Chapter08rw-final

Chapter08rw-final - 8 8.1 (a) (b) (c) 8.2 Basic Concepts of...

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Unformatted text preview: 8 8.1 (a) (b) (c) 8.2 Basic Concepts of Chemical Bonding Visualizing Concepts Analyze/Plan. Count the number of electrons in the Lewis symbol. This corresponds to the ‘A’-group number of the family. Solve. Group 14 or 4A Group 2 or 2A Group 15 or 5A (These are the appropriate groups in the s and p blocks, where Lewis symbols are most useful.) Analyze. Given the size and charge of four different ions, determine their ionic bonding characteristics. Plan. The magnitude of lattice energy is directly proportional to the charges of the two ions and inversely proportional to their separation. Eel = κQ1Q2/d. Apply these concepts to A, B, X and Y. (a) AY and BX have a 1:1 ratio of cations and anions. In an ionic compound, the total positive and negative charges must be equal. In order to form a 1:1 compound, the magnitude of positive charge on the cation must equal the magnitude of negative charge on the anion. A2+ combines with Y2− and B+ combines with X− to form 1:1 compounds. AY has the larger lattice energy. The A−Y and B−X separations are nearly equal. (A is smaller than B, but X is smaller than Y, so the differences in cation and anion radii approximately cancel.) In AY, Q1Q2 = (2)(2) = 4, while in BX, Q1Q2 = (1)(1) = 1. BX has the smaller lattice energy. To recap the arguments in part (b), the d values in the two compounds are similar and BX has the smaller Q1Q2, so it has the smaller lattice energy. (b) (c) 8.3 Analyze/Plan. Count the valence electrons in the orbital diagram, take ion charge into account, and find the element with this orbital electron count on the periodic chart. Write the complete electron configuration for the ion. Solve. (a) This ion has seven 3d electrons. Transition metals, or d-block elements, have valence electrons in d-orbitals. Transition metal ions first lose electrons from the 4s orbital, then from 3d if required by the charge. This 2+ ion has lost two electrons from 4s, none from 3d. The transition metal with seven 3d-electrons is cobalt, Co. The electron configuration of Co is [Ar]4s 2 3d 7 . (The configuration of Co 2 + is [Ar]3d 7 ). (b) 190 8 Chemical Bonding 8.4 A: E: D: Q: X: Z: Solutions to Exercises Analyze/Plan. This question is a ”reverse” Lewis structure. Count the valence electrons shown in the Lewis structure. For each atom, assume zero formal charge and determine the number of valence electrons an unbound atom has. Name the element. Solve. 1 shared e – pair = 1 valence electron + 3 unshared pairs = 7 valence electrons, F 2 shared pairs = 2 valence electrons + 2 unshared pairs = 6 valence electrons, O 4 shared pairs = 4 valence electrons, C 3 shared pairs = 3 valence electrons + 1 unshared pair = 5 valence electrons, N 1 shared pair = 1 valence electron, no unshared pairs, H same as X, H Check. Count the valence electrons in the Lewis structure. Does the number correspond to the molecular formula CH 2 ONF? 12 e – pair in the Lewis structure. CH 2 ONF = 4 + 2 + 6 + 5 + 7 = 24 e – , 12 e – pair. The molecular formula we derived matches the Lewis structure. 8.5 Analyze/Plan. Since there are no unshared pairs in the molecule, we use single bonds to H to complete the octet of each C atom. For the same pair of bonded atoms, the greater the bond order, the shorter and stronger the bond. Solve. (a) Moving from left to right along the molecule, the first C needs two H atoms, the second needs one, the third needs none, and the fourth needs one. The complete molecule is: (b) (c) 8.6 In order of increasing bond length: 3 < 1 < 2 In order of increasing bond enthalpy (strength): 2 < 1 < 3 Analyze/Plan. Given an oxyanion of the type XO4n–, find the identity of X from elements in the third period. Use the generic Lewis structure to determine the identity of X, and to draw the ion-specific Lewis structures. Use the definition of formal charge, [# of valence electrons − # of nonbonding electrons – (# bonding electrons/2)], to draw Lewis structures where X has a formal charge of zero. Solve. (a) According to the generic Lewis structure, each anion has 12 nonbonding and 4 bonding electron pairs, for a total of 32 electrons. Of these 32 electrons, the 4 O atoms contribute (4 × 6) = 24, and the overall negative charges contribute 1, 2 or 3. # X electrons = 32 – 24 – n. For n = 1−, X has (32 – 24 –1) = 7 valence electrons. X is Cl, and the ion is ClO4−. For n = 2–, X has (32 – 24 – 2) = 6 valence electrons. X is S, and the ion is SO42–. For n = 3–, X has (32 – 24 – 3) = 5 valence electrons. X is P, and the ion is PO43–. Check. The identity of the ions is confirmed in Figure 2.27. (b) In the generic Lewis structure, X has 0 nonbonding electrons and (8/2) = 4 bonding electrons. Differences in formal charge are due to difference in the number of valence electrons on X. 191 8 Chemical Bonding For PO43–, formal charge of P is (5 – 4) = +1. For SO42–, formal charge of S is (6 – 4) = +2. For ClO4−, formal charge of Cl is (7 – 4) = +3. (c) Solutions to Exercises In order to reduce the formal charge of X to zero, X must have more bonding electrons. This is accomplished by changing the appropriate number of lone pairs on O to multiple bonds between X and O. O O P O O 3− O O S O O 2− O O Cl O O − (d) In part (c) the Lewis structures that cause the formal charge on X to be zero all violate the octet rule. According to Section 8.7, the best single Lewis structure for an anion is the one that obeys the octet rule. Lewis Symbols 8.7 (a) Valence electrons are those that take part in chemical bonding, those in the outermost electron shell of the atom. This usually means the electrons beyond the core noble-gas configuration of the atom, although it is sometimes only the outer shell electrons. N : [He] 2s 2 2 p 3 A nitrogen atom has 5 valence electrons. Valence electrons (b) (c) 1s 2 2s 2 2 p 6 [Ne] 3s 2 3 p 2 The atom (Si) has 4 valence electrons. valence electrons 8.8 (a) Atoms will gain, lose or share electrons to achieve the nearest noble-gas electron configuration. Except for H and He, this corresponds to eight electrons in the valence shell, thus the term octet rule. S: [Ne]3s 2 3p 4 A sulfur atom has six valence electrons, so it must gain two electrons to achieve an octet. 1s 2 2s 2 2p 3 = [He]2s 2 2p 3 The atom (N) has five valence electrons and must gain three electrons to achieve an octet. (b) (c) 8.9 P: 1s 2 2s 2 2p 6 3s 2 3p 3 . The 3s and 3p electrons are valence electrons; the 1s, 2s and 2p electrons are nonvalence or core electrons. Valence electrons are involved in chemical bonding, while nonvalence or core electrons are not. (a) Ti: [Ar]4s23d2. Ti has four (4) valence electrons. These valence electrons are available for chemical bonding, while core electons do not participate in chemical bonding. 8.10 192 8 Chemical Bonding (b) (c) Hf: [Xe]6s24f145d2 Solutions to Exercises If Hf and Ti both behave as if they have four (4) valence electrons, the 6s and 5d orbitals in Hf behave as valence orbitals and the 4f behaves as a core orbital. This is reasonable because 4f is complete and 4f electrons are, on average, closer to the nucleus than 5d or 6s electrons. 8.11 (a) (a) Al Ca (b) (b) Br P (c) (c) Ar 2+ Mg or Mg 2+ (d) (d) Sr 2− S or S2− 8.12 Ionic Bonding 8.13 8.14 8.15 8.16 8.17 Ca + F + F Ca2+ + 2 F (a) (a) (a) (b) (c) (d) (e) (f) AlF 3 BaF2 (b) (b) K2S CsCl (c) Y 2 O 3 (c) Li 3 N (d) Mg 3 N 2 (d) Al2O3 Sr 2 +: [Kr], noble-gas configuration Ti 2 +: [Ar]3d 2 Se 2 –: [Ar]4s 2 3d 1 04p 6 = [Kr], noble-gas configuration Ni 2 +: [Ar]3d 8 Br – : [Ar]4s 2 3d 1 04p 6 = [Kr], noble-gas configuration Mn 3 +: [Ar]3d 4 Zn 2 +: [Ar]3d 1 0 Te 2 –: [Kr]5s 2 4d 1 05p 6 = [Xe], noble-gas configuration Sc 3 +: [Ar], noble-gas configuration Rh 3 +: [Kr]4d 6 Tl + : [Xe]6s 2 4f 1 45d 1 0 Bi3+: [Xe]6s2 4f1 4 5d 1 0 Lattice energy is the energy required to totally separate one mole of solid ionic compound into its gaseous ions. The magnitude of the lattice energy depends on the magnitudes of the charges of the two ions, their radii, and the arrangement of ions in the lattice. The main factor is the charges, because the radii of ions do not vary over a wide range. 