CHAPTER 9 MODELS OF CHEMICAL BONDING

Transcription

1 CAPTER 9 MODELS OF CEMICAL BONDING 9.1 a) Larger inizatin energy decreases metallic character. b) Larger atmic radius increases metallic character. c) Larger number f uter electrns decreases metallic character. d) Larger effective nuclear charge decreases metallic character. 9.2 A has cvalent bnding, B has inic bnding and C has metallic bnding. 9. The tendency f main-grup elements t frm catins decreases frm Grup 1A(1) t 4A(14), and the tendency t frm anins increases frm Grup 4A(14) t 7A(17). 1A(1) and 2A(2) elements frm mn- and divalent catins, respectively, while 6A(16) and 7A(17) elements frm di- and mnvalent anins, respectively. 9.4 Metallic behavir increases t the left and dwn n the peridic table. a) Cs is mre metallic since it is further dwn the alkali metal grup than Na. b) Rb is mre metallic since it is bth t the left and dwn frm Mg. c) As is mre metallic since it is further dwn Grup 5A than N. 9.5 a) O b) Be c) Se 9.6 Inic bnding ccurs between metals and nnmetals, cvalent bnding between nnmetals, and metallic bnds between metals. a) Bnd in CsF is inic because Cs is a metal and F is a nnmetal. b) Bnding in N 2 is cvalent because N is a nnmetal. c) Bnding in Na(s) is metallic because this is a mnatmic, metal slid. 9.7 a) cvalent b) cvalent c) inic 9.8 Inic bnding ccurs between metals and nnmetals, cvalent bnding between nnmetals, and metallic bnds between metals. a) Bnding in O wuld be cvalent since O is a nnmetal. b) Bnding in MgCl 2 wuld be inic since Mg is a metal and Cl is a nnmetal. c) Bnding in BrO 2 wuld be cvalent since bth Br and O are nnmetals. 9.9 a) metallic b) cvalent c) inic 9.10 Lewis electrn-dt symbls shw valence electrns as dts. Place ne dt at a time n the fur sides (this methd explains the structure in b) and then pair up dts until all valence electrns are used. The grup number f the main grup elements (Grups 1A-8A) gives the number f valence electrns. Rb is Grup 1A, Si is Grup 4A, I is Grup 7A a) Rb b) Si c) I a) Ba b) Kr c) Br 9-1

2 9.12 a) Sr b) P c) S 9.1 a) As b) Se c) Ga 9.14 a) Assuming X is an A grup element, the number f dts (valence electrns) equals the grup number. Therefre, X is a 6A(16) element with 6 valence electrns. Its general electrn cnfiguratin is [nble gas]ns 2 np 4, where n is the energy level. b) X has three valence electrns and is a A(1) element with general e cnfiguratin [nble gas]ns 2 np a) 5A(15); ns 2 np b) 4A(14); ns 2 np Energy is required t frm the catins and anins in inic cmpunds but energy is released when the ppsitely charged ins cme tgether t frm the cmpund. This energy is the lattice energy and mre than cmpensates fr the required energy t frm ins frm metals and nnmetals a) Because the lattice energy is the result f electrstatic attractins amng the ppsitely charged ins, its magnitude depends n several factrs, including inic size, inic charge, and the arrangement f ins in the slid. Fr a particular arrangement f ins, the lattice energy increases as the charges n the ins increase and as their radii decrease. b) Increasing lattice energy: A < B < C 9.18 The lattice energy releases even mre energy when the gas is cnverted t the slid The lattice energy drives the energetically unfavrable electrn transfer resulting in slid frmatin a) Barium is a metal and lses 2 electrns t achieve a nble gas cnfiguratin: Ba ([Xe]6s 2 ) Ba 2+ ([Xe]) + 2 e 2+ Ba Ba + 2 e - Chlrine is a nnmetal and gains 1 electrn t achieve a nble gas cnfiguratin: Cl ([Ne]s 2 p 5 ) + 1 e Cl ([Ne]s 2 p 6 ) Cl + 1 e - Cl - Tw Cl atms gain the tw electrns lst by Ba. The inic cmpund frmed is BaCl 2. Ba + Cl Cl - Ba 2+ Cl - Cl 9-2

