9-1

CHAPTER 9 MODELS OF CHEMICAL BONDING 9.1 a) Larger ionization energy decreases metallic character. b) Larger atomic radius increases metallic character. c) Larger number of outer electrons decreases metallic character. d) Larger effective nuclear charge decreases metallic character. 9.2 Bismuth is more metallic than nitrogen. Metallic properties increase from top to bottom in a group. 9.3 The tendency of main-group elements to form cations decreases from Group 1A(1) to 4A(14), and the tendency to form anions increases from Group 4A(14) to 7A(17). 1A(1) and 2A(2) elements form mono- and divalent cations, respectively, while 6A(16) and 7A(17) elements form di- and monovalent anions, respectively. 9.4 Metallic behavior increases to the left and down on the periodic table. a) Cs is more metallic since it is further down the alkali metal group than Na. b) Rb is more metallic since it is both to the left and down from Mg. c) As is more metallic since it is further down Group 5A than N. 9.5 a) O b) Be c) Se 9.6 Ionic bonding occurs between metals and nonmetals, covalent bonding between nonmetals, and metallic bonds between metals. a) Bond in CsF is ionic because Cs is a metal and F is a nonmetal. b) Bonding in N2 is covalent because N is a nonmetal. c) Bonding in Na(s) is metallic because this is a monatomic, metal solid. 9.7 a) covalent b) covalent c) ionic 9.8 a) Bonding in O3 would be covalent since O is a nonmetal. b) Bonding in MgCl2 would be ionic since Mg is a metal and Cl is a nonmetal. c) Bonding in BrO2 would be covalent since both Br and O are nonmetals. 9.9 a) metallic b) covalent c) ionic 9.10 Lewis electron-dot symbols show valence electrons as dots. Place one dot at a time on the four sides (this method explains the structure in b) and then pair up dots until all valence electrons are used.

Rb Si Ia) b) c)

9.11

a) b) c)KrBa Br

9.12

Sa) b) c)Sr P

9-2

9.13

a) b) c)As Se Ga

9.14 a) Assuming X is an A-group element, the number of dots (valence electrons) equals the group number. Therefore, X is a 6A(16) element. Its general electron configuration is [noble gas]ns2np4, where n is the energy level. b) X has three valence electrons and is a 3A(13) element with general e- configuration [noble gas]ns2np1. 9.15 a) 5A(15); ns2np3 b) 4A(14); ns2np2 9.16 The lattice energy is an important indication of the strength of ionic interactions and more than compensates for the required energy to form ions from metals and non-metals. 9.17 Because the lattice energy is the result of electrostatic attractions among the oppositely charged ions, its magnitude depends on several factors, including ionic size, ionic charge, and the arrangement of ions in the solid. For a particular arrangement of ions, the lattice energy increases as the charges on the ions increase and as their radii decrease. 9.18 The lattice energy releases even more energy when the gas is converted to the solid. 9.19 The lattice energy drives the energetically unfavorable electron transfer resulting in solid formation. 9.20 a) Barium is a metal and loses 2 electrons to achieve a noble gas configuration: Ba ([Xe]6s2) → Ba2+ ([Xe]) + 2 e-

Ba Ba

2++ 2 e-

Chlorine is a nonmetal and gains 1 electron to achieve a noble gas configuration: Cl ([Ne]3s23p5) + 1 e- → Cl- ([Ne]3s23p6)

Cl + 1 e- Cl

-

Two Cl atoms gain the two electrons lost by Ba. The ionic compound formed is BaCl2.

Ba +

Cl

Cl

Ba Cl-2+

Cl-

b) Sr ([Kr]5s2) → Sr2+ ([Kr]) + 2 e- O ([He]2s22p4) + 2 e- → O2- ([He]2s22p6) The ionic compound formed is SrO.

Sr O+ O2-

Sr2+

9-3

c) Al ([Ne]3s23p1) → Al3+ ([Ne]) + 3 e- F ([He]2s22p5) + 1 e- → F- ([He]2s22p6)

AlF

F

FAl

3+ F-

F-

F-

The ionic compound formed is AlF3. d) Rb ([Kr]5s1) → Rb+ ([Kr]) + 1 e- O ([He]2s22p4) + 2 e- → O2- ([He]2s22p6)

Rb

RbO O

2-Rb

+Rb

+

The ionic compound formed is Rb2O. 9.21 a) 2 Cs + S → Cs2S 2 Cs ([Xe]6s1) + S ([Ne]3s23p4) → 2 Cs+ ([Xe]) + S2- ([Ne]3s23p6)

2 Cs + S 2 Cs + S

+ 2-

b) 3 O + 2 Ga → Ga2O3 3 O ([He]2s22p4) + 2 Ga ([Ar]3d104s24p1) → 3 O2- ([He]2s22p6) + 2 Ga3+ ([Ar]3d10)

