1. Chapter 9 Ionic and Covalent Bonding

2. Copyright © Cengage Learning. All rights reserved.9 | 2 Contents and Concepts Ionic Bonds Molten salts and aqueous solutions of salts are electrically conducting. This conductivity results from the motion of ions in the liquids. It suggests the possibility that ions exist in certain solids, held together by the attraction of ions of opposite charges. 1. Describing Ionic Bonds 2. Electron Configurations of Ions 3. Ionic Radii

3. Copyright © Cengage Learning. All rights reserved.9 | 3 Covalent Bonds Not all bonds can be ionic. Hydrogen, H 2, is a clear example in which there is a strong bond between two like atoms. The bonding in the hydrogen molecule is covalent. A covalent bond forms between atoms by the sharing of a pair of electrons. 4. Describing Covalent Bonds 5. Polar Covalent Bonds; Electronegativity 6. Writing Lewis Electron-Dot Formulas 7. Delocalized Bonding: Resonance 8. Exceptions to the Octet Rule 9. Formal Charge and Lewis Formulas 10. Bond Length and Bond Order 11. Bond Energy

4. Copyright © Cengage Learning. All rights reserved.9 | 4 Learning Objectives Ionic Bonds 1.Describing Ionic Bonds a.Define ionic bond. b.Explain the Lewis electron-dot symbol of an atom. c.Describe the energetics of ionic bonding. d.Define lattice energy. e.Describe the Born–Haber cycle to obtain a lattice energy from thermodynamic data. f.Describe some general properties of ionic substances.

5. Copyright © Cengage Learning. All rights reserved.9 | 5 2.Electron Configurations of Ions a.State the three categories of monatomic ions of main-group elements. b.Write the electron configuration and Lewis symbol for a main-group ion. c.Note the polyatomic ions given earlier in Table 2.5. d.Note the formation of +2 and +3 transition- metal ions. e.Write the electron configurations of transition-metal ions.

6. Copyright © Cengage Learning. All rights reserved.9 | 6 3.Ionic Radii a.Define ionic radius. b.Define isoelectronic ions. c.Use periodic trends to obtain relative ionic radii.

7. Copyright © Cengage Learning. All rights reserved.9 | 7 Covalent Bonds 4.Describing Covalent Bonds a.Describe the formation of a covalent bond between two atoms. b.Define Lewis electron-dot formula. c.Define bonding pair and lone (nonbonding) pair of electrons. d.Define coordinate covalent bond. e.State the octet rule. f.Define single, double, and triple bond.

8. Copyright © Cengage Learning. All rights reserved.9 | 8 5.Polar Covalent Bonds; Electronegativity a.Define polar covalent bond. b.Define electronegativity. c.State the general periodic trends in electronegativity. d.Use electronegativity to obtain relative bond polarity. 6.Writing Lewis Electron-Dot Formulas a.Write Lewis formulas having single bonds. b.Write Lewis formulas having multiple bonds. c.Write Lewis formulas for ionic species.

9. Copyright © Cengage Learning. All rights reserved.9 | 9 7.Delocalized Bonding: Resonance a.Define localized bonding. b.Define resonance description. c.Write resonance forms. 8.Exceptions to the Octet Rule a.Write Lewis formulas (exceptions to the octet rule). b.Note exceptions to the octet rule in Group IIA and Group IIIA.

10. Copyright © Cengage Learning. All rights reserved.9 | 10 9.Formal Charge and Lewis Formulas a.Define formal charge. b.State the rules for obtaining the formal charge. c.State two rules useful in writing Lewis formulas. d.Use formal charges to determine the best Lewis formula.

11. Copyright © Cengage Learning. All rights reserved.9 | 11 10. Bond Length and Bond Order a.Define bond length (bond distance). b.Define covalent radii. c.Define bond order. d.Explain how bond order and bond length are related. 11. Bond Energy a.Define bond energy. b.Estimate  H from bond energies.

12. Copyright © Cengage Learning. All rights reserved.9 | 12 A chemical bond is a strong attractive force that exists between certain atoms in a substance. There are three types of chemical bonds: Ionic bonds Covalent bonds Metallic bonds

13. Copyright © Cengage Learning. All rights reserved.9 | 13 An ionic bond is a chemical bond formed by the electrostatic attraction between positive and negative ions.

14. Copyright © Cengage Learning. All rights reserved.9 | 14 An ionic bond forms when one or more electrons are transferred from the valence shell of one atom to the valence shell of another atom. Na ([Ne]3s 1 ) + Cl ([Ne]3s 2 3p 5 )  Na + ([Ne]) + Cl − ([Ne]3s 2 3p 6 ) The atom that transferred the electron(s) becomes a cation. The atom that gained the electron(s) becomes an anion.

