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THIRTEEN CHAPTER BONDING: GENERAL CONCEPTS Chemical Bonds and Electronegativity 11. a. Electronegativity: The ability of an atom in a molecule to attract electrons to itself. Electron affinity: The energy change for M(g) + e- M-(g). EA deals with isolated atoms in the gas phase. b. Covalent bond: Sharing of electron pair(s); Polar covalent bond: Unequal sharing of electron pair(s). c. Ionic bond: Electrons are no longer shared, i.e., complete transfer of electron(s) from one atom to another. 12. a. There are two attractions of the form (+1) (-1) , where r = 1 10-10 m = 0.1 nm. r V = 2 (2.31 10-19 J nm) = -4.62 10-18 J 0.1 nm b. There are 4 attractions of +1 and -1 charges at a distance of 0.1 nm from each other. The two negative charges and the two positive charges repel each other across the diagonal of the square. This is at a distance of V = 4 (2.31 10-19) (+1) (-1) 2 0.1 nm. (+1) (+1) (+1) (-1) -19 -19 + 2.31 10 2 (0.1) + 2.31 10 0.1 (-1) (-1) 2 (0.1) V = -9.24 10-18 J + 1.63 10-18 J + 1.63 10-18 J = -5.98 10-18 J 371 Note: 13. There is a greater net attraction in arrangement b than in a. Using the periodic table we expect the general trend for electronegativity to be: 1) increase as we go from left to right across a period 2) decrease as we go down a group a. C < N < O d. Tl < Ge < S b. Se < S < Cl e. Rb < K < Na f. c. Sn < Ge < Si Ga < B < O 372 CHAPTER 13 14. BONDING: GENERAL CONCEPTS 373 The most polar bond will have the greatest difference in electronegativity between the two atoms. From positions in the periodic table, we would predict: a. Ge-F b. P-Cl c. S-F d. Ti-Cl e. Sn-H f. Tl-Br 15. The general trends in electronegativity used on Exercises 13.13 and 13.14 are only rules of thumb. In this exercise we use experimental values of electronegativities and can begin to see several exceptions. The order of EN using Figure 13.3 is: a. C (2.6) < N (3.0) < O (3.4) b. Se (2.6) = S (2.6) < Cl (3.2) same as predicted different d. Tl (2.0) = Ge (2.0) < S (2.6) different f. Ga (1.8) < B (2.0) < O (3.4) same c. Si (1.9) < Ge (2.0) = Sn (2.0) different e. Rb (0.8) = K (0.8) < Na (0.9) different Most polar bonds using actual EN values: a. Si-F (Ge-F predicted) c. S-F (same as predicted) e. C-H (Sn-H predicted) 16. F Cl Br I (IE - EA) 2006 kJ/mol 1604 1463 1302 (IE - EA)/502 4.0 3.2 2.9 2.6 b. P-Cl (same as predicted) d. Ti-Cl (same as predicted) f. Al-Br (Tl-Br predicted) 2006/502 = 4.0 EN (text) 4.0 3.2 3.0 2.7 The values calculated from IE and EA show the same trend (and agree fairly closely) to the values given in the text. Ionic Compounds 17. a. Cu > Cu+ > Cu2+ d. La3+ > Eu3+ > Gd3+ > Yb3+ b. Pt2+ > Pd2+ > Ni2+ e. Te2- > I- > Cs+ > Ba2+ > La3+ c. O2- > O- > O For answer a, as electrons are removed from an atom, size decreases. Answers b and d follow the radii 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 most protons. 