3 b) Sr ([Kr]5s 2 ) Sr 2+ ([Kr]) + 2 e O ([e]2s 2 2p 4 ) + 2 e O 2 ([e]2s 2 2p 6 ) The inic cmpund frmed is SrO Sr Sr + O O c) Al ([Ne]s 2 p 1 ) Al + ([Ne]) + e F ([e]2s 2 2p 5 ) + 1 e F ([e]2s 2 2p 6 ) - F + F Al F Al - F F F - The inic cmpund frmed is AlF. d) Rb ([Kr]5s 1 ) Rb + ([Kr]) + 1 e O ([e]2s 2 2p 4 ) + 2 e O 2 ([e]2s 2 2p 6 ) + 2- Rb Rb O Rb O Rb The inic cmpund frmed is Rb 2 O a) 2 Cs + S Cs 2 S 2 Cs ([Xe]6s 1 ) + S ([Ne]s 2 p 4 ) 2 Cs + ([Xe]) + S 2 ([Ne]s 2 p 6 ) Cs + S 2 Cs + S b) O + 2 Ga Ga 2 O O ([e]2s 2 2p 4 ) + 2 Ga ([Ar]d 10 4s 2 4p 1 ) O 2 ([e]2s 2 2p 6 ) + 2 Ga + ([Ar]d 10 ) + 2- O + 2 Ga 2 Ga + O c) 2 N + Mg Mg N 2 2 N ([e]2s 2 2p ) + Mg ([Ne]s 2 ) 2 N ([e]2s 2 2p 6 ) + Mg 2+ ([Ne]) 2+ - Mg + 2 N Mg + 2 N d) Br + Li LiBr Br ([Ar]d 10 4s 2 4p 5 ) + Li ([e]2s 1 ) Br ([Ar]d 10 4s 2 4p 6 ) + Li + ([e]) + - Li + Br Li + Br 9.22 a) X in XF 2 is a catin with +2 charge since the anin is F and there are tw fluride ins in the cmpund. Grup 2A(2) metals frm +2 ins. b) X in MgX is an anin with 2 charge since Mg 2+ is the catin. Elements in Grup 6A(16) frm -2 ins. c) X in X 2 SO 4 must be a catin with +1 charge since the plyatmic sulfate in has a charge f 2. X cmes frm Grup 1A(1). 9.2 a) 1A(1) b) A(1) c) 2A(2) 9.24 a) X in X 2 O is a catin with + charge. The xygen in this cmpund has a 2 charge. T prduce an electrically neutral cmpund, 2 catins with + charge bnd with anins with 2 charge: 2(+) + ( 2) = 0. Elements in Grup A(1) frm + ins. b) The carbnate in, CO 2, has a 2 charge, s X has a +2 charge. Grup 2A(2) elements frm +2 ins. 9-

4 c) X in Na 2 X has a 2 charge, balanced with the +2 verall charge frm the tw Na + ins. Grup 6A(16) elements gain 2 electrns t frm 2 ins with a nble gas cnfiguratin a) 7A(17) b) 6A(16) c) A(1) 9.26 a) BaS wuld have the higher lattice energy since the charge n each in is twice the charge n the ins in CsCl and lattice energy is greater when inic charges are larger. b) LiCl wuld have the higher lattice energy since the inic radius f Li + is smaller than that f Cs + and lattice energy is greater when the distance between ins is smaller a) CaO; O has a smaller radius than S. b) SrO; Sr has a smaller radius than Ba a) BaS has the lwer lattice energy because the inic radius f Ba 2+ is larger than Ca 2+. A larger inic radius results in a greater distance between ins. The lattice energy decreases with increasing distance between ins. b) NaF has the lwer lattice energy since the charge n each in (+1, 1) is half the charge n the Mg 2+ and O 2 ins. Lattice energy increases with increasing in charge a) NaCl; Cl has a larger radius than F. b) K 2 S; S has a larger radius than O. 9.0 Lattice energy: NaCl(s) Na + (aq) + Cl (g) Use ess Law: º Na(s) Na(g) 109 kj 1/2Cl2(g) Cl(g) 24/2 = kj Na(g) Na + (g) + e 496 kj Cl(g) + e Cl (g) 49 kj NaCl(s) Na(s) + 1/2Cl (g) 411 kj (Reactin is reversed and the sign f º changed). NaCl(s) Na + (aq) + Cl (g) = 788 kj The lattice energy fr NaCl is less than that f LiF, which is expected since lithium and fluride ins are smaller than sdium and chlride ins. 9.1 Lattice energy: MgF 2 (s) Mg 2+ (aq) + 2 F (g) Use ess Law: º Mg(s) Mg(g) 148 kj F 2 (g) 2 F(g) 159 kj Mg(g) Mg + (g) + e 78 kj Mg+(g) Mg 2+ (g) + e 1450 kj 2 F(g) + e 2 F (g) 2( 28) = 656 kj MgF 2 (s) Mg(s) + F 2 (g) 112 kj (Reactin is reversed and the sign f º changed). MgF 2 (s) Mg 2+ (aq) + 2 F (g) 2962 kj The lattice energy fr MgF 2 is greater than that f LiF and NaCl, which is expected since magnesium ins have twice the charge f lithium and sdium ins. Lattice energy increases with increasing in charge. 9.2 The lattice energy in an inic slid is directly prprtinal t the prduct f the in charges and inversely prprtinal t the sum f the in radii. The strng interactins between ins cause mst inic materials t be hard. A very large lattice energy implies a very hard material. The lattice energy is predicted t be high fr Al 2 O since the ins invlved, Al + and O 2, have fairly large charges and are relatively small ins. 9-4

