Chapter 8 Basic Concepts of Chemical Bonding

1. Chapter 8Basic Concepts of Chemical Bonding

2. Chemical Bonds • Three basic types of bonds • Ionic • Electrostatic attraction between ions. • Covalent • Sharing of electrons. • Metallic • Metal atoms bonded to several other atoms.

3. A. Metallic B. Ionic C. Covalent

4. Lewis Symbols • G.N. Lewis pioneered the use of chemical symbols surrounded with dots to symbolize the valence electrons around an atom. • When forming compounds, atoms tend to add or subtract electrons until they are surrounded by eight valence electrons (the octet rule).

5. Yes, all three are correct. • No, the first two are different Lewis structures because of differing placement of unpaired electron. • No, because the Lewis structure on the right has only six valence electrons and elemental Cl has seven valence electrons.

6. Energetics of Ionic Bonding As we saw in the last chapter, it takes 496 kJ/mol to remove an electron from sodium. We get 349 kJ/mol back by giving an electron to chlorine.

7. Energetics of Ionic Bonding But these numbers don’t explain why the reaction of sodium metal and chlorine gas to form sodium chloride is so exothermic!

8. No pair of an alkali metal with a halogen gives a sum of I1 and EA that is negative, exothermic. • Na and F • K and Cl • Rb and I

9. Energetics of Ionic Bonding • There must be a third piece to the puzzle. • What is as yet unaccounted for is the electrostatic attraction between the newly formed sodium cation and chloride anion.

10. A. No, because we must know the names of each element assigned to a sphere to make a conclusion. B. Yes, because Na and Cl are in the same period. Atomic size generally increases from left to right in a period—a cation is smaller than its neutral atom, and an anion is greater than its neutral atom. C. No, because we need actual ion sizes to make a conclusion. D. Yes, because Na and Cl are in the same period. Atomic size generally decreases from left to right in a period —a cation is larger than its neutral atom, and an anion is smaller than its neutral atom.

11. A. Each calcium atom loses one electron and each fluorine atom gains two electrons. B. Each calcium atom loses two electrons and each fluorine atom gains one electron. C. Each calcium atom gains one electron and each fluorine atom loses two electrons. D. Each calcium atom gains two electrons and each fluorine atom loses one electron.

12. Q1Q2 d Eel =  Lattice Energy • This third piece of the puzzle is the lattice energy: • The energy required to completely separate a mole of a solid ionic compound into its gaseous ions. • The energy associated with electrostatic interactions is governed by Coulomb’s law:

13. Lattice Energy • Lattice energy, then, increases with the charge on the ions. • It also increases with decreasing size of ions.

14. A. Yes, the relative sizes of the reactants are similar. B. No, the physical properties of the reactants are different. C. Yes, the metals are alkali metals and the nonmetals are elemental halogens. D. No, the metals and nonmetals are each from different families of elements.

15. Sample Exercise 8.1Magnitudes of Lattice Energies Without consulting Table 8.2, arrange the ionic compounds NaF, CsI, and CaO in order of increasing lattice energy. Practice Exercise Which substance do you expect to have the greatest lattice energy, MgF2, CaF2, or ZrO2?

16. A. Lattice energy range of KF cannot be estimated unless we know ion sizes exactly and the type of crystal structure. B. Between 691 kJ/mol and 788 kJ/mol estimated from the range between the lattice energies of NaCl and RbCl because KCl is between these two lattice energies. C. Between 910 kJ/mol and 788 kJ/mol estimated from the range between the lattice energies of NaF and NaCl because a fluoride ion is present in the pair. D. Between 701 kJ/mol and 910 kJ./mole estimated by recognizing that the distance between adjacent ions of opposite charge should be larger than that in NaF and smaller than in KCl.

17. Energetics of Ionic Bonding By accounting for all three energies (ionization energy, electron affinity, and lattice energy), we can get a good idea of the energetics involved in such a process.

18. Sample Exercise 8.2Charges on Ions Predict the ion generally formed by (a) Sr, (b) S, (c) Al. Practice Exercise Predict the charges on the ions formed when magnesium reacts with nitrogen.

