CHAPTER 13 BONDING: GENERAL CONCEPTS

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CHAPTER 13BONDING: GENERAL CONCEPTSChemical Bonds and Electronegativity11.Electronegativity is the ability of an atom in a molecule to attract electrons to itself.Electronegativity is a bonding term. Electron affinity is the energy change when an electron isadded to a substance. Electron affinity deals with isolated atoms in the gas phase.A covalent bond is a sharing of electron pair(s) in a bond between two atoms. An ionic bondis a complete transfer of electrons from one atom to another to form ions. The electrostaticattraction of the oppositely charged ions is the ionic bond.A pure covalent bond is an equal sharing of shared electron pair(s) in a bond. A polarcovalent bond is an unequal sharing.Ionic bonds form when there is a large difference in electronegativity between the two atomsbonding together. This usually occurs when a metal with a small electronegativity is bondedto a nonmetal having a large electronegativity. A pure covalent bond forms between atomshaving identical or nearly identical eletronegativities. A polar covalent bond forms whenthere is an intermediate electronegativity difference. In general, nonmetals bond together byforming covalent bonds, either pure covalent or polar covalent.Ionic bonds form due to the strong electrostatic attraction between two oppositely chargedions. Covalent bonds form because the shared electrons in the bond are attracted to twodifferent nuclei, unlike the isolated atoms where electrons are only attracted to one nuclei.The attraction to another nuclei overrides the added electron-electron repulsions.12.a. There are two attractions of the form( 1)( 1), where r 1 10 10 m 0.1 nm.r ( 1)( 1) 18 18V 2 (2.31 10 19 J nm) 4.62 10 J 5 10 J 0.1 nm b. There are 4 attractions of 1 and 1 charges at a distance of 0.1 nm from each other. Thetwo negative charges and the two positive charges repel each other across the diagonal ofthe square. This is at a distance of 2 0.1 nm.490

CHAPTER 13BONDING: GENERAL CONCEPTS ( 1)( 1) 19V 4 (2.31 10 19) 2.31 10 0 .1 491 ( 1)( 1) 2 (0.1) ( 1)( 1) 2.31 10 19 2 (0.1) V 9.24 10 18 J 1.63 10 18 J 1.63 10 18 J 5.98 10 18 J 6 10 18 JNote: There is a greater net attraction in arrangement b than in a.13.Using the periodic table, we expect the general trend for electronegativity to be:1. Increase as we go from left to right across a period2. Decrease as we go down a group14.a. C N Ob. Se S Clc. Sn Ge Sid. Tl Ge Se. Rb K Naf.The most polar bond will have the greatest difference in electronegativity between the twoatoms. From positions in the periodic table, we would predict:a. Ge F15.Ga B Ob. P Clc. S Fd. Ti Cle. Sn Hf.Tl BrThe general trends in electronegativity used in Exercises 13.13 and 13.14 are only rules ofthumb. In this exercise we use experimental values of electronegativities and can begin tosee several exceptions. The order of EN using Figure 13.3 is:a. C (2.6) N (3.0) O (3.4)same as predictedb. Se (2.6) S (2.6) Cl (3.2) differentc. Si (1.9) Ge (2.0) Sn (2.0) differentd. Tl (2.0) Ge (2.0) S (2.6) differente. Rb (0.8) K (0.8) Na (0.9) differentf. Ga (1.8) B (2.0) O (3.4) sameMost polar bonds using actual EN values:16.a. Si F (Ge F predicted)b. P Cl (same as predicted)c. S F (same as predicted)d. Ti Cl (same as predicted)e. C H (Sn H predicted)f.Al Br (Tl Br predicted)Electronegativity values increase from left to right across the periodic table. The order ofelectronegativities for the atoms from smallest to largest electronegativity will be H P C N O F. The most polar bond will be F‒H since it will have the largest difference in