5 9. An analgus Brn-aber cycle has been described in Figure 9.6 fr LiF. Use ess s Law and slve fr the EA f flurine: Δº K(s) K(g) 90 kj K(g) K + (g) + e 419 kj 1/2 F 2 (g) F(g) 1/2(159) = 79.5 kj F(g) + e F (g)? = EA KF(s) K(s) + 1/2 F 2 (g) 569 kj (Reverse the reactin and change the sign f Δº) KF(s) K + + F (g) 821 kj (Reverse the reactin and change the sign f Δº ) 821 kj = 90 kj kj kj + EA kj EA = 821 kj ( )kJ EA = 6.5 = 6 kj 9.4 When tw chlrine atms are far apart, there is n interactin between them. Once the tw atms mve clser tgether, the nucleus f each atm attracts the electrns n the ther atm. As the atms mve clser this attractin increases, but the repulsin f the tw nuclei als increases. When the atms are very clse tgether the repulsin between nuclei is much strnger than the attractin between nuclei and electrns. The final internuclear distance fr the chlrine mlecule is the distance at which maximum attractin is achieved in spite f the repulsin. At this distance, the energy f the mlecule is at its lwest value. 9.5 The bnd energy is the energy required t vercme the attractin between atms and Cl atms in ne mle f Cl mlecules in the gaseus state. Energy input is needed t break bnds, s bnd energy is always absrbed (endthermic) and Δ bnd breaking is psitive. The same amunt f energy needed t break the bnd is released upn its frmatin, s Δ bnd frming has the same magnitude as Δ bnd breaking, but ppsite in sign (always exthermic and negative). 9.6 The strength f the cvalent bnd is generally inversely related t the size f the bnded atms. Fr larger atms, their bnding rbitals are mre diffuse, s they frm weaker bnds. 9.7 Bnd strength increases with bnd rder, s C C > C=C > C C. Tw nuclei are mre strngly attracted t tw shared electrn pairs than t ne shared electrn pair and t three shared electrn pairs than t tw. The atms are drawn clser tgether with mre electrn pairs in the bnd and the bnd is strnger. 9.8 When benzene bils, the gas cnsists f C 6 6 mlecules. The energy supplied disrupts the intermlecular attractins between mlecules but nt the intramlecular frces f bnding within the mlecule. 9.9 a) I I < Br Br < Cl Cl. Bnd strength increases as the atmic radii f atms in the bnd decreases. Atmic radii decrease up a grup in the peridic table, s I is the largest and Cl is the smallest f the three. b) S Br < S Cl < S. Radius f is the smallest and Br is the largest, s the bnd strength fr S is the greatest and that fr S Br is the weakest. c) C N < C=N < C N. Bnd strength increases as the number f electrns in the bnd increases. The triple bnd is the strngest and the single bnd is the weakest a) F < Cl < I b) C=O < C O < C S c) N < N O < N S 9.41 a) Fr given pair f atms, in this case carbn and xygen, bnd strength increases with increasing bnd rder. The C=O bnd (bnd rder = 2) is strnger than the C O bnd (bnd rder = 1). b) O is smaller than C s the O bnd is shrter and strnger than the C bnd. 9-5

6 9.42 C=C wuld shw absrptin f IR at shrter wavelength (higher energy) due t it being a strnger bnd than C C. The C C bnd wuld shw absrptin at a shrter wavelength (higher energy) than the C=C bnd since the triple bnd has a larger bnd energy than the duble bnd. Bnd strength increases as the number f electrns in the bnd increases (g) + O 2 (g) O O (g) Δ rxn = Δ bnds brken + Δ bnds frmed Δ rxn = BE + BE Use negative values fr the bnd energies f 2 O= O + [2 (BE O) + BE O O] the prducts Reactin between mlecules requires the breaking f existing bnds and the frmatin f new bnds. Substances with weak bnds are mre reactive than are thse with strng bnds because less energy is required t break weak bnds Bnd energies are average values fr a particular bnd in a variety f cmpunds. eats f frmatin are specific fr a cmpund Fr methane: C 4 (g) + 2 O 2 (g) CO 2 (g) O(l) which requires that 4 C bnds and 2 O=O bnds be brken and 2 C=O bnds and 4 O bnds be frmed. Fr frmaldehyde: C 2 O(g) + O 2 (g) CO 2 (g) + 2 O(l) which requires that 2 C bnds, 1 C=O bnd and 1 O=O bnd be brken and 2 C=O bnds and 2 O bnds be frmed. Methane cntains mre C bnds and fewer C=O bnds than frmaldehyde. Since C bnds take less energy t break than C=O bnds, mre energy is released in the cmbustin f methane than f frmaldehyde Methanl has the greater heat f reactin per mle since C bnds must be brken in methanl while 5 C bnds are brken in the cmbustin f ethanl. Less energy is required t break the bnds in the reactant mlecules in the cmbustin f methanl T find the heat f reactin, add the energy required t break all the bnds in the reactants t the energy released t frm all bnds in the prduct. Remember t use a negative sign fr the energy f the bnds frmed since bnd frmatin is exthermic. The bnd energy values are fund in Table 9.2. Reactant bnds brken: 1(C=C) + 4(C ) + 1(Cl Cl) = (1 ml) (614 kj/ml) + (4 ml) (41 kj/ml) + (1 ml) (24 kj/ml) = 2509 kj Prduct bnds frmed: 1(C C) + 4(C ) + 2(C Cl) = (1 ml) ( 47 kj/ml) + (4 ml) ( 41 kj/ml) + (2 ml) ( 9 kj/ml) = 2677 kj Δ rxn = Δ bnds brken + Δ bnds frmed = 2509 kj + ( 2677 kj) = 168 kj Nte: It is crrect t reprt the answer in kj r kj/ml as lng as the value refers t a reactant r prduct with a mlar cefficient f CO N ( 2 N) 2 CO + 2 O Δ rxn = Δ bnds brken + Δ bnds frmed Δ rxn = [(2 ml BE C=O + 6 BE N ] + [4 (BE N )+ (BE C=O ) + 2 (BE C N ) + 2 ( BE O )] = [2(799) + 6(91)] + [4( 91) + ( 745) + 2( 05) + 2( 467)] = = 91 kj 9.50 T find the heat f reactin, add the energy required t break all the bnds in the reactants t the energy released t frm all bnds in the prduct. Remember t use a negative sign fr the energy f the bnds frmed since bnd frmatin is exthermic. The bnd energy values are fund in Table

7 The reactin: O C O + C O C C O Δ bnds brken = 1 C O= 1 ml (58 kj/ml) Δ bnds frmed = C = ml ( 41 kj/ml) C = ml (41 kj/ml) 1 C C = 1 ml ( 47 kj/ml) 1 O = 1 ml (467 kj/ml) 1 C=O = 1 ml ( 745 kj/ml) 1 C O = 1 ml (1070 kj/ml) 1 C O = 1 ml ( 58 kj/ml) = 14 kj 1 O = 1 ml ( 467 kj/ml) = 156 kj Δ rxn = Δ bnds brken + Δ bnds frmed = 14 kj + ( 156 kj) = 22 kj 9.51 Add the energy required t break all the bnds in the reactants t the energy released t frm all bnds in the prduct. Remember t use a negative sign fr the energy f the bnds frmed since bnd frmatin is exthermic. The bnd energy values are fund in Table 9.2. C C + Br C C Br Bnds brken: 1(C=C) + 4(C ) + 1( Br) = (41) + 6 = kj Bnds frmed: 1(C C) + 5(C ) + 1(C Br) = ( 47) + 5( 41) + ( 276) = 2688 kj Δ rxn = Δ bnds brken + Δ bnds frmed = ( 2688) = 59 kj 9.52 Electrnegativity increases frm left t right acrss a perid (except fr the nble gases) and increases frm bttm t tp within a grup. Flurine (F) and xygen (O) are the tw mst electrnegative elements. Cesium (Cs) and francium (Fr) are the tw least electrnegative elements. 9.5 In general, electrnegativity and inizatin energies are directly related. Electrnegativity relates the strength with which an atm attracts bnding electrns and the inizatin energy measures the energy needed t remve an electrn. Atms that d nt require much energy t have an electrn remved d nt have much attractin fr bnding electrns The O bnd in water is plar cvalent, because the tw atms differ in electrnegativity which results in an unequal sharing f the bnding electrns. A nnplar cvalent bnd ccurs between tw atms with identical electrnegativity values where the sharing f bnding electrns is equal. Althugh electrn sharing ccurs t a very small extent in sme inic bnds, the primary frce in inic bnds is attractin f ppsite charges resulting frm electrn transfer between the atms Electrnegativity is the tendency f a bnded atm t hld the bnding electrns mre strngly. Electrn affinity is the energy invlved when an atm acquires an electrn The difference in E.N. is a reflectin f hw strngly ne atm attracts bnding electrns. The greater this difference is, the mre likely the bnd will be inic; the smaller the E.N. difference, the mre cvalent the bnd. 9-7

8 9.57 Electrnegativity increases frm left t right acrss a perid (except fr the nble gases) and increases frm bttm t tp within a grup. a) Si < S < O, sulfur is mre electrnegative than silicn since it is lcated further t the right in the table. Oxygen is mre electrnegative than sulfur since it is lcated nearer the tp f the table. b) Mg < As < P, magnesium is the least electrnegative because it lies n the left side f the peridic table and phsphrus and arsenic n the right side. Phsphrus is mre electrnegative than arsenic because it is higher in the table a) I < Br < N b) Ca < < F 9.59 Electrnegativity generally increases up a grup and left t right acrss a perid. a) N > P > Si, nitrgen is abve P in Grup 5(A)15 and P is t the right f Si in perid. b) As > Ga > Ca, all three elements are in Perid 4, with As the right-mst element a) Cl > Br > P b) F > O > I 9.61 The arrw pints tward the mre electrnegative atm. nne a) N B b) N O c) C S d) S O e) N f) Cl O 9.62 The mre electrnegative element is partially negative (δ ) and the less electrnegative element is partially psitive (δ + ). δ + δ δ δ + δ + δ a) Br Cl b) F Cl c) O δ δ + δ + δ δ + δ d) Se e) As f) S N 9.6 The mre plar bnd will have a greater difference in electrnegativity, ΔEN. a) N: EN =.0; B: EN = 2.0; ΔEN a = 1.0 b) N: EN =.0; O: EN =.5; ΔEN b = 0.5 c) C: EN = 2.5; S: EN = 2.5; ΔEN c = 0 d) S: EN = 2.5; O: EN =.5; ΔEN d = 1.0 e) N: EN =.0; : EN = 2.1;ΔEN e = 0.9 f) Cl: EN =.0; O: EN =.5; ΔEN f = 0.5 (a), (d), and (e) have greater bnd plarity b) is mre plar; ΔEN is 1.0 fr F Cl and 0.2 fr Br Cl. c) is mre plar; ΔEN is 1.4 fr O and 0. fr Se. f) is mre plar; ΔEN is 0.5 fr S N and 0.1 fr As a) Bnds in S 8 are nnplar cvalent. All the atms are nnmetals s the substance is cvalent and bnds are nnplar because all the atms are f the same element and thus have the same electrnegativity value. ΔEN = 0. b) Bnds in RbCl are inic because Rb is a metal and Cl is a nnmetal. ΔEN is large. c) Bnds in PF are plar cvalent. All the atms are nnmetals s the substance is cvalent. The bnds between P and F are plar because their electrnegativity differs. (By 1.9 units fr P F) d) Bnds in SCl 2 are plar cvalent. S and Cl are nnmetals and differ in electrnegativity. (By 0.5 unit fr S Cl) e) Bnds in F 2 are nnplar cvalent. F is a nnmetal. Bnds between tw atms f the same element are nnplar since ΔEN =

9 f) Bnds in SF 2 are plar cvalent. S and F are nnmetals that differ in electrnegativity. (By 1.5 units fr S F) Increasing bnd plarity: SCl 2 < SF 2 < PF 9.66 a) KCl inic b) P 4 nnplar cvalent c) BF plar cvalent d) SO 2 plar cvalent e) Br 2 nnplar cvalent f) NO 2 plar cvalent NO 2 < SO 2 < BF 9.67 Increasing inic character ccurs with increasing ΔEN. a) ΔEN Br = 0.7, ΔEN Cl = 0.9, ΔEN I = 0.4 b) ΔEN O = 1.4, ΔEN C = 0.4, ΔEN F = 1.9 c) ΔEN SCl = 0.5, ΔEN PCl = 0.9, ΔEN SiCl = 1.2 a) I < Br < Cl b) C < O < F c) S Cl < P Cl < Si Cl 9.68 Increasing inic character ccurs with increasing ΔEN. a) ΔEN PCl = 0.9, ΔEN PBr = 0.7, ΔEN PF = 1.9 P F > P Cl > P Br δ+ δ δ+ δ δ+ δ b) ΔEN BF = 2.0, ΔEN NF = 1.0, ΔEN CF = 1.5 B F > C F > N F δ+ δ δ+ δ δ+ δ c) ΔEN SeF = 1.6, ΔEN TeF = 1.9, ΔEN BrF = 1.2 Te F > Se F > Br F δ+ δ δ+ δ δ+ δ 9.69 C C + Cl Cl 2 C Cl 47 kj/ml 24 kj/ml d) The value shuld be greater than the average f the tw bnd energies given. This is due t the electrnegativity difference a) A slid metal is a shiny slid that cnducts heat, is malleable, and melts at high temperatures. (Other answers include relatively high biling pint and gd cnductr f electricity.) b) Metals lse electrns t frm psitive ins and metals frm basic xides a) Ptassium is a larger atm than sdium, s its electrns are held mre lsely and thus its metallic bnd strength is weaker. b) Be has tw valence electrns per atm cmpared with Li, which has ne. The metallic bnd strength is strnger fr the Be. c) The biling pint is high due t the large amunt f energy needed t separate the metal ins frm each ther in the electrn sea When metallic magnesium is defrmed, the atms are displaced and pass ver ne anther while still being tightly held by the attractin f the sea f electrns. When inic MgF 2 is defrmed, the ins are displaced s that repulsive frces between neighbring ins f like charge cause shattering f the crystals. 9.7 Mlten rck cls frm tp t bttm. The mst stable cmpund (the ne with the largest lattice energy) will slidify first near the tp. The less stable cmpunds will remain in the mlten state at the bttm and eventually crystallize there later. 9-9

