Chapter9ISM

Chapter9ISM - CHAPTER 9 CHEMICAL BONDING I: BASIC CONCEPTS...

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CHAPTER 9 CHEMICAL BONDING I: BASIC CONCEPTS Problem Categories Biological : 9.81, 9.125. Conceptual : 9.62, 9.84, 9.87, 9.96, 9.100, 9.108, 9.111, 9.115, 9.116, 9.126. Descriptive : 9.19, 9.20, 9.35, 9.36, 9.37, 9.38, 9.39, 9.40, 9.73, 9.74, 9.76, 9.78, 9.92, 9.94, 9.127. Environmental : 9.99, 9.102, 9.119. Industrial : 9.98, 9.127. Organic : 9.91, 9.95, 9.101, 9.103, 9.105, 9.122. Difficulty Level Easy : 9.15, 9.16, 9.17, 9.18, 9.19, 9.20, 9.35, 9.36, 9.37, 9.38, 9.39, 9.40, 9.48, 9.62, 9.64, 9.65, 9.69, 9.71, 9.73, 9.74, 9.75, 9.76, 9.79, 9.81, 9.86, 9.89, 9.91, 9.94, 9.99, 9.101, 9.102, 9.103, 9.104, 9.106, 9.108, 9.112. Medium : 9.43, 9.44, 9.45, 9.46, 9.47, 9.51, 9.52, 9.53, 9.54, 9.55, 9.56, 9.61, 9.63, 9.66, 9.70, 9.72, 9.77, 9.78, 9.80, 9.82, 9.83, 9.85, 9.87, 9.88, 9.90, 9.92, 9.93, 9.95, 9.96, 9.98, 9.100, 9.105, 9.107, 9.109, 9.111, 9.113, 9.114, 9.116, 9.117, 9.118, 9.120, 9.121, 9.124, 9.125, 9.127, 9.128, 9.130. Difficult : 9.25, 9.26, 9.84, 9.97, 9.110, 9.115, 9.119, 9.122, 9.123, 9.126, 9.129, 9.131, 9.132, 9.133, 9.134. 9.15 We use Coulomb’s law to answer this question: cation anion = QQ Ek r (a) Doubling the radius of the cation would increase the distance, r , between the centers of the ions. A larger value of r results in a smaller energy, E , of the ionic bond. Is it possible to say how much smaller E will be? (b) Tripling the charge on the cation will result in tripling of the energy, E , of the ionic bond, since the energy of the bond is directly proportional to the charge on the cation, Q cation . (c) Doubling the charge on both the cation and anion will result in quadrupling the energy, E , of the ionic bond. (d) Decreasing the radius of both the cation and the anion to half of their original values is the same as halving the distance, r , between the centers of the ions. Halving the distance results in doubling the energy. 9.16 (a) RbI, rubidium iodide (b) Cs 2 SO 4 , cesium sulfate (c) Sr 3 N 2 , strontium nitride (d) Al 2 S 3 , aluminum sulfide 9.17 Lewis representations for the ionic reactions are as follows. 2K S Na F Ba O Al N + (a) 2 + (b) 2 2+ (c) Al + N 3 3+ (d) Ba + O Na + F 2K + S
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CHAPTER 9: CHEMICAL BONDING I: BASIC CONCEPTS 247 9.18 The Lewis representations for the reactions are: Ca 2H Sr Se 3Li N 2Al 3 S Sr + Se 2 2+ (a) Ca + 2 H 2+ (b) 3Li + N 3 + (c) 2Al + 3 S 2 3+ (d) 9.19 (a) I and Cl should form a molecular compound; both elements are nonmetals. One possibility would be ICl, iodine chloride. (b) Mg and F will form an ionic compound; Mg is a metal while F is a nonmetal. The substance will be MgF 2 , magnesium fluoride. 9.20 (a) Covalent (BF 3 , boron trifluoride) (b) ionic (KBr, potassium bromide) 9.25 (1) Na( s ) Na( g ) 1 108 kJ/mol Δ= H D (2) 1 2 Cl 2 ( g ) Cl( g ) 2 121.4 kJ/mol H D (3) Na( g ) Na + ( g ) + e 3 495.9 kJ/mol H D (4) Cl( g ) + e Cl ( g ) 4 349 kJ/mol H D (5) Na + ( g ) + Cl ( g ) NaCl( s ) 5 ? Δ = H D Na( s ) + 1 2 Cl 2 ( g ) NaCl( s ) overall 411 kJ/mol H D 5o v e r a l l 1 2 3 4 ( 411) (108) (121.4) (495.9) ( 349) 787 kJ/mol Δ = Δ Δ− Δ =− − = HH H H H H D D DDDD The lattice energy of NaCl is 787 kJ/mol . 9.26 (1) Ca(s) Ca(g) 1 121 kJ/mol H D (2) Cl 2 ( g ) 2Cl( g ) 2 242.8 kJ/mol H D (3) Ca( g ) Ca + ( g ) + e ' 3 589.5 kJ/mol H D Ca + ( g ) Ca 2 + ( g ) + e " 3 1145 kJ/mol H D (4) 2[Cl( g ) + e Cl ( g )] 4 2( 349 kJ/mol) 698 kJ/mol = H D (5) Ca 2 + ( g ) + 2Cl ( g ) CaCl 2 ( s ) 5 ? Δ = H D Ca( s ) + Cl 2 ( g ) CaCl 2 ( s ) overall 795 kJ/mol H D
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CHAPTER 9: CHEMICAL BONDING I: BASIC CONCEPTS 248 Thus we write: '" overall 1 2 3 3 4 5 Δ = Δ+ Δ + Δ HH H H H H H DD D D D D D 5 ( 795 121 242.8 589.5 1145 698)kJ/mol 2195 kJ / mol Δ= + = H D The lattice energy is represented by the reverse of equation (5); therefore, the lattice energy is + 2195 kJ/mol .
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Chapter9ISM - CHAPTER 9 CHEMICAL BONDING I: BASIC CONCEPTS...

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