15. Copyright © Cengage Learning. All rights reserved.9 | 15 A Lewis electron-dot symbol is a notation in which the electrons in the valence shell of an atom or ion are represented by dots placed around the chemical symbol of the element. Note: Dots are placed one to a side, until all four sides are occupied.

16. Copyright © Cengage Learning. All rights reserved.9 | 16 Table 9.1 illustrates the Lewis electron-dot symbols for second- and third-period atoms.

17. ? Copyright © Cengage Learning. All rights reserved.9 | 17 Represent the transfer of electrons in forming calcium oxide, CaO, from atoms. + Ca 2+ Ca O + O 2− ] [

18. Copyright © Cengage Learning. All rights reserved.9 | 18 Let’s look next at the energy involved in forming ionic compounds. The energy to remove an electron is the ionization energy. The energy to add an electron is the electron affinity.

19. Copyright © Cengage Learning. All rights reserved.9 | 19 The combination of ionization energy and electron affinity is still endothermic; the process requires energy. However, when the two ions bond, more than enough energy is released, making the overall process exothermic.

20. Copyright © Cengage Learning. All rights reserved.9 | 20 The lattice energy is the change in energy that occurs when an ionic solid is separated into gas- phase ions. It is very difficult to measure lattice energy directly. It can be found, however, by using the energy changes for steps that give the same result.

21. Copyright © Cengage Learning. All rights reserved.9 | 21 For example, to find the lattice energy for NaCl, we can use the following steps.

22. Copyright © Cengage Learning. All rights reserved.9 | 22 The process of finding the lattice energy indirectly from other thermochemical reactions is called the Born–Haber cycle.

23. Copyright © Cengage Learning. All rights reserved.9 | 23 Ionic substances are typically high-melting solids. There are two factors that affect the strength of the ionic bond. They are given by Coulomb’s law: The higher the ionic charge, the stronger the force; the smaller the ion, the stronger the force.

24. Copyright © Cengage Learning. All rights reserved.9 | 24 Based on this relationship, we can predict the relative melting points of NaCl and MgO. The charge on the ions of MgO is double the charge on the ions of NaCl. Because the charge is double, the force will be four times stronger. The size of Na + is larger than that of Mg 2+ ; the size of Cl - is larger than that of O 2-. Because the distance between Mg 2+ and O 2- is smaller than the distance between Na + and Cl -, the force between Mg 2+ and O 2- will be greater.

25. Copyright © Cengage Learning. All rights reserved.9 | 25 Based on the higher charge and the smaller distance for MgO, its melting point MgO should be significantly higher than the melting point of NaCl. The actual melting point of NaCl is 801°C; the melting point of MgO is 2800°C.

26. Copyright © Cengage Learning. All rights reserved.9 | 26 When we examine the electron configuration of main-group ions, we find that each element gains or loses electrons to attain a noble-gas configuration.

27. ? Copyright © Cengage Learning. All rights reserved.9 | 27 Give the electron configuration and the Lewis symbol for the chloride ion, Cl -. Cl − ] [ For chlorine, Cl, Z = 17, so the Cl - ion has 18 electrons. The electron configuration for Cl - is 1s21s2 2s22s2 2p62p6 3s23s2 3p63p6 The Lewis symbol for Cl − is

28. Copyright © Cengage Learning. All rights reserved.9 | 28 Group IIIA to VA metals often exhibit two different ionic charges: one that is equal to the group number and one that is 2 less than the group number. The higher charge is due to the loss of both the s subshell electrons and the p subshells electron(s). The lower charge is due to the loss of only the p subshell electron(s). For example, in Group IVA, tin and lead each form both +4 and +2 ions. In Group VA, bismuth forms +5 and +3 ions.

29. Copyright © Cengage Learning. All rights reserved.9 | 29 Polyatomic ions are atoms held together by covalent bonds as a group and that, as a group, have gained or lost one or more electron.

30. Copyright © Cengage Learning. All rights reserved.9 | 30 Transition metals form several ions. The atoms generally lose the ns electrons before losing the (n – 1)d electrons. As a result, most transition metals form the +2 ion.

