CHAPTER 2 Bonding and Molecular Structure 23 Attractive forces betv such molecules can mutually mix and solution is easy.The attractive forces between polarH.orC.H.OH mol. ecules are strong H-bonds.Most nonpolar molecules cannot overcome these H-bonds and therefore do not dis- solve in such polar protic solvents. Problem 2.19 Explain why CH,CH,OH is much more soluble in water than is CH (CH,),CH,OH. Problem 2.20 Explain why NaCl dissolves in water Water.a protie solvent,helps separate the strongly attracting ions of the solid salt by solvation.Several negative ends o Hrcmdn CI-- Nat. and H Problem 2.21 Compare the ways in which NaCl dissolves in water and in dimethyl sulfoxide. also solvates positive ions by an ion-dipole attraction;the Oo the group is d to is surrounded hy the methyl o and cannot get close enough to solvate the anion The bare negative ions discussed in Problem 2.21 have a greatly enhanced reactivity.The small amounts of salts that dissolve in nonpolar or weakly polar solvents exist mainly as ion-pairs or ion-clusters.where the oppo sitely charged on ose to eacn other and move a as units.I etween ion-pairs are separated by a small number of s hloanpiteheenosolemtmolecules 2.8 Resonance and Delocalized x Electrons Resonance theory describes species for which a single Lewis electron structure cannot be written.As an example,consider dinitrogen oxide,N,O: 放==0： resonance N=N-0: 0.1200.115 0.1100.147 Observed 0.1120.119 0.1120.119 Bond Length on of the calculated and observed bond leng s show that neithe ture is tributing (resonance s tell us th ance hybrid ha character between Nand and some triple-bond character between Nand N.This state of affairs is described by the non-Lewis structure: :=i6
Attractive forces between nonpolar molecules such as mineral oil and n-hexane are very weak. Therefore, such molecules can mutually mix and solution is easy. The attractive forces between polar H2 O or C2 H5 OH molecules are strong H-bonds. Most nonpolar molecules cannot overcome these H-bonds and therefore do not dissolve in such polar protic solvents. Problem 2.19 Explain why CH3CH2OH is much more soluble in water than is CH3(CH2)3CH2OH. The OH portion of an alcohol molecule tends to interact with water—it is hydrophilic. The hydrocarbon portion does not interact. Rather, it is repelled—it is hydrophobic. The larger the hydrophobic portion, the less soluble in water is the molecule. Problem 2.20 Explain why NaCl dissolves in water. Water, a protic solvent, helps separate the strongly attracting ions of the solid salt by solvation. Several water molecules surround each positive ion (Na+) by an ion-dipole attraction. The O atoms, which are the negative ends of the molecular dipole, are attracted to the cation. H2O typically forms an H-bond with the negative ion (in this case Cl). CHAPTER 2 Bonding and Molecular Structure 23 Problem 2.21 Compare the ways in which NaCl dissolves in water and in dimethyl sulfoxide. The way in which NaCl, a typical salt, dissolves in water, a typical protic solvent, was discussed in Problem 2.20. Dimethyl sulfoxide also solvates positive ions by an ion-dipole attraction; the O of the SO group is attracted to the cation. However, since this is an aprotic solvent, there is no way for an H-bond to be formed and the negative ions are not solvated when salts dissolve in aprotic solvents. The S, the positive pole, is surrounded by the methyl groups and cannot get close enough to solvate the anion. The bare negative ions discussed in Problem 2.21 have a greatly enhanced reactivity. The small amounts of salts that dissolve in nonpolar or weakly polar solvents exist mainly as ion-pairs or ion-clusters, where the oppositely charged ions are close to each other and move about as units. Tight ion-pairs have no solvent molecules between the ions; loose ion-pairs are separated by a small number of solvent molecules. 2.8 Resonance and Delocalized π Electrons Resonance theory describes species for which a single Lewis electron structure cannot be written. As an example, consider dinitrogen oxide, N2 O: A comparison of the calculated and observed bond lengths show that neither structure is correct. Nevertheless, these contributing (resonance) structures tell us that the actual resonance hybrid has some double-bond character between N and O, and some triple-bond character between N and N. This state of affairs is described by the non-Lewis structure:
24 CHAPTER 2 Bonding and Molecular Structure in which broken lines stand for the partial bonds in which there a ons in an tended abond created from overlap of porbitals on each atom.See also the orbital diagram in Fig.2.10.The symboldenotes resonance,not equilibrium. 0 Figure 2.10 hetical E,=E-E ter the res the hybrid looks most like the lowest-energy structure. Contributing structures (a)differ only in positions of electro ons (atomic nuclei must have the same must have the same number of paired electrons.Relative energies of contributing structures 1.Structures with the test number of covalent bonds are most stable.However,for second-period elements (C..N).the octet rule must be observed. With a few exceptions.structures with the least amount of formal charges are most stable. ive form and+。 charge,the most stable (lowest energy)one hason the more electronegative lectropo h lik 5.Resonance positively charged atomhave very high enery and are usually ignored. Problem 2.22 Write contributing structures,showing formal charges when necessary,for(a)ozone,O @）6于i9:→9i=:((qual--enery re).The hybrid is:6-6-高 65c6: →0-c=→：6=c-: (I) (2) 3) (1)is most stable;it has no forma charge.(2)and (3)have equal energy and are least stable because they (ges.In a ition,in harge.S :
