Chemical Bonding

1. Chemical Bonding • Chapter 8

2. Chemical Bonding & Structure • Molecular bonding and structure play the central role in determining the course of chemical reactions.

3. Bonds • Forces that hold groups of atoms together and make them function as a unit.

4. Bond Energy • It is the energy required to break a bond. • It gives us information about the strength of a bonding interaction. • The stronger the bond, the higher the bond energy.

5. Bond Energies • Bond breaking requires energy (endothermic). • Bond formation releases energy (exothermic).

6. Chemical Bonds • Chemical Bond • Ionic Covalent • Cation Anion Molecule

7. Ionic Bonds • Formed from electrostatic attractions of closely packed, oppositely charged ions. • Formed when an atom that easily loses electrons (metal) reacts with one that has a high electron affinity(nonmetal). • 2Na(s) +Cl2(g) ----> 2Na+(aq) + 2Cl-(aq)

8. Figure 11.8: The structure of lithium fluoride

9. Covalent Bonding • Covalent bonds are formed by sharing electrons between nuclei. • H.+ .H ----> H-H 2 hydrogen atoms hydrogen molecule

10. Figure 11.1: The formation of a bond between two hydrogen atoms

11. Types of Covalent Bonds • Polar covalent bond -- covalent bond in which the electrons are not shared equally because one atom attracts them more strongly than the other. A dipole moment exists. HOH, HCl, & CO • Nonpolar covalent bond -- covalent bond in which the electrons are shared equally between both atoms. No dipole moment exists. CO2, CH4, & Cl2

12. Electronegativity • The ability of an atom in a molecule to attract shared electrons to itself. • As electronegativity increases, the attraction for electrons increases. Fluorine has the highest value at 4.0 and cesium and francium are lowest at 0.7.

13. Pauling Electronegativity Values

14. Electronegativity values for selected elements. See Figure 8.3 on page 334 in Zumdahl.

15. Homework #24 • a. Rb < K < Na • b. Ga < B < O • c. Br < Cl < F • d. S < O < F

16. Three Possible Types of Bonds Nonpolar Covalent (Electrons equally shared.) Polar Covalent (Electrons shared unequally.) Ionic (Electrons are transferred.)

17. Percent Ionic Character where xA is the larger electronegativity and xB is the smaller value. Watch significant figures!!! Ionic Bond % IC > 50 % Polar Covalent % IC 5 - 50 % Nonpolar Covalent % IC < 5 %

18. Percent Ionic Character What type of bonding & % ionic character does KCl have? Ionic

19. Percent Ionic Character What type of bonding & % ionic character does HOH have? Polar covalent

20. Percent Ionic Character What type of bonding & % ionic character does N2 have? Nonpolar covalent

21. Polarity • A molecule, such as HF, that has a center of positive charge and a center of negative charge is said to be polar, or to have a dipole moment. partial positive charge partial negative charge

22. Figure 11.2: Probability representations of the electron sharing in HF

23. The Effect of an electric field on hydrogen fluoride molecules.

24. Dipole Moment for the water molecule.

25. Polar Water Molecule • The polarity of water allows it to dissolve ionic materials which are essential for life. • The polarity of the water molecule allows water molecules to attract each other strongly (hydrogen bonds). Because of this fact water remains as a liquid at room temperatures and allows the existence of life as we know it.

26. Dipole moment for the ammonia molecule.

27. Homework #30 • a. • PH is a pure covalent (nonpolar) bond since P and H have identical electronegativities.

28. Nonpolar molecule--zero dipole moment.

30. Cation Size • Cations are always smaller than the parent atom because they have lost an entire electron shell. As well, the number of protons is greater than the number of electrons so the electrons are held tighter.

31. Anion Size • Anions are always larger than the parent atom because they have added electrons which repel each other. As well, the number of protons is less than the number of electrons so they are not held as tightly.

32. Relative sizes of some ions and their parent atoms.

33. Energy Changes & Chemical Bonding • Most chemical reactions can be explained in terms of the rearrangements of bonds. • Energy must be added to break existing bonds in the reacting substances. • Energy is released when new bonds are formed in the products. • For the reactions of covalent compounds, the net energy of the reaction, ∆Hrxn, = (sum of energy added to break bonds (reactants) - (sum of energy released when bonds form(products).

34. Energy Changes & Chemical Bonding • Consider the reaction for the formation of water from elemental hydrogen and oxygen, 2H2(g) + O2(g) ---> 2H2O(g). • First, write the Lewis dot structures for the reactants and products, and determine which bonds must be broken and which bonds must be formed. • Then use the average bond energy values found in Table 8.4 p351 of the book. to calculate the energy needed to break the bonds in the reactants and the energy released when the bonds form in the products.

35. Energy Changes & Chemical Bonding • To break the H-H bond requires 432 kJ/mole H2(g). Since there are two moles of H2(g) in the reaction, this value is doubled to 864 kJ. • The oxygen atoms in the O2(g) are held together by a double bond which requires 495 kJ/mole to break. • For each water molecule produced, there are two O-H bonds formed. For the 2H2O(g), the total energy released when four moles of O-H bonds form is: 4 moles O-H x 467 kJ/mole = 1868 kJ. • 2H2(g) + O2(g) ---> 2H2O(g).

36. Energy Changes & Chemical Bonding • Therefore, the ∆Hrxn= (864 kJ to break H-H bonds + 495 kJ to break O=O bonds) – (1868 kJ released when O-H bonds formed) = -509 kJ. • The negative value indicates that the reaction releases energy and is exothermic. • 2H2(g) + O2(g) ---> 2H2O(g). (Note that bond energies are average values, and can vary depending upon what other elements are bonded to the atoms which are being broken apart.)

