND MOLECULAR STRUCTURE 1.10 (a) CH3CH CH + CH OH CH3CH (b) CH2 CH CH + CH δ+ CH CH + CH2 CH2 δ+ CH CH2 CH OH CH CH δ+ OH CH2 δ+ OH CH CH3CH CH CH + δ+ CH3CH + CH CH CH CH CH CH2 + CH2 δ+ CH CH2 + (c) + + δ+ δ+ (d) CH2 CH δ+ − Br δ− CH2 CH Br + δ+ Br CH CH2+ CH2 + CH2+ CH2 CH2 CH2 (e) + δ+ δ+ (f ) CH3 δ− H2C (g) CH3 S − C C − H2C CH2 δ+ O O + δ+ O δ− CH3 C CH3 CH2+ CH3 H2C CH3 + S CH2 δ+ δ+ S CH2 CONFIRMING PAGES 20:10 JWCL234-01 P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Bind Rite JWCL234-Solomons-v1 December 8, 2009 THE BASICS: BONDING AND MOLECULAR STRUCTURE (h) CH3 O + N O O + CH3 N − − CH3 O CH3 + O 2+ O − O 5 − N (minor) δ− N − Oδ 1.11 (a) CH2 N(CH3)2 because all atoms have a complete octet (rule 3), and there are more covalent bonds (rule 1). + •• •• P1: PBU/OVY O (b) CH3 because it has no charge separation (rule 2). C O (c) NH2 C H N because it has no charge separation (rule 2). 1.12 (a) In its ground state, the valence electrons of carbon might be disposed as shown in the following figure. The electronic configuration of a ground state carbon atom: The p orbitals are designated 2 px , 2 p y , and 2 pz to indicate their respective orientations along the x , y , and z axes. The 2s 2py assignment of the unpaired electrons to the 2 p y and 2 px orbitals is arbitrary. They could also have been placed in the 2 px C 2px and 2 pz or 2 p y and 2 pz orbitals. (To have placed them both 2pz in the same orbital would not have been correct, however, for this would have violated Hund’s rule.) (Section 1.10A) The formation of the covalent bonds of methane from individual atoms requires that the carbon atom overlap its orbitals containing single electrons with 1s orbitals of hydrogen atoms (which also contain a single electron). If a ground state carbon atom were to combine with hydrogen atoms in this way, the result would be that depicted below. Only two carbon-hydrogen bonds would be formed, and these would be at right angles to each other. The hypothetical formation of CH2 from a carbon atom in its ground state: C +2 H H C H CONFIRMING PAGES 20:10 P1: PBU/OVY JWCL234-01 P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Bind Rite JWCL234-Solomons-v1 6 December 8, 2009 THE BASICS: BONDING AND MOLECULAR STRUCTURE (b) An excited-state carbon atom might combine with four hydrogen atoms as shown in the figure above. The promotion of an electron from the 2s orbital to the 2 pz orbital requires energy. The amount of energy required has been determined and is equal to 400 kJ mol−1 . This expenditure of energy can be rationalized by arguing that the energy released when two additional covalent bonds form would more than compensate for that required to excite the electron. No doubt this is true, but it solves only one problem. The problems that cannot be solved by using an excited-state carbon as a basis for a model of methane are the problems of the carbon-hydrogen bond angles and the apparent equivalence of all four carbon-hydrogen bonds. Three of the hydrogens—those overlapping their 1s orbitals with the three p orbitals—would, in this model, be at angles of 90◦ with respect to each other; the fourth hydrogen, the one overlapping its 1s orbital with the 2s orbital of carbon, would be at some other angle, probably as far from the other bonds as the confines of the molecule would allow. Basing our model of methane on this excited state of carbon gives us a carbon that is tetravalent but one that is not tetrahedral, and it predicts a structure for methane in which