8.18 (a) (b) (c) (d) (e) (f) 8.19 (a) (b) 193 8 Chemical Bonding 8.20 (a) NaCl, 788 kJ/mol; KF, 808 kJ/mol Solutions to Exercises The two factors that affect lattice energies are ionic charge and radius. The ionic charges, 1+ and 1−, are the same in the two compounds. Since lattice energy is inversely proportional to the ion separation (d), we expect the compound with the smaller lattice energy, NaCl, to have the larger ion separation. That is, the K−F distance should be shorter than the Na−Cl distance. (b) Na−Cl, 1.16 Å + 1.67 Å = 2.83 Å K−F, 1.52 Å + 1.19 Å = 2.71 Å This estimate of the relative ion separations agrees with the estimate from lattice energies. Ionic radii indicate that the K−F distance is shorter than the Na−Cl distance. 8.21 KF, 808 kJ/mol; CaO, 3414 kJ/mol; ScN, 7547 kJ/mol The sizes of the ions vary as follows: Sc 3 + < Ca 2 + < K + and F – < O 2 – < N 3 –. Therefore, the inter-ionic distances are similar. According to Coulomb’s law for compounds with similar ionic separations, the lattice energies should be related as the product of the charges of the ions. The lattice energies above are approximately related as (1)(1): (2)(2): (3)(3) or 1:4:9. Slight variations are due to the small differences in ionic separations. 8.22 (a) According to Equation 8.4, electrostatic attraction increases with increasing charges of the ions and decreases with increasing radius of the ions. Thus, lattice energy (i) increases as the charges of the ions increase and (ii) decreases as the sizes of the ions increase. KBr < NaF < MgO < ScN. This order is confirmed by the lattice energies given in Table 8.2. ScN has the highest lattice energy, because its ions have 3+ and 3− charges. Na+ is smaller than K+, and F− is smaller than Br−. The ion separation is smaller in NaF, so it has the larger lattice energy. (b) 8.23 Since the ionic charges are the same in the two compounds, the K–Br and Cs–Cl separations must be approximately equal. Since the radii are related as Cs + > K + and Br – > Cl – , the difference between Cs + and K + must be approximately equal to the difference between Br – and Cl – . This is somewhat surprising, since K + and Cs + are two rows apart and Cl – and Br – are only one row apart. (a) The ion charges in CaF2 and BaF2 are the same, 2+ for the cations and 1− for the anions. Ba2+ is a larger cation than Ca2+, so the ion separation, d, is greater in BaF2 and the lattice energy is smaller than that of CaF2. The ions have 1+ and 1– charges in all three compounds. In NaCl the cationic and anionic radii are smaller than in the other two compounds, so it has the largest lattice energy. In RbBr and CsBr, the anion is the same, but the Cs cation is larger, so CsBr has the smaller lattice energy. In BaO, the magnitude of the charges of both ions is 2; in KF, the magnitudes are 1. Charge considerations alone predict that BaO will have the higher lattice energy. The distance effect is less clear; O 2 – and F – are isoelectronic, so F – , with the larger Z, has a slightly smaller radius. Ba 2 + is two rows lower on the periodic 8.24 (b) (c) 194 8 Chemical Bonding 8.25 Solutions to Exercises chart than K + , but it has a greater positive charge, so the radii are probably similar. In any case, the ionic separations in the two compounds are not very different, and the charge effect dominates. Equation 8.4 predicts that as the oppositely charged ions approach each other, the energy of interaction will be large and negative. This more than compensates for the energy required to form Ca 2 + and O 2 – from the neutral atoms (see Figure 8.4 for the formation of NaCl). By analogy to the Born-Haber cycle for NaCl(s), Figure 8.4, the enthalpy of formation for NaCl2(s) is ΔH o NaCl2(s) = −ΔHlattNaCl2 + ΔH o Na(g) + 2 ΔH o Cl(g) + I1(Na) + I2(Na) + 2E(Cl) f f f 8.26 (a) ΔH o NaCl2(s) = −ΔHlattNaCl2 + 107.7 kJ + 2(121.7 kJ) + 496 kJ + 4562 kJ f + 2(−349 kJ) ΔH o f NaCl2(s) = −ΔHlattNaCl2 + 4711 kJ The collective energy of the “other” steps in the cycle (vaporization and ionization of Na2+, dissociation of Cl2 and electron affinity of Cl) is +4711 kJ. In order for the sign of ΔH o NaCl2 to be negative, the lattice energy would have to be f greater than 4711 kJ. (b) ΔH o NaCl2(s) = −(2326 kJ) + 4711 kJ = 2385 kJ f This value is large and positive. 8.27 RbCl(s) → Rb + (g) + Cl – (g) ΔH (lattice energy) = ? By analogy to NaCl, Figure 8.4, the lattice energy is ΔH latt = −ΔH ° RbCl(s) + ΔH ° Rb(g) + ΔH ° Cl(g) + I 1 (Rb) + E (Cl) f f f = −(−430.5 kJ) + 85.8 kJ + 121.7 kJ + 403 kJ + (−349 kJ) = +692 kJ This value is smaller than that for NaCl (+788 kJ) because Rb + has a larger ionic radius than Na + . This means that the value of d in the denominator of Equation 8.4 is larger for RbCl, and the potential energy of the electrostatic attraction is smaller. 8.28 (a) MgCl2, 2326 kJ; SrCl2, 2127 kJ. Since the ionic radius of Ca2+ is greater than that of Mg2+, but less than that of Sr2+, the ion separation (d) in CaCl2 will be intermediate as well. We expect the lattice energy of CaCl2 to be in the range 2200-2250 kJ. By analogy to Figure 8.4: ΔH latt = −ΔH ° CaCl 2 + ΔH ° Ca(g) + 2ΔH ° Cl(g) + I 1 (Ca) + I 2 (Ca) + 2E (Cl) f f f = −( −795.8 kJ) + 179.3 kJ + 2(121.7 kJ) + 590 kJ + 1145 kJ + 2( −349 kJ) = +2256 kJ (b) This value is near the range predicted in part (a). Covalent Bonding, Electronegativity, and Bond Polarity 8.29 (a) A covalent bond is the bond formed when two atoms share one or more pairs of electrons. 195 8 Chemical Bonding (b) (c) 8.30 Solutions to Exercises Any simple compound whose component atoms are nonmetals, such as H 2 , SO 2 , and CCl 4 , are molecular and have covalent bonds between atoms. Covalent because it is a gas even below room temperature. K and Ar. K is an active metal with one valence electron. It is most likely to achieve an octet by losing this single electron and to participate in ionic bonding. Ar has a stable octet of valence electrons; it is not likely to form chemical bonds of any type. Analyze/Plan. Follow the logic in Sample Exercise 8.3. Solve. 8.31 Check. Each pair of shared electrons in SiCl 4 is shown as a line; each atom is surrounded by an octet of electrons. 8.32 8.33 (a) (b) (c) A double bond is required because there are not enough electrons to satisfy the octet rule with single bonds and unshared pairs. The greater the number of shared electron pairs between two atoms, the shorter the distance between the atoms. If O 2 has a double bond, the O–O distance will be shorter than the O–O single bond distance. The H atoms must be terminal because H can form only one bond. 14 e−, 7 e− pairs H O O H 8.34 (a) (b) 8.35 (a) (b) (c) (d) 8.36 (a) (b) (c) From Solution 8.33, O2 has a double bond. The O−O bond in H2O2 is a single bond, and thus longer than the O−O bond in O2. Electronegativity is the ability of an atom in a molecule (a bonded atom) to attract electrons to itself. The range of electronegativities on the Pauling scale is 0.7–4.0. Fluorine, F, is the most electronegative element. Cesium, Cs, is the least electronegative element that is not radioactive. The electronegativity of the elements increases going from left to right across a row of the periodic chart. Electronegativity decreases going down a family of the periodic chart. Generally, the trends in electronegativity are the same as those in ionization energy and opposite those in electron affinity. That is, the more positive the 196 8 Chemical Bonding 8.37 Solutions to Exercises ionization energy and the more negative the electron affinity (ignoring a few exceptions), the greater the electronegativity of an element. Plan. Electronegativity increases going up and to the right in the periodic table. Solve. (a) O (b) C (c) P (d) Be Check. The electronegativity values in Figure 8.6 confirm these selections. 