Fall 07 HW9 solution

# Fall 07 HW9 solution - Homework Assignment#9 Name...

• Notes
• 10

This preview shows pages 1–3. Sign up to view the full content.

Homework Assignment #9 Name____________________________ Laboratory Section (circle one) MW1.30 MW6.30 TTh7 1. Chapter 9, Problem 9, 13, 15, 23, 25, 27, 29, 31 [Show your work for Problem 9.30 (you will need to do this).], 40, 49, 58, 62, 68, 76 9.9 a) metallic b) covalent c) ionic 9.13 a) b) c) As Se Ga 9.15 a) 5A(15); n s 2 n p 3 b) 4A(14); n s 2 n p 2 9.23 a) 1A(1) b) 3A(13) c) 2A(2) 9.25 a) 7A(17) b) 6A(16) c) 3A(13) 9.27 a) CaO; O has a smaller radius than S. b) SrO; Sr has a smaller radius than Ba. 9.29 a) NaCl; Cl has a larger radius than F. b) K 2 S; S has a larger radius than O. 9.30 Lattice energy can be found from the fact that the sum of the energies from each step which equals the heat of formation for solid sodium chloride. o f H Δ + ) g ( Na ) s ( Na H Δ + ) g ( Cl ) g ( Cl H 2 2 2 1 Δ + - + + Δ e ) g ( Na ) g ( Na H + ) g ( Cl e ) g ( Cl H - - + Δ = lattice H Δ lattice H Δ = -41 kJ + 109 kJ + (1/2) (243 kJ) + 496 kJ + (-349 kJ) = 336.5 = 336 kJ The lattice energy for NaCl is less than that for LiF, which is expected since lithium and fluoride ions are smaller than sodium and chloride ions. 9.31 2 MgF H ° Δ = f H ° Δ - ( 1 H ° Δ + 2 H ° Δ + 3 H ° Δ + 4 H ° Δ + 2 5 H ° Δ ) 2 MgF H ° Δ = -1123 - (148 + 159 + 738 + 1450 + 2(-328)) kJ 2 MgF H ° Δ = -2962 kJ This value is larger due to greater ionic charge on Mg 2+ and greater numbers of ions, compared with those of NaCl. 9.40 a) H-F < H-Cl < H-I b) C=O < C-O < C-S c) N-H < N-O < N-S 9.49 CO 2 + 2 NH 3 (H 2 N) 2 CO + H 2 O Δ H rxn = Σ BE reactants - Σ BE products Δ H rxn = (2 BE C=O + 6 BE N-H ) - (4 BE N-H + BE C=O + 2 BE C-N + 2 BE O-H ) = [2(799) + 6(391)] - [4(391) + 745 + 2(305) + 2(467)] = 3944 - 3853 = 91 kJ

This preview has intentionally blurred sections. Sign up to view the full version.

9.58 a) I < Br < N b) Ca < H < F 9.62 a) Br Cl b) F Cl c) H O d) Se H e) As H f) S N 9.68 Increasing ionic character occurs with increasing Δ EN. a) Δ EN PCl = 0.9, Δ EN PBr = 0.7, Δ EN PF = 1.9 P-F > P-Cl > P-Br δ + δ - δ + δ - δ + δ - b) Δ EN BF = 2.0, Δ EN NF = 1.0, Δ EN CF = 1.9 B-F > C-F > N-F δ + δ - δ + δ - δ + δ - c) Δ EN SeF = 1.6, Δ EN TeF = 1.9, Δ EN BrF = 1.2 Te-F > Se-F > Br-F 9.76 a) 1) Mg( s ) Mg( g ) o H 1 Δ = 148 kJ 2) 1/2 Cl 2 ( g ) Cl( g ) o H 2 Δ = 1/2 (243 kJ) 3) Mg( g ) Mg + ( g ) + e - o H 3 Δ = 738 kJ 4) Cl( g ) + e - Cl - ( g ) o H 4 Δ = -349 kJ 5) Mg + ( g ) + Cl - ( g ) MgCl( s ) o H 5 Δ = -783.5 kJ (= o lattice H Δ (MgCl)) 6) Mg( s ) + 1/2 Cl 2 ( g ) MgCl( s ) o f H Δ (MgCl) = ? o f H Δ (MgCl) = o H 1 Δ + o H 2 Δ + o H 3 Δ + o H 4 Δ + o H 5 Δ = 148 kJ + 1/2 (243 kJ) + 738 kJ - 349 kJ - 783.5 kJ = -125 kJ b) Yes , since f H ° Δ for MgCl is negative, MgCl( s ) is stable relative to its elements. c) 2 MgCl( s ) MgCl 2 ( s ) + Mg( s ) ° Δ H = 1 mol ( o f H Δ , MgCl 2 ( s )) + 1 mol ( o f H Δ , Mg( s )) - 2 mol ( o f H Δ , MgCl( s )) = 1 mol (-641.6 kJ/mol) + 1 mol (0) - 2 mol (-125 kJ/mol) = -391.6 = -392 kJ d) No, o f H Δ for MgCl 2 is much more negative than that for MgCl. This makes the ° Δ H value for the above reaction very negative, and the formation of MgCl 2 would be favored. 2. Chapter 10, Problem 3, 5, 6, 9, 10, 14, 16, 18, 20, 27, 31, 30, 35, 37, 40, 41, 52, 53, 56, 73 10.3 For an element to obey the octet rule it must be surrounded by 8 electrons. To determine the number of electrons present (1) count the individual electrons actually shown adjacent to a particular atom, and (2) add two times the number of bonds to that atom. Using this method the structures shown give: (a) 0 + 2(4) = 8; (b) 2 + 2(3) = 8; (c) 0 + 2(5) = 10; (d) 2 + 2(3) = 8; (e) 0 + 2(4) = 8; (f) 2 + 2(3) = 8; (g) 0 + 2(3) = 6; (h) 8 + 2(0) = 8. All the structures obey the octet rule except: c and g.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern