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Unformatted text preview: CHAPTER 2 ATOMIC STRUCTURE AND INTERATOMIC BONDING PROBLEM SOLUTIONS 2.7 The electron configurations of the ions are determined using Table 2.2. Fe 2+ 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6 Fe 3+ 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 Cu + 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 Ba 2+ 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 4d 10 5s 2 5p 6 Br 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 S 2 1s 2 2s 2 2p 6 3s 2 3p 6 2.8 The Na + ion is just a sodium atom that has lost one electron; therefore, it has an electron configuration the same as neon (Figure 2.6). The Cl ion is a chlorine atom that has acquired one extra electron; therefore, it has an electron configuration the same as argon. 2.10 (a) The 1s 2 2s 2 2p 6 3s 2 3p 6 3d 7 4s 2 electron configuration is that of a transition metal because of an incomplete d subshell. (b) The 1s 2 2s 2 2p 6 3s 2 3p 6 electron configuration is that of an inert gas because of filled 3 s and 3 p subshells. (c) The 1s 2 2s 2 2p 5 electron configuration is that of a halogen because it is one electron deficient from having a filled L shell. (d) The 1s 2 2s 2 2p 6 3s 2 electron configuration is that of an alkaline earth metal because of two s electrons. (e) The 1s 2 2s 2 2p 6 3s 2 3p 6 3d 2 4s 2 electron configuration is that of a transition metal because of an incomplete d subshell. (f) The 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 electron configuration is that of an alkali metal because of a single s electron. 2.22 For brass, the bonding is metallic since it is a metal alloy. For rubber, the bonding is covalent with some van der Waals. (Rubber is composed primarily of carbon and hydrogen atoms.) For BaS, the bonding is predominantly ionic (but with some covalent character) on the basis of the relative positions of Ba and S in the periodic table. For solid xenon, the bonding is van der Waals since xenon is an inert gas. For bronze, the bonding is metallic since it is a metal alloy (composed of copper and tin). For nylon, the bonding is covalent with perhaps some van der Waals. (Nylon is composed primarily of carbon and hydrogen.) For AlP the bonding is predominantly covalent (but with some ionic character) on the basis of the relative positions of Al and P in the periodic table. CHAPTER 3 THE STRUCTURE OF CRYSTALLINE SOLIDS PROBLEM SOLUTIONS 3.10 This problem asks for us to calculate the radius of a vanadium atom. For BCC, n = 2 atoms/ unit cell, and V C = 4R 3 3 = 64R 3 3 3 Since, ρ = nA V V C N A and solving for R R = 3 n 3 A V 64 ρ N A 1/3 = 3 3 ( 29 2 atoms/unit cell ( 29 50.9 g/mol ( 29 64 ( 29 5.96 g/cm 3 ( 29 6.023 x 10 23 atoms/mol ( 29 1/3 = 1.32 x 108 cm = 0.132 nm 3.12. (a) The volume of the Ti unit cell may be computed using Equation (3.5) as V C = nA Ti ρ N A Now, for HCP, n = 6 atoms/unit cell, and for Ti, A Ti = 47.9 g/mol. Thus, V C = (6 atoms/unit cell)(47.9 g/mol)(6 atoms/unit cell)(47....
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This note was uploaded on 05/03/2010 for the course ME 250750 taught by Professor Signer during the Summer '10 term at Wichita State.
 Summer '10
 Signer

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