Chap13_Soln_even_problems

Chap13_Soln_even_problems - CHAPTER 13 BONDING: GENERAL...

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CHAPTER 13 BONDING: GENERAL CONCEPTS Chemical Bonds and Electronegativity 12. a. V = −5 × 10 −18 J b. V = −6 × 10 −18 J Note : There is a greater net attraction in arrangement b than in a. 14. The most polar bond will have the greatest difference in electronegativity between the two atoms. From positions in the periodic table, we would predict: a. Ge F b. P Cl c. S F d. Ti Cl e. Sn H f. Tl Br 16. Electronegativity values increase from left to right across the periodic table. The order of electronegativities for the atoms from smallest to largest electronegativity will be H = P < C < N < O < F. The most polar bond will be F-H since it will have the largest difference in electronegativities, and the least polar bond will be P-H since it will have the smallest difference in electronegativities (ΔEN = 0). The order of the bonds in decreasing polarity will be F-H > O-H > N-H > C-H > P-H. 18. Ionic Compounds 20.
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22. 24. a. Sc 3+ b. Te 2− c. Ce 4+ and Ti 4+ d. Ba 2+ All of these have the number of electrons of a noble gas. 26. a. LiF; Li + is smaller than Cs + . b. NaBr; Br - is smaller than I . c. BaO; O 2− has a greater charge than Cl - . d. CaSO 4 ; Ca 2+ has a greater charge than Na + . 28. Ionic solids can be characterized as being held together by strong omnidirectional forces. i. For electrical conductivity, charged species must be free to move. In ionic solids the charged ions are held rigidly in place. Once the forces are disrupted (melting or dissolution), the ions can move about (conduct). ii. Melting and boiling disrupts the attractions of the ions for each other. If the forces are strong, it will take a lot of energy (high temperature) to accomplish this. iii. If we try to bend a piece of material, the atoms/ions must slide across each other. For an ionic solid, the following might happen: Just as the layers begin to slide, there will be very strong repulsions causing the solid
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to snap across a fairly clean plane. 30. ΔH f = 1085 kJ/mol 32. Two other factors that must be considered are the ionization energy needed to produce more positively charged ions and the electron affinity needed to produce more negatively charged ions. The favorable lattice energy more than compensates for the unfavorable ionization energy of the metal and for the unfavorable electron affinity of the nonmetal, as long as electrons are added to or removed from the valence shell. Once the valence shell is full, the ionization energy required to remove another electron is extremely unfavorable; the same is true for electron affinity when an electron is added to a higher n shell. These two quantities are so unfavorable after the valence shell is complete that they overshadow the favorable lattice energy, and the higher-charged ionic compounds do not form. 34. EA
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This note was uploaded on 02/28/2010 for the course CHEM 27 taught by Professor Pandey during the Spring '10 term at Pacific.

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Chap13_Soln_even_problems - CHAPTER 13 BONDING: GENERAL...

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