492CHAPTER 13BONDING: GENERAL CONCEPTSelectronegativities, and the least polar bond will be P‒H since it will have the smallestdifference in electronegativities (ΔEN 0). The order of the bonds in decreasing polarity willbe F‒H O‒H N‒H C‒H P‒H.17.Ionic character is proportional to the difference in electronegativity values between the twoelements forming the bond. Using the trend in electronegativity, the order will be:Br‒Br N‒O C‒F Ca‒O K‒Fleastmostionic characterionic characterNote that Br‒Br, N‒O and C‒F bonds are all covalent bonds since the elements are all nonmetals. The Ca‒O and K‒F bonds are ionic, as is generally the case when a metal forms abond with a nonmetal.(IE EA)18.FClBrI(IE EA)/5022006 kJ/mol1604146313024.03.22.92.6EN (text)2006/502 4.04.03.23.02.7The values calculated from IE and EA show the same trend as (and agree fairly closely) withthe values given in the text.Ionic Compounds19.Anions are larger than the neutral atom, and cations are smaller than the neutral atom. Foranions, the added electrons increase the electron-electron repulsions. To counteract this, thesize of the electron cloud increases, placing the electrons further apart from one another. Forcations, as electrons are removed, there are fewer electron-electron repulsions, and theelectron cloud can be pulled closer to the nucleus.Isoelectronic: same number of electrons. Two variables, the number of protons and thenumber of electrons, determine the size of an ion. Keeping the number of electrons constant,we only have to consider the number of protons to predict trends in size. The ion with themost protons attracts the same number of electrons most strongly, resulting in a smaller size.20.All of these ions have 18 e ; the smallest ion (Sc3 ) has the most protons attracting the 18 e ,and the largest ion has the fewest protons (S2 ). The order in terms of increasing size is Sc3 Ca2 K Cl S2 . In terms of the atom size indicated in the question:K Ca2 Sc3 S2-Cl-

CHAPTER 1321.BONDING: GENERAL CONCEPTS493c. O2 O Oa. Cu Cu Cu2 b. Pt2 Pd2 Ni2 d. La3 Eu3 Gd3 Yb3 e. Te2 I Cs Ba2 La3 For answer a, as electrons are removed from an atom, size decreases. Answers b and dfollow the radius trend. For answer c, as electrons are added to an atom, size increases.Answer e follows the trend for an isoelectronic series, i.e., the smallest ion has the mostprotons.22.23.a. Mg2 : 1s22s22p6Sn2 :[Kr]5s24d10K :1s22s22p63s23p6Al3 :1s22s22p6Tl :[Xe]6s24f145d10As3 :[Ar]4s23d10b. N3 , O2 and F : 1s22s22p6Te2-:[Kr]5s24d105p6c. Be2 :1s2Rb :[Ar]4s23d104p6Ba2 :[Kr]5s24d105p6Se2 :[Ar]4s23d104p6I :[Kr]5s24d105p6a. Cs2S is composed of Cs and S2 . Cs has the same electron configuration as Xe, and S2 has the same configuration as Ar.b. SrF2; Sr2 has the Kr electron configuration, and F has the Ne configuration.c. Ca3N2; Ca2 has the Ar electron configuration, and N3 has the Ne configuration.d. AlBr3; Al3 has the Ne electron configuration, and Br has the Kr configuration.24.a. Sc3 b. Te2 c. Ce4 and Ti4 d. Ba2 All of these have the number of electrons of a noble gas.25.Se2 , Br-, Rb , Sr2 , Y3 , and Zr4 are some ions that are isoelectronic with Kr (36 electrons).In terms of size, the ion with the most protons will hold the electrons tightest and will be thesmallest. The size trend is:Zr4 Y3 Sr2 Rb Br- Se2 smallestlargest26.Lattice energy is proportional to Q1Q2/r, where Q is the charge of the ions and r is thedistance between the ions. In general, charge effects on lattice energy are greater than sizeeffects.a. LiF; Li is smaller than Cs .b. NaBr; Br- is smaller than I .c. BaO; O2 has a greater charge than Cl-. d. CaSO4; Ca2 has a greater charge than Na .

494CHAPTER 13e. K2O; O2 has a greater charge than F .27.f.BONDING: GENERAL CONCEPTSLi2O; The ions are smaller in Li2O.a. Al3 and S2 are the expected ions. The formula of the compound would be Al2S3(aluminum sulfide).b. K and N3 ; K3N, potassium nitridec. Mg2 and Cl ; MgCl2, magnesium chlorided. Cs and Br ; CsBr, cesium bromide28.Ionic solids can be characterized as being held together by strong omnidirectional forces.i.For electrical conductivity, charged species must be free to move. In ionic solids thecharged ions are held rigidly in place. Once the forces are disrupted (melting ordissolution), the ions can move about (conduct).ii. Melting and boiling disrupts the attractions of the ions for each other. If the forcesare strong, it will take a lot of energy (high temperature) to accomplish this.iii. If we try to bend a piece of material, the atoms/ions must slide across each other. Foran ionic solid, the following might happen: Strong repulsionStrong attractionJust as the layers begin to slide, there will be very strong repulsions causing the solidto snap across a fairly clean plane.These properties and their correlation to chemical forces will be discussed in detail in Chapter16.K(s) K(g)29. ΔH 64 kJ (sublimation) K(g) K (g) e1/2 Cl2(g) Cl(g) ΔH 419 kJ (ionization energy)ΔH 239/2 kJ (bond energy)Cl(g) e Cl (g)ΔH !349 kJ (electron affinity) ΔH !690. kJ (lattice energy)K (g) Cl (g) KCl(s)ΔH f !437 kJ/molK(s) 1/2 Cl2(g) KCl(s)

CHAPTER 1330.BONDING: GENERAL CONCEPTSMg(s) Mg(g)Mg(g) Mg (g) e Mg (g) Mg2 (g) e F2(g) 2 F(g)2 F(g) 2 e 2 F (g)Mg2 (g) 2 F (g) MgF2(s)ΔH 150. kJΔH 735 kJΔH 1445 kJΔH 154 kJΔH 2(-328) kJΔH 2913 kJ495(sublimation)(IE1)(IE2)(BE)(EA)(LE)Mg(s) F2(g) MgF2(s)31.ΔH of 1085 kJ/molUse Figure 13.11 as a template for this problem.Li(s) Li(g) 1/2 I2(g) I(g) e Li (g) I (g) Li(g)Li (g) e I(g)I (g)LiI(s)ΔHsub ?ΔH 520. kJΔH 151/2 kJΔH !295 kJΔH !753 kJLi(s) 1/2 I2(g) LiI(s)ΔH !272 kJΔHsub 520. 151/2 ! 295 ! 753 !272, ΔHsub 181 kJ32.Two other factors that must be considered are the ionization energy needed to produce morepositively charged ions and the electron affinity needed to produce more negatively chargedions. The favorable lattice energy more than compensates for the unfavorable ionizationenergy of the metal and for the unfavorable electron affinity of the nonmetal, as long aselectrons are added to or removed from the valence shell. Once the valence shell is full, theionization energy required to remove another electron is extremely unfavorable; the same istrue for electron affinity when an electron is added to a higher n shell. These two quantitiesare so unfavorable after the valence shell is complete that they overshadow the favorablelattice energy, and the higher-charged ionic compounds do not form.33.a. From the data given, less energy is required to produce Mg (g) O (g) than to produceMg2 (g) O2 (g). However, the lattice energy for Mg2 O2 will be much more exothermicthan for Mg O (due to the greater charges in Mg2 O2 ). The favorable lattice energyterm will dominate and Mg2 O2 forms.b. Mg and O both have unpaired electrons. In Mg2 and O2 there are no unpairedelectrons. Hence Mg O would be paramagnetic; Mg2 O2 would be diamagnetic.Paramagnetism can be detected by measuring the mass of a sample in the presence andabsence of a magnetic field. The apparent mass of a paramagnetic substance will belarger in a magnetic field because of the force between the unpaired electrons and thefield.