O + Ga 2 Ga + O

3+ 2-3 2 3

c) 2 N + 3 Mg → Mg3N2 2 N ([He]2s22p3) + 3 Mg ([Ne]3s2) → 2 N3- ([He]2s22p6) + 3 Mg2+ ([Ne])

Mg + N 3 Mg + N

2+ 3-3 2 2

d) Br + Li → LiBr Br ([Ar]3d104s24p5) + Li ([He]2s1) → Br- ([Ar]3d104s24p6) + Li+ ([He])

Li + Br Li + Br

+ -

9.22 a) X in XF2 is a cation with +2 charge since the anion is F- and there are two fluoride ions in the compound. Group 2A(2) metals form +2 ions. b) X in MgX is an anion with -2 charge since Mg2+ is the cation. Elements in Group 6A(16) form -2 ions. c) X in X2SO4 must be a cation with +1 charge since the polyatomic sulfate ion has a charge of -2. X comes from Group 1A(1). 9.23 a) 1A(1) b) 3A(13) c) 2A(2) 9.24 a) X in X2O3 is a cation with +3 charge. The oxygen in this compound has a -2 charge. To produce an electrically neutral compound, 2 cations with +3 charge bond with 3 anions with -2 charge: 2(+3) + 3(-2) = 0. Elements in Group 3A(13) form +3 ions. b) The carbonate ion, CO3

2-, has a -2 charge, so X has a +2 charge. Group 2A(2) elements form +2 ions. c) X in Na2X has a -2 charge, balanced with the +2 overall charge from the two Na atoms. Group 6A(16) elements gain 2 electrons to form -2 ions with a noble gas configuration. 9.25 a) 7A(17) b) 6A(16) c) 3A(13) 9.26 a) BaS would have the higher lattice energy since the charge on each ion is twice the charge on the ions in CsCl and lattice energy is greater when ionic charges are larger. b) LiCl would have the higher lattice energy since the ionic radius of Li+ is smaller than that of Cs+ and lattice energy is greater when the distance between ions is smaller.

9-4

9.27 a) CaO; O has a smaller radius than S. b) SrO; Sr has a smaller radius than Ba. 9.28 a) BaS has the lower lattice energy because the ionic radius of Ba2+ is larger than Ca2+. A larger ionic radius results in a greater distance between ions. The lattice energy decreases with increasing distance between ions. b) NaF has the lower lattice energy since the charge on each ion (+1, -1) is half the charge on the Mg2+ and O2- ions. 9.29 a) NaCl; Cl has a larger radius than F. b) K2S; 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.

ofH∆ + )g(Na)s(NaH →∆ + )g(Cl)g(ClH 222

1→∆ + −+ +→∆ e)g(Na)g(NaH + )g(Cle)g(ClH −− →+∆ = latticeH∆

latticeH∆ = -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

2MgFH°∆ = fH°∆ - ( 1H°∆ + 2H°∆ + 3H°∆ + 4H°∆ + 2 5H°∆ )

2MgFH°∆ = -1123 - (148 + 159 + 738 + 1450 + 2(-328)) kJ

2MgFH°∆ = -2962 kJ This value is larger due to greater ionic charge on Mg2+ and greater numbers of ions, compared with those of NaCl. 9.32 The lattice energy in an ionic solid is directly proportional to the product of the ion charges and inversely proportional to the sum of the ion radii. The strong interactions between ions cause most ionic materials to be hard. A very large lattice energy implies a very hard material. 9.33 An analogous Born-Haber cycle has been described in Figure 9.6 for LiF. Use Figure 9.6 as a basis for this diagram and solve for step 4, the electron affinity of F. 1) K(s) → K(g) ∆H1 = 90 kJ 2) K(g) → K+(g) + e- ∆H2 = 419 kJ 3) 1/2 F2(g) → F(g) ∆H3 = (1/2) (159) kJ 4) F(g) + e- → F-(g) ∆H4 = ? = EA 5) K+ + F-(g) → KF(s) ∆H5 = -821 kJ 6) K(s) + 1/2 F2(g) → KF(s) ∆Hf = -569 kJ ∆H4 = EA = ∆Hf - (∆H1 + ∆H2 + ∆H3 + ∆H5)

EA = -569 - (90 + 419 + 2

159 + (-821))

EA = -336 kJ 9.34 When two chlorine atoms are far apart, there is no interaction between them. Once the two atoms move closer together, the nucleus of each atom attracts the electrons on the other atom. As the atoms move closer this attraction increases, but the repulsion of the two nuclei also increases. When the atoms are very close together the repulsion between nuclei is much stronger than the attraction between nuclei and electrons. The final internuclear distance for the chlorine molecule is the distance at which maximum attraction is achieved in spite of the repulsion. At this distance, the energy of the molecule is at its lowest value.