37. Homework #54 • Bonds broken: Bonds formed: • 1 C ≡ N (891 kJ/mol) 1 C - N (305 kJ/mol) • 2 H - H (432 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 kJ

38. Part B • Bonds broken: Bonds formed: • 1 N - N (160. kJ/mol) 4 H - F (565 kJ/mol) • 4 N - H (391 kJ/mol) 1 N ≡ N (941 kJ/mol) • 2 F - F (154 kJ/mol) • ΔH = 160. kJ + 4(391 kJ) + 2(154 kJ) - [4(565 kJ) + 941 kJ] = -1169 kJ

39. Energy Changes & Chemical Bonding • For theformation of ionic compounds, the ∆Hrxn is based upon all of the energy changes involved in the transfer of electrons between the metal and nonmetal. • Consider the reaction for the formation of sodium chloride from elemental sodium and chlorine, • 2Na(s) + Cl2(g) ---> 2NaCl(s).

40. Energy Changes & Chemical Bonding • Ionization energy must be added to completely remove electrons from the metal in the gaseous state. • Since most metals are solids at room temperature, energy must be added to vaporize the metal before this ionization occurs. This energy value is called the heat of formation, ∆Hºf. • Then energy is released when the nonmetal gains electrons, as measured by the electron affinity value. • Energy is also released when the ions bond to form a crystal lattice structure, called the lattice energy.

41. Ionic Compounds • 2Na(s) ---> 2Na(g) • add ∆Hºf = (108 kJ/mole Na x 2 moles Na) = 216 kJ • 2Na(g) ---> 2Na1+(g) + 2e1- • addionization energy = (496 kJ/mole Na x 2 moles Na) = 992 kJ • Cl2(g) ---> 2Cl(g) • add bond energy = 243 kJ/mole Cl2 x 1 mole Cl2) = 243 kJ • 2Cl(g) + 2e1- ---> 2Cl1-(g) • release electron affinity = 349 kJ/mole Cl1- x 2 moles Cl1- = -698 kJ • 2Na1+(g) + 2Cl1-(g) ---> 2NaCl(s) • releaselattice energy = 788 kJ/mole NaCl x 2 moles NaCl = -1576kJ • ∆Hrxn = 216 kJ + 992 kJ + 243 Kj + -698 kJ + -1576 kJ = -823 kJper mole of reaction • 2Na(s) + Cl2(g) ---> 2NaCl(s)

42. Homework #50 • 50. Let us look at the complete cycle for Na2S. • 2 Na(s) → 2 Na(g) 2 ΔHsub, Na = 2(109) kJ • 2 Na(g) → 2 Na+(g) + 2 e 2 IE = 2(495) kJ • S(s) → S(g) ΔHsub, S = 277 kJ • S(g) + e → S (g) EA1 = -200. kJ • S-(g) + e → S2 (g) EA2 = ? • 2 Na+(g) + S2(g) → Na2S LE = -2203 kJ • _____________________________________________________ • 2 Na(s) + S(s) → Na2S(s) ΔH= -365 kJ • ΔH= 2 + 2 IE + ΔHsubS + EA1 + EA2 + LE, • -365 = -918 + EA2, • EA2 = 553kJ • For each salt: ΔH= 2 ΔHsub, M + 2 IE + 277 - 200. + LE + EA2

43. # 50 continued • 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: = 538 kJ

44. Energy Changes & Chemical Bonding • In your textbook, • Ionization energies for some of the elements can be found in Table 7.5 p.310 & Figure 7.6 p.310. • Electron affinity values are listed in Figure 7.7, page 313. • Lattice energies for ionic crystals are found in the problems or given data • Values for the enthalpy of formation, ∆Hºf, are listed in Figure Appendix 4 pp.A19-A22. • Values for the enthalpy of vaporization, ∆Hºvap, are listed in The problems or given appendix. • The accepted values for the various energy changes that occur during bonding reactions have been carefully calculated using calorimeter experiments.

45. Achieving Noble Gas Electron Configurations (NGEC) • Two nonmetals react: They share electrons to achieve NGEC. • A nonmetal and a representative group metal react (ionic compound): The valence orbitals of the metal are emptied to achieve NGEC. The valence electron configuration of the nonmetal achieves NGEC.

46. Noble Gas Configuration • When a Group I, II, or III metal reacts with a nonmetal to form a binary ionic compound, the nonmetal gains electrons to obtain the configuration of the next noble gas. The metal loses electrons to gain the configuration of the previous noble gas. • Na ----> Na+ + e- configuration of Ne • Cl + e- ----> Cl- configuration of Ar

47. Lewis Structure • Shows how valence electrons are arranged among atoms in a molecule. • Reflects central idea that stability of a compound relates to noble gas electron configuration. • Developed by G.N. Lewis in 1902.

48. Lewis Structures • Na. sodium atom [Na]+ sodium ion • sulfur atom [ ] sulfide ion

49. Drawing Lewis structures • Write the electron dot diagrams for each element in the compound. • Check the electronegativity difference between the elements to determine if electrons are transferred or shared. • If the electronegativitydifference > 1.67, the reaction forms ions. Remove the electrons from the metal and add them to the nonmetal.

50. Drawing Lewis Structures Write the charges of the ions formed and use coefficientsto show how many of each ion are needed to balance the overall charge. 2- + [ ] 2Na , O Ionic sodium oxide