49634.CHAPTER 13BONDING: GENERAL CONCEPTSLet us look at the complete cycle for Na2S.2 Na(s) 2 Na(g)2 Na(g) 2 Na (g) 2 e S(s) S(g)S(g) e S (g)S (g) e S2 (g) 2 Na (g) S2 (g) Na2S(s)2ΔHsub, Na 2(109) kJ2IE 2(495) kJΔHsub, S 277 kJEA1 !200. kJEA2 ?LE !2203 kJΔH f !365 kJ2 Na(s) S(s) Na2S(s)ΔH f 2ΔH sub , Na 2IE ΔHsub, S EA1 EA2 LE, !365 !918 EA2, EA2 553 kJFor each salt: ΔH f 2ΔHsub, M 2IE 277 ! 200. LE EA2K2S: !381 2(90.) 2(419) 277 ! 200. ! 2052 EA2, EA2 576 kJRb2S: !361 2(82) 2(409) 277 ! 200. ! 1949 EA2, EA2 529 kJCs2S: !360. 2(78) 2(382) 277 ! 200. ! 1850. EA2, EA2 493 kJWe get values from 493 to 576 kJ.The mean value is:540 50 kJ.35.553 576 529 493 538 kJ. We can represent the results as EA2 4Ca2 has a greater charge than Na , and Se2 is smaller than Te2 . The effect of charge on thelattice energy is greater than the effect of size. We expect the trend from most exothermic toleast exothermic to be:CaSe CaTe Na2Se Na2Te( 2862)36.( 2721)( 2130)( 2095 kJ/mol)Lattice energy is proportional to the charge of the cation times the charge of the anion, Q1Q2.CompoundQ1Q2Lattice EnergyFeCl2( 2)( 1) 2 2631 kJ/molFeCl3( 3)( 1) 3 5339 kJ/molFe2O3( 3)( 2) 6 14,744 kJ/molBond Energies37.This is what we observe.a.HH ClCl2HCl

CHAPTER 13BONDING: GENERAL CONCEPTSBonds broken:497Bonds formed:1 H‒H (432 kJ/mol)1 Cl‒Cl (239 kJ/mol)2 H‒Cl (427 kJ/mol)ΔH ΣDbroken ! ΣDformed, ΔH 432 kJ 239 kJ ! 2(427) kJ !183 kJb.N 3 HN2HHNHHBonds broken:Bonds formed:6 N‒H (391 kJ/mol)1 N N (941 kJ/mol)3 H‒H (432 kJ/mol)ΔH 941 kJ 3(432) kJ ‒ 6(391) kJ ‒109 kJc. Sometimes some of the bonds remain the same between reactants and products. To savetime, only break and form bonds that are involved in the reaction.HN 2 HCHHHHCNHHBonds broken:Bonds formed:1 C N (891 kJ/mol)2 H H (432 kJ/mol)1 C N (305 kJ/mol)2 C H (413 kJ/mol)2 N H (391 kJ/mol)ΔH 891 kJ 2(432 kJ) [305 kJ 2(413 kJ) 2(391 kJ)] 158 kJHHNd. 2 FNHF4 HF NNHBonds broken:1 N N (160. kJ/mol)4 N H (391 kJ/mol)2 F F (154 kJ/mol)Bonds formed:4 H F (565 kJ/mol)1 N N (941 kJ/mol)ΔH 160. kJ 4(391 kJ) 2(154 kJ) [4(565 kJ) 941 kJ] 1169 kJ38.a. ΔH 2ΔH fo, HCl 2 mol( 92 kJ/mol) 184 kJ ( 183 kJ from bond energies)

498CHAPTER 13BONDING: GENERAL CONCEPTSb. ΔH 2ΔH of , NH 3 2 mol(-46 kJ/mol) 92 kJ ( 109 kJ from bond energies)Comparing the values for each reaction, bond energies seem to give a reasonably goodestimate for the enthalpy change of a reaction. The estimate is especially good for gas phasereactions.HH39.HC NCH C C NHHBonds broken: 1 C‒N (305 kJ/mol)Bonds formed: 1 C‒C (347 kJ/mol)ΔH ΣDbroken ! ΣDformed, ΔH 305 ! 347 !42 kJNote: Sometimes some of the bonds remain the same between reactants and products.To save time, only break and form bonds that are involved in the reaction.40.HHHCCHHOH 3OBonds broken:O2OCO 3 HOHBonds formed:5 C H (413 kJ/mol)1 C C (347 kJ/mol)1 C O (358 kJ/mol)1 O H (467 kJ/mol)3 O O (495 kJ/mol)2 2 C O (799 kJ/mol)3 2 O H (467 kJ/mol)ΔH 5(413 kJ) 347 kJ 358 kJ 467 kJ 3(495 kJ) – [4(799 kJ) 6(467 kJ)] 1276 kJ41.H C/C H 5/2 O O 2 O C O H O HBonds broken:2 C H (413 kJ/mol)1 C/C (839 kJ/mol)5/2 O O (495 kJ/mol)Bonds formed:2 2 C O (799 kJ/mol)2 O H (467 kJ/mol)ΔH 2(413 kJ) 839 kJ 5/2 (495 kJ) – [4(799 kJ) 2(467 kJ)] !1228 kJ42.Let x bond energy for A2, and then 2x bond energy for AB.ΔH !285 kJ x 432 kJ – [2(2x)], 3x 717, x 239 kJ/mol bond energy for A2

