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CHAPTER Chemical Bonding 12 Section 12.1 The Chemical Bond Concept 2. Atom H Br Valence Electrons 1 valence e– 7 valence e– 4. Atom H Br Covalent Bond 2 valence e– 8 valence e– 6. 8. 10. (a) (c) Compound ZnBr2 IF7 Bond ionic covalent Compound (b) CO2 (d) PbSO4 Bond covalent ionic (a) (c) Substance CH3 OH CH3 COCH3 Particle molecule molecule Substance (b) CoCl2 (d) SnCO3 Particle formula unit formula unit (a) (c) Substance Xe XeO3 Particle atom molecule Substance (b) P4 (d) Fe3 O4 Particle molecule formula unit 2014 © Pearson Education, Inc. Chemical Bonding 89 Section 12.2 Ionic Bonds 12. 14. 16. 18. 20. 22. 24. 90 (a) (c) Ion Li ion Al ion Ionic Charge 1+ (Group 1) 3+ (Group 13) Ion (b) Sr ion (d) Pb ion Ionic Charge 2+ (Group 2) 4+ (Group 14) (a) (c) Ion Cl ion S ion Ionic Charge 1– (Group 17) 2– (Group 16) Ion (b) I ion (d) P ion Ionic Charge 1– (Group 17) 3– (Group 15) (a) (b) (c) (d) Ion Sc3+ K+ Ti4+ Ba2+ Electron Configuration 1s2 2s2 2p6 3s2 3p6 1s2 2s2 2p6 3s2 3p6 1s2 2s2 2p6 3s2 3p6 1s2 2s2 2p6 3s2 3p6 4s2 3d1 0 4p6 5s2 4d1 0 5p6 Ion (a) Br– (b) O2 – (c) Se2 – (d) N3 – Electron Configuration 1s2 2s2 2p6 3s2 3p6 4s2 3d1 0 4p6 1s2 2s2 2p6 1s2 2s2 2p6 3s2 3p6 4s2 3d1 0 4p6 1s2 2s2 2p6 (a) (b) (c) (d) Ion Se2 – Br– Rb+ Sr2+ Isoelectronic Noble Gas Kr Kr Kr Kr (a) (b) (c) (d) Ion Sc3+ K+ Ti4+ Ba2+ Isoelectronic Noble Gas Ar Ar Ar Xe (a) (b) (c) (d) Ion Br– O2 – Se2 – N3 – Isoelectronic Noble Gas Kr Ne Kr Ne Chapter 12 2014 © Pearson Education, Inc. 26. (a) (b) (c) (d) Al atom radius > Al ion radius Pb atom radius > Pb ion radius Se atom radius < Se ion radius N atom radius < N ion radius 28. The true statements are (b) and (c). The corrected statements are: (a) Aluminum and nitride ions bond by electrostatic attraction. (d) Breaking an ionic bond between aluminum and nitrogen requires energy. Section 12.3 Covalent Bonds 30. (a) (b) The bond length in H–Br is less than the sum of the atomic radii. The bond length in S=O is less than the sum of the atomic radii. 32. The true statements are (b) and (c). The corrected statements are: (a) Valence electrons are shared between carbon and oxygen atoms. (d) Energy is required to break a C—O covalent bond. Section 12.4 Electron Dot Formulas of Molecules 34. Molecule Valence Electrons (a) Cl2 7 + 7 = 14 e– (b) O2 6 + 6 = 12 e– (c) HI 1 + 7 = 8 e– (d) PH 3 5 + 3(1) = 8 e– Electron Dot Structural Formula Cl Cl Cl O O O H I H H P H 2014 © Pearson Education, Inc. Cl O H I H H P H Chemical Bonding 91 36. 38. Molecule Valence Electrons (a) 2(5) = 10 e– N2 (b) PI3 5 + 3(7) = 26 e– (c) CS 2 4 + 2(6) = 16 e– (d) HOCl 1 + 6 + 7 = 14 e– Molecule Valence Electrons (a) (b) (c) CCl4 HONO2 CH 3 OH (d) HOCN 92 Chapter 12 4 + 4(7) = 32 e– 1 + 3(6) + 5 = 24 e– 4 + 4(1) + 6 = 14 e– 1 + 6 + 4 + 5 = 16 e– Electron Dot N Structural Formula N N I I P I S C N I I P I S C S H O Cl S H O Cl Electron Dot Cl Cl Cl C Cl Cl H O N O Cl 2014 © Pearson Education, Inc. C Cl Cl O H O N O O H H H C O H H H O C Structural Formula H C O H C N H N H O Section 12.5 Electron Dot Formulas of Polyatomic Ions 40. Polyatomic Ion Valence Electrons (a) IO– 7 + 6 + 1 = 14 e– (b) IO2 – 7 + 2(6) + 1 = 20 e– (c) IO3 – 7 + 3(6) + 1 = 26 e– Electron Dot I Structural Formula O – O I O – [O I O]– O I O – O [O I O]– O O O – O I O (d) IO4 – 42. 7 + 4(6) + 1 = 32 e– Polyatomic Ion Valence Electrons [ O O PO4 3 – 5 + 4(6) + 3 = 32 e– Electron Dot HPO4 2 – 1 + 5 + 4(6) + 2 = 32 e– O PO3 3 – (d) HPO3 2 – 5 + 3(6) + 3 = 26 e– 1 + 5 + 3(6) + 2 = 26 e– [ O H O P O O 2– O ] 3– O [H–O 3– H O P O O P O ] 2– O [ O O 2014 © Pearson Education, Inc. P O O P O (c) O ]– Structural Formula O O (b) I O O 3– O P O (a) O]– [I P O ] 3– O 2– [H–O P O ] 2– O Chemical Bonding 93 44. Polyatomic Ion Valence Electrons Electron Dot H (a) (b) PH 4 + SeO3 2 – H P H H 5 + 4(1) – 1 = 8 e– Structural Formula H [ H – P – H ]+ + H [ O – Se – O ] 2– O Se O 2– O 6 + 3(6) + 2 = 26 e– O [ O – C – O ] 2– O C O 2– (c) CO 3 2 – (d) BO3 3 – 4 + 3(6) + 2 = 24 e– O O [ O – B – O ] 3– O B O 3– O 5 + 3(6) + 3 = 26 e– O Section 12.6 Polar Covalent Bonds 46. Within a period of elements, the electronegativity increases from left to right in the periodic table. 