8.38 Electronegativity increases going up and to the right in the periodic table. (a) 8.39 O (b) Al (c) Cl (d) F The bonds in (a), (c) and (d) are polar because the atoms involved differ in electronegativity. The more electronegative element in each polar bond is: (a) F (c) O (d) I 8.40 The more different the electronegativity values of the two elements, the more polar the bond. (a) (b) O–F < C–F < Be–F. This order is clear from the periodic trend. S–Br < C–P < O–Cl. Refer to the electronegativity values in Figure 8.6 to confirm the order of bond polarity. The 3 pairs of elements all have the same positional relationship on the periodic chart. The more electronegative element is one row above and one column to the left of the less electronegative element. This leads us to conclude that ΔEN is similar for the 3 bonds, which is confirmed by values in Figure 8.6. The most polar bond, O–Cl, involves the most electronegative element, O. Generally, the largest electronegativity differences tend to be between row 2 and row 3 elements. The 2 bonds in this exercise involving elements in row 2 and row 3 do have slightly greater ΔEN than the S–Br bond, between elements in rows 3 and 4. C–S < N–O < B–F. You might predict that N–O is least polar since the elements are adjacent on the table. However, the big decrease going from the second row to the third means that the electronegativity of S is not only less than that of O, but essentially the same as that of C. C–S is the least polar. Analyze/Plan. Q is the charge at either end of the dipole. Q = μ/r. The values for OH are μ = 1.78 D and r = 0.98 Å. Change Å to m and use the definition of debyes and the charge of an electron to calculate effective charge in units of e. Solve. Q= 1Å 3.34 × 10 −30 C - m 1e μ 1.78 D = × × × = 0.38 e −10 r 0.98 Å 1D 1 × 10 m 1.60 × 10 −19 C (c) 8.41 (a) (b) From Sample Exercise 8.5, the effective charges on H and Cl in the HCl molecule are +0.178 e and −0.178 e, respectively. These effective charges can be thought of as the amount of charge “transferred” from H to Cl. From part (a), 0.38 e is transferred from H to O in OH. There is larger charge separation in OH than in HCl, so OH is more polar. The dipole moments, 1.78D for OH and 1.08 for HCl, reinforce this conclusion. 197 8 Chemical Bonding (c) Solutions to Exercises The greater polarity of OH is also predicted by electronegtativities. According to electronegativity trends (and values in Figure 8.6), O is more electronegative than Cl. The electronegativity difference between O and H is greater than that between Cl and H, so OH is more polar. The more electronegative element, Br, will have a stronger attraction for the shared electrons and adopt a partial negative charge. Q is the charge at either end of the dipole. Q= 1Å 3.34 × 10 −30 C - m 1e μ 1.21 D = × × × −10 r 2.49 Å 1D 1 × 10 m 1.60 × 10 −19 C = 0.1014 = 0.101 e 8.42 (a) (b) The charges on I and Br are 0.101 e. 8.43 Analyze/Plan. Generally, compounds formed by a metal and a nonmetal are described as ionic, while compounds formed from two or more nonmetals are covalent. However, substances with metals in a high oxidation state often have properties of molecular compounds. In this exercise we know that one substance in each pair is molecular and one is ionic; we may need to distinguish by comparison. Solve. (a) SiF4, metalloid and nonmetal, molecular, silicon tetrafluoride. LaF3, metal and nonmetal, ionic, lanthanum(III) fluoride (b) FeCl2, metal and nonmetal, ionic, iron(II) chloride ReCl6, metal in high oxidation state, Re(VI), molecular, rhenium hexachloride (c) PbCl4, metal and nonmetal, Pb(IV) is relatively high oxidation state, molecular (by contrast with RbCl, which is definitely ionic), lead tetrachloride RbCl, metal and nonmetal, ionic, rubidium chloride 8.44 Generally, compounds formed by a metal and a nonmetal are described as ionic, while compounds formed from two or more nonmetals are covalent. However, substances with metals in a high oxidation states often have properties of molecular compounds. (a) TiCl4, metal and nonmetal, Ti(IV) is a relatively high oxidation state, molecular (by contrast with CaF2, which is definitely ionic), titanium tetrachloride CaF2, metal and nonmetal, ionic, calcium fluoride (b) ClF3, two nonmetals, molecular, chlorine trifluoride VF3, metal and nonmetal, ionic, vanadium(III) fluoride (c) SbCl5, metalloid and nonmetal, molecular, antimony pentachloride AlF3, metal and nonmetal, ionic, aluminum fluroide Lewis Structures; Resonance Structures 8.45 Analyze. Counting the correct number of valence electrons is the foundation of every Lewis structure. Plan/Solve. 198 8 Chemical Bonding (a) Solutions to Exercises Count valence electrons: 4 + (4 × 1) = 8 e – , 4 e – pairs. Follow the procedure in Sample Exercise 8.6. (b) (c) Valence electrons: 4 + 6 = 10 e – , 5 e – pairs Valence electrons: [6 + (2 × 7)] = 20 e – , 10 e – pairs i. ii. iii. (d) Place the S atom in the middle and connect each F atom with a single bond; this requires 2 e – pairs. Complete the octets of the F atoms with nonbonded pairs of electrons; this requires an additional 6 e – pairs. The remaining 2 e – pairs complete the octet of the central S atom. 32 valence e – , 16 e – pairs (Choose the Lewis structure that obeys the octet rule, Section 8.7.) (e) Follow Sample Exercise 8.8. 20 valence e – , 10 e – pairs (f) 14 valence e – , 7 e – pairs Check. In each molecule, bonding e – pairs are shown as lines, and each atom is surrounded by an octet of electrons (duet for H). 8.46 (a) 12 e – , 6 e – pairs (b) 14 valence e – , 7 e – pairs (c) 50 valence e – , 25 e – pairs (d) 26 valence e – , 13 e – pairs (Choose the Lewis structure that obeys the octet rule, Section 8.7) 199 8 Chemical Bonding (e) 26 valence e – , 13 e – pairs (f) Solutions to Exercises 10 e – , 5 e – pairs (Choose the Lewis structure that obeys they octet rule, Section 8.7.) 8.47 (a) (b) (c) Formal charge is the charge on each atom in a molecule, assuming all atoms have the same electronegativity. Formal charges are not actual charges. They assume perfect covalency, one extreme for the possible electron distribution in a molecule. The other extreme is represented by oxidation numbers, which assume that the more electronegative element holds all electrons in a bond. The true electron distribution is some composite of the two extremes. 26 e – , 13 e – pairs 8.48 (a) The octet rule is satisfied for all atoms in the structure. (b) (c) (d) F is more electronegative than P. Assuming F atoms hold all shared electrons, the oxidation number of each F is –1. The oxidation number of P is +3. Assuming perfect sharing, the formal charges on all F and P atoms are 0. The oxidation number on P is +3; the formal charge is 0. These represent extremes in the possible electron distribution, not the best picture. By virtue of their greater electronegativity, the F atoms carry a partial negative charge, and the P atom a partial positive charge. 8.49 Analyze/Plan. Draw the correct Lewis structure: count valence electrons in each atom, total valence electrons and electron pairs in the molecule or ion; connect bonded atoms with a line, place the remaining e – pairs as needed, in nonbonded pairs or multiple bonds, so that each atom is surrounded by an octet (or duet for H). Calculate formal charges: assign electrons to individual atoms [nonbonding e – + 1/2 (bonding e – )]; formal charge = valence electrons – assigned electrons. Assign oxidation numbers, assuming that the more electronegative element holds all electrons in a bond. Solve. Formal charges are shown near the atoms, oxidation numbers (ox. #) are listed below the structures. (a) 10 e – , 5 e – pairs (b) 32 valence e – , 16 e – pairs ox. #: N, +3; O, –2 ox #: P, +5; Cl, –1; O, –2 200 8 Chemical Bonding (c) 32 valence e – , 16 e – pairs pairs Solutions to Exercises (d) 26 valence e – , 13 e – ox. #: Cl, +5; H, +1; O, –2 ox. #: Cl, +7; O, –2 Check. Each atom is surrounded by an octet (or duet) and the sum of the formal charges and oxidation numbers is the charge on the particle. 8.50 Formal charges are given near the atoms, oxidation numbers are listed below the structures. (a) 18 e – , 9 e – pairs (b) 24 e – , 12 e – pairs ox. #: S, +4; O, –2 ox. #: S, +6; O, –2 (c) 26 e , 13 e pairs – – ox. #: S, +4; O, –2 (d) SO2 < SO3 < SO32− Double bonds are shorter than single bonds. SO2 has two resonance structures with alternating single and double bonds, for an approximate average ‘’one-anda-half” bond. SO3 has three resonance structures with one double and two single bonds, for an approximately, ‘’one-and-a-third” bond. SO32− has all single bonds. The order of increasing bond length is the order of decreasing bond type. SO2 (1.5) < SO3 (1.3) < SO32− (1.0). 8.51 (a) Plan. Count valence electrons, draw all possible correct Lewis structures, taking note of alternate placements for multiple bonds. Solve. 18 e – , 9 e – pairs Check. The octet rule is satisfied. 