19. Fe • Co • Pd • Rh

20. Covalent Bonding • In covalent bonds, atoms share electrons. • There are several electrostatic interactions in these bonds: • Attractions between electrons and nuclei, • Repulsions between electrons, • Repulsions between nuclei.

21. Repulsions:Electron-ElectronNuclei-NucleiAttractions:Electron-NucleiRepulsions:Electron-ElectronNuclei-NucleiAttractions:Electron-Nuclei A. Unaffected Decrease Decrease B. Unaffected Decrease Increase C. Increase Increase Increase D. Increase Increase Decrease

22. Stronger, because a H-H covalent bond in H2+ has one less electron than in H2. • Stronger, because a H-H covalent bond in H2+ has one more electron than in H2. • Weaker, because a H-H covalent bond in H2+ has one less electron than in H2. • Weaker, because a H-H covalent bond in H2+ has one more electron than in H2.

23. Lewis Structures Lewis structures are representations of molecules showing all electrons, bonding and nonbonding.

24. Sample Exercise 8.3Lewis Structure of a Compound Given the Lewis symbols for nitrogen and fluorine in Table 8.1, predict the formula of the stable binary compound (a compound composed of two elements) formed when nitrogen reacts with fluorine and draw its Lewis structure. Practice Exercise Compare the Lewis symbol for neon with the Lewis structure for methane, CH4. In what important way are the electron arrangements about neon and carbon alike? In what important way are they different?

25. Polar Covalent Bonds • Though atoms often form compounds by sharing electrons, the electrons are not always shared equally. • Fluorine pulls harder on the electrons it shares with hydrogen than hydrogen does. • Therefore, the fluorine end of the molecule has more electron density than the hydrogen end.

26. Electronegativity • Electronegativity is the ability of atoms in a molecule to attract electrons to themselves in a bond. • On the periodic chart, electronegativity increases as you go… • …from left to right across a row. • …from the bottom to the top of a column.

27. A. EN and EA are the same; they both measure the same characteristic. B. EA measures the energy released when an isolated atom/ion gains an electron; EN measures the ability of an atom in a molecule to attract electrons to itself. C. EN values of neutral atoms are just the negative of EA values of neutral atoms. D. EN measures the energy released when an isolated atom/ion gains an electron; EA measures the ability measures the ability of an atom in a molecule to attract electrons to itself.

28. A. Increasing • Decreasing • constant

29. Polar Covalent Bonds • When two atoms share electrons unequally, a bond dipole results. • The dipole moment, , produced by two equal but opposite charges separated by a distance, r, is calculated:  = Qr • It is measured in debyes (D).

30. A. Increase, because r increases B. Decrease, because r decreases C. Stays the same because the values of Q and r change in opposite but equal directions.

31. Sample Exercise 8.5Dipole Moments of Diatomic Molecules The bond length in the HCl molecule is 1.27 Å. (a) Calculate the dipole moment, in debyes, that results if the charges on the H and Cl atoms were and 1+ and 1–, respectively. (b) The experimentally measured dipole moment of HCl(g) is 1.08 D. What magnitude of charge, in units of e, on the H and Cl atoms leads to this dipole moment? Practice Exercise The dipole moment of chlorine monofluoride, ClF(g), is 0.88 D. The bond length of the molecule is 1.63 Å. (a) Which atom is expected to have the partial negative charge? (b) What is the charge on that atom in units of e?

32. A. IF B. ClF C. They have the same dipole moment. D. Neither has a dipole moment; they are both nonpolar.

33. H—C has the same bond dipole moment as H—I, because the magnitude of Q and r inare about the same for both. • H—C has the larger bond dipole moment, because Q and r are larger in H—C than in H—I. • H—I has the smaller bond dipole moment, because Q is about the same for both, but r is larger in H—C than in H—I. • H—I has the larger bond dipole moment, because Q is about the same for both, but r is larger in H—I than in H—C.