10 9.74 a) C /2 O 2 2 CO O Δ rxn = 1259 kj/ml C C O=O 2 O=C=O + O 9.75 a) 5 Δ rxn = [2 BE C + BE C C + BE O ] + [4 ( BE 2 C=O ) + 2 ( BE O )] kj = [2(41) + BE C C + 5/2 (498)] + [4( 799) + 2( 467)] 1259 kj = [826 + BE C C ] + [ 410)] kj BE C C = 800. kj/ml Table 9.2 lists the value as 89 kj/ml 1mlC kj b) heat (kj) = ( g C22) g C22 1 ml C22 = x 10 4 = x 10 4 kj 1 ml C22 2 ml CO g CO2 c) CO 2 prduced (g) = ( g C22) g C22 1 ml C22 1 ml CO2 = = g CO 2 1mlC2 2 (5/2)mlO2 d) ml O 2 = ( g C22) = ml O 2 (unrunded) g C22 1 ml C22 L atm ( ml O2 ) ( 298 K) ml K V = nrt / P = = = 65.2 L O atm Br + F F Br F b) Al + F + Al + F F a) 1) Mg(s) Mg(g) Δ 1 = 148 kj 2) 1/2 Cl 2 (g) Cl(g) Δ 2 = 1/2 (24 kj) ) Mg(g) Mg+(g) + e Δ = 78 kj 4) Cl(g) + e Cl (g) Δ 4 = 49 kj 5) Mg+(g) + Cl (g) MgCl(s) Δ = 78.5 kj (= Δ (MgCl)) 5 lattice 6) Mg(s) + 1/2 Cl 2 (g) MgCl(s) Δ f (MgCl) =? Δ f (MgCl) = Δ 1 + Δ 2 + Δ + Δ 4 + Δ 5 = 148 kj + 1/2 (24 kj) + 78 kj + ( 49 kj) + ( 78.5 kj) = 125 kj b) Yes, since Δ f fr MgCl is negative, MgCl(s) is stable relative t its elements. c) 2 MgCl(s) MgCl 2 (s) + Mg(s) Δ = [1 ml ( Δ f, MgCl 2 (s)) + 1 ml ( Δ f, Mg(s))] [2 ml ( Δ f, MgCl(s))] = [1 ml ( kj/ml) + 1 ml (0)] [2 ml ( 125 kj/ml)] = 91.6 = 92 kj d) N, Δ f fr MgCl 2 is much mre negative than that fr MgCl. This makes the Δ value fr the abve reactin very negative, and the frmatin f MgCl 2 wuld be favred. 9-10

11 9.77 a) Δ = Σ(n)BE reactants + Σ(m)BE prducts = [1 ml (BE - ) + 1 ml (BE Cl-Cl )] + [2 ml (BE -Cl )] = [1 ml (42 kj/ml) + 1 ml (24 kj/ml)] + [2 ml ( 427 kj/ml)] = 179 kj b) Δ = Σ(n)BE reactants + Σ(m)BE prducts = [1 ml (BE - ) + 1 ml (BE I-I )] + [2 ml (BE -I )] = [1 ml (42 kj/ml) + 1 ml (151 kj/ml)] + [2 ml ( 295 kj/ml)] = 7 kj c) Δ = Σ(n)BE reactants + Σ(m)BE prducts = [2 ml (BE - ) + 1 ml (BE O=O )] + [4 ml (BE -O )] = [2 ml (42 kj/ml) + 1 ml (498 kj/ml)] +[4 ml ( 467 kj/ml)] = 506 kj Reactins (a) and (c) are strngly exthermic and are a ptential explsive hazard. Reactin (c) shuld ccur mst explsively a) Find the bnd energy fr an I bnd frm Table 9.2. Calculate wavelength frm this energy using the relatinship frm Chapter 7: E = hc / λ. Bnd energy fr ( I is 295 kj/ml (Table 9.2). 4 )( x 10 Jis.00 x 10 m/s ) 1nm λ = hc / E = 9 = = 406 nm kj 10 J ml 10 m 295 ml 1 kj x 10 2 b) Calculate the energy fr a wavelength f 254 nm and then subtract the energy frm part a) t get the excess energy. kj 10 J ml E (I) = 295 ml 1 kj x 10 2 = x J (unrunded) ( 4 )( 8 ) E (254 nm) = hc / λ. = x 10 J s.00 x 10 m/s 1nm 254 nm 9 10 m = x J Excess energy = x J x J = x = 2.9 x J c) Speed can be calculated frm the excess energy since E k = 1/2 mu 2. E k = 1/2mu 2 thus, u = u = 2E m m = g ml 1 kg 2 ml x 10 = x g 27 kg 2( x 10 J) kg m /s x 10 kg J = x 104 = 1.87 x 10 4 m/s 9.79 a) Vibratinal mtins have frequencies which are in the IR regin f the electrmagnetic spectrum. b)e = hv = (6.626 x 10 4 J s) (4.02 x 10 1 s 1 ) = x = 2.66 x J (symmetric stretch) E = (6.626 x 10 4 J s) (2.00 x 10 1 s 1 ) = x = 1. x J (bending) E = (6.626 x 10 4 J s) (7.05 x 10 1 s 1 ) = x = 4.67 x J (asymmetrical stretch) Bending requires the least amunt f energy Excess bnd energy refers t the difference between the actual bnd energy fr an X Y bnd and the average f the energies fr the X X and the Y Y bnds. Excess bnd energy = BE X Y 1/2 (BE X X + BE Y Y ). The excess bnd energy is zer when the atms X and Y are identical r have the same electrnegativity, as in (a), (b), and (e). ΔEN P = 0, ΔEN CS = 0, ΔEN BrCl = 0.2, ΔEN B = 0.1, ΔEN SeSe =