9-5

9.35 The bond energy is the energy required to overcome the attraction between H atoms and Cl atoms in one mole of HCl molecules in the gaseous state. Energy input is needed to break bonds, so bond energy is always endothermic and ∆H°bond breaking is positive. The same amount of energy needed to break the bond is released upon its formation, so ∆H°bond forming has the same magnitude as ∆H°bond breaking, but opposite in sign (always exothermic and negative). 9.36 The strength of the covalent bond is generally inversely related to the size of the bonded atoms. For larger atoms, their bonding orbitals are more diffuse, so they form weaker bonds. 9.37 Bond strength increases with bond order, so C≡C > C=C > C-C. 9.38 When benzene boils, the gas consists of C6H6 molecules. The energy supplied disrupts the intermolecular attractions and not the intramolecular forces of bonding. 9.39 a) I-I < Br-Br < Cl-Cl. Bond strength increases as the atomic radii of atoms in the bond decreases. Atomic radii decrease up a group in the periodic table, so I is the largest and Cl is the smallest of the three. b) S-Br < S-Cl < S-H. Radius of H is the smallest and Br is the largest, so the bond strength for S-H is the greatest and that for S-Br is the weakest. c) C-N < C=N < C≡N. Bond strength increases as the number of electrons in the bond increases. The triple bond is the strongest and the single bond is the weakest. 9.40 a) H-F < H-Cl < H-I b) C=O < C-O < C-S c) N-H < N-O < N-S 9.41 a) For given pair of atoms, in this case carbon and oxygen, bond strength increases with increasing bond order. The C=O bond (bond order = 2) is stronger than the C-O bond (bond order = 1). These covalent bonds occur between atoms in the formic acid molecule and are much greater than the intermolecular forces that occur between formic acid molecules. b) O is smaller than C so the O-H bond is shorter and stronger than the C-H bond. 9.42 C≡C would show absorption of IR at shorter wavelength (higher energy) due to it being a stronger bond. 9.43 H2(g) + O2(g) → H-O-O-H(g) ∆H ]Be + )BE( [2 BE + BE = OOOHOOH 2 −= − 9.44 Reaction between molecules requires the breaking of existing bonds and the formation of new bonds. Substances with weak bonds are more reactive than are those with strong bonds because less energy is required to break weak bonds. 9.45 Bond energies are average values for a particular bond in a variety of compounds. Heats of formation are specific for a compound. 9.46 For methane: CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) which requires that 4 C-H bonds and 2 O=O bonds be broken and 2 C=O bonds and 4 O-H bonds be formed. For formaldehyde: CH2O(g) + O2(g) → CO2(g) + H2O(l) which requires that 2 C-H bonds, 1 C=O bond and 1 O=O bond be broken and 2 C=O bonds and 2 O-H bonds be formed. The fact that methane contains more C-H bonds and fewer C-O bonds than formaldehyde suggests that more energy is released in the combustion of methane than of formaldehyde. 9.47 With more C-H and C-C bonds, ethanol will have the greater heat of combustion per mole.

9-6

9.48 To find the heat of reaction, add the energy required to break all the bonds in the reactants to the energy released to form all bonds in the product. Reactant bonds broken: 1(C=C) + 4(C-H) + 1(Cl-Cl) = (1 mol) (614 kJ/mol) + (4 mol) (413 kJ/mol) + (1 mol) (243 kJ/mol) = 2509 kJ Product bonds formed: 1(C-C) + 4(C-H) + 2(C-Cl) = (1 mol) (-347 kJ/mol) + (4 mol) (-413 kJ/mol) + (2 mol) (-339 kJ/mol) = -2677 kJ ∆H°rxn = ∆H°bonds broken + ∆H°bonds formed = 2509 kJ + (-2677 kJ) = -168 kJ Note: It is correct to report the answer in kJ or kJ/mol as long as the value refers to a reactant or product with a molar coefficient of 1. 9.49 CO2 + 2 NH3 → (H2N)2CO + H2O ∆Hrxn = ΣBEreactants - ΣBEproducts ∆Hrxn = (2 BEC=O + 6 BEN-H) - (4 BEN-H + BEC=O + 2 BEC-N + 2 BEO-H) = [2(799) + 6(391)] - [4(391) + 745 + 2(305) + 2(467)] = 3944 - 3853 = 91 kJ 9.50 The reaction:

C O H

H

H

H + C O C

H

H

H C

O

O H

∆H°bonds broken = 1 C-O = 1 mol (358 kJ/mol) ∆H°bonds formed = 3 C-H = 3 mol (-413 kJ/mol) 3 C-H = 3 mol (413 kJ/mol) 1 C-C = 1 mol (-347 kJ/mol) 1 O-H = 1 mol (467 kJ/mol) 1 C=O = 1 mol (-745 kJ/mol) 1 C ≡ O = 1 mol (1070 kJ/mol) 1 C-O = 1 mol (-358 kJ/mol) = 3134 kJ 1 O-H = 1 mol (-467 kJ/mol) = -3156 kJ ∆H°rxn = ∆H°bonds broken + ∆H°bonds formed = 3134 kJ + (-3156 kJ) = -22 kJ 9.51 Plan: Examine the structures for all substances and assume that all reactant bonds are broken and all product bonds are formed. Solution:

C C

H

H

H

H

H Br C C

H

H

H

H

Br

H

+

Bonds broken: 1(C=C) + 4(C-H) + 1(H-Br) = 614 + 4(413) + 363 = +2629 kJ Bonds formed: 1(C-C) + 5(C-H) + 1(C-Br) = (-347) + 5(-413) + (-276) = -2688 kJ ∆H°rxn = ∆H°bonds broken + ∆H°bonds formed = 2629 - 2688 = -59 kJ 9.52 Electronegativity increases from left to right across a period (except for the noble gases) and increases from bottom to top within a group. Fluorine (F) and oxygen (O) are the two most electronegative elements. Cesium (Cs) and francium (Fr) are the two least electronegative elements. 9.53 In general, electronegativity and ionization energies are directly related. Electronegativity relates the strength with which an atom attracts bonding electrons and the ionization energy measures the energy needed to remove an electron.

9-7

9.54 The H-O bond in water is polar covalent, because the two atoms differ in electronegativity which results in an unequal sharing of the bonding electrons. A nonpolar covalent bond occurs between two atoms with identical electronegativities where the sharing of bonding electrons is equal. Although electron sharing occurs to a very small extent in some ionic bonds, the primary force in ionic bonds is attraction of opposite charges resulting from electron transfer between the atoms. 9.55 E.N. is the tendency of a bonded atom to hold the bonding electrons more strongly. E.A. is the energy involved when an atom acquires an electron. 9.56 The difference in E.N. is a reflection of how strongly one atom attracts bonding electrons. The greater this difference is, the more likely the bond will be ionic; the smaller the E.N. difference, the more covalent the bond. 9.57 a) Si < S < O, sulfur is more electronegative than silicon since it is located further to the right on the table. Oxygen is more electronegative than sulfur since it is located nearer the top of the table. b) Mg < As < P, magnesium is the least electronegative because it lies on the left side of the periodic table and phosphorus and arsenic on the right side. Phosphorus is more electronegative than arsenic because it is higher on the table. 9.58 a) I < Br < N b) Ca < H < F 9.59 Electronegativity generally increases up a group and left to right across a period. a) N > P > Si, nitrogen is above P in Group 5(A)15 and P is to the right of Si in period 3. b) As > Ga > Ca, all three elements are in Period 4, with As the right-most element. 9.60 a) Cl > Br > P b) F > O > I 9.61

a) N B b) N O c) C S

d) S O e) N H f) Cl O

none

9.62

a) Br Cl b) F Cl c) H O

d) Se H e) As H f) S N 9.63 The more polar bond will have a greater difference in electronegativity, ∆EN. a) ∆ENa = 1.0 b) ∆ENb = 0.5 c) ∆ENc = 0 d) ∆ENd = 1.0 e) ∆ENe = 0.9 f) ∆ENf = 0.5 (a), (d), and (e) have greater bond polarity. 9.64 b) is more polar; ∆EN is 1.0 for F-Cl and 0.2 for Br-Cl. c) is more polar; ∆EN is 1.4 for H-O and 0.3 for Se-H. f) is more polar; ∆EN is 0.5 for S-N and 0.1 for As-H.

9-8

9.65 a) Bonds in S8 are nonpolar covalent. All the atoms are nonmetals so the substance is covalent and bonds are nonpolar because all the atoms are of the same element. b) Bonds in RbCl are ionic because Rb is a metal and Cl is a nonmetal. c) Bonds in PF3 are polar covalent. All the atoms are nonmetals so the substance is covalent. The bonds between P and F are polar because their electronegativity differs. (By 1.9 units for P-F) d) Bonds in SCl2 are polar covalent. S and Cl are nonmetals and differ in electronegativity. (By 0.5 unit for S-Cl) e) Bonds in F2 are nonpolar covalent. F is a nonmetal. Bonds between two atoms of the same element are nonpolar. f) Bonds in SF2 are polar covalent. S and F are nonmetals that differ in electronegativity. (By 1.5 units for S-F) Increasing bond polarity: SCl2 < SF2 < PF3 9.66 a) KCl ionic b) P4 nonpolar covalent c) BF3 polar covalent d) SO2 polar covalent e) Br2 nonpolar covalent f) NO2 polar covalent NO2 < SO2 < BF3 9.67 Increasing ionic character occurs with increasing ∆EN. a) ∆ENHBr = 0.7, ∆ENHCl = 0.9, ∆ENHI = 0.4 b) ∆ENHO = 1.4, ∆ENCH = 0.4, ∆ENHF = 1.9 c) ∆ENSCl = 0.5, ∆ENPCl = 0.9, ∆ENSiCl = 1.2