34. Polar Covalent Bonds • \ The greater the difference in electronegativity, the more polar the bond.

35. Nonpolar, because both S and O are in the same family. • Polar covalent, because the difference in electronegativity values is 1.0. • Ionic, because the difference in electronegativity values is –1.0. • Ionic, because the difference in electronegativity values is >0.9.

36. A. Polarity of HBr and HI bonds are insufficiently polar to cause significant excess electron density (shown by red shading) on the halogen atoms. B. Polarity of HBr and HI bonds are sufficiently polar to cause excess electron density (shown by dark green shading) on the halogen atoms. C. Polarity of HBr and HI bonds are insufficiently polar to cause significant excess electron density (shown by yellow shading) on the halogen atoms. D. Polarity of HBr and HI bonds are sufficiently polar to be represented by significant yellow shading between the bonded atoms

37. A. OsO4 B. MoO3 • Melting points are not useful in distinguishing between types of compounds. • More information about types of bonds within structures is needed to answer this question.

38. Sample Exercise 8.4 Bond Polarity In each case, which bond is more polar: (a) B—Cl or C—Cl, (b) P—F or P—Cl ? Indicate in each case which atom has the partial negative charge. Practice Exercise Which of the following bonds is most polar: S—Cl, S—Br, Se—Cl, or Se—Br ?

39. Find the sum of valence electrons of all atoms in the polyatomic ion or molecule. If it is an anion, add one electron for each negative charge. If it is a cation, subtract one electron for each positive charge. The central atom is the least electronegative element that isn’t hydrogen. Connect the outer atoms to it by single bonds. Fill the octets of the outer atoms. Fill the octet of the central atom. Writing Lewis Structures

40. Sample Exercise 8.6Drawing a Lewis Structure Practice Exercise (a) How many valence electrons should appear in the Lewis structure for CH2Cl2? (b) Draw the Lewis structure.

41. Writing Lewis Structures • If you run out of electrons before the central atom has an octet… …form multiple bonds until it does.

42. A. Single covalent bond B. Double covalent bond C. Triple covalent bond

43. Sample Exercise 8.7Lewis Structure with a Multiple Bond Practice Exercise Draw the Lewis structure for (a) NO+ ion, (b) C2H4.

44. Sample Exercise 8.8 Lewis Structure for a Polyatomic Ion Draw the Lewis structure for the BrO3– ion. Practice Exercise Draw the Lewis structure for (a) ClO2–, (b) PO43–

45. Writing Lewis Structures • Then assign formal charges. • For each atom, count the electrons in lone pairs and half the electrons it shares with other atoms. • Subtract that from the number of valence electrons for that atom: the difference is its formal charge. • The best Lewis structure… • …is the one with the fewest charges. • …puts a negative charge on the most electronegative atom.

46. Sample Exercise 8.9Lewis Structures and Formal Charges Three possible Lewis structures for the thiocyanate ion, NCS– , are (a) Determine the formal charges in each structure. (b) Based on the formal charges, which Lewis structure is the dominant one? Practice Exercise The cyanate ion, NCO, has three possible Lewis structures. (a) Draw these three structures and assign formal charges in each. (b) Which Lewis structure is dominant?

47. A. The structure actually represents an ion. B. The F atom in the structure must have four covalent bonds attached to it. C. There must be another F atom in the structure carrying a –1 formal charge, since F is the most electronegative element and it should carry a negative formal charge. D. There must be a better Lewis structure, since F is the most electronegative element and it should carry a negative formal charge.

48. A. The colors of the two O atoms are the same. B. The substance is bent. C. The substance has a bond angle of 117°. D. The two O—O bond lengths are equivalent.

49. Resonance • This is the Lewis structure we would draw for ozone, O3. • But this is at odds with the true, observed structure of ozone, in which… • …both O—O bonds are the same length. • …both outer oxygens have a charge of −1/2. • One Lewis structure cannot accurately depict a molecule like ozone. • We use multiple structures, or resonance structures, to describe the molecule.

50. Resonance Just as green is a synthesis of blue and yellow… …ozone is a synthesis of these two resonance structures.