CHAPTER 13BONDING: GENERAL CONCEPTS49943.H4NHO 5NCHOHNOH12 HNOH 9 NN 4 OCOHBonds broken:Bonds formed:9 N‒N (160. kJ/mol)4 N‒C (305 kJ/mol)12 C‒H (413 kJ/mol)12 N‒H (391 kJ/mol)10 N O (607 kJ/mol)10 N‒O (201 kJ/mol)24 O‒H (467 kJ/mol)9 N N (941 kJ/mol)8 C O (799 kJ/mol)ΔH 9(160.) 4(305) 12(413) 12(391) 10(607) 10(201) [24(467) 9(941) 8(799)]ΔH 20,388 kJ 26,069 kJ 5681 kJ44.a. I.HH * OHCH C* HOBonds broken (*):H * CNHCNC*CHHHBonds formed (*):1 C‒O (358 kJ)1 C‒H (413 kJ)1 O‒H (467 kJ)1 C‒C (347 kJ)ΔHI 358 kJ 413 kJ (467 kJ 347 kJ) 43 kJII.OH HH*C * C*HHHCNHCH* CCNH * OHO

500CHAPTER 13Bonds broken (*):BONDING: GENERAL CONCEPTSBonds formed (*):1 C‒O (358 kJ/mol)1 C‒H (413 kJ/mol)1 C‒C (347 kJ/mol)1 H‒O (467 kJ/mol)1 C C (614 kJ/mol)ΔHII 358 kJ 413 kJ 347 kJ [467 kJ 614 kJ] 37 kJΔHoverall ΔHI ΔHII 43 kJ 37 kJ 6 kJb.HHHC4CCHHH4 6 NOHCC 6 HCHBonds broken:NOH NNHBonds formed:4 3 C‒H (413 kJ/mol)4 C N (891 kJ/mol)6 N O (630. kJ/mol)6 2 H‒O (467 kJ/mol)1 N N (941 kJ/mol)ΔH 12(413) 6(630.) [4(891) 12(467) 941] 1373 kJc.HH2CHHHCCH 2HNHH 3 O2C2HC 6CHBonds broken:NHOHHBonds formed:2 3 C‒H (413 kJ/mol)2 3 N‒H (391 kJ/mol)3 O O (495 kJ/mol)2 C N (891 kJ/mol)6 2 O‒H (467 kJ/mol)ΔH 6(413) 6(391) 3(495) [2(891) 12(467)] 1077 kJ45.Because both reactions are highly exothermic, the high temperature is not needed to provideenergy. It must be necessary for some other reason. The reason is to increase the speed ofthe reaction. This will be discussed in Chapter 15 on kinetics.46.HHC O H C OHHHOC C O HH