in which broken lines stand for the partial bonds in which there are delocalized p electrons in an extended π bond created from overlap of p orbitals on each atom. See also the orbital diagram in Fig. 2.10. The symbol ↔ denotes resonance, not equilibrium. 24 CHAPTER 2 Bonding and Molecular Structure Figure 2.10 The energy of the hybrid, Eh, is always less than the calculated energy of any hypothetical contributing structure, Ec . The difference between these energies is the resonance (delocalization) energy, Er : Er Ec – Eh The more nearly equal in energy the contributing structures, the greater the resonance energy and the less the hybrid looks like any of the contributing structures. When contributing structures have dissimilar energies, the hybrid looks most like the lowest-energy structure. Contributing structures (a) differ only in positions of electrons (atomic nuclei must have the same positions) and (b) must have the same number of paired electrons. Relative energies of contributing structures are assessed by the following rules: 1. Structures with the greatest number of covalent bonds are most stable. However, for second-period elements (C, O, N), the octet rule must be observed. 2. With a few exceptions, structures with the least amount of formal charges are most stable. 3. If all structures have formal charge, the most stable (lowest energy) one has on the more electronegative atom and on the more electropositive atom. 4. Structures with like formal charges on adjacent atoms have very high energies. 5. Resonance structures with electron-deficient, positively charged atoms have very high energy and are usually ignored. Problem 2.22 Write contributing structures, showing formal charges when necessary, for (a) ozone, O3; (b) CO2; (c) hydrazoic acid, HN3; (d) isocyanic acid, HNCO. Indicate the most and least stable structures and give reasons for your choices. Give the structure of the hybrid. (a) (b) (1) is most stable; it has no formal charge. (2) and (3) have equal energy and are least stable because they have formal charges. In addition, in both (2) and (3), one O, an electronegative element, bears a formal charge. Since (1) is so much more stable than (2) and (3), the hybrid is which is just (1).
CHAPTER 2 Bonding and Molecular Structure 25 H-N=衣=：《H--=N:→H一衣=衣-：→H一对-- (1) (2 (3) (4 (1)and (2)ha the leas charge about the y high sme on adiacent atoms and.in terms a total formal charge of 4.(4)has a very high energy because the N bonded to H has only six electrons The hybrid,composed of (1)and (2),is as follows: H-=N d H-i-c=0:→H-8-C=0→H-=C-: 3 (3) (1)has no formal charge and is most stable.(2)is least stable since the-charge is on N rather than on the more electronegatve as in (3).The nybnd is H:which is the same as (D).the mo ontributing structure Problem 2.23 (a)Write contributing structures and the delocalized structure for (i)NO and (ii)NO The-is delocalized over both O's so that each can be assumed to have acharge.Each N-O bond has the same bond length 0 h)pad oe c sharge (c)We can use resonance theory to compare the stability of these two ions because they differ in only one fea ture- he number of 'son chN.which is related to the oxidation numbers of the N's.We could not.fo O;an s0ncoheymoehhom0Oys delocalized(number of Os and since NOhasa more extended bond system. Indicate which one of the following pairs of resonance structures is the less stable and is an
(c) (1) and (2) have about the same energy and are the most stable, since they have the least amount of formal charge. (3) has a very high energy, since it has charge on adjacent atoms and, in terms of absolute value, a total formal charge of 4. (4) has a very high energy because the N bonded to H has only six electrons. The hybrid, composed of (1) and (2), is as follows: (d) CHAPTER 2 Bonding and Molecular Structure 25 (1) has no formal charge and is most stable. (2) is least stable since the charge is on N rather than on the more electronegative O as in (3). The hybrid is which is the same as (1), the most stable contributing structure. Problem 2.23 (a) Write contributing structures and the delocalized structure for (i) NO 2 and (ii) NO 3. (b) Use p AO’s to draw a structure showing the delocalization of the p electrons in an extended π bond for (i) and (ii). (c) Compare the stability of the hybrids of each. (a) The is delocalized over both O’s so that each can be assumed to have a 1 –2 charge. Each N—O bond has the same bond length. The – charges are delocalized over three O’s so that each has a 2 –3 charge. (b) See Fig. 2.11. (c) We can use resonance theory to compare the stability of these two ions because they differ in only one feature—the number of O’s on each N, which is related to the oxidation numbers of the N’s. We could not, for example, compare NO 3 and HSO 3 , since they differ in more than one way; N and S are in different groups and periods of the periodic table. NO 3 is more stable than NO 2 since the charge on NO 3 is delocalized (dispersed) over a greater number of O’s and since NO 3 has a more extended π bond system. Problem 2.24 Indicate which one of the following pairs of resonance structures is the less stable and is an unlikely contributing structure. Give reasons in each case. (ii) (i)