a) H I H Br H Cl

b) H C H O H F

< <

c) S Cl P Cl Si Cl

< <

< <

9.68 Increasing ionic character occurs with increasing ∆EN. a) ∆ENPCl = 0.9, ∆ENPBr = 0.7, ∆ENPF = 1.9 P-F > P-Cl > P-Br δ+ δ- δ+ δ- δ+ δ- b) ∆ENBF = 2.0, ∆ENNF = 1.0, ∆ENCF = 1.9 B-F > C-F > N-F δ+ δ- δ+ δ- δ+ δ- c) ∆ENSeF = 1.6, ∆ENTeF = 1.9, ∆ENBrF = 1.2 Te-F > Se-F > Br-F δ+ δ- δ+ δ- δ+ δ- 9.69 C-C + Cl-Cl → 2 C-Cl 347 kJ/mol 243 kJ/mol d) The value should be greater than the sum of the averages for the bonds. This is due to the electronegativity difference. 9.70 a) A solid metal is a shiny solid that conducts heat, is malleable, and melts at high temperatures. (Other answers include relatively high boiling point and good conductor of electricity.) b) Metals lose electrons to form positive ions and metals form basic oxides. 9.71 a) Potassium is a larger atom than sodium, so its electrons are held more loosely and thus its metallic bond strength is weaker. b) Be has two valence electrons per atom compared with Li, which has one. The metallic bond strength is stronger for the Be. c) The boiling point is high due to the large amount of energy needed to separate the metal ions from each other in the electron sea.

9-9

9.72 When metallic magnesium is deformed, the atoms are displaced and pass over one another while still being tightly held by the attraction of the “sea of electrons.” When ionic MgF2 is deformed, the ions are displaced so that repulsive forces between neighboring ions of like charge cause shattering of the crystals. 9.73 Molten rock cools from top to bottom. The most stable compound (the one with the largest lattice energy) will solidify first near the top. The less stable compounds will remain in the molten state at the bottom and eventually crystallize there later. 9.74 a) C2H2 + 5/2 O2 → 2 CO2 + H2O ∆H = -1259 kJ/mol

H-C ≡ C-H + 52

O=O → H-O-H + 2 O=C=O

∆Hrxn = (2 BEC-H + BEC≡C + 2OBE

25 ) - (4 BEC=O + 2 BEO-H)

-1259 kJ = [(2(413) + BEC≡C + (5/2) (498)] - [(4(799) + 2(467)] -1259 kJ = [(826 + BEC≡C + 1245) - (4130)] kJ BEC≡C = 800. kJ/mol Table 9.2 lists the value as 839 kJ/mol

b) heat (kJ) = ( ) 2 22 2

2 2 2 2

1 mol C H 1259 kJ500.0 g C H26.04 g C H 1 mol C H

−

= -2.4174347 x 104 = -2.417 x 104 kJ

c) CO2 produced (g) = ( ) 2 2 2 22 2

2 2 2 2 2

1 mol C H 2 mol CO 44.01 g CO500.0 g C H

26.04 g C H 1 mol C H 1 mol CO

= 1690.092 = 1690. g CO2

d) mol O2 = ( ) 2 2 22 2

2 2 2 2

1 mol C H (5/2) mol O500.0 g C H

26.04 g C H 1 mol C H

= 48.00307 mol O2 (unrounded)

V = nRT / P = ( ) ( )2

L • atm48.00307 mol O 0.08206 298 Kmol • K

18.0 atm

= 65.2145 = 65.2 L O2

9.75 a)

b)

Br + 3 F Br

F

F F

+ 3 FAl + 3Al F

-3+

9.76 a) 1) Mg(s) → Mg(g) oH 1∆ = 148 kJ 2) 1/2 Cl2(g) → Cl(g) oH 2∆ = 1/2 (243 kJ) 3) Mg(g) → Mg+(g) + e- oH 3∆ = 738 kJ 4) Cl(g) + e- → Cl-(g) oH 4∆ = -349 kJ 5) Mg+(g) + Cl-(g) → MgCl(s) oH 5∆ = -783.5 kJ (= o

latticeH∆ (MgCl)) 6) Mg(s) + 1/2 Cl2(g) → MgCl(s) o

fH∆ (MgCl) = ?

ofH∆ (MgCl) = oH 1∆ + oH 2∆ + oH 3∆ + oH 4∆ + oH 5∆

= 148 kJ + 1/2 (243 kJ) + 738 kJ - 349 kJ - 783.5 kJ = -125 kJ b) Yes, since fH°∆ for MgCl is negative, MgCl(s) is stable relative to its elements.