one carbon-hydrogen bond differs from the other three. The hypothetical formation of CH4 from an excited-state carbon atom: C +4 H H H C H H 1.13 (a) Cis-trans isomers are not possible. CH3 (b) CH3 C and C H H CH3 C C CH3 H H (c) Cis-trans isomers are not possible. Cl (d) CH3CH2 C H and C H H CH3CH2 C H C Cl CONFIRMING PAGES 20:10 P1: PBU/OVY JWCL234-01 P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Bind Rite JWCL234-Solomons-v1 December 8, 2009 THE BASICS: BONDING AND MOLECULAR STRUCTURE 7 1.14 sp 3 1.15 sp 3 1.16 sp 2 1.17 sp 1.18 (a) − H H H (b) There are four bonding pairs. The geometry is tetrahedral. B H Be F (c) There are two bonding pairs about the central atom. The geometry is linear. F + H There are four bonding pairs. The geometry is tetrahedral. N H H H (d) There are two bonding pairs and two nonbonding pairs. The geometry is angular. S H H H (e) There are three bonding pairs. The geometry is trigonal planar. B H H F (f) There are four bonding pairs around the central atom. The geometry is tetrahedral. C F F F F (g) Si F F F There are four bonding pairs around the central atom. The geometry is tetrahedral. − (h) Cl C Cl Cl There are three bonding pairs and one nonbonding pair around the central atom. The geometry is trigonal pyramidal. CONFIRMING PAGES 20:10 P1: PBU/OVY JWCL234-01 P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Bind Rite JWCL234-Solomons-v1 8 December 8, 2009 THE BASICS: BONDING AND MOLECULAR STRUCTURE 120° F F C 1.19 (a) C 120° F F (b) CH3 180° CC CH3 linear H O H C H C (c) H 180° CN linear H C H 1.20 H trigonal planar at each carbon atom H H CH3 1.21 CH3CHCHCHCH3 or (CH3)2CHCH(CH3)CH(CH3)2 CH3 CH3 CH3 1.22 (a) CH3 CH3 = CH2 CH = CH2 CH3 (b) CH CH3 CH3 (c) CH3 (d) CH 3 CH2 ΟΗ OH H C C = CH3 CH2 CH2 CH2 CH2 CH2 (e) CH3 CH3 = CH2 CH CH3 = ΟΗ OH CH2 CH3 CH2 (f ) CH2 CH3 = C O (g) C CH3 CH2 CH2 Ο CH3 CH CH3 = CH2 Cl (h) CH3 CH CH2 = Cl CH3 CONFIRMING PAGES 20:10 P1: PBU/OVY JWCL234-01 P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Bind Rite JWCL234-Solomons-v1 December 8, 2009 THE BASICS: BONDING AND MOLECULAR STRUCTURE 9 1.23 (a) and (d) are constitutional isomers with the molecular formula C5 H12 . (b) and (e) are constitutional isomers with the molecular formula C5 H12 O. (c) and (f) are constitutional isomers with the molecular formula C6 H12 . H H 1.24 (a) H Cl H H O H C C C C C C H H H H H H C H H H (b) H H C H H H C C H H H C C C H H C C H H H C C H H C H H C H O C C C H H H C (c) H H H HH Cl 1.25 (a) C H (Note that the Cl atom and the three H atoms may be written at any of the four positions.) H H Cl Cl (b) or C Cl H H H and so on C H Cl Cl (c) Br H and others C H H (d) C H and others Cl C H H H Problems Electron Configuration 1.26 (a) Na+ has the electronic configuration, 1s 2 2s 2 2 p 6 , of Ne. (b) Cl− has the electronic configuration, 1s 2 2s 2 2 p 6 3s 2 3 p 6 , of Ar. CONFIRMING PAGES 20:10 P1: PBU/OVY JWCL234-01 P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Bind Rite JWCL234-Solomons-v1 10 December 8, 2009 THE BASICS: BONDING AND MOLECULAR STRUCTURE (c) F+ and (h) Br+ do not have the electronic configuration of a noble gas. (d) H− has the electronic configuration, 1s 2 , of He. (e) Ca2+ has the electronic configuration, 1s 2 2s 2 2 p 6 3s 2 3 p 6 , of Ar. (f) S2− has the electronic configuration, 1s 2 2s 2 2 p 6 3s 2 3 p 6 , of Ar. (g) O2− has the electronic configuration, 1s 2 2s 2 2 p 6 , of Ne. Lewis Structures 1.27 (a) Cl Cl O O (b) S Cl P Cl (c) Cl P Cl Cl Cl Cl (d) H O O − (c) − O (b) CH3 S + − S O − O O O N O S O O O Cl 1.28 (a) CH3 O + − O CH3 (d) CH3 S O − O Structural Formulas and Isomerism 1.29 (a) (CH3)2CHCH2OH (c) H2C HC CH2 CH O (b) (CH3)2CHCCH(CH3)2 1.30 (a) C4H10O (b) C7H14O (d) (CH3)2CHCH2CH2OH (c) C4H6 (d) C5H12O 1.31 (a) Different compounds, not isomeric (d) Same compound (b) Constitutional isomers (e) Same compound (c) Same compound (f) Constitutional isomers CONFIRMING PAGES 20:10 P1: PBU/OVY JWCL234-01 P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Bind Rite JWCL234-Solomons-v1 December 8, 2009 THE BASICS: BONDING AND MOLECULAR STRUCTURE (g) Different compounds, not isomeric (l) Different compounds, not isomeric (h) Same compound (m) Same compound (i) Different compounds, not isomeric (n) Same compound (j) Same compound (o) Same compound (k) Constitutional isomers 11 (p) Constitutional isomers Ο 1.32 (a) ΟΗ (d) Ο (e) (b) or Ο ΟΗ (c) (f ) 1.33 H 1.34 H C H O + N H O H − C H O − + N O O H O C N H H H H C H O N O (Other structures are possible.) CONFIRMING PAGES 20:10 JWCL234-01 P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Bind Rite JWCL234-Solomons-v1 12 December 8, 2009 THE BASICS: BONDING AND MOLECULAR STRUCTURE Resonance Structures − P1: PBU/OVY O 1.35 (a) Ο+ Ο + + + NH2 − Ο Ο − − + Ο (g) Ο Ο (h) + Ν − Ο − + Ο Ο Ο − − Ο+ Ο − Ο − + Ν Ν Ο Ο 1.36 − + + Ο Ο Ο − + (f ) + + (d) (e) NH2 NH2 − − + H2N H2N Ν 1.37 − + (b) (c) − Ο Ο Ν Ο N Ο − N + 1.38 (a) While the structures differ in the position of their electrons, they also differ in the positions of their nuclei and thus they are not resonance structures. (In cyanic acid the hydrogen nucleus is bonded to oxygen; in isocyanic acid it is bonded to nitrogen.) (b) The anion obtained from either acid is a − resonance hybrid of the following structures: O C N O C N − CONFIRMING PAGES 20:10 P1: PBU/OVY JWCL234-01 P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Bind Rite JWCL234-Solomons-v1 December 8, 2009 THE BASICS: BONDING AND MOLECULAR STRUCTURE 13 H 1.39 H C H (a) A +1 charge. ( F = 4 − 6 /2 − 2 = +1) (b) A +1 charge. (It is called a methyl cation.) (c) Trigonal planar, that is, H C + H H (d) sp 2 H 1.40 H C H (a) A −1 charge. ( F = 4 − 6/2 − 2 = −1) (b) A −1 charge. (It is called a methyl anion.) (c) Trigonal pyramidal, that is H C− H H (d) sp 3 H 1.41 H C H (a) No formal charge. ( F = 4 − 6/2 − 1 = 0) (b) No charge. (c) sp 2 , that is, H C H H CONFIRMING PAGES 20:10 P1: PBU/OVY JWCL234-01 P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Bind Rite JWCL234-Solomons-v1 14 December 8, 2009 THE BASICS: BONDING AND MOLECULAR STRUCTURE 1.42 (a) and (b) + − O O O + O O O − (c) Because the two resonance structures are equivalent, they should make equal contributions to the hybrid and, therefore, the bonds should be the same length. (d) Yes. We consider the central atom to have two groups or units of bonding electrons and one unshared pair. 1.43 N + N N 2− − + N N N − 2− + N N N B A C Structures A and C are equivalent and, therefore, make equal contributions to the hybrid. The bonds of the hybrid, therefore, have the same length. 