12 9.81 Rb ([Kr]5s 1 ) has ne valence electrn, s the metallic bnding wuld be fairly weak, resulting in a sft, lw-melting material. Cd ([Kr]5s 2 4d 10 ) has tw valence electrns s the metallic bnding is strnger. V ([Ar]4s 2 d ) has five valence electrns, s its metallic bnding is the strngest, that is, its hardness, melting pint and ther metallic prperties wuld be greatest Find the apprpriate bnd energies in Table 9.2. Calculate the wavelengths using E = hc / λ. C Cl bnd energy = 9 kj/ml λ = hc ( 4 )( 8 ) E = x 10 Jis.00 x 10 m/s =.511 x 10 7 =.5 x 10 7 m kj 10 J ml 9 ml 1 kj x 10 2 O 2 bnd energy = 498 kj/ml λ = hc ( 4 )( 8 ) E = x 10 Jis.00 x 10 m/s = x 10 7 = 2.40 x 10 7 m kj 10 J ml 498 ml 1 kj x Write balanced chemical equatins fr the frmatin f each f the cmpunds. Obtain the bnd energy f flurine frm Table 9.2 (159 kj/ml). Determine the average bnd energy frm Δ = bnds brken + bnds frmed. Remember that the bnds frmed (Xe F) have negative values since bnd frmatin is exthermic. XeF 2 Xe(g) + F 2 (g) XeF 2 (g) Δ = [1 ml F 2 (159 kj/ml)] + [2 ( Xe F)] = 105 kj/ml Xe F = 12 kj/ml XeF 4 Xe(g) + 2 F 2 (g) XeF 4 (g) Δ = [2 ml F 2 (159 kj/ml)] + [4 ( Xe F)] = 284 kj/ml Xe F = = 150. kj/ml XeF 6 Xe(g) + F 2 (g) XeF 6 (g) Δ = [ ml F 2 (159 kj/ml)] + [6 ( Xe F)] = 402 kj/ml Xe F = = 146 kj/ml 9.84 The difference in electrnegativity prduces a greater than expected verlap f rbitals, which shrtens the bnd. As ΔEN becmes smaller (i.e., as yu prceed frm F t I), this effect lessens and the bnd lengths becme mre predictable a) The presence f the very electrnegative flurine atms bnded t ne f the carbns makes the C C bnd plar. This plar bnd will tend t underg heterlytic rather than hmlytic cleavage. Mre energy is required t frce heterlytic cleavage. b) Since ne atm gets bth f the bnding electrns in heterlytic bnd breakage, this results in the frmatin f ins. In heterlytic cleavage a catin is frmed, invlving inizatin energy; an anin is als frmed, invlving electrn affinity. The bnd energy f the O 2 bnd is 498 kj/ml. Δ = (hmlytic cleavage + electrn affinity + first inizatin energy) Δ = (498/2 kj/ml + ( 141 kj/ml) kj/ml) = 1422 = 1420 kj/ml It wuld require 1420 kj t heterlytically cleave 1 ml f O The bnd energies are needed frm Table 9.2. N 2 = 945 kj/ml; O 2 = 498 kj/ml; F 2 = 159 kj/ml N 2 : λ = hc ( 4 )( 8 ) E = x 10 Jis.00 x 10 m/s = x 10 7 = 1.27 x 10 7 m kj 10 J ml 945 ml 1 kj x

13 O 2 : F 2 : ( x 10 4 Jis )(.00 x 10 8 m/s ) λ = kj 10 J ml ml 1 kj x 10 ( x 10 4 Jis )(.00 x 10 8 m/s ) λ = kj 10 J ml ml 1 kj x 10 = x 10 7 = 2.40 x 10 7 m = x 10 7 = 7.5 x 10 7 m 9.87 a) T cmpare the tw energies, the inizatin energy must be cnverted t the energy t remve an electrn frm an atm. The energy needed t remve an electrn frm a single gaseus Ag atm (J) = 71 kj 10 J ml ml 1 kj x 10 = x = 1.21 x J > 7.59 x J It requires less energy t remve an electrn frm the surface f slid silver. b) The electrns in slid silver are held less tightly than the electrns in gaseus silver because the electrns in metals are delcalized, meaning they are shared amng all the metal nuclei. The delcalized attractin f many nuclei t an electrn (slid silver) is weaker than the lcalized attractin f ne nucleus t an electrn (gaseus silver) Use ess Law. Δ f f SiO 2 is fund in Appendix B. 1) Si(s) Si(g) Δ 1 = 454 kj 2) Si(g) Si 4+ (g) + 4 e Δ 2 = 9949 kj ) O 2 (g) 2 O(g) Δ = 498 kj 4) 2 O(g) + 4 e 2 O 2 (g) Δ 4 = 2(77) kj 5) Si 4+ (g) + 2 O 2 (g) SiO 2 (s) Δ 5 = Δ lattice (SiO 2 ) =? 6) Si(s) + O 2 (g) SiO 2 (s) Δ f (SiO 2 ) = kj Δ 6 = - ( Δ 1 + Δ 2 + Δ + Δ 4 + Δ lattice ) kj = (454 kj kj kj + 2(77) kj + Δ lattice ) Δ lattice = = 1286 kj 9.89 Δ rxn = Σ(n) BE reactants + Σ BE prducts Fr ethane: Δ rxn = [1 ml (BE C C ) + 6 ml (BE C ) + 1 ml (BE )] + [8 ml (BE C )] kj = [1 ml (47 kj/ml) + 6 ml (BE C ) + 1 ml (42 kj/ml)] + [8( 415 kj/ml)] ( ) kj BE C = = = 41 kj/ml 6ml Fr ethene: Δ rxn =[ 1 ml (BE C =C ) + 4 ml (BE C ) + 2 ml (BE )] + [8 ml (BE C )] kj =[ 1 ml (614 kj/ml) + 4 ml (BE C ) + 2 ml (42 kj/ml]) + [8 ml ( 415 kj/ml)] ( ) kj BE C = = = 410. kj/ml 4ml 9-1