201 8 Chemical Bonding (b) Solutions to Exercises Plan. Isoelectronic species have the same number of valence electrons and the same electron configuration. Solve. A single O atom has 6 valence electrons, so the neutral ozone molecule O 3 is isoelectronic with NO 2 – . Check. The octet rule is satisfied. (c) Since each N–O bond has partial double bond character, the N–O bond length in NO 2 – should be shorter than in species with formal N–O single bonds. 16 e – , 8 e – pairs 8.52 (a) (b) (c) More than one correct Lewis structure can be drawn, so resonance structures are needed to accurately describe the structure. NO 2 + has 16 valence electrons. Consider other triatomic molecules involving second-row nonmetallic elements. O 3 2 + or C 3 4 – are not “common” (or stable). CO 2 is common and matches the description (as does N 3 – , azide ion). 8.53 Plan/Solve. The Lewis structures are as follows: 5 e – pairs 8 e – pairs 12 e – pairs The more pairs of electrons shared by two atoms, the shorter the bond between the atoms. The average number of electron pairs shared by C and O in the three species is 3 for CO, 2 for CO 2 , and 1.33 for CO 3 2 –. This is also the order of increasing bond length: CO < CO 2 < CO 3 2 –. 8.54 The Lewis structures are as follows: 5 e – pairs 9 e – pairs 12 e – pairs 202 8 Chemical Bonding Solutions to Exercises The average number of electron pairs in the N–O bond is 3.0 for NO + , 1.5 for NO 2 – , and 1.33 for NO 3 – . The more electron pairs shared between two atoms, the shorter the bond. Thus the N–O bond lengths vary in the order NO + < NO 2 – < NO 3 – . 8.55 (a) Two equally valid Lewis structures can be drawn for benzene. Each structure consists of alternating single and double C–C bonds; a particular bond is single in one structure and double in the other. The concept of resonance dictates that the true description of bonding is some hybrid or blend of the two Lewis structures. The most obvious blend of these two resonance structures is a molecule with six equivalent C–C bonds, each with some but not total doublebond character. If the molecule has six equivalent C–C bonds, the lengths of these bonds should be equal. (b) The resonance model described in (a) has six equivalent C–C bonds, each with some double bond character. That is, more than one pair but less than two pairs of electrons is involved in each C–C bond. This model predicts a uniform C–C bond length that is shorter than a single bond but longer than a double bond. 8.56 (a) (b) The resonance model of this molecule has bonds that are neither single nor double, but somewhere in between. This results in bond lengths that are intermediate between C–C single and C = C double bond lengths. (c) Exceptions to the Octet Rule 8.57 (a) (b) The octet rule states that atoms will gain, lose, or share electrons until they are surrounded by eight valence electrons. The octet rule applies to the individual ions in an ionic compound. That is, the cation has lost electrons to achieve an octet and the anion has gained electrons to achieve an octet. For example, in MgCl 2 , Mg loses 2 e – to become Mg 2 + with the electron configuration of Ne. Each Cl atom gains one electron to form Cl – with the electron configuration of Ar. 203 8 Chemical Bonding 8.58 Solutions to Exercises Carbon, in group 14, needs to form four single bonds to achieve an octet, as in CH 4 . Nitrogen, in group 15, needs to form three, as in NH 3 . If G = group number and n = the number of single bonds, G + n = 18 is a general relationship for the representative nonmetals. Check: O as in H 2 O (G = 16) + (n = 2 bonds) = 18 8.59 The most common exceptions to the octet rule are molecules with more than eight electrons around one or more atoms, usually the central atom. Examples: SF 6 , PF 5 In the third period, atoms have the space and available orbitals to accommodate extra electrons. Since atomic radius increases going down a family, elements in the third period and beyond are less subject to destabilization from additional electron-electron repulsions. Also, the third shell contains d orbitals that are relatively close in energy to 3s and 3p orbitals (the ones that accommodate the octet) and provide an allowed energy state for the extra electrons. (a) 26 e – , 13 e – pairs 8.60 8.61 Other resonance structures with one, two, or three double bonds can be drawn. While a structure with three double bonds minimizes formal charges, all structures with double bonds violate the octet rule. Theoretical calculations show that the single best Lewis structure is the one that doesn’t violate the octet rule. Such a structure is shown above. (b) 6 e – , 3 e – pairs 6 electrons around Al; impossible to satisfy octet rule with only 6 valence electrons. (c) 16 e – , 8 e – pairs 3 resonance structures; all obey octet rule. (d) 20 e – , 10 e – pairs Obeys octet rule. 204 8 Chemical Bonding (e) 40 e – , 20 e – pairs Solutions to Exercises Does not obey octet rule; 10 e – around central Sb 8.62 (a) 8 e – , 4 e – pairs H H N H H + (b) 16 e – , 8 e – pairs S C N − S C N − S C N − Three resonance structures, all obey the octet rule. The middle structure is probably the largest contributor to the actual structure. (See Sample Exercise 8.9.) (c) 26 e – , 13 e – pairs Cl P Cl Cl (d) 34 e – , 17 e – pairs F F Te F F Does not obey the octet rule. (e) 22 e – , 11 e – pair Does not obey the octet rule. 8.63 (a) 16 e – , 8 e – pairs This structure violates the octet rule; Be has only 4 e – around it. (b) (c) The formal charges on each of the atoms in the four resonance structures are: 205 8 Chemical Bonding Solutions to Exercises Formal charges are minimized on the structure that violates the octet rule; this form is probably most important. Note that this is a different conclusion than for molecules that have resonance structures with expanded octets that minimize formal charge. 8.64 (a) 19 e – , 9.5 e – pairs, odd electron molecule (b) None of the structures satisfies the octet rule. In each structure, one atom has only 7 e – around it. If a molecule has an odd number of electrons in the valence shell, no Lewis structure can satisfy the octet rule. (c) Formal charge arguments predict that the two resonance structures with the odd electron on O are most important. This contradicts electronegativity arguments, which would predict that the less electronegative atom, Cl, would be more likely to have fewer than 8 e – around it. Bond Enthalpies 8.65 Analyze. Given: structural formulas. Find: enthalpy of reaction. Plan. Count the number and kinds of bonds that are broken and formed by the reaction. Use bond enthalpies from Table 8.4 and Equation 8.12 to calculate the overall enthalpy of reaction, ΔH. Solve. (a) ΔH = 2D(O–H) + D(O–O) + 4D(C–H) + D(C=C) –2D(O–H) – 2D(O–C) – 4D(C–H) – D(C–C) ΔH = D(O–O) + D(C=C) – 2D(O–C) – D(C–C) = 146 + 614 – 2(358) – 348 = –304 kJ (b) ΔH = 5D(C–H) + D(C ≡ N) + D(C=C) – 5D(C–H) – D(C ≡ N) – 2D(C–C) = D(C=C) – 2D(C–C) = 614 – 2(348) = –82 kJ (c) ΔH = 6D(N–Cl) – 3D(Cl–Cl) – D(N ≡ N) = 6(200) – 3(242) – 941 = –467 kJ 8.66 (a) ΔH = 3D(C–Br) + D(C–H) + D(Cl–Cl) – 3D(C–Br) – (C–Cl) – D(H–Cl) = D(C–H) + D(Cl–Cl) – D(C–Cl) – D(H–Cl) ΔH = 413 + 242 – 328 – 431 = –104 kJ (b) ΔH = 4D(C–H) + 2D(C–S) + 2D(S–H) + D(C–C) + 2D(H–Br) –4D(S–H) – D(C–C) – 2D(C–Br) – 4D(C–H) = 2D(C–S) + 2D(H–Br) – 2D(S–H) – 2D(C–Br) 206 8 Chemical Bonding ΔH = 2(259) + 2(366) – 2(339) – 2(276) = 20 kJ (c) = D(N–N) + D(Cl–Cl) – 2D(N–Cl) ΔH = 163 + 242 – 2(200) = 5 kJ 8.67 Solutions to Exercises ΔH = 4D(N–H) + D(N–N) + D(Cl–Cl) – 4D(N–H) – 2D(N–Cl) Plan. Draw structural formulas so bonds can be visualized. Then use Table 8.4 and Equation 8.12. Solve. (a) ΔH = 8D(C–H) + D(O=O) – 6D(C–H) – 2D(C–O) – 2D(O–H) = 2D(C–H) + D(O=O) – 2D(C–O) – 2D(O–H) = 2(413) + (495) – 2(358) – 2(463) = –321 kJ (b) H–H + Br–Br → 2 H–Br ΔH = D(H–H) + D(Br–Br) – 2D(H–Br) = (436) + (193) – 2(366) = –103 kJ (c) a) 8.68 2 H–O–O–H → 2 H–O–H + O = O ΔH = 4D(O–H) + 2D(O–O) – 4D(O–H) – D(O=O) ΔH = 2D(O–O) – D(O=O) = 2(146) – (495) = –203 kJ Plan. Draw structural formulas so bonds can be visualized. (a) H H C H H C H H C H H + 5O O 3 O C O+ 4 H O H ( Solve. ΔH = 2D(C–C) + 8D(C–H) + 5D(O=O) – 6D(C=O) – 8D(O–H) = 2(348) + 8(413) + 5(495) – 6(799) – 8(463) = −2023 kJ (b) H H C H H C H O H + 3O O 2O C O+3 H O H ΔH = D(C–C) + 5D(C–H) + D(C–O) + D(O–H) + 3D(O=O) – 4D(C=O) – 6D(O–H) =348 + 5(413) + 358 + 3(495) – 4(799) – 5(463) = –1255 KJ (c) Plan. Use bond enthalpies to calculate ΔH for the reaction