# ch38 - CHAPTER 38 Molecules and Solids 1*Would you expect...

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CHAPTER 38 Molecules and Solids 1*· Would you expect the NaCl molecule to be polar or nonpolar? NaCl is a polar molecule. 2 · Would you expect the N 2 molecule to be polar or nonpolar? N 2 is a non-polar molecule. 3 · Does neon occur naturally as Ne or Ne 2 ? Why? Neon occurs naturally as Ne, not Ne 2 . Neon is a rare gas atom with a closed shell electron configuration. 4 · What type of bonding mechanism would you expect for ( a ) HF, ( b ) KBr, ( c ) N 2 ? ( a ) HF - ionic bonding. ( b ) KBr - ionic bonding. ( c ) N 2 - covalent bonding. 5* · What kind of bonding mechanism would you expect for ( a ) the N 2 molecule, ( b ) the KF molecule, ( c ) Ag atoms in a solid? ( a ) N 2 - covalent bonding. ( b ) KF - ionic bonding. ( c ) Ag (solid) - metallic bonding. 6 · Calculate the separation of Na + and Cl - ions for which the potential energy is -1.52 eV. U e = - ke 2 / r ; solve for r (see Problem 37-70) r = (1.44/1.52) nm = 0.947 nm 7 · The dissociation energy of Cl 2 is 2.48 eV. Consider the formation of an NaCl molecule according to the reaction Na + 1/2 Cl 2 NaCl Does this reaction absorb energy or release energy? How much energy per molecule is absorbed or released? Find the net energy change, E . Note that the E < 0, therefore energy is released E = 1.24 eV - (Binding energy of NaCl - see p. 1205) = (1.24 - 4.27) eV = -3.03 eV Reaction is exothermic 8 · The dissociation energy is sometimes expressed in kilocalories per mole (kcal/mol). ( a ) Find the relation between the units eV/molecule and kcal/mol. ( b ) Find the dissociation energy of molecular NaCl in kcal/mol. ( a ) 1 eV/molecule = (1 eV/molecule)(1 kcal/4184 J)(6.02 × 10 23 molecules/mole)(1.6 × 10 -19 J/eV) = 23.0 kcal/mol ( b ) Dissociation energy of NaCl = (4.27 × 23) kcal/mol = 98.3 kcal/mol 9* · The equilibrium separation of the HF molecule is 0.0917 nm and its measured electric dipole moment is 6.40 × 10 -30 C m. What percentage of the bonding is ionic ? 1. Find the dipole moment for 100% ionic bonding p 100 = er = 1.6 × 10 -19 × 9.17 × 10 -11 C . m = 1.47 × 10 -29 C . m

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Chapter 38 Molecules and Solids 2. Percent ionic bonding = 100( p meas / p 100 ) Ionic bonding = 43.6% 10 · Do Problem 9 for CsCl, for which the equilibrium separation is 0.291 nm and the measured electric dipole moment is 3.48 × 10 -29 C m. 1. Find the dipole moment for 100% ionic bonding 2. Percent ionic bonding = 100( p meas / p 100 ) p 100 = 1.6 × 10 -19 × 2.91 × 10 -10 C . m = 4.656 × 10 -29 C . m Ionic bonding = 74.7% 11 ·· The dissociation energy of LiCl is 4.86 nm and the equilibrium separation is 0.202 nm. The electron affinity of chlorine is 3.62 eV, and the ionization energy of lithium is 5.39 eV. Determine the core-repulsion energy of LiCl. 1. Determine
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## This note was uploaded on 01/23/2011 for the course PHYS 122 taught by Professor Cramer during the Winter '08 term at University of Washington.

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ch38 - CHAPTER 38 Molecules and Solids 1*Would you expect...

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