31. ? Copyright © Cengage Learning. All rights reserved.9 | 31 Give the electron configurations of Mn and Mn 2+. Manganese, Z = 25, has 25 electrons. Its electron configuration is 1s21s2 2s22s2 2p62p6 3s23s2 3p63p6 4s24s2 3d53d5 Mn 2+ has 23 electrons. When ionized, Mn loses the 4s electrons first; the electron configuration for Mn 2+ is 1s21s2 2s22s2 2p62p6 3s23s2 3p63p6 3d53d5

32. Copyright © Cengage Learning. All rights reserved.9 | 32 Concept Check 9.1 The following are electron configurations for some ions. Which ones would you expect to see in chemical compounds? State the concept or rule you used to decide for or against any ion. a. Fe 2+ [Ar]3d 4 4s 2 b. N 2− [He]2s 2 2p 5 c. Zn 2+ [Ar]3d 10 d. Na 2+ [He]2s 2− p 5 e. Ca 2+ [Ne]3s 2 3p 6

33. Copyright © Cengage Learning. All rights reserved.9 | 33 a.Fe 2+ : [Ar]3d 4 4s 2 No. The 4s 2 electrons would be lost before the 3d electrons. b.N 2- : [He]2s 2 2p 5 No. Nitrogen will gain three electrons to fill the shell, forming N 3-. c.Zn 2+ : [Ar]3d 10 Yes! d.Na 2+ : [He]2s 2 2p 5 No. Sodium will lose only its one valence electron, forming Na +. e.Ca 2+ : [Ne]3s 2 3p 6 Yes!

34. Copyright © Cengage Learning. All rights reserved.9 | 34 Ionic radius is a measure of the size of the spherical region around the nucleus of an ion within which the electrons are most likely to be found. While ionic radius, like atomic radius, can be somewhat arbitrary, it can be measured in ionic compounds.

35. Copyright © Cengage Learning. All rights reserved.9 | 35 A cation is always smaller than its neutral atom. An anion is always larger than its neutral atom.

36. Copyright © Cengage Learning. All rights reserved.9 | 36 The term isoelectronic refers to different species having the same number and configuration of electrons. For example, Ne, Na +, and F - are isoelectronic. Ionic radius for an isoelectronic series decreases with increasing atomic number.

37. ? Copyright © Cengage Learning. All rights reserved.9 | 37 Using the periodic table only, arrange the following ions in order of increasing ionic radius: Br -, Se 2-, Sr 2+. 35 Br 34 Se 38 Sr These ions are isoelectronic, so their size decreases with increasing atomic number: Sr 2+ < Br - < Se 2-

38. Copyright © Cengage Learning. All rights reserved.9 | 38 A covalent bond is a chemical bond formed by sharing a pair of electrons.

39. Copyright © Cengage Learning. All rights reserved.9 | 39 To consider how a covalent bond forms, we can monitor the energy of two isolated hydrogen atoms as they move closer together. The energy decreases—first gradually, and then more steeply—to a minimum. As the atoms continue to move closer, it increases dramatically. The distance between the atoms when energy is at a minimum is called the bond length. This is illustrated on the following graph, from right to left.

40. Copyright © Cengage Learning. All rights reserved.9 | 40

41. Copyright © Cengage Learning. All rights reserved.9 | 41 As the hydrogen atoms move closer together, the electron of each atom is attracted to both its own nucleus and the nucleus of the second atom. The electron probability distribution illustrates this relationship.

42. Copyright © Cengage Learning. All rights reserved.9 | 42 A formula using dots to represent valence electrons is called a Lewis electron-dot formula. An electron pair is represented by two dots. A electron pair that is between two atoms is a bonding pair. It can also be represented by one line for each bonding pair. Electron pairs that are not bonding are nonbonding, or lone pair electrons.

43. Copyright © Cengage Learning. All rights reserved.9 | 43 A coordinate covalent bond is formed when both electrons of the bond are donated by one atom. The two electrons forming the bond with the hydrogen on the left were both donated by the nitrogen. Once shared, they are indistinguishable from the other N—H bonds.

44. Copyright © Cengage Learning. All rights reserved.9 | 44 In forming covalent bonds, atoms tend toward having a full eight electrons in their valence shell. This tendency is called the octet rule. Hydrogen is an exception to the octet rule: it has two electrons in its valence shell (a duet).

45. Copyright © Cengage Learning. All rights reserved.9 | 45 A single bond is a covalent bond in which one pair of electrons is shared by two atoms. A double bond is a covalent bond in which two pairs of electrons are shared by two atoms. A triple bond is a covalent bond in which three pairs of electrons are shared by two atoms. Double bonds form primarily with C, N, and O. Triple bonds form primarily with C and N.