with S 8 (g) as a product. Then, 207 8 Chemical Bonding 8H 2 S(g) → 8H 2 (g) + S 8 (g) S 8 (g) → S 8 (s) 8H 2 S(g) → 8H 2 (g) + S 8 (s) ΔH Solutions to Exercises º − ΔH f for S 8 (g ) º ΔH rxn = [ΔH − ΔH f S 8 (g )] ΔH = 16D(S–H) – 8(H–H) – 8(S–S) = 16(339) – 8(436) – 8(266) = –192 kJ ΔH rxn = ΔH − ΔH ° S 8 (g) = −192 kJ − 102.3 kJ = −294.3 = −294 kJ f 8.69 Plan. Draw structural formulas so bonds can be visualized. Then use Table 8.4 and Equation 8.12. Solve. (a) ΔH = D(N ≡ N) + 3D(H–H) – 6(N–H) = 941 kJ + 3(436 kJ) – 6(391 kJ) = –97 kJ/2 mol NH 3 ; exothermic (b) Plan. Use Equation 5.31 to calculate ΔH rxn from ΔH ° values. f ΔH ° = Σn ΔH ° (products) − Σn ΔH ° (reactants). ΔH ° NH 3 (g) = −46.19 kJ. rxn f f f Solve. ΔH ° = 2 ΔH ° NH 3 (g) − 3 ΔH ° H 2 (g) − ΔH ° N 2 (g) rxn f f f ΔH ° = 2( −46.19) − 3(0) − 0 = −92.38 kJ/2 mol NH 3 rxn The ΔH calculated from bond enthalpies is slightly more exothermic (more negative) than that obtained using ΔH ° values. f 8.70 (a) ΔH = 4D(C–H) + D(C=C) + D(H–H) – 6D(C–H) – D(C–C) = D(C=C) + D(H–H) – 2D(C–H) – D(C–C) ΔH = 614 + 436 – 2(413) – 348 = –124 kJ (b) ΔH ° = ΔH ° C 2 H 6 (g) − ΔH ° C 2 H 4 (g) − ΔH ° H 2 (g) f f f = –84.68 – 52.30 – 0 = –136.98 kJ The values of ΔH for the reaction differ because the bond enthalpies used in part (a) are average values that can differ from one compound to another. For example, the exact enthalpy of a C–H bond in C 2 H 4 is probably not equal to the enthalpy of a C–H bond in C 2 H 6 . Thus, reaction enthalpies calculated from average bond enthalpies are estimates. On the other hand, standard enthalpies of formation are measured quantities and should lead to accurate reaction enthalpies. The advantage of average bond enthalpies is that they can be used for reactions where no measured enthalpies of formation are available. 208 8 Chemical Bonding 8.71 8.72 (a) (i) Solutions to Exercises The average Ti–Cl bond enthalpy is just the average of the four values listed. 430 kJ/mol. ΔH = 2D(F–F) – 4D(C–F) = 2(155) – 4(485) = –1630 kJ (ii) ΔH = D(C≡O) + 3D(F–F) – 4D(C–F) – 2D(O–F) = 1072 + 3(155) – 4(485) – 2(190) = –783 kJ (iii) ΔH = 2D(C=O) + 4D(F–F) – 4D(C–F) – 4D(O–F) = 2(799) + 4(155) – 4(485) – 4(190) = –482 kJ Reaction (i) is most exothermic. (b) The more oxygen atoms bound to carbon, the less exothermic the reaction in this series. Additional Exercises 8.73 Six nonradioactive elements in the periodic table have Lewis symbols with single dots. Yes, they are in the same family, assuming H is placed with the alkali metals, as it is on the inside cover of the text. This is because the Lewis symbol represents the number of valence electrons of an element, and all elements in the same family have the same number of valence electrons. By definition of a family, all elements with the same Lewis symbol must be in the same family. (a) Lattice energy is proportional to Q 1 Q 2 /d. For each of these compounds, Q 1 Q 2 is the same. The anion H – is present in each compound, but the ionic radius of the cation increases going from Be to Ba. Thus, the value of d (the cation-anion separation) increases and the ratio Q 1 Q 2 /d decreases. This is reflected in the decrease in lattice energy going from BeH 2 to BaH 2 . Again, Q 1 Q 2 for ZnH 2 is the same as that for the other compounds in the series and the anion is H – . The lattice energy of ZnH 2 , 2870 kJ, is closest to that of MgH 2 , 2791 kJ. The ionic radius of Zn 2 + is similar to that of Mg 2 +. 8.74 (b) 8.75 (a) 209 8 Chemical Bonding Solutions to Exercises The difference in lattice energy between LiCl and LiI is 104 kJ. The difference between NaCl and NaI is 106 kJ; the difference between NaCl and NaBr is 56 kJ, or 53% of the difference between NaCl and NaI. Applying this relationship to the Li salts, 0.53(104 kJ) = 55 kJ difference between LiCl and LiBr. The approximate lattice energy of LiBr is (834 – 55) kJ = 779 kJ. (b) By analogy to the Na salts, the difference between lattice energies of CsCl and CsBr should be approximately 53% of the difference between CsCl and CsI. The lattice energy of CsBr is approximately 627 kJ. (c) By analogy to the oxides, the difference between the lattice energies of MgCl 2 and CaCl 2 should be approximately 66% of the difference between MgCl 2 and SrCl 2 . That is, 0.66(199 kJ) = 131 kJ. The lattice energy of CaCl 2 is approximately (2326 – 131) kJ = 2195 kJ. 8.76 AlN and ScN have the same charges on cations (3+) and anions (3−), so differences in lattice energy are due to differences in ion separation (d). With the same anion in both compounds, any separation difference depends on the difference in ionic radii of Al3+ and Sc3+. The Al3+ cation is isoelectronic with Ne in the second row, while Sc3+ is isoelectronic with Ar in the third row. Size increases going down a family, so the ionic radius of Al3+ is smaller than that of Sc3+. Lattice energy is inversely proportional to ion separation, so AlN, with the smaller separation, has the larger lattice energy. E= − 8.99 × 10 9 J - m C2 × 4(1.60 × 10 −19 C) 2 (1.14 + 1.26) × 10 −10 m = −3.836 × 10 −18 = −3.84 × 10 −18 J 8.77 On a molar basis: (–3.836 × 10 – 18 J)(6.022 × 10 2 3) = –2.310 × 10 6 J = –2310 kJ Note that the absolute value of this potential energy is less than the lattice energy of CaO, 3414 kJ/mol. The difference represents the added energy of putting all the Ca 2 +O 2 – ion pairs together in a three-dimensional array, similar to the one in Figure 8.3. 8.78 E = Q 1 Q 2 /d; k = 8.99 × 10 9 J-m/coul 2 (a) Na + , Br − : E = − 8.99 × 10 9 J - m C2 × (1 × 1.60 × 10 −19 C) 2 (1.16 + 1.82) × 10 −10 m = −7.7230 × 10 −19 = −7.72 × 10 −19 J The sign of E is negative because one of the interacting ions is an anion; this is an attractive interaction. On a molar basis: –7.723 × 10 – 19 × 6.022 × 10 2 3 = –4.65 × 10 5 J = –465 kJ 210 8 Chemical Bonding (b) Rb + , Br − : E = − 8.99 × 10 9 J - m C2 × Solutions to Exercises (1 × 1.60 × 10 −19 C) 2 (1.66 + 1.82) × 10 −10 m = −6.61 × 10 −19 J On a molar basis: –3.98 × 10 5 J = –398 kJ (c) Sr 2 + , S 2 − : E = − 8.99 × 10 9 J - m C 2 × ( 2 × 1.60 × 10 −19 C) 2 (1.32 + 1.70) × 10 −10 m = −3.05 × 10 −18 J On a molar basis: –1.84 × 10 6 J = –1.84 × 10 3 kJ 8.79 (a) A polar molecule has a measurable dipole moment; its centers of positive and negative charge do not coincide. A nonpolar molecule has a zero net dipole moment; its centers of positive and negative charge do coincide. Yes. If X and Y have different electronegativities, they have different attractions for the electrons in the molecule. The electron density around the more electronegative atom will be greater, producing a charge separation or dipole in the molecule. μ = Qr. The dipole moment, μ, is the product of the magnitude of the separated charges, Q, and the distance between them, r. (b) (c) 8.80 Molecule (b) H 2 S and ion (c) NO 2 – contain polar bonds. The atoms that form the bonds (H–S) and N–O) have different electronegativity values. (a) (b) 2NaAlH4(s) → 2NaH(s) + 2Al(s) + 3H2(g) Hydrogen is the only nonmetal in NaAlH4, so we expect it to be most electronegative. (The position of H on the periodic table is problematic. Its electronegativity does not fit the typical trend for Gp 1A elements.) For the two metals, Na and Al, electronegativity increases moving up and to the right on the periodic table, so Al is more electronegative. The least electronegative element in the compound is Na. Covalent bonds hold polyatomic anions together; elements involved in covalent bonding have smaller electronegativity differences than those that are involved in ionic bonds. Possible covalent bonds in NaAlH4 are Na−H and Al−H. Al and H have a smaller electronegativity difference than Na and H and are more likely to form covalent bonds. The anion has an overall 1− charge, so it can be thought of as four hydride ions and one Al3+ ion. The formula is AlH4−. For the purpose of counting valence electrons, assume neutral atoms. 