2. No simple, and yet complete, way to define chemical bonds. A possible definition could be, “forces that hold groups of atoms together and make them function as a unit.” A bond will form if the energy of the bonded atoms is lower than that of the separated atoms. – Everything will the lowest energy state. – Stable, bonded compounds have less energy than if their atoms were “floating about alone”. 8.1 Types of Chemical Bonds

3. 8.1 Types of Chemical Bonds The Interaction of Two Hydrogen Atoms: -The dots represent probable electron location (density).

4. 8.1 Types of Chemical Bonds The Interaction of Two Hydrogen Atoms:

5. 8.1 Types of Chemical Bonds Key Ideas Ionic Bonding – electrons are transferred Covalent Bonding – electrons are shared equally What about intermediate cases? – Unequal sharing of electrons between atoms in a molecule. – Results in a charge separation in the bond (partial positive and partial negative charge).

6. 8.1 Types of Chemical Bonds indicates a positive or negative fractional charge. The Effect of an Electric Field on Hydrogen Fluoride Molecules

7. The ability of an atom in