1.44 (a) ΟΗ OH OH O O O OH (b) (CH3)2NH CH3CH2NH2 (c) (CH3)3N CH3CH2NHCH3 CH3CH2CH2NH2 CH3CHCH3 NH2 (d) 1.45 (a) constitutional isomers (b) the same (c) resonance forms (d) constitutional isomers (e) resonance forms (f) the same Challenge Problems 1.46 (a) O + N O (b) Linear (c) Carbon dioxide CONFIRMING PAGES 20:10 P1: PBU/OVY JWCL234-01 P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Bind Rite JWCL234-Solomons-v1 December 8, 2009 THE BASICS: BONDING AND MOLECULAR STRUCTURE 15 Br 1.47 Set A: Br Br Br Br Br Br Set B: H2N H2N OH OH OH N Set C: OH H H C b a H OH O O O N H H O O N O H NH2 O NH2 OH N Br H O c O [and unstable enol forms of a, b, and c] NH3 H + H − Set E: − O CH3 Dimethylacetylene CH3 − (i.e., CH3CH2CH2 and CH3CHCH3) − 1.48 (a) Dimethyl ether CH3 N + Set D: C cis-1,2-Dichloro-1,2-difluoroethene Cl CH3 C Cl C C F F Cl Cl F (b) F O H (c) H C H O C H H H H H H H C C C C Cl C H H Cl F C F Cl Cl or F C C 1.49 The large lobes centered above and below the boron atom represent the 2 p orbital that was not involved in hybridization to form the three 2sp 2 hybrid orbitals needed for the three boron-fluorine covalent bonds. This orbital is not a pure 2 p atomic orbital, since it is not an isolated atomic p orbital but rather part of a molecular orbital. Some of the other lobes in this molecular orbital can be seen near each fluorine atom. CONFIRMING PAGES F 20:10 P1: PBU/OVY JWCL234-01 P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Bind Rite JWCL234-Solomons-v1 16 December 8, 2009 THE BASICS: BONDING AND MOLECULAR STRUCTURE − CH2 CH O and CH2 CH O − . 1.50 The two resonance forms for this anion are The MEP indicates that the resonance contributor where the negative charge on the anion is on the oxygen is more important, which is what we would predict based on the fact that oxygen is more electronegative than carbon. δ− δ− Resonance hybrid, CH2 CH O QUIZ 1.1 Which of the following is a valid Lewis dot formula for the nitrite ion (NO− )? 2 (a) − O N (b) O O N O − (c) O N − (d) Two of these O (e) None of the above 1.2 What is the hybridization state of the boron atom in BF3 ? (a) s (b) p (c) sp (d) sp 2 (e) sp 3 F 1.3 BF3 reacts with NH3 to produce a compound, F of B is (a) s (b) p (c) sp (d) sp 2 (e) sp 3 H B N F H H . The hybridization state 1.4 The formal charge on N in the compound given in Problem 1.3 is (a) −2 (b) −1 (c) 0 (d) +1 (e) +2 1.5 The correct bond-line formula of the compound whose condensed formula is CH3 CHClCH2 CH(CH3 )CH(CH3 )2 is Cl Cl Cl (a) (b) Cl (c) Cl (d) (e) 1.6 Write another resonance structure for the acetate ion. Ο − O Acetate ion CONFIRMING PAGES 20:10 P1: PBU/OVY JWCL234-01 P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Bind Rite JWCL234-Solomons-v1 December 8, 2009 THE BASICS: BONDING AND MOLECULAR STRUCTURE 17 1.7 In the boxes below write condensed structural formulas for constitutional isomers of CH3 (CH2 )3 CH3 . 1.8 Write a three-dimensional formula for a constitutional isomer of compound A given below. Complete the partial structure shown. H C Cl H C H3C C H H HC H3C H H A Constitutional isomer of A 1.9 Consider the molecule (CH3 )3 B and give the following: (a) Hybridization state of boron (b) Hybridization state of carbon atoms (c) Formal charge on boron (d) Orientation of groups around boron (e) Dipole moment of (CH3)3B 1.10 Give the formal charge on oxygen in each compound. (a) CH3 O CH3 (c) CH3 O − O (b) 1.11 Write another resonance structure in which all of the atoms have a formal charge of zero. O − + H N H H CONFIRMING PAGES 20:10 P1: PBU/OVY JWCL234-01 P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Bind Rite JWCL234-Solomons-v1 18 December 8, 2009 THE BASICS: BONDING AND MOLECULAR STRUCTURE 1.12 Indicate the direction of the net dipole moment of the following molecule. Cl H3C C H3C F 1.13 Write bond-line formulas for all compounds with the formula C3 H6 O. CONFIRMING PAGES 20:10 ...
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