8 e−, 4 e− pairs H H Al H H − 8.81 (c) 8.82 (a) B–O. The most polar bond will be formed by the two elements with the greatest difference in electronegativity. Since electronegativity increases moving right and up on the periodic chart, the possibilities are B–O and Te–O. These two bonds are likely to have similar electronegativitiy differences (3 columns apart vs. 3 rows 211 8 Chemical Bonding (b) (c) Solutions to Exercises apart). Values from Figure 8.6 confirm the similarity, and show that B–O is slightly more polar. Te–I. Both are in the fifth row of the periodic chart and have the two largest covalent radii among this group of elements. TeI 2 . Te needs to participate in two covalent bonds to satisfy the octet rule, and each I atom needs to participate in one bond, so by forming a TeI 2 molecule, the octet rule can be satisfied for all three atoms. (d) B 2 O 3 . Although this is probably not a purely ionic compound, it can be understood in terms of gaining and losing electrons to achieve a noble-gas configuration. If each B atom were to lose 3 e – and each O atom were to gain 2 e – , charge balance and the octet rule would be satisfied. P 2 O 3 . Each P atom needs to share 3 e – and each O atom 2 e – to achieve an octet. Although the correct number of electrons seem to be available, a correct Lewis structure is difficult to imagine. In fact, phosphorus (III) oxide exists as P 4 O 6 rather than P 2 O 3 (Chapter 22). 8.83 To calculate empirical formulas, assume 100 g of sample. (a) 76.0 g Ru 101.07 g / mol 24.0 g O 15.9994 g / mol = 0.752 mol Ru; 0.752/0.752 = 1 Ru = 1.50 mol O; 1.50/0.762 = 2 O The empirical formula of compound 1 is RuO2. (b) 61.2 g Ru = 0.6055 mol Ru; 0.6055/0.6055 = 1 Ru 101.07 g / mol 38.8 g O 15.9994 g / mol = 2.425 mol O; 2.425/0.6055 = 4 O The empirical formula of compound 2 is RuO4. (c) Ionic compounds have very high melting points, while the melting points of molecular compounds are lower and variable. Clearly the black powder, m.p. > 1200°C, is ionic and the yellow substance, m.p. = 25°C, is molecular. Substances with metals in high oxidation states are often molecular. RuO4 contains Ru(VIII), while RuO2 contains Ru(IV), so RuO4 is more likely to be molecular. The yellow compound is RuO4, ruthenium tetroxide. The black compound is RuO2, ruthenium(IV) oxide. 8.84 To calculate empirical formulas, assume 100 g of sample. (a) 47.7 g Cr 51.9961g / mol = 0.9174 mol Cr; 0.9174/0.9174 = 1 Cr 212 8 Chemical Bonding 52.3 g F 18.9984 g / mol Solutions to Exercises = 2.753 mol F; 2.753/0.9174 = 3 F The empirical formula of compound 1 is CrF3. (b) 45.7 g Mo 95.94 g / mol 54.3 g F 18.9984 g / mol = 0.4763 mol Mo; 0.4763/0.4763 = 1 Mo = 2.858 mol F; 2858/0.4763 = 6 F The empirical formula of Compound 2 is MoF6. (c) The high-melting green powder is ionic and contains the group 6B metal in the lower oxidation state, Cr(III). The green powder is CrF3, chromium(III) fluoride. The colorless liquid is molecular and contains the group 6B metal in the higher oxidation state, Mo(VI). The colorless liquid is MoF6, molybdenum hexafluoride. 8.85 (a) 12 + 3 + 15 = 30 valence e – , 15 e – pairs. Structures with H bound to N and nonbonded electron pairs on C can be drawn, but the structures above minimize formal charges on the atoms. (b) The resonance structures indicate that triazine will have six equal C–N bond lengths, intermediate between C–N single and C–N double bond lengths. (See Solutions 8.55 and 8.56.) From Table 8.5, an average C–N length is 1.43 Å, a C=N length is 1.38 Å. The average of these two lengths is 1.405 Å. The C–N bond length in triazine should be in the range 1.40–1.41 Å. 8.86 Use the method detailed in Section 8.5, A Closer Look, to estimate partial charges from electronegativity values. From Figure 8.6, the electronegativity of Br is 2.8 and of Cl is 3.0. Br has 2.8/(3.0 + 2.8) = 0.48 of the charge of the bonding e – pair. Cl has 3.0/(3.0 + 2.8) = 0.52 of the charge of the bonding e – pair. This amounts to 0.52 × 2e = 1.04e on Cl or 0.04e more than a neutral Cl atom. This implies a –0.04 charge on Cl and +0.04 charge on Br. From Figure 7.7, the covalent radius of Br is 1.14 Å and of Cl is 0.99 Å. The Br–Cl separation is 2.13 Å. μ = Qr = 0.04e × 1.60 × 10 −19 C 1 × 10 −10 m 1D × 2.13 Å × × = 0.41 D e Å 3.34 × 10 − 30 C - m Clearly, this method is approximate. The estimated dipole moment of 0.41 D is within 28% of the measured value of 0.57 D. 213 8 Chemical Bonding 8.87 Solutions to Exercises I 3 – has a Lewis structure with an expanded octet of electrons around the central I. F cannot accommodate an expanded octet because it is too small and has no available d orbitals in its valence shell. 8.88 Formal charge (FC) = # valence e – – (# nonbonding e – + 1/2 # bonding e – ) (a) 18 e – , 9 e – pairs FC for the central O = 6 – [2 + 1/2 (6)] = +1 (b) 48 e – , 24 e – pairs FC for P = 5 – [0 + 1/2 (12)] = –1 The three nonbonded pairs on each F have been omitted. (c) 17 e – ; 8 e – pairs, 1 odd e – The odd electron is probably on N because it is less electronegative than O. Assuming the odd electron is on N, FC for N = 5 – [1 + 1/2 (6)] = +1. If the odd electron is on O, FC for N = 5 – [2 + 1/2 (6)] = 0. (d) 28 e – , 14 e – pairs (e) 32 e – , 16 e – pairs FC for I = 7 – [4 + 1/2 (6)] = 0 8.89 (a) 14e – , 7 e – pairs FC for Cl = 7 – [0 + 1/2 (8)] = +3 32 e – , 16 e – pairs FC on Cl = 7 – [6 + 1/2(2)] = 0 FC on Cl = 7 – [0 + 1/2(8)] = +3 (b) (c) The oxidation number of Cl is +1 in ClO – and +7 in ClO 4 – . The definition of formal charge assumes that all bonding pairs of electrons are equally shared by the two bonded atoms, that all bonds are purely covalent. The definition of oxidation number assumes that the more electronegative element in the bond gets all of the bonding electrons, that the bonds are purely ionic. These two definitions represent the two extremes of how electron density is distributed between bonded atoms. 214 8 Chemical Bonding Solutions to Exercises In ClO – and ClO 4 – , Cl is the less electronegative element, so the oxidation numbers have a higher positive value than the formal charges. The true description of the electron density distribution is somewhere between the extremes indicated by formal charge and oxidation number. (d) Oxidizing power is the tendency of a substance to be reduced, to gain electrons. Oxidation numbers show the maximum electron deficiency (or excess) of a substance. The higher the oxidation number of the central atom in an oxyanion, the greater its electron deficiency and oxidizing power. Formal charges can also be used to show oxidizing (or reducing) power, but trends are less obvious because the magnitudes are smaller. 8.90 (a) In the leftmost structure, the more electronegative O atom has the negative formal charge, so this structure is likely to be most important. (b) In general, the more shared pairs of electrons between two atoms, the shorter the bond, and vice versa. That the N–N bond length in N 2 O is slightly longer than the typical N≡N indicates that the middle and right resonance structures where the N atoms share less than three electron pairs are contributors to the true structure. That the N–O bond length is slightly shorter than a typical N=O indicates that the middle structure, where N and O share more than two electron pairs, does contribute to the true structure. This physical data indicates that while formal charge can be used to predict which resonance form will be more important to the observed structure, the influence of minor contributors on the true structure cannot be ignored. 8.91 ΔH = 8D(C–H) – D(C–C) – 6D(C–H) – D(H–H) = 2D(C–H) – D(C–C) – D(H–H) = 2(413) – 348 – 436 = +42 kJ ΔH = 8D(C–H) + 1/2 D(O=O) – D(C–C) – 6D(C–H) – 2D(O–H) = 2D(C–H) + 1/2 D(O=O) – D(C–C) – 2D(O–H) = 2(413) + 1/2 (495) – 348 – 2(463) = –200 kJ The fundamental difference in the two reactions is the formation of 1 mol of H–H bonds versus the formation of 2 mol of O–H bonds. The latter is much more exothermic, so the reaction involving oxygen is more exothermic. 