46. Copyright © Cengage Learning. All rights reserved.9 | 46 A polar covalent bond (or polar bond) is a covalent bond in which the bonding electrons spend more time near one atom than near the other atom. Electronegativity, X, is a measure of the ability of an atom in a molecule to draw bonding electrons to itself. Electronegativity is related to ionization energy and electron affinity.

47. Copyright © Cengage Learning. All rights reserved.9 | 47 Electronegativity increases from left to right and from bottom to top in the periodic table. F, O, N, and Cl have the highest electronegativity values.

48. Copyright © Cengage Learning. All rights reserved.9 | 48 The difference in electronegativity between the two atoms in a bond is a rough measure of bond polarity. When the difference is very large, an ionic bond forms. When the difference is large, the bond is polar. When the difference is small, the bond is nonpolar.

49. ? Copyright © Cengage Learning. All rights reserved.9 | 49 Using electronegativities, arrange the following bonds in order by increasing polarity: C—N, Na—F, O—H. For Na—F, the difference is 4.0 (F) – 0.9 (Na) = 3.1. For C—N, the difference is 3.0 (N) – 2.5 (C) = 0.5. For O—H, the difference is 3.5 (O) – 2.1 (H) = 1.4. C—N < Bond polarities: O—H < Na—F

50. Copyright © Cengage Learning. All rights reserved.9 | 50 Writing Lewis Electron-Dot Formulas 1.Calculate the number of valence electrons. 2.Write the skeleton structure of the molecule or ion. 3.Distribute electrons to the atoms surrounding the central atom or atoms to satisfy the octet rule. 4.Distribute the remaining electrons as pairs to the central atom or atoms.

51. ? Copyright © Cengage Learning. All rights reserved.9 | 51 Write the electron dot formulas for the following: a.OF 2 b.NF 3 c.NH 2 OH, hydroxylamine

52. Copyright © Cengage Learning. All rights reserved.9 | 52 Count the valence electrons in OF 2 : O1(6) F2(7) 20 valence electrons O is the central atom (it is less electronegative). Now, we distribute the remaining 16 electrons, beginning with the outer atoms. The last four electrons go on O.

53. Copyright © Cengage Learning. All rights reserved.9 | 53 Count the valence electrons in NF 3 : N1(5) F3(7) 26 valence electrons N is the central atom (it is less electronegative). Now, we distribute the remaining 20 electrons, beginning with the outer atoms. The last two electrons go on N.

54. Copyright © Cengage Learning. All rights reserved.9 | 54 Count the electrons in NH 2 OH: N1(5) H3(1) O1(6) 14 valence electrons N is the central atom. Now, we distribute the remaining six electrons, beginning with the outer atoms. The last two electrons go on N.

55. ? Copyright © Cengage Learning. All rights reserved.9 | 55 Write electron-dot formulas for the following: a. CO 2 b. HCN

56. Copyright © Cengage Learning. All rights reserved.9 | 56 Count the electrons in CO 2 : C1(4) O2(6) 16 valence electrons C is the central atom. Now, we distribute the remaining 12 electrons, beginning with the outer atoms. Carbon does not have an octet, so two of the lone pairs shift to become a bonding pair, forming double bonds.

57. Copyright © Cengage Learning. All rights reserved.9 | 57 Count the electrons in HCN: H1(1) C1(4) N1(5) 10 valence electrons C is the central atom. The remaining electrons go on N. Carbon does not have an octet, so two of the lone pairs shift to become a bonding pair, forming a triple bond.

58. ? Copyright © Cengage Learning. All rights reserved.9 | 58 Phosphorus pentachloride exists in solid state as the ionic compound [PCl 4 ] + [PCl 6 ] - ; it exists in the gas phase as the PCl 5 molecule. Write the Lewis formula of the PCl 4 + ion.

59. Copyright © Cengage Learning. All rights reserved.9 | 59 Count the valence electrons in PCl 4 + : P1(5) Cl4(7) −1 −1 32 valence electrons P is the central atom. The remaining 24 nonbonding electrons are placed on Cl atoms. Add square brackets with the charge around the ion.

60. Copyright © Cengage Learning. All rights reserved.9 | 60 Delocalized bonding is a type of bonding in which a bonding pair of electrons is spread over a number of atoms rather than being localized between two atoms.

61. Copyright © Cengage Learning. All rights reserved.9 | 61 A single electron-dot diagram cannot properly describe delocalized bonding. Using the resonance description, the electron structure of a molecule or ion having delocalized bonding is given by writing all possible electron-dot formulas. They are connected with a double-headed arrow.