48. Nonmetals are more electronegative than semimetals. 50. (a) (c) More Electronegative Se < Br Te < S More Electronegative (b) C > B (d) Ba < Be (Note: The more electronegative element is in bold.) 52. 54. (a) (c) Bond H—Cl N—O Polarity 3.0 – 2.1 = 0.9 3.5 – 3.0 = 0.5 Bond (b) H—Br (d) C—O (a) (b) (c) (d) Polar Bonds Using Delta Notation δ– C—H δ+ δ+ Se—O δ– δ+ P—I δ– δ+ H—Br δ– Polarity 2.8 – 2.1 = 0.7 3.5 – 2.5 = 1.0 (Note: δ– indicates the more electronegative atom and δ+ indicates the more electropositive atom.) 94 Chapter 12 2014 © Pearson Education, Inc. Section 12.7 Nonpolar Covalent Bonds 56. (a) (b) (c) (d) Bond I—C C—S S—H H—Br Polarity 2.5 – 2.5 2.5 – 2.5 2.5 – 2.1 2.8 – 2.1 = = = = 0 0 0.4 0.7 Classification nonpolar nonpolar polar polar Thus, (a) and (b) are nonpolar. 58. O2 , F2 , and I2 occur naturally as diatomic molecules. Section 12.8 Coordinate Covalent Bonds (Note: Coordinate covalent bonds are indicated by a dash, —.) 60. 62. Molecule Valence Electrons HIO 1 + 7 + 6 = 14 e– Molecule Valence Electrons Electron Dot H 64. 1 + 7 + 2(6) = 20 e– Polyatomic Ion Valence Electrons O Electron Dot H HIO2 I I O 66. 5 + 4(1) – 1 = 8 e– Polyatomic Ion Valence Electrons 5 + 4(6) + 3 = 32 e– I O Coord. Cov. Bond H I O O Electron Dot Coord. Cov. Bond H P H H H + Electron Dot O 3– O P O PO4 3 – H O H PH 4 + Coord. Cov. Bond O 2014 © Pearson Education, Inc. H P H + H Coord. Cov. Bond O O P O 3– O Chemical Bonding 95 Section 12.9 Hydrogen Bonds 68. H–F: – – – H–F: 70. A hydrogen bond is ~50% longer than a polar covalent bond. Section 12.10 Shapes of Molecules 72. Formula CBr4 Electron Pair tetrahedral Molecular Shape tetrahedral Bond Angle 109.5° 74. Formula PH 3 Electron Pair tetrahedral Molecular Shape trigonal pyramidal Bond Angle 107° 76. Formula Cl2 O Electron Pair tetrahedral Molecular Shape bent Bond Angle 104.5° 78. In CF 4 each of the four C–F bonds is polar, but the four fluorine atoms pull equally in opposite directions to give a nonpolar molecule. The four fluorine atoms are at the corners of a tetrahedron, and the molecular shape is described as tetrahedral. General Exercises 80. 82. 84. 96 (a) (b) (c) (d) Substance Pu O2 Pu2 O3 H2 O2 Particle atom molecule formula unit molecule (a) (b) (c) (d) Ions 1 Sc3+ and 1 N3 – 1 Ti4+ and 2 O2 – 2 NH4 + and 1 CO3 2 – 3 Hg2 2+ and 2 PO4 3 – Chemical Formula ScN TiO2 (NH4 )2 CO3 (Hg2 )3 (PO4 )2 A chloride ion has one more electron than a chlorine atom. As a result, the negative electrons in the 3p subenergy level repel each other, and the ion becomes larger. Chapter 12 2014 © Pearson Education, Inc. 86. Bond Sb—Cl Polarity 3.0 – 1.9 = 1.1 (polar) 88. Bond S—I Polarity 2.5 – 2.5 = 0 (nonpolar) 90. Polar Bond and Delta Notation δ+ 92. Molecule Electron Dot Valence Electrons As—Cl δ– H Sb H SbH3 94. 5 + 3(1) = 8 e– Polyatomic Ion Valence Electrons 4 + 3(6) + 2 = 24 e– H H Electron Dot O Si O 2– SiO3 2 – Structural Formula H Sb H Structural Formula [ O Si O ] 2– O O 96. Formula SiO2 Electron Pair linear Molecular Shape linear Bond Angle 180° 98. Formula Electron Pair Molecular Shape Bond Angle CO 3 2 – trigonal planar trigonal planar 120° Electron Dot Structural Formula O Xe O Challenge Exercises 100. Molecule Valence Electrons XeO3 8 + 3(6) = 26 e– O Xe O O 2014 © Pearson Education, Inc. O Chemical Bonding 97 Online Exercises 102. In addition to water and ammonia, hydrogen fluoride, HF, can also form strong hydrogen bonds between molecules. A hydrogen atom in one HF molecule can bond to a nonbonding electron pair in another HF molecule. 104. The two strands of a DNA molecule that form the double helix are held together by two hydrogen bonds between the DNA bases adenine(A) and thymine(T). 98 Chapter 12 2014 © Pearson Education, Inc.