8.92 (a) ΔH = 5D(C–H) + D(C–C) + D(C–O) + D(O–H) – 6D(C–H) – 2D(C–O) = D(C–C) + D(O–H) – D(C–H) – D(C–O) = 348 kJ + 463 kJ – 413 kJ – 358 kJ ΔH = +40 kJ; ethanol has the lower enthalpy 215 8 Chemical Bonding (b) = 2D(C–O) – D(C=O) = 2(358 kJ) – 799 kJ Solutions to Exercises ΔH = 4D(C–H) + D(C–C) + 2D(C–O) – 4D(C–H) – D(C–C) – D(C=O) ΔH = –83 kJ; acetaldehyde has the lower enthalpy (c) ΔH = 8D(C–H) + 4D(C–C) + D(C=C) – 8D(C–H) – 2D(C–C) – 2D(C=C) = 2D(C–C) – D(C=C) = 2(348 kJ) – 614 kJ ΔH = +82 kJ; cyclopentene has the lower enthalpy (d) ΔH = 3D(C–H) + D(C–N) + D(C ≡ N) – 3D(C–H) – D(C–C) – D(C ≡ N) = D(C–N) – D(C–C) = 293 kJ – 348 kJ ΔH = –55 kJ; acetonitrile has the lower enthalpy 8.93 (a) ΔH = 20D(C–H) + 8D(C–C) + 12D(C–O) + 24D(O–N) + 12D(N=O) – [6D(N ≡ N) + 24D(C=O) + 20D(H–O) + D(O=O)] ΔH = 20(413) + 8(348) + 12(358) + 24(201) + 12(607) – [6(941) + 24(799) + 20(463) + 495] = –7129 kJ 1.00 g C 3 H 5 N 3 O 9 × −7129 kJ 1 mol C 3 H 5 N 3 O 9 × = 7.85 kJ/g C 3 H 5 N 3 O 9 4 mol C 3 H 5 N 3 O 9 227 .1 g C 3 H 5 N 3 O 9 (b) 8.94 (a) 4C 7 H 5 N 3 O 6 (s) → 6N 2 (g) + 7CO 2 (g) + 10H 2 O(g) + 21C(s) C3H6N6O6 12 + 6 + 30 + 36 = 84 e – , 42 e – pairs 216 8 Chemical Bonding Solutions to Exercises 42 e – pairs – 24 shared e – pairs 18 unshared (lone) e – pairs Use unshared pairs to complete octets on terminal O atoms (15 unshared pairs) and ring N atoms (3 unshared pairs). (b) No C=N bonds in the 6-membered ring are possible, because all C octets are complete with 4 bonds to other atoms. N=N are possible, as shown below. There are 8 possibilities involving some combination of N–N and N=N groups [1 with 0 N=N, 3 with 1 N=N, 3 with 2N=N, 1 with 3N=N]. A resonance structure with 1 N=N is shown below. Each terminal O=N–O group has two possible placements for the N=O. This generates 8 structures with 0 N=N groups (and 3 O = N–O groups), 4 with 1 N=N and 2 O=N–O, 2 with 2 N=N and 1 O=N–O, and 1 with 3 N=N and no O=N–O. This sums to a total of 15 resonance structures (that I can visualize). Can you find others? (c) (d) (e) C 3 H 6 N 6 O 6 (s) → 3CO(g) + 3N 2 (g) + 3H 2 O(g) The molecule contains N=O, N=N, C–H, C–N, N–O, and N–N bonds. According to Table 8.4, N–N bonds have the smallest bond enthalpy and are weakest. Calculate the enthalpy of decomposition for the resonance structure drawn in part (a). ΔH = 3D(N=O) + 3D(N–O) + 3D(N–N) + 6D(N–C) + 6D(C–H) – 3D(C≡O) – 3D(N≡N) – 6D(O–H) = 3(607) + 3(201) + 3(163) + 6(293) + 6(413) – 3(1072) – 3(941) – 6(463) = –1668 kJ/mol C 3 H 6 N 6 O 6 5.0 g C 3 H 6 N 6 O 6 × −1668 kJ 1 mol C 3 H 6 N 6 O 6 × = 37.55 = 38 kJ 222.1 g C 3 H 6 N 6 O 6 mol C 3 H 6 N 6 O 6 While exchanging N=O and N–O bonds has no effect on the enthalpy calculation, structures with N=N and 2 N–O do have different enthalpy of decomposition. For the resonance structure with 3 N=N and 6 N–O bonds instead of 3 N–N, 3 N–O and 3 N=O, ΔH = –2121 kJ/mol. The actual enthalpy of decomposition is probably somewhere between –1668 and –2121 kJ/mol. The enthalpy charge for the decomposition of 5.0 g RDX is then in the range 38–48 kJ. 217 8 Chemical Bonding 8.95 Solutions to Exercises When comparing the same pair of bonded atoms (C–N vs. C=N vs. C≡N), the shorter the bond the greater the bond energy, but the two quantities are not necessarily directly proportional. The plot clearly shows that there are no simple length/strength correlations for single bonds alone, double bonds alone, triple bonds alone, or among different pairs of bonded atoms (all C–C bonds vs. all C–N bonds, etc.). 8.96 (a) (b) S–N ≈ 1.77 Å (sum of the bonding atomic radii from Figure 7.7). S–O ≈ 1.75 Å (the sum of the bonding atomic radii from Figure 7.7.) Alternatively, half of the S–S distance in S 8 (1.02) plus half of the O–O distance from Table 8.5 (0.74) is 1.76 Å. Owing to the resonance structures for SO 2 , we assume that the S–O bond in SO 2 is intermediate between a double and single bond, so the distance of 1.43 Å should be significantly shorter than an S–O single bond distance, 1.75 Å. 54 e – , 27 e – pair (c) (d) The observed S–O bond distance, 1.48 Å, is similar to that in SO 2 , 1.43 Å, which can be described by resonance structures showing both single and double S–O bonds. Thus, S 8 O must have resonance structures with both single and double S–O bonds. The structure with the S=O bond has 5 e – pairs about this S atom. To the extent that this resonance form contributes to the true structure, the S atom bound to O has more than an octet of electrons around it. 218 8 Chemical Bonding Integrative Exercises 8.97 (a) (b) Solutions to Exercises Ti 2 + : [Ar]3d 2 ; Ca : [Ar]4s 2 . Yes. The two valence electrons in Ti 2 + and Ca are in different principle quantum levels and different subshells. According to the Aufbau Principle, valence electrons will occupy the lowest energy empty orbital. Thus, in Ca the 4s is lower in energy than the 3d, while in Ti 2 +, the 3d is lower in energy than the 4s. Since there is only one 4s orbital, the two valence electrons in Ca are paired. There are five degenerate 3d orbitals, so the two valence electrons in Ti 2 + are unpaired. Ca has no unpaired electrons, Ti 2 + has two. Sr(s) → Sr(g) Sr(g) → Sr + (g) + 1 e – Sr + (g) → Sr 2 +(g) + 1 e – Cl 2 (g) → 2Cl(g) 2Cl(g) + 2 e – → 2Cl – (g) SrCl 2 (s) → Sr(s) + Cl 2 (g) ° ΔH ° Sr(g) [ΔH sub Sr(s)] f (c) 8.98 (a) I 1 Sr I 2 Sr 2 ΔH ° Cl(g) [D(Cl 2 )] f 2E 1 Cl − ΔH ° SrCl 2 f SrCl 2 (s) → Sr 2 +(g) + 2Cl – (g) (b) ΔH latt ΔH ° SrCl 2 (s) = ΔH ° Sr(g) + I 1 (Sr) + I 2 (Sr) + 2 ΔH ° Cl(g) + 2E(Cl) − ΔH latt SrCl 2 f f f ΔH ° SrCl 2 (s) = 164.4 kJ + 549 kJ + 1064 kJ + 2(121.7) kJ + 2( −349) kJ − 2127 kJ f = −804 kJ 8.99 The pathway to the formation of K 2 O can be written: 2K(s) → 2K(g) 2K(g) → 2K + (g) + 2 e – 1/2 O 2 (g) → O(g) O(g) + 1 e – → O – (g) O – (g) + 1 e – → O 2 –(g) 2K + (g ) + O 2 − (g ) → K 2 O(s) 2ΔH ° K(g) f 2 I 1 (K) ΔH ° O(g) f E 1 (O) E 2 (O) − ΔH latt K 2 O(s) ΔH ° K 2 O(s) f 2K(s) + 1/2 O 2 (g) → K 2 O(s) ΔH ° K 2 O(s) = 2 ΔH ° K(g) + 2 I 1 (K) + ΔH ° O(g) + E 1 (O) + E 2 (O) − ΔH latt K 2 O(s) f f f E 2 (O) = ΔH ° K 2 O(s) + ΔH latt K 2 O(s) − 2 ΔH ° K(g) − 2 I t (K) − ΔH ° O(g) − E 1 (O) f f f E 2 (O) = –363.2 kJ + 2238 kJ – 2(89.99) kJ – 2(419) kJ – 247.5 kJ – (–141) kJ = +750 kJ 8.100 (a) Assume 100 g. A: 87.7 g In/114.82 = 0.764 mol In; 0.764/0.384 ≈ 2 12.3 g S/32.07 B: = 0.384 mol S; 0.384/0.384 = 1 − 78.2 g In/114.82 = 0.681 mol In; 0.681/0.68 0 ~ 1 21.8 g S/32.07 = 0.680 mol S; 0.680/0.680 = 1 219 8 Chemical Bonding C: 29.5 g S/32.07 A: (b) (c) (d) In 2 S; B: InS; C: In 2 S 3 Solutions to Exercises = 0.920 mol S; 0.920/0.614 = 1.5 70.5 g In/114.82 = 0.614 mol In; 0.614/0.614 = 1 A: In(I); B: In(II); C: In(III) In(I) : [Kr]5s 2 4d 1 0; In(II) : [Kr]5s 1 4d 1 0; In(III) : [Kr]4d 1 0 None of these is a noble-gas configuration. The ionic radius of In 3 + in compound C will be smallest. Removing successive electrons from an atom reduces electron repulsion, increases the effective nuclear charge experienced by the valence electrons and decreases the ionic radius. The higher the charge on a cation, the smaller the radius. Lattice energy is directly related to the charge on the ions and inversely related to the interionic distance. Only the charge and size of the In varies in the three compounds. In(I) in compound A has the smallest charge and the largest ionic radius, so compound A has the smallest lattice energy and the lowest melting point. In(III) in compound C has the greatest charge and the smallest ionic radius, so compound C has the largest lattice energy and highest melting point. Even though Cl has the greater (more negative) electron affinity, F has a much larger ionization energy, so the electronegativity of F is greater. F: k(IE–EA) = k(1681 – (–328)) = k(2009) Cl: k(IE–EA) = k(1251 – (–349)) = k(1600) (e) 8.101 (a) (b) Electronegativiy is the ability of an atom in a molecule to attract electrons to itself. It can be thought of as the ability to hold its own electrons (as measured by ionization energy) and the capacity to attract the electrons of other atoms (as measured by electron affinity). Thus, both properties are relevant to the concept of electronegativity. EN = k(IE – EA). For F: 4.0 = k(2009), k = 4.0/2009 = 2.0 × 10 – 3 Cl: EN = 2.0 × 10 – 3 (1600) = 3.2 O: EN = 2.0 × 10 – 3 (1314 – (–141)) = 2.9 These values do not follow the trend on Figure 8.6. The Pauling scale on the figure shows O to be second only to F in electronegativity, more electronegative than Cl. The simple definition EN = k(IE – EA) that employs thermochemical properties of isolated, gas phase atoms does not take into account the complex bonding environment of molecules. (c) (d) 8.102 (a) Assume 100 g. 14.52 g C × 1.83 g H × 1 mol = 1.209 mol C; 1.209 / 1.209 = 1 12.011 g C 1 mol = 1.816 mol H; 1.816 / 1.209 = 1.5 1.008 g H 220 8 Chemical Bonding 64.30 g Cl × 19.35 g O × Solutions to Exercises 1 mol = 1.814 mol Cl; 1.814 / 1.209 = 1.5 35.453 g Cl 1 mol = 1.209 mol O; 1.209 / 1.209 = 1.0 15.9994 g O Multiplying by 2 to obtain an integer ratio, the empirical formula is C 2 H 3 Cl 3 O 2 . (b) The empirical formula weight is 2(12.0) + 3(1.0) + 3(35.5) + 2(16) = 165.5. The empirical formula is the molecular formula. 