374 18. a. Mg2+: 1s22s22p6 K+: 1s22s22p63s23p6 CHAPTER 13 Sn2+: Al3+: As3+: BONDING: GENERAL CONCEPTS [Kr]5s24d10 1s22s22p6 Tl+: [Xe]6s24f145d10 b. N3-, O2- and F-: 1s22s22p6 c. Be2+: Ba2+: I-: 19. a. Sc3+ 1s2 [Kr]5s24d105p6 [Kr]5s24d105p6 b. Te2- [Ar]4s23d10 [Kr]5s24d105p6 [Ar]4s23d104p6 [Ar]4s23d104p6 Te2-: Rb+: Se2-: c. Ce4+ and Ti4+ d. Ba2+ All of these have the number of electrons of a noble gas. 20. Isoelectronic: Same number of electrons. There are two variables, number of protons and number of electrons, that will 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 smallest ion has the most protons. Se2-, Br-, Rb+, Sr2+, Y3+, Zr4+ are some ions which are isoelectronic with Kr (36 electrons). In terms of size, the ion with the most protons will hold the electrons tightest and will be the smallest. The size trend is: Zr4+ < Y3+ < Sr2+ < Rb+ < Br- < Se2smallest largest 22. Lattice energy is proportional to Q1Q2/r where Q is the charge of the ions and r is the distance between the ions. In general, charge effects on lattice energy are greater than size effects. a. NaCl; Na+ is smaller than K+. b. LiF; F- is smaller than Cl-. 21. c. MgO; O2- has a greater charge than OH-.d. Fe(OH)3; Fe3+ has a greater charge than Fe2+. e. Na2O; O2- has a greater charge than Cl-. 23. f. MgO; The ions are smaller in MgO. a. Li+ and N3- are the expected ions. The formula of the compound would be Li3N and the name is lithium nitride. b. Ga3+ and O2-; Ga2O3, gallium oxide c. Rb+ and Cl-; RbCl, rubidium chloride d. Ba2+ and S2-; BaS, barium sulfide CHAPTER 13 24. BONDING: GENERAL CONCEPTS 375 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 the charged ions are held rigidly in place. Once the forces are disrupted (melting or dissolution), the ions can move about (conduct). ii. Melting and boiling disrupts the attractions of the ions for each other. If the forces are strong, it will take a lot of energy (high temp.) to accomplish this. iii. If we try to bend a piece of material, the atoms/ions must slide across each other. For an ionic solid the following might happen: strong attraction strong repulsion Just as the layers begin to slide, there will be very strong repulsions causing the solid to snap across a fairly clean plane. These properties and their correlation to chemical forces will be discussed in detail in Chapter 16. 25. Na(s) Na(g) Na(g) Na+(g) + e1/2 Cl2(g) Cl(g) Cl(g) + e- Cl-(g) Na+(g) + Cl-(g) NaCl(s) Na(s) + 1/2 Cl2(g) NaCl(s) H = 109 kJ (sublimation) H = 495 kJ (ionization energy) H = 239/2 kJ (bond energy) H = -349 kJ (electron affinity) H = -786 kJ (lattice energy) Hf = -412 kJ/mol H = 150. kJ H = 735 kJ H = 1445 kJ H = 154 kJ H = 2(-328) kJ H = -3916 kJ (sublimation) (IE1) (IE2) (BE) (EA) (LE) 26. Mg(s) Mg(g) Mg(g) Mg+(g) + eMg+(g) Mg2+(g) + eF2(g) 2 F(g) 2 F(g) + 2 e- 2 F-(g) Mg2+(g) + 2 F-(g) MgF2(s) Mg(s) + F2(g) MgF2(s) Hf = -2088 kJ/mol 376 CHAPTER 13 BONDING: GENERAL CONCEPTS 27. a. From the data given, less energy is required to produce Mg+(g) + O-(g) than to produce Mg2+(g) + O2-(g). However, the lattice energy for Mg2+O2- will be much more exothermic than for Mg+O- (due to the greater charges in Mg2+O2-). The favorable lattice energy term will dominate and Mg2+O2- forms. b. Mg+ and O- both have unpaired electrons. In Mg2+ and O2-, there are no unpaired electrons. 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 and absence of a magnetic field. The apparent mass of a paramagnetic substance will be larger in a magnetic field because of the force between the unpaired electrons and the field. 