9-10

c) 2 MgCl(s) → MgCl2(s) + Mg(s) °∆H = 1 mol ( o

fH∆ , MgCl2(s)) + 1 mol ( ofH∆ , Mg(s)) - 2 mol ( o

fH∆ , MgCl(s)) = 1 mol (-641.6 kJ/mol) + 1 mol (0) - 2 mol (-125 kJ/mol) = -391.6 = -392 kJ d) No, o

fH∆ for MgCl2 is much more negative than that for MgCl. This makes the °∆H value for the above reaction very negative, and the formation of MgCl2 would be favored. 9.77 a) b) The general trend is for ionization energy to increase with each successive electron removed. The increase in ionization energy is small between the first and second ionization energies as electrons are removed from the 4s orbital. Ionization energies increase by an amount greater than expected when (1) the first electron is removed from a 3d orbital, (2) the first unpaired electron is removed from a 3d orbital, (3) the first electron is removed from a 3p orbital, (4) the first unpaired electron is removed from a 3p orbital, and (5) the first electron is removed from a 3s orbital. There is a large increase in ionization energy when the first electron is removed from a 2p orbital. 9.78 a) ∆H° = Σ(n)BEreactants - Σ(m)BEproducts = 1 mol (BEH-H) + 1 mol (BECl-Cl) - 2 mol (BEH-Cl) = 1 mol (432 kJ/mol) + 1 mol (243 kJ/mol) - 2 mol (427 kJ/mol) = -179 kJ b) ∆H° = Σ(n)BEreactants - Σ(m)BEproducts = 1 mol (BEH-H) + 1 mol (BEI-I) - 2 mol (BEH-I) = 1 mol (432 kJ/mol) + 1 mol (151 kJ/mol) - 2 mol (295 kJ/mol) = -7 kJ c) ∆H° = Σ(n)BEreactants - Σ(m)BEproducts = 2 mol (BEH-H) + 1 mol (BEO=O) - 4 mol (BEH-O) = 2 mol (432 kJ/mol) + 1 mol (498 kJ/mol) - 4 mol (467 kJ/mol) = -506 kJ Reactions (a) and (c) are strongly exothermic and are a potential explosive hazard. 9.79 Plan: Find the bond energy for an H-I bond from Table 9.2. For part a) calculate wavelength from this energy using the relationship from Chapter 7: E = hc / λ. For part b) calculate the energy for a wavelength of 254 nm and then subtract the energy from part a) to get the excess energy. Speed can be calculated from the excess energy since Ek = 1/2 mu2.

9-11

Solution: Bond energy for H-I is 295 kJ/mol (Table 9.2).

a) λ = hc / E = ( )( )34 8

93

23

6.626 x 10 J • s 3.00 x 10 m/s 1 nm10 mkJ 10 J mol295

mol 1 kJ 6.022 x 10

−

−

= 405.7807 = 406 nm

b) E (HI) = 3

23kJ 10 J mol295

mol 1 kJ 6.022 x 10

= 4.8987 x 10-19 J (unrounded)

E (254 nm) = hc / λ. = ( )( )34 8

9

6.626 x 10 J • s 3.00 x 10 m/s 1 nm254 nm 10 m

−

−

= 7.82598 x 10-19 J (unrounded)

Excess energy = 7.82598 x 10-19 J - 4.8987 x 10-19 J = 2.92728 x 10-19 = 2.93 x 10-19 J

c) Ek = 1/2mu2 thus, u = 2Em

m = 23 31.008 g H mol 1 kg

mol 6.022 x 10 10 g

= 1.67386 x 10-27 kg (unrounded)

u = 19 2 2

272(2.92728 x 10 J) kg • m /s

J1.67386 x 10 kg

−

−

= 1.8701965 x 104 = 1.87 x 104 m/s

Check: The energy difference and speed are similar in magnitude to those calculated in Chapter 7 for atoms, so the answers appear reasonable. 9.80 a) Vibrational motions have frequencies, which are in the IR region of the electromagnetic spectrum. b) E = hv = (6.626 x 10-34 J•s) (4.02 x 1013 s-1) = 2.66365 x 10-20 = 2.66 x 10-20 J (symmetric stretch) E = (6.626 x 10-34 J•s) (2.00 x 1013 s-1) = 1.3252 x 10-20 = 1.33 x 10-20 J (bending) E = (6.626 x 10-34 J•s) (7.05 x 1013 s-1) = 4.6713 x 10-20 = 4.67 x 10-20 J (asymmetrical stretch) Bending requires the least amount of energy. 9.81 “Excess bond energy” refers to the difference between the actual bond energy for an A-B bond and the average of the energies for the A-A and the B-B bonds, i.e., excess bond energy = BEA-B - 1/2 (BEA-A + BEB-B). The excess bond energy is zero when the atoms A and B are identical or have the same electronegativity, as in (a), (b), and (e). 9.82 Rb ([Kr]5s1) has one valence electron, so the metallic bonding would be fairly weak, resulting in a soft, low- melting material. Cd ([Kr]5s24d10) has two valence electrons so the metallic bonding is stronger. V ([Ar]4s23d3) has five valence electrons, so its metallic bonding is the strongest, that is, its hardness, melting point and other metallic properties would be greatest. 9.83 Plan: Find the appropriate bond energies in Table 9.2. Calculate the wavelengths using E = hc / λ. Solution: C-Cl bond energy = 339 kJ/mol

λ = ( )( )34 8

3

23

6.626 x 10 J • s 3.00 x 10 m/s

kJ 10 J mol339mol 1 kJ 6.022 x 10

−

= 3.53113 x 10-7 = 3.53 x 10-7 m

O2 bond energy = 498 kJ/mol

λ = ( )( )34 8

3

23

6.626 x 10 J • s 3.00 x 10 m/s

kJ 10 J mol498mol 1 kJ 6.022 x 10

−

= 2.40372 x 10-7 = 2.40 x 10-7 m

9.84 Plan: Write balanced chemical equations for the formation of each of the compounds. Obtain the bond energy of fluorine from Table 9.2 (159 kJ/mol). Determine the average bond energy from ∆H = bonds broken - bonds formed.