CHAPTER 13BONDING: GENERAL CONCEPTSBonds broken:501Bonds formed:1 C‒C (347 kJ/mol)1 C O (745 kJ/mol)1 C‒O (358 kJ/mol)1 C O (1072 kJ/mol)1 C‒O (358 kJ/mol)ΔH 1072 358 (347 745 358) 20. kJCH3OH(g) CO(g) CH3COOH(l)ΔH 484 kJ [( 201 kJ) ( 110.5 kJ)] 173 kJUsing bond energies, ΔH 20. kJ. For this reaction, bond energies give a much poorerestimate for ΔH as compared with gas phase reactions. The major reason for the largediscrepancy is that not all species are gases in this exercise. Bond energies do not account forthe energy changes that occur when liquids and solids form instead of gases. These energychanges are due to intermolecular forces and will be discussed in Chapter 16.47.a.HF(g) H(g) F(g)H(g) H (g) e F(g) e F (g)ΔH 565 kJΔH 1312 kJΔH 327.8 kJHF(g) H (g) F (g)b.HCl(g) H(g) Cl(g)H(g) H (g) e Cl(g) e Cl (g)ΔH 1549 kJΔH 427 kJΔH 1312 kJΔH 348.7 kJHCl(g) H (g) Cl (g)c.HI(g) H(g) I(g)H(g) H (g) e I(g) e I (g)ΔH 1390. kJΔH 295 kJΔH 1312 kJΔH 295.2 kJd.HI(g) H (g) I (g)ΔH 1312 kJH2O(g) OH(g) H(g)H(g) H (g) e OH(g) e OH (g)ΔH 467 kJΔH 1312 kJΔH 180. kJH2O(g) H (g) OH (g)48.ΔH 1599 kJa. Using SF4 data: SF4(g) S(g) 4 F(g)ΔH 4DSF 278.8 kJ 4(79.0 kJ) ( 775 kJ) 1370. kJDSF 1370. kJ 342.5 kJ/mol4 mol SF bonds

502CHAPTER 13BONDING: GENERAL CONCEPTSUsing SF6 data: SF6(g) S(g) 6 F(g)ΔH 6DSF 278.8 kJ 6(79.0 kJ) ( 1209 kJ) 1962 kJDSF 1962 kJ 327.0 kJ/mol6b. The S‒F bond energy in Table 13.6 is 327 kJ/mol. The value in the table was based onthe S‒F bond in SF6.c. S(g) and F(g) are not the most stable form of the element at 25 C and 1 atm. The moststable forms are S8(s) and F2(g); ΔH of 0 for these two species.49.NH3(g) N(g) 3 H(g)ΔH 3DNH 472.7 kJ 3(216.0 kJ) ( 46.1 kJ) 1166.8 kJDNH 1166. 8 kJ 388.93 kJ/mol3 mol NH bondsDcalc 389 kJ/mol compared with 391 kJ/mol in the table. There is good agreement.50.ΔH of for H(g) is ΔH for the reaction: 1/2 H2(g) H(g); ΔH of for H(g) equals one-half theH‒H bond energy.Lewis Structures and Resonance51.Drawing Lewis structures is mostly trial and error. However, the first two steps are alwaysthe same. These steps are (1) count the valence electrons available in the molecule/ion, and(2) attach all atoms to each other with single bonds (called the skeletal structure). Unlessnoted otherwise, the atom listed first is assumed to be the atom in the middle, called thecentral atom, and all other atoms in the formula are attached to this atom. The most notableexceptions to the rule are formulas that begin with H, e.g., H2O, H2CO, etc. Hydrogen cannever be a central atom since this would require H to have more than two electrons. In thesecompounds, the atom listed second is assumed to be the central atom.After counting valence electrons and drawing the skeletal structure, the rest is trial and error.We place the remaining electrons around the various atoms in an attempt to satisfy the octetrule (or duet rule for H). Keep in mind that practice makes perfect. After practicing, you can(and will) become very adept at drawing Lewis structures.

CHAPTER 13BONDING: GENERAL CONCEPTSa. HCN has 1 4 5 10 valenceelectrons.HCNHC503b. PH3 has 5 3(1) 8 valence electrons.HNPHHHSkeletalstructureSkeletal structure uses 4 e ; 6 e remainc.ClCd. NH4 has 5 4(1) ! 1 8 valenceelectrons.HHClClClCHClNLewisstructuree. H2CO has 2(1) 4 6 12 valenceelectrons. HHClSkeletalstructureLewisstructureSkeletal structures uses 6 e ; 2 e remainCHCl3 has 4 1 3(7) 26 valencee

CHAPTER 13 BONDING: GENERAL CONCEPTS Chemical Bonds and Electronegativity 11. Electronegativity is the ability of an atom in a molecule to attract electrons to itself. Electronegativity is a bonding term. Electron affinity is the energy change when an electron is added to a substance. Electron affinity deals with isolated atoms in the gas phase.

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