28. Let us look at the complete cycle for Li2S. 2 Li(s) 2 Li(g) 2 Li(g) 2 Li+(g) + 2 eS(s) S(g) S(g) + e- S-(g) S-(g) + e- S2-(g) + 2 Li (g) + S2-(g) Li2S 2 Li(s) + S(s) Li2S(s) 2 Hsub, Li = 2(161) kJ 2 IE = 2(520.) kJ Hsub, S = 277 kJ EA1 = -200. kJ EA2 = ? LE = -2472 kJ Hf = -500. kJ Hf = 2 Hsub, Li + 2 IE + Hsub, S + EA1 + EA2 + LE, -500. = -1033 + EA2, EA2 = 533 kJ For each salt: Hf = 2 Hsub, M + 2 IE + 277 - 200. + LE + EA2 Na2S: -365 = 2(109) + 2(495) + 277 - 200. - 2203 + EA2; EA2 = 553 kJ K2S: -381 = 2(90.) + 2(419) + 277 - 200. - 2052 + EA2; EA2 = 576 kJ Rb2S: -361 = 2(82) + 2(409) + 277 - 200. - 1949 + EA2; EA2 = 529 kJ Cs2S: -360. = 2(78) + 2(382) + 277 - 200. - 1850. + EA2; EA2 = 493 kJ We get values from 493 to 576 kJ. The mean value is: 533 + 553 + 576 + 529 + 493 = 537 kJ 5 We can represent the results as EA2 = 540 50 kJ. 29. Ca2+ has a greater charge than Na+, and Se2- is smaller than Te2-. The effect of charge on the lattice CHAPTER 13 BONDING: GENERAL CONCEPTS 377 energy is greater than the effect of size. We expect the trend from most exothermic to least exothermic to be: CaSe > CaTe > Na2Se > Na2Te (-2862) (-2721) (-2130) (-2095 kJ/mol) This is what we observe. 30. Lattice energy is proportional to the charge of the cation times the charge of the anion, Q1Q2. Compound FeCl2 FeCl3 Fe2O3 Q1Q2 (+2)(-1) = -2 (+3)(-1) = -3 (+3)(-2) = -6 Lattice Energy -2631 kJ/mol -5339 kJ/mol -14,744 kJ/mol Bond Energies 31. a. H - H + Cl - Cl 2 H - Cl Bonds broken: 1 H - H (432 kJ/mol) 1 Cl - Cl (239 kJ/mol) Bonds formed: 2 H - Cl (427 kJ/mol) H = Dbroken - Dformed, H = 432 kJ + 239 kJ - 2(427) kJ = -183 kJ b. N N + 3H H Bonds broken: 1 N N (941 kJ/mol) 3 H - H (432 kJ/mol) 2 H N H Bonds formed: 6 N - H (391 kJ/mol) H H = 941 kJ + 3(432) kJ - 6(391) kJ = -109 kJ c. 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. 378 CHAPTER 13 BONDING: GENERAL CONCEPTS Bonds broken: 1 C N (891 kJ/mol) 2 H - H (432 kJ/mol) Bonds formed: 1 C - N (305 kJ/mol) 2 C - H (413 kJ/mol) 2 N - H (391 kJ/mol) H = 891 kJ + 2(432 kJ) - [305 kJ + 2(413 kJ) + 2(391 kJ)] = -158 kJ d. Bonds 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) H = 160. kJ + 4(391 kJ) + 2(154 kJ) - [4(565 kJ) + 941 kJ] = -1169 kJ 32. a. H = 2 Hf, HCl = 2 mol(-92 kJ/mol) = -184 kJ (-183 kJ from bond energies) b. H = 2 Hf, NH3 = 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 good estimate for the enthalpy change of a reaction. The estimate is especially good for gas phase reactions. 33. H H C N C H Bonds broken: 1 C - N (305 kJ/mol) H H C H Bonds formed: 1 C - C (347 kJ/mol) C N H = Dbroken - Dformed, H = 305 - 347 = -42 kJ Note: Some bonds usually remain the same between reactants and products. To save time, only break and form bonds that are involved in the reaction. 