26 CHAPTER 2 Bonding and Molecular Structure @访 0- @H-:→H8- 6: Q: (H.C-CH-EHs Hc-H-H(d)::C N: m (e)H,C-d:→H,c=t: Figure 2.11 elciron-deficienN (d)VII has fewer covalent bonds and a+on the more electronegative N.which is also electron-deficient. (e)CinXhas the second period SUPPLEMENTARY PROBLEMS Problem 2.25 Distinguish between an AO.an HO.an MO and a localized MO An AO is a region of space in an atom in which an electron may exist.An HO is mathematically fabricated maeA2oorg atoms in which the bonding electrons are assumed to be present
(a) I has fewer covalent bonds, more formal charge, and an electron-deficient N. (b) IV has on the more electronegative O. (c) VI has similar charges on adjacent C’s, fewer covalent bonds, more formal charge, and an electrondeficient C. (d) VII has fewer covalent bonds and a on the more electronegative N, which is also electron-deficient. (e) C in X has 10 electrons; this is not possible with the elements of the second period. SUPPLEMENTARY PROBLEMS Problem 2.25 Distinguish between an AO, an HO, an MO and a localized MO. An AO is a region of space in an atom in which an electron may exist. An HO is mathematically fabricated from some number of AO’s to explain equivalency of bonds. An MO is a region of space about the entire molecule capable of accommodating electrons. A localized MO is a region of space between a pair of bonded atoms in which the bonding electrons are assumed to be present. 26 CHAPTER 2 Bonding and Molecular Structure Figure 2.11
CHAPTER 2 Bonding and Molecular Structure 27 回法 日可家 回 山器封 Note that since the energy difference between hybrid and p orbitals is so small,Hund's rule prevails over the Aufbau principle. Problem 2.27 ()NOis linear,(b)NO is bent.Explain in terms of the hybrid orbitals used by N (a)NO,:--6:.N has two bonds,no unshared pairs of electrons and therefore needs two hybrid orbitals.N uses sp hybrid orbitals and the bonds are linear.The geometry is controlled by the arrange ment of the sigma bonds. ()NON:Nhas twobonds and one unshared pair ofelectrons and,therefore.needs three hybrid orbitals.N uses sphybrid HO's,and,the bond angle is about 120 Problem2.8 Draw an orbital representation of the cyanide ion.C-N: See Fig.2.12.The C and N each have one o bond and one unshared pair of electrons.and therefore each needs two sp hybrid HO's.On each atom.one sp hybrid orbital formsa bond,while the other has the unshared pair.Each atom hasap AOandap.AO.The twop,orbitals overlap to formabond in the xy-plane:the two ist betwecr thotood Nond he zpane.hus.two bond a ngnt angles to each other and ao bond ead"only Figure 2.12 (@HF,BrCI.CH.CHCl,CHOH. (b)HF.BrCl,CHCl CH,OH.The symmetrical individual bond moments in CH cancel. Problem 2 30 Conside ronegativity between O and S.would H,O or H,S exhibit (a)H,0(b)H,0
CHAPTER 2 Bonding and Molecular Structure 27 (a) 1s 2s 2p 1s 2sp3 1s 2sp2 1s 2sp2 2p 2p (b) (d) (c) Problem 2.26 Show the orbital population of electrons for unbonded N in (a) ground state, and for (b) sp3 , (c) sp2 , and (d) sp hybrid states. Note that since the energy difference between hybrid and p orbitals is so small, Hund’s rule prevails over the Aufbau principle. Problem 2.27 (a) NO+ 2 is linear, (b) NO 2 is bent. Explain in terms of the hybrid orbitals used by N. (a) . N has two σ bonds, no unshared pairs of electrons and therefore needs two hybrid orbitals. N uses sp hybrid orbitals and the σ bonds are linear. The geometry is controlled by the arrangement of the sigma bonds. (b) . N has two σ bonds and one unshared pair of electrons and, therefore, needs three hybrid orbitals. N uses sp2 hybrid HO’s, and, the bond angle is about 120º. Problem 2.28 Draw an orbital representation of the cyanide ion, :C ⎯ N: . See Fig. 2.12. The C and N each have one σ bond and one unshared pair of electrons, and therefore each needs two sp hybrid HO’s. On each atom, one sp hybrid orbital forms a σ bond, while the other has the unshared pair. Each atom has a py AO and a pz AO. The two py orbitals overlap to form a πy bond in the xy-plane; the two pz orbitals overlap to form a πz bond in the xz-plane. Thus, two π bonds at right angles to each other and a σ bond exist between the C and N atoms. Figure 2.12 Problem 2.29 (a) Which of the following molecules possess polar bonds: F2 , HF, BrCl, CH4 , CHCl3 , CH3 OH? (b) Which are polar molecules? (a) HF, BrCl, CH4, CHCl3, CH3OH. (b) HF, BrCl, CHCl3, CH3OH. The symmetrical individual bond moments in CH4 cancel. Problem 2.30 Considering the difference in electronegativity between O and S, would H2 O or H2 S exhibit greater (a) dipole-dipole attraction, (b) H-bonding? (a) H2 O (b) H2O.