62. ? Copyright © Cengage Learning. All rights reserved.9 | 62 Draw the resonance formulas of the acetate ion, CH 3 COO -.

63. Copyright © Cengage Learning. All rights reserved.9 | 63 CH 3 COO - Valence electrons: 2(4) + 3(1) + 2(6) + 1 = 24 C is the central atom. A double bond is needed between C—O. There are two equivalent places for it, so two resonance structures are required.

64. Copyright © Cengage Learning. All rights reserved.9 | 64 Some molecules have electron-dot structures that do not satisfy the octet rule. Some have an odd number of electrons, such as NO. Other molecules either have too few or too many electrons around the central atom.

65. Copyright © Cengage Learning. All rights reserved.9 | 65 Elements that have too few electrons are in Groups IIA and IIIA. Be, B, and Al exhibit too few electrons around the central atom.

66. Copyright © Cengage Learning. All rights reserved.9 | 66 There are many more examples of central atoms with more than an octet. Because elements of the third period and beyond have a d subshell, they can expand their valence electron configurations. S, P, Cl (as a central atom), and other elements in period 3 are examples of atoms in this situation. Elements in the second period, having only s and p subshells, are unable to do this.

67. ? Copyright © Cengage Learning. All rights reserved.9 | 67 Give the Lewis formula of the IF 5 molecule.

68. Copyright © Cengage Learning. All rights reserved.9 | 68 Count the valence electrons in IF 5 : I1(7) F5(7) 42 valence electrons I is the central atom. Thirty-two electrons remain; they first complete F octets. The remaining electrons go on I.

69. Copyright © Cengage Learning. All rights reserved.9 | 69 The formal charge on an atom in the Lewis formula is the hypothetical charge you obtain by assuming that bonding electrons are equally shared between bonded atoms and that the electrons of each lone pair belong completely to one atom.

70. Copyright © Cengage Learning. All rights reserved.9 | 70 Formal charge = valence electrons on free atom – ½ (number of electrons in bonds) – (number of lone-pair electrons) The sum of the formal charges on the atoms equals the charge on the formula.

71. Copyright © Cengage Learning. All rights reserved.9 | 71 Formal charges can help to determine the most likely electron-dot formula using three rules: 1.Whenever you can write several Lewis formulas for a molecule, choose the one having the lowest magnitudes of formal charges. 2.When two proposed Lewis formulas have the same magnitudes of formal charges, choose the one having the negative formal charge on the more electronegative atom. 3.When possible, choose Lewis formulas that do not have like charges on adjacent atoms.

72. ? Copyright © Cengage Learning. All rights reserved.9 | 72 Compare the formal charges for the following electron-dot formulas of CO 2. For the left structure:For the right structure: O:6 – 2 – 4 = 0 C:4 – 4 – 0 = 0 The left structure is better. Formal charge = group number – (number of bond pairs) – (number of nonbonding electrons) C:4 – 4 – 0 = 0 O:6 – 1 – 6 = –1 O:6 – 3 – 2 = +1

73. Copyright © Cengage Learning. All rights reserved.9 | 73 Bond length (or bond distance) is the distance between nuclei in a bond. Bond order is, defined in terms of the Lewis formula, the number of pairs of electrons in a bond. Bond length decreases as bond order increases.

74. ? Copyright © Cengage Learning. All rights reserved.9 | 74 Consider the propylene molecule: 134 pm 150 pm The shorter bond is the double bond; the longer bond is the single bond. One of the carbon–carbon bonds has a length of 150 pm; the other 134 pm. Identify each bond with a bond length.

75. Copyright © Cengage Learning. All rights reserved.9 | 75 Bond energy is the average enthalpy change for breaking the A—B bond in a molecule in the gas phase. Bond energy is a measure of bond strength: the larger the bond energy, the stronger the bond.

76. Copyright © Cengage Learning. All rights reserved.9 | 76 Bond energies can be used to estimate the enthalpy change,  H, for a reaction. To do so, we imagine the reaction in two steps: breaking bonds and forming new bonds.  H = sum of the bond energies for bonds broken – sum of the bond energies for bonds formed When  H is negative, heat is released. When  H is positive, heat is absorbed.

77. ? Copyright © Cengage Learning. All rights reserved.9 | 77 Estimate the enthalpy change for the following reaction, using bond energies: Bonds Broken: 1 C=C602 kJ 1 Cl—Cl 240 kJ Absorbed842 kJ Bonds Formed: 1 C—C 346 kJ 2 C—Cl 654 kJ Released1000 kJ  H = 842 kJ – 1000 kJ  H = –158 kJ