34. CHAPTER 13 BONDING: GENERAL CONCEPTS 379 Bonds broken: 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) Bonds formed: 2 2 C = O (799 kJ/mol) 3 2 O - H (467 kJ/mol) H = 5(413 kJ) + 347 + kJ 358 kJ + 467 kJ + 3(495 kJ) - [4(799 kJ) + 6(467 kJ)] = -1276 kJ 35. H C C H H H + O O O H H O C C H H + O O Bonds broken: 1 C _ C (614 kJ/mol) 1 O - O (146 kJ/mol) 1 C - H (413 kJ/mol) Bonds formed: 1 C - C (347 kJ/mol) 1 C _ O (745 kJ/mol) 1 C - H (413 kJ/mol) H = 614 + 146 + 413 - (347 + 745 + 413) = -332 kJ 36. a. I. H C H H C * H O + H * C N H* O H C H C * C H N H Bonds broken (*): 1 C - O (358 kJ) 1 C - H (413 kJ) Bonds formed (*): 1 O - H (467 kJ) 1 C - C (347 kJ) HI = 358 kJ + 413 kJ - [467 kJ + 347 kJ] = -43 kJ II. H * OH H C * C * H H C N H H C * C C N H + H * O H Bonds broken (*): 1 C - O (358 kJ/mol) 1 C - H (413 kJ/mol) 1 C - C (347 kJ/mol) Bonds formed (*): 1 H - O (467 kJ/mol) 1 C _ C (614 kJ/mol) 380 CHAPTER 13 BONDING: GENERAL CONCEPTS HII = 358 kJ + 413 kJ + 347 kJ - [467 kJ + 614 kJ] = +37 kJ Hoverall = HI + HII = -43 kJ + 37 kJ = -6 kJ b. H H 4 H C C H C H + 6 NO 4 H H H C C H C N + 6 H O H + N N Bonds broken: 4 3 C - H (413 kJ/mol) 6 N _ O (630. kJ/mol) Bonds formed: 4 C N (891 kJ/mol) 6 2 H - O (467 kJ/mol) 1 N N (941 kJ/mol) H = 12(413) + 6(630.) - [4(891) + 12(467) + 941] = -1373 kJ c. H H 2 H C C H C H + 2 H H H N H + 3 O2 2 H H C C H C N + 6 H O H Bonds broken: 2 3 C - H (413 kJ/mol) 2 3 N - H (391 kJ/mol) 3 O _ O (495 kJ/mol) Bonds formed: 2 C N (891 kJ/mol) 6 2 O - H (467 kJ/mol) H = 6(413) + 6(391) + 3(495) - [2(891) + 12(467)] = -1077 kJ 37. Since both reactions are highly exothermic, the high temperature is not needed to provide energy. It must be necessary for some other reason. The reason is to increase the speed of the reaction. This will be discussed in Chapter 15 on kinetics. H H C O H + C O H H H O C C O H H 38. Bonds broken: 1 C O (1072 kJ/mol) 1 C - O (358 kJ/mol) Bonds formed: 1 C - C (347 kJ/mol) 1 C _ O (745 kJ/mol) CHAPTER 13 BONDING: GENERAL CONCEPTS 1 C - O (358 kJ/mol) 381 H = 1072 + 358 - [347 + 745 + 358] = -20. kJ CH3OH(g) + CO(g) CH3COOH(l) H = -484 kJ - [(-201 kJ) + (-110.5 kJ)] = -173 kJ Using bond energies, H = -20. kJ. For this reaction, bond energies give a much poorer estimate for H as compared to gas phase reactions. The major reason for the large discrepancy is that not all species are gases in this exercise. Bond energies do not account for the energy changes that occur when liquids and solids form instead of gases. These energy changes are due to intermolecular forces and will be discussed in Chapter 16. 39. a. HF(g) H(g) + F(g) H(g) H+(g) + eF(g) + e- F-(g) HF(g) H+(g) + F-(g) b. HCl(g) H(g) + Cl(g) H(g) H+(g) + eCl(g) + e- Cl-(g) HCl(g) H+(g) + Cl-(g) c. HI(g) H(g) + I(g) H(g) H+(g) + eI(g) + e- I-(g) HI(g) H+(g) + I-(g) d. H2O(g) OH(g) + H(g) H(g) H+(g) + eOH(g) + e- OH-(g) H2O(g) H+(g) + OH-(g) 40. H = 565 kJ H = 1312 kJ H = -327.8 kJ H = 1549 kJ H = 427 kJ H = 1312 kJ H = -348.7 kJ H = 1390. kJ H = 295 kJ H = 1312 kJ H = -295.2 kJ H = 1312 kJ H = 467 kJ H = 1312 kJ H = -180. kJ H = 1599 kJ a. Using SF4 data: SF4(g) S(g) + 4 F(g) H = 4 DSF = 278.8 kJ + 4(79.0 kJ) - (-775 kJ) = 1370. kJ DSF = 1370. kJ = 342.5 kJ/mol 4 mol SF bonds Using SF6 data: SF6(g) S(g) + 6 F(g) 382 CHAPTER 13 BONDING: GENERAL CONCEPTS H = 6 DSF = 278.8 kJ + 6(79.0 kJ) - (-1209 kJ) = 1962 kJ DSF = 1962 kJ = 327.0 kJ/mol 6 b. The S - F bond energy in Table 13.6 is 327 kJ/mol. The value in the table was based on the 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 most stable forms are S8(s) and F2(g); Hf = 0 for these two species. 