9-12

Solution: XeF2 Xe(g) + F2(g) → XeF2(g) ∆H = 1 mol F2 (159 kJ/mol) - 2 (Xe-F) = -105 kJ/mol Xe-F = 132 kJ/mol XeF4 Xe(g) + 2 F2(g) → XeF4(g) ∆H = 2 mol F2 (159 kJ/mol) - 4 (Xe-F) = -284 kJ/mol Xe-F = 150.5 = 150. kJ/mol XeF6 Xe(g) + 3 F2(g) → XeF6(g) ∆H = 3 mol F2 (159 kJ/mol) - 6 (Xe-F) = -402 kJ/mol Xe-F = 146.5 = 146 kJ/mol 9.85 The difference in electronegativity produces a greater than expected overlap of orbitals, which shortens the bond. As ∆EN becomes smaller (i.e., as you proceed from HF to HI), this effect lessens and the bond lengths become more predictable. 9.86 a) The presence of the very electronegative fluorine atoms bonded to one of the carbons makes the C-C bond polar. This polar bond will tend to undergo heterolytic rather than homolytic cleavage. More energy is required to force homolytic cleavage. b) ∆H = (homolytic cleavage + electron affinity + first ionization energy) ∆H = (249.2 kJ/mol + (-141 kJ/mol) + 1.31 x 103 kJ/mol) = 1420 kJ/mol It would require 1420 kJ to heterolytically cleave 1 mol of O2. Since one atom gets both of the bonding electrons in heterolytic bond breakage, this results in the formation of ions. 9.87 The bond energies are needed from Table 9.2. N2 = 945 kJ/mol; O2 = 498 kJ/mol; F2 = 159 kJ/mol N2:

λ = ( ) ( )34 8

3

23

6.626 x 10 J • s 3.00 x 10 m/s

kJ 10 J mol945mol 1 kJ 6.022 x 10

−

= 1.26672 x 10-7 = 1.27 x 10-7 m

O2:

λ = ( )( )34 8

3

23

6.626 x 10 J • s 3.00 x 10 m/s

kJ 10 J mol498mol 1 kJ 6.022 x 10

−

= 2.40372 x 10-7 = 2.40 x 10-7 m

F2:

λ = ( ) ( )34 8

3

23

6.626 x 10 J • s 3.00 x 10 m/s

kJ 10 J mol159mol 1 kJ 6.022 x 10

−

= 7.528636 x 10-7 = 7.53 x 10-7 m

9.88 a) To compare the two energies, the ionization energy must be converted to the energy to remove an electron from an atom. The energy needed to remove an electron from a single gaseous Ag atom (J) =

3

23731 kJ 10 J mol

mol 1 kJ 6.022 x 10

= 1.21388 x 10-18 = 1.21 x 10-18 J > 7.59 x 10-19 J

It requires less energy to remove an electron from the surface of solid silver. b) The electrons in solid silver are held less tightly than the electrons in gaseous silver because the electrons in metals are delocalized, meaning they are shared among all the metal nuclei. The delocalized attraction of many nuclei to an electron (solid silver) is weaker than the localized attraction of one nucleus to an electron (gaseous silver).

9-13

9.89 1) Si(s) → Si(g) oH 1∆ = 454 kJ 2) Si(g) → Si4+(g) + 4 e- oH 2∆ = 9949 kJ 3) O2(g) → 2 O(g) oH 3∆ = 498 kJ 4) 2 O(g) + 4 e- → 2 O2-(g) oH 4∆ = 2(737) kJ 5) Si4+(g) + 2 O2-(g) → SiO2(s) oH 5∆ = o

latticeH∆ (SiO2) = ? 6) Si(s) + O2(g) → SiO2(s) o

latticeH∆ (SiO2) = -910.9 kJ oH 5∆ = o

latticeH∆ - ( oH 1∆ + oH 2∆ + oH 3∆ + oH 4∆ ) = -910.9 kJ - (454 kJ + 9949 kJ + 498 kJ + 2(737) kJ) = -13285.9 = -13286 kJ 9.90 o

rxnH∆ = Σ(n) BEreactants - Σ BEproducts For ethane: o

rxnH∆ = 1 mol (BEC - C) + 6 mol (BEC - H) + 1 mol (BEH - H) - 8 mol (BEC - H) -65.07 kJ = 1 mol (347 kJ/mol) + 6 mol (BEC - H) + 1 mol (432 kJ/mol) - 8(415 kJ/mol)