14 Fr ethyne: Δ rxn = [1 ml (BE C C ) + 2 ml (BE C ) + ml (BE )] + [8 ml (BE C )] kj = [1 ml (89 kj/ml) + 2 ml (BE C ) + ml (42 kj/ml)] + [8 ml ( 415 kj/ml)] ( ) kj BE C = = = 404 kj/ml 2ml 9.90 Use the equatins E = hν, and E = hc / λ. kj 10 J ml ν = E 47 ml 1 kj x 10 2 h = 4 = x = 8.70 x s x 10 Jis ( x 10 4 Jis )(.00 x 10 8 m/s ) λ = hc / E = = x 10 7 =.45 x 10 7 m kj 10 J ml 47 ml 1 kj x 10 2 This is in the ultravilet regin f the electrmagnetic spectrum ν = E h = kj 10 J ml ml 1 kj x x 10 Jis ( x 10 4 Jis )(.00 x 10 8 m/s ) λ = hc / E = kj 10 J ml ml 1 kj x 10 = x = 1.17 x s 1 = x 10 7 = 2.56 x 10 7 m kj 1 ml E phtn = 467 ml x 10 2 phtns = x = 7.75 x kj/phtn 9.92 a) 2 C 4 (g) + O 2 (g) C OC (g) + 2 O(g) Δ rxn = ΣBE reactants + ΣBE prducts Δ rxn =[ 8 x (BE C ) + (BE O=O )] + [6 (BE C ) + 2 (BE C O ) + 2 (BE O )] Δ rxn = [8 (41 kj) kj] + [6 ( 41 kj) + 2 ( 58 kj) + 2 ( 467)] Δ rxn = 26 kj 2 C 4 (g) + O 2 (g) C C 2 O(g) + 2 O(g) Δ rxn = ΣBE reactants + ΣBE prducts Δ rxn = [8 x (BE C ) + (BE O=O )] + [5( BE C ) + (BE C C ) + (BE C O ) + ( BE O )] Δ rxn = [8 (41 kj) k]+ [5 ( 41 kj) + ( 47 kj) + ( 58 kj) + ( 467)] Δ rxn = 69 kj b) The frmatin f gaseus ethanl is mre exthermic. c) Δ rxn = 26 kj ( 69 kj) = 4 kj 9.9 a) C 2 =C 2 (g) + 2 O(g) C C 2 O(g) Using bnd energies: Δ rxn = ΣBE reactants + ΣBE prducts Δ rxn =[ 4 x (BE C ) + (BE C=C ) + 2 (BE O )] + [5 (BE C ) + (BE C C ) + (BE C O ) + (BE O )] Δ rxn = [4 (41 kj) kj+ 2(467 kj)] + [5 ( 41 kj) + ( 47 kj) + ( 58) + ( 467)] Δ rxn = 7 kj Using heats f frmatin: Δ = Σ Δ f prducts Σ Δ f reactants Δ = [1 ml ( Δ f C C 2 O(g))] [ 1ml ( Δ f C 2 =C 2 (g)) + 1ml ( Δ f 2 O (g))] Δ = [ 25.1 kj] [52.47 kj kj] Δ = = 45.7 kj 9-14

15 b) C 2 4 O(l) + 2 O(l) C 2 6 O 2 (l) Δ rxn = ΣBE reactants + ΣBE prducts Δ rxn =[ 4 x (BE C ) + (BE C-C ) + 2(BE C O ) + 2 (BE O )] + [4 (BE C ) + (BE C C ) + 2(BE C O ) + 2(BE O )] Δ rxn =[4(41 kj) + 47 kj + 2(58 kj) + 2(467 kj)] + [4( 41 kj) + ( 47 kj) + 2( 58 kj) + 2( 467 kj)] Δ rxn = 0 kj c) In the hydrlysis in part b), the Δ rxn appears t be 0 kj using bnd energies since the number and types f bnds brken and the number and types f bnds frmed are the same. Since the bnd energy values used are average values, this methd des nt differentiate between an O- bnd in water, fr example and an O- bnd in ethylene glycl. 9-15