41. NH3(g) N(g) + 3 H(g) H = 3 DNH = 472.7 kJ + 3(216.0 kJ) - (-46.1 kJ) = 1166.8 kJ DNH = 1166.8 kJ = 388.93 kJ/mol 3 mol NH bonds Dcalc = 389 kJ/mol as compared to 391 kJ/mol in the table. There is good agreement. 42. N2 + 3 H2 2 NH3; H = D N2 + 3 DH2 - 6 D NH ; H = 2(-46 kJ) = -92 kJ -92 kJ = 941 kJ + 3(432 kJ) - (6 DNH), 6 DNH = 2329 kJ, DNH = 388.2 kJ/mol Exercise 13.41: 389 kJ/mol; Table in text: 391 kJ/mol; There is good agreement between all three values. 43. H N H N H H 2 N(g) + 4 H(g) From Exercise 13.41, the N-H bond energy is 388.9 kJ/mol. H = DN-N + 4 DN-H = DN-N + 4(388.9) H = 2 Hf, N + 4 Hf, H - Hf, N2 H4 = 2(472.7 kJ) + 4(216.0 kJ) - 95.4 kJ CHAPTER 13 BONDING: GENERAL CONCEPTS 383 H = 1714.0 kJ = DN-N + 4(388.9), DN-N = 158.4 kJ/mol (160. kJ/mol in Table 13.6) 44. Hf for H(g) is H for the reaction: 1/2 H2(g) H(g); Hf for H(g) equals one-half the H-H bond energy. Lewis Structures and Resonance 45. Drawing Lewis structures is mostly trial and error. However, the first two steps are always the same. These steps are 1) count the valence electrons available in the molecule or ion and 2) attach all atoms to each other with single bonds (called the skeletal structure). Generally, the atom listed first is assumed to be the atom in the middle (called the central atom) and all other atoms in the formulas are attached to this atom. The most notable exceptions to the rule are formulas which begin with H, e.g., H2O, H2CO, etc. Hydrogen can never be a central atom since this would require H to have more than two electrons. 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 octet rule (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. a. HCN has 1 + 4 + 5 = 10 valence electrons. b. PH3 has 5 + 3(1) = 8 valence electrons. Skeletal structure Lewis structure Skeletal structure Lewis structure Skeletal structures uses 4 e- ; 6 e- remain c. CHCl3 has 4 + 1 + 3(7) = 26 valence electrons. Skeletal structure uses 6 e-; 2 e- remain d. NH4+ has 5 + 4(1) - 1 = 8 valence electrons. Note: Subtract valence electrons for positive charged ions. Skeletal structure Lewis structure Lewis structure 384 CHAPTER 13 Skeletal structure uses 8 e-; 18 e- remain e. H2CO has 2(1) + 4 + 6 = 12 valence electrons. O C H H BONDING: GENERAL CONCEPTS f. SeF2 has 6 + 2(7) = 20 valence electrons. F Se F g. CO2 has 4 + 2(6) = 16 valence electrons. O C O h. O2 has 2(6) = 12 valence electrons. O O i. HBr has 1 + 7 = 8 valence electrons. H Br 46. a. POCl3 has 5 + 6 + 3(7) = 32 valence electrons. This structure uses all 32 e- while satisfying the octet rule for all atoms. This is a valid Lewis structure. SO42- has 6 + 4(6) + 2 = 32 valence electrons. Note: A negatively charged ion will have additional electrons to those that come from the valence shell of the atoms. XeO4, 8 + 4(6) = 32 e- PO43-, 5 + 4(6) + 3 = 32 e- CHAPTER 13 BONDING: GENERAL CONCEPTS 385 O O Xe O O O O P O O 3- ClO4- has 7 + 4(6) + 1 = 32 valence electrons. O O Cl O O - Note: All of these species have the same number of atoms and the same number of valence electrons. They also have the same Lewis structure. b. NF3 has 5 + 3(7) = 26 valence electrons. F N F F F N F F 