44 e – , 22 e – pairs (c) 8.103 (a) Assume 100 g. 62.04 g Ba × 37.96 g N × 1 mol = 0.4518 mol Ba; 0.4518 / 0.4518 = 1.0 137.33 g Ba 1 mol = 2.710 mol N; 2.710 / 0.4518 = 6.0 14.007 g N The empirical formula is BaN 6 . Ba has an ionic charge of 2+, so there must be two 1– azide ions to balance the charge. The formula of each azide ion is N 3 – . (b) 16 e – , 8 e – pairs (c) (d) The left structure minimizes formal charges and is probably the main contributor. The two N–N bond lengths will be equal. The two minor contributors would individually cause unequal N–N distances, but collectively they contribute equally to the lengthening and shortening of each bond. The N–N distance will be approximately 1.24 Å, the average N=N distance. C 2 H 2 : 10 e – , 5 e – pair N 2 : 10 e – , 5 e – pair 8.104 (a) (b) N 2 is an extremely stable, unreactive compound. Under appropriate conditions, it can be either oxidized (Section 22.7) or reduced (Sections 14.7 and 15.2). C 2 H 2 is a reactive gas, used in combination with O 2 for welding and as starting material for organic synthesis (Section 25.3). 2N 2 (g) + 5O 2 (g) → 2N 2 O 5 (g) 2C 2 H 2 (g) + 5O 2 (g) → 4CO 2 (g) + 2H 2 O(g) (c) 221 8 Chemical Bonding (d) = 2(11.30) – 2(0) – 5(0) = 22.60 kJ ΔH ° = 11.30 kJ/mol N 2 ox Solutions to Exercises ΔH ° (N 2 ) = 2 ΔH ° N 2 O 5 (g) − 2 ΔH ° N 2 (g) − 5ΔH ° O 2 (g) rxn f f f ΔH ° (C 2 H 2 ) = 4 ΔH ° CO 2 (g) + 2ΔH ° H 2 O(g) − 2 ΔH ° C 2 H 2 (g) − 5ΔH ° O 2 (g) rxn f f f f = 4(–393.5 kJ) + 2(–241.82 kJ) – 2(226.7 kJ) – 5(0) = –2511.0 kJ ° ΔH ox (C 2 H 2 ) = −1255.5 kJ/mol C 2 H 2 The oxidation of C 2 H 2 is highly exothermic, which means that the energy state of the combined products is much lower than that of the reactants. The reaction is “downhill” in an energy sense, and occurs readily. The oxidation of N 2 is mildly endothermic (energy of products higher than reactants) and the reaction does not readily occur. This is in agreement with the general reactivities from part (b). Referring to bond enthalpies in Table 8.4, when the C–H bonds are taken into account, even more energy is required for bond breaking in the oxidation of C 2 H 2 than in the oxidation of N 2 . The difference seems to be in the enthalpies of formation of the products. CO 2 (g) and H 2 O(g) have extremely exothermic ΔH ° f values, which cause the oxidation of C 2 H 2 to be energetically favorable. N 2 O 5 (g) has an endothermic ΔH ° value, which causes the oxidation of N 2 to be f energetically unfavorable. 8.105 (a) Assume 100 g of compound 69.6 g S × 30.4 g N × 1 mol S = 2.17 mol S 32.07 g 1 mol N = 2.17 mol N 14.01 g S and N are present in a 1:1 mol ratio, so the empirical formula is SN. The empirical formula weight is 46. MM/FW = 184.3/46 = 4 The molecular formula is S 4 N 4. (b) 44 e – , 22 e – pairs. Because of its small radius, N is unlikely to have an expanded octet. Begin with alternating S and N atoms in the ring. Try to satisfy the octet rule with single bonds and lone pairs. At least two double bonds somewhere in the ring are required. S N S N S N S N N S N S S N S N N S N S S N S N N S N S S N S N These structures carry formal charges on S and N atoms as shown. Other possibilities include: 222 8 Chemical Bonding S N S N S N S N S Solutions to Exercises S N N S N N S These structures have zero formal charges on all atoms and are likely to contribute to the true structure. Note that the S atoms that are shown with two double bonds are not necessarily linear, because S has an expanded octet. Other resonance structures with four double bonds are. S N S N S N S N N S N S S N S N In either resonance structure, the two ‘extra’ electron pairs can be placed on any pair of S atoms in ring, leading to a total of 10 resonance structures. The sulfur atoms alternately carry formal charges of +1 and –1. Without further structural information, it is not possible to eliminate any of the above structures. Clearly, the S 4 N 4 molecule stretches the limits of the Lewis model of chemical bonding. (c) Each resonance structure has 8 total bonds and more than 8 but less than 16 bonding e – pairs, so an “average” bond will be intermediate between a S–N single and double bond. We estimate an average S–N single bond length to be 1.77 Å (sum of bonding atomic radii from Figure 7.7). We do not have a direct value for a S–N double bond length. Comparing double and single bond lengths for C–C (1.34 Å, 1.54 Å), N–N (1.24 Å, 1.47 Å) and O–O (1.21 Å, 1.48 Å) bonds from Table 8.5, we see that, on average, a double bond is approximately 0.23 Å shorter than a single bond. Applying this difference to the S–N single bond length, we estimate the S–N double bond length as 1.54 Å. Finally, the intermediate S–N bond length in S 4 N 4 should be between these two values, approximately 1.60–165 Å. (The measured bond length is 1.62 Å.) S 4 N 4 → 4S(g) + 4N(g) ΔH = 4 ΔH ° S(g) + 4 ΔH ° N(g) − ΔH ° S 4 N 4 f f f (d) ΔH = 4(222.8 kJ) + 4(472.7 kJ) – 480 kJ = 2302 kJ This energy, 2302 kJ, represents the dissociation of 8 S–N bonds in the molecule; the average dissociation energy of one S–N bond in S 4 N 4 is then 2302 kJ/8 bonds = 287.8 kJ. 8.106 (a) Yes. In the structure shown in the exercise, each P atom needs 1 unshared pair to complete its octet. This is confirmed by noting that only 6 of the 10 valence e – pairs are bonding pairs. There are six P–P bonds in P 4. 20 e−, 10 e− pr P P P P (b) (c) 223 8 Chemical Bonding Solutions to Exercises In this Lewis structure, the octet rule is satisfied for all atoms. However, it requires P=P, which are uncommon because P has a covalent radius that is too large to accommodate the side-to-side π overlap of parallel p orbitals required for double bond formation. (d) From left to right, the formal charges are on the P atoms in the linear structure are −1, +1, +1, −1. In the tetrahedral structure, all formal charges are zero. Clearly the linear structure does not minimize formal charge and is probably less stable than the tetrahedral structure, owing to the difficulty of P=P bond formation (see above). C6 H 6 (g) → 6H(g) + 6C(g) ΔH ° = 6 ΔH ° H(g) + 6 ΔH ° C(g) − ΔH ° C 6 H 6 (g ) f f f 8.107 (a) ΔH ° = 6(217.94) kJ + 6(718.4) kJ – 82.9 kJ = 5535 kJ (b) (c) C 6 H 6 (g) → 6CH(g) C 6 H 6 (g) → 6H(g) + 6C(g) 6H(g) + 6C(g) → 6CH(g) C 6 H 6 (g) → 6CH(g) ΔH ° − 6 D(C −H) 5535 kJ − 6( 413) kJ 3057 kJ 3057 kJ is the energy required to break the six C–C bonds in C 6 H 6 (g). The average bond dissociation energy for one carbon-carbon bond in C 6 H 6 (g) is 3057 kJ = 509.5 kJ. 6 C − C bonds (d) The value of 509.5 kJ is between the average value for a C–C single bond (348 kJ) and a C=C double bond (614 kJ). It is somewhat greater than the average of these two values, indicating that the carbon-carbon bond in benzene is a bit stronger than we might expect. Br 2 (l) → 2Br(g) CCl 4 (l) → C(g) + 4Cl(g) ΔH ° = ΔH ° C(g) + 4 ΔH ° Cl(g) − ΔH ° CCl 4 (l) f f f ΔH ° = 2 ΔH ° Br(g) = 2(111.8) kJ = 223.6 kJ f 8.108 (a) (b) = 718.4 kJ + 4(121.7) kJ – (–139.3) kJ = 1344.5 1344.5 kJ 4 C − Cl bonds = 336.1 kJ (c) H 2 O 2 (l) → 2H(g) + 2O(g) 2H(g) + 2O(g) → 2OH(g) H 2 O 2 (l) → 2OH(g) D(O − O)(l) = 2ΔH ° H(g) + 2ΔH ° O(g) − ΔH ° H 2 O 2 (l ) − 2 D(O − H)(g) f f f = 2(217.94) kJ + 2(247.5) kJ – (–187.8) kJ – 2(463) kJ = 193 kJ 224 8 Chemical Bonding (d) The data are listed below. bond Br–Br C–Cl O–O D gas kJ/mol 193 328 146 Solutions to Exercises D liquid kJ/mol 223.6 336.1 192.7 Breaking bonds in the liquid requires more energy than breaking bonds in the gas phase. For simple molecules, bond dissociation from the liquid phase can be thought of in two steps: molecule (l) → molecule (g) molecule (g) → atoms (g) The first step is evaporation or vaporization of the liquid and the second is bond dissociation in the gas phase. Average bond enthalpy in the liquid phase is then the sum of the enthalpy of vaporization for the molecule and the gas phase bond dissociation enthalpies, divided by the number of bonds dissociated. This is greater than the gas phase bond dissociation enthalpy owing to the contribution from the enthalpy of vaporization. 225 ...
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This note was uploaded on 04/04/2009 for the course CLASSICS 222 taught by Professor Lopez during the Summer '07 term at Ohio State.

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