Chapter 9 - CHAPTER 9 CHEMICAL BONDING I: THE COVALENT BOND...

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CHAPTER 9 CHEMICAL BONDING I: THE COVALENT BOND 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 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.
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CHAPTER 9: CHEMICAL BONDING I: THE COVALENT BOND 183 (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 o (2) 1 2 Cl 2 ( g ) Cl( g ) 2 121.4 kJ/mol H o (3) Na( g ) Na + ( g ) + e 3 495.9 kJ/mol H o (4) Cl( g ) + e Cl ( g ) 4 349 kJ/mol H o (5) Na + ( g ) + Cl ( g ) NaCl( s ) 5 ? Δ = H o Na( s ) + 1 2 Cl 2 ( g ) NaCl( s ) overall 411 kJ/mol H o 5o v e r a l l 1 2 3 4 ( 411) (108) (121.4) (495.9) ( 349) 787 kJ/mol Δ = Δ Δ− − − = H H HHHH o o oooo The lattice energy of NaCl is 787 kJ/mol . 9.26 (1) Ca(s) Ca(g) 1 121 kJ/mol H o (2) Cl 2 ( g ) 2Cl( g ) 2 242.8 kJ/mol H o (3) Ca( g ) Ca + ( g ) + e ' 3 589.5 kJ/mol H o Ca + ( g ) Ca 2 + ( g ) + e " 3 1145 kJ/mol H o (4) 2[Cl( g ) + e Cl ( g )] 4 2( 349 kJ/mol) 698 kJ/mol = H o (5) Ca 2 + ( g ) + 2Cl ( g ) CaCl 2 ( s ) 5 ? Δ = H o Ca( s ) + Cl 2 ( g ) CaCl 2 ( s ) overall 795 kJ/mol H o Thus we write: '" overall 1 2 3 3 4 5 Δ = Δ+ Δ + Δ H HHH H HH oo o o o o o 5 ( 795 121 242.8 589.5 1145 698)kJ/mol 2195 kJ / mol + = H o The lattice energy is represented by the reverse of equation (5); therefore, the lattice energy is + 2195 kJ/mol . 9.33 The degree of ionic character in a bond is a function of the difference in electronegativity between the two bonded atoms. Figure 9.5 lists electronegativity values of the elements. The bonds in order of increasing ionic character are: N N (zero difference in electronegativity) < S O (difference 1.0) = Cl F (difference 1.0) < K O (difference 2.7) < Li F (difference 3.0).
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CHAPTER 9: CHEMICAL BONDING I: THE COVALENT BOND 184 9.34 Strategy: We can look up electronegativity values in Figure 9.5 of the text.
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This note was uploaded on 10/11/2009 for the course CHE 102 taught by Professor Maggie during the Fall '09 term at Wisc Whitewater.

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Chapter 9 - CHAPTER 9 CHEMICAL BONDING I: THE COVALENT BOND...

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