2- SO32-, 6 + 3 O S O O Ske letal Lewis structure structure 26 e + 1 = 26 e- PO33-, 5 + O Cl O O 3- - O P O O 3(6) + 3 = ClO3-, 7 + 3(6) Note: Species with the same number of atoms and valence electrons have similar Lewis structures. c. ClO2- has 7 + 2(6) + 1 = 20 valence electrons. O Cl O O Cl O Skeletal structure Lewis structure 386 CHAPTER 13 BONDING: GENERAL CONCEPTS SCl2, 6 + 2(7) = 20 eCl S Cl PCl2-, 5 + 2(7) + 1 = 20 eCl P Cl Note: Species with the same number of atoms and valence electrons have similar Lewis structures. 47. Molecules/ions that have the same number of valence electrons and the same number of atoms will have similar Lewis structures. a. NO2- has 5 + 2(6) + 1 = 18 valence electrons. The skeletal structure is: O - N - O To get an octet about the nitrogen and only use 18 e- , we must form a double bond to one of the oxygen atoms. O N O O N O 48. Since there is no reason to have the double bond to a particular oxygen atom, we can draw two resonance structures. Each Lewis structure uses the correct number of electrons and satisfies the octet rules so each is a valid Lewis structure. Resonance structures occur when you have multiple bonds that can be in various positions. We say the actual structure is an average of these two resonance structures. NO3- has 5 + 3(6) + 1 = 24 valence electrons. We can draw three resonance structures for NO3-, with the double bond rotating between the three oxygen atoms. O N O O O O N O O O N O b. OCN- has 6 + 4 + 5 + 1 = 16 valence electrons. We can draw three resonance structures for OCN-. O C N O C N O C N CHAPTER 13 BONDING: GENERAL CONCEPTS 387 SCN- has 6 + 4 + 5 + 1 = 16 valence electrons. Three resonance structures can be drawn. S C N S C N S C N N3- has 3(5) + 1 = 16 valence electrons. As with OCN- and SCN-, three different resonance structures can be drawn. N N N N N N N N N 49. Ozone: O3 has 3(6) = 18 valence electrons. Two resonance structures can be drawn. O O O O O O Sulfur dioxide: SO2 has 6 + 2(6) = 18 valence electrons. Two resonance structures are possible. O S O O S O Sulfur trioxide: SO3 has 6 + 3(6) = 24 valence electrons. Three resonance structures are possible. O S O O O O S O O O S O 50. PAN (H3C2NO5) has 3(1) + 2(4) + 5 + 5(6) = 46 valence electrons. Skeletal structure with complete octets about 388 CHAPTER 13 BONDING: GENERAL CONCEPTS oxygen atoms (46 electrons used). This structure has used all 46 electrons, but there are only six electrons around one of the carbon atoms and the nitrogen atom. Two unshared pairs must become shared, that is we form two double bonds. 51. CH3NCO has 4 + 3(1) + 5 + 4 + 6 = 22 valence electrons. The order of the elements in the formula give the skeletal structure. CHAPTER 13 52. BONDING: GENERAL CONCEPTS SCl2 has 6 + 2(7) = 20 valence electrons. electrons. 389 S2Cl2 has 2(6) + 2(7) = 26 valence 53. Benzene has 6(4) + 6(1) = 30 valence electrons. Two resonance structures can be drawn for benzene. The actual structure of benzene is an average of these two resonance structures, that is, all carbon-carbon bonds are equivalent with a bond length and bond strength somewhere between a single and a double bond.

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