BEC - H = ( )65.07 347 432 3320 kJ

6 mol− − − +

= 412.655 = 413 kJ/mol

For ethene: orxnH∆ = 1 mol (BEC - C) + 4 mol (BEC - H) + 2 mol (BEH - H) - 8 mol (BEC - H)

-202.21 kJ = 1 mol (614 kJ/mol) + 4 mol (BEC - H) + 2 mol (432 kJ/mol) - 8 mol (415 kJ/mol)

BEC - H = ( )202.21 614 862 3320 kJ

4 mol− − − +

= 410.4475 = 410. kJ/mol

For ethyne: rxnH°∆ = 1 mol (BEC≡C) + 2 mol (BEC - H) + 3 mol (BEH - H) - 8 mol (BEC - H) -376.74 kJ = 1 mol (839 kJ/mol) + 2 mol (BEC - H) + 3 mol (432 kJ/mol) - 8 mol (415 kJ/mol)

BEC - H = ( )376.74 839 1296 3320 kJ

2 mol− − − +

= 404.13 = 404 kJ/mol

9.91 Plan: Use the equations E = hν, and E = hc / λ. Solution:

ν = Eh

=

3

23

34

kJ 10 J mol347mol 1 kJ 6.022 x 10

6.626 x 10 J • s−

= 8.6963556 x 1014 = 8.70 x 1014 s-1

λ = hc / E = ( )( )34 8

3

23

6.626 x 10 J • s 3.00 x 10 m/s

kJ 10 J mol347mol 1 kJ 6.022 x 10

−

= 3.44972 x 10-7 = 3.45 x 10-7 m

This is in the ultraviolet region of the electromagnetic spectrum. 9.92

ν = Eh

=

3

23

34

kJ 10 J mol467mol 1 kJ 6.022 x 10

6.626 x 10 J • s−

= 1.170374 x 1015 = 1.17 x 1015 s-1

λ = hc / E = ( ) ( )34 8

3

23

6.626 x 10 J • s 3.00 x 10 m/s

kJ 10 J mol467mol 1 kJ 6.022 x 10

−

= 2.56328 x 10-7 = 2.56 x 10-7 m

9-14

Ephoton = 23kJ 1 mol467

mol 6.022 x 10 photons

= 7.7548987 x 10-22 = 7.75 x 10-22 kJ/photon

9.93 a) E = hc / λ

E575 = ( )( )

( )

34 8

9

6.626 x 10 J • s 3.00 x 10 m/s 1 nm575 nm 10 m

−

−

= 3.45704 x 10-19 J (unrounded)

E300 = ( )( )

( )

34 8

9

6.626 x 10 J • s 3.00 x 10 m/s 1 nm300 nm 10 m

−

−

= 6.626 x 10-19 J (unrounded)

Ratio = (6.626 x 10-19 J) / (3.45704 x 10-19 J) = 1.9166686 = 1.92 b) E = hc / λ

E1000 = ( )( )

( )

34 8

9

6.626 x 10 J • s 3.00 x 10 m/s 1 nm1000 nm 10 m

−

−

= 1.9878 x 10-19 J (unrounded)

Ratio = (1.9878 x 10-19 J) / (3.45704 x 10-19 J) = 0.57500 = 0.575 c) Using Table 9.2, it takes 347 kJ/mol to break a C-C bond. Converting this to Joules/photon:

E = 3

23kJ 10 J 1 mol347

mol 1 kJ 6.022 x 10 photons

= 5.7622 x 10-19 = 5.77 x 10-19 J/photon

Only the ultraviolet photons (300 nm) have sufficient energy. d) Using Table 9.2, it takes 413 kJ/mol to break a C-H bond. Converting this to Joules/photon:

E = 3

23kJ 10 J 1 mol413

mol 1 kJ 6.022 x 10 photons

= 6.8581866 x 10-19 = 6.86 x 10-19 J/photon

None of the photons has sufficient energy. 9.94 a) 2 CH4(g) + O2(g) → CH3OCH3(g) + H2O(g) ∆Hrxn = ΣBEreactants - ΣBEproducts ∆Hrxn = (2 x 4 BEC-H + BEO=O) - (6 BEC-H + 2 BEC-O + 2 BEO-H) ∆Hrxn = (8 (413 kJ) + 498 kJ) - (6 (413 kJ) + 2 (358 kJ) + 2 (467)) ∆Hrxn = -326 kJ 2 CH4(g) + O2(g) → CH3CH2OH(g) + H2O(g) ∆Hrxn = ΣBEreactants - ΣBEproducts ∆Hrxn = (2 x 4 BEC-H + BEO=O) - (5 BEC-H + BEC-C + BEC-O + 3 BEO-H) ∆Hrxn = (8 (413 kJ) + 498 kJ) - (5 (413 kJ) + (347 kJ) + (358 kJ) + 3 (467)) ∆Hrxn = -369 kJ b) The formation of gaseous ethanol is more exothermic. c) ∆Hrxn = -326 kJ - (-369 kJ) = 43 kJ