Solutions: Chapter 8

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Part B True-False 10. NT 12. ST 14. AT 11. AT 13. ST 15. AT Part C Matching 16. b 18. e 20. a 17. d 19. c Part D Questions and Problems 21. dispersion forces, dipole interactions, hydrogen bonds 22. a. ionic b. polar covalent bonds c. polar covalent bonds d. nonpolar covalent bonds Practice Problems 8 Section 8.1 1. a. atom d. molecule b. molecule e. atom c. molecule 2. a. not diatomic b. diatomic c. diatomic d. not diatomic e. diatomic 3. Molecular compounds are usually composed from two or more nonmetallic elements. 4. A molecular structure gives information about the kinds and numbers of atoms present in a molecule. 5. Molecular compounds tend to have lower melting and boiling points that that of ionic compounds Section 8.2 1. The two atoms share a pair of electrons in order to form a single covalent bond. H F 2. Phosphorous needs 3 more electrons to fill the 3p orbitals. Fluorine needs one more electron to fill its second energy level. Since each fluorine atom only needs one electron and phosphorus needs 3 electrons, three fluorine atoms are required to bond with phosphorus. F P F 786 Core Teaching Resources F 3. Nitrogen needs 3 more electrons to fill its second energy level. Chlorine needs one more electron to achieve a noble gas configuration. Because each chlorine atom needs only one electron and nitrogen needs 3 electrons, three chlorine atoms are required to bond with nitrogen. 4. Because carbon can form four single covalent bonds, there is an apparent shortage of atoms with which to bond. This is a clue that a carbon-carbon multiple bond exists in this compound. Each carbon atom shares one electron with one of the two hydrogen atoms. The remaining three electrons for each carbon atom form a triple covalent bond. The electron dot structure is: HCCH 5. Carbon has 4 valence electrons and each of the oxygens has 6 valence electrons. Two additional electrons are added to account for the ion having a 2 charge. The carbon and oxygen can satisfy the octet rule by having the oxygens bonded to a central carbon. There is one double covalent bond between a carbon and oxygen, which can shift to any one of the carbon-oxygen bonds giving rise to three resonance structures. O C O O Section 8.3 Cl N Cl Cl 2 O C O O 1. The four fluorine atoms are covalently bonded to the central carbon atom. The four shared pairs of electrons repel each other to the corners of a tetrahedron. All four bond angles are 109.5°. 2. The four valence electron pairs repel each other, but the unshared pair is held closer to the phosphorus than the three bonding pairs. The unshared pair repels the shared pairs more strongly. Thus, the angle between bonds is expected to be slightly smaller than the tetrahedral bond angle of 109.5. The actual bond angle for NH3 , a similar molecule, is 107. 3. Boron forms three sp2 orbitals by mixing one 2s orbital and two 2p orbitals. The three sp2 orbitals lie in the same plane, 120 apart from one another. Each sp2 orbital overlaps with an 2 O C O O 2 © Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved.
© Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. atomic orbital of chlorine to form three equivalent sigma bonds. 4. To form four equivalent bonds, silicon mixes one s orbital and all three of the p orbitals. The hybridization in SiF4 is sp3 . 5. Oxygen is the central atom in this molecule. It has 6 valence electrons, two of which are bonding electrons. The other 4 electrons are unshared pairs. These 2 unshared pairs repel the two bonding pairs and prevent F2O from being linear. The molecule is a bent triatomic molecule, with a bond angle of approximately 104.5°. This angle is slightly smaller than the tetrahedral bond angle because the two unshared pairs repel each other more strongly than the two shared pairs. 6. Because each carbon can form single covalent bonds with four other atoms, there exists in this compound an apparent shortage of atoms with which to bond. This is a clue that CH2CF2 contains a carbon-carbon multiple bond. The two hydrogen atoms bond with one carbon atom while the two fluorine atoms bond with the other carbon. Each carbon atom has two electrons left over. These electrons form a carbon-carbon double covalent bond. The molecule looks very much like the ethene molecule (C2H4 ). H2C2H and F2C2F bond angles of 120°, the hybridization involved in the carbon-carbon bond is sp2 . H F C C H F 7. Carbon 1 mixes one s orbital and three p orbitals to form four sp 3 hybrid orbitals, which form 4 sigma bonds. Carbon 2 mixes one s and two p orbitals to form three sp 2 hybrid orbitals, which overlap with the hybrid orbitals of the carbon and oxygen atoms to form three equivalent sigma bonds. The non-hybridized p carbon orbital overlaps with an oxygen p orbital to form one pi bonding orbital. Section 8.4 1. a. The difference in electronegativity between Na and O is about 2.4 and the bond is ionic. b. With like atoms, the difference is zero and the bond is nonpolar covalent. c. The electronegativity difference between P and O is about 1.4 and the bond is polar covalent. 2. For a bond to be classified as nonpolar covalent, like atoms must bond, as in diatomic molecules. Most bonds are between unlike atoms; therefore, they must be ionic or polar covalent. 3. Both carbon dioxide and carbon monoxide contain polar bonds. However, the effect of the polar bond on the polarity of the entire molecule depends on the shape of the molecule. In carbon monoxide, there is a partial positive pole and a partial negative pole. Therefore, the molecule is a dipole. In carbon dioxide, the carbon and oxygens lie along the same axis. The bond polarities cancel, producing a nonpolar molecule. 4. The more electronegative atom in a covalent bond will have the symbol and the less electronegative atom the symbol. a. H H N H δ+ δ+ δ+ δ– b. δ– F Fδ– δ+ C δ– F Fδ– 5. CaO is an ionic compound and CS2 is a polar covalent compound. Generally, ionic compounds have much higher melting points than molecular compounds. Interpreting Graphics 8 1. O 3 C 3 O ; linear; 180°; none 2. H H ; tetrahedral; 109.5°; none C H H 3. O O ; trigonal planar; 120°; S O O S O O 4. F Be F ; linear; 180°; none O S O O O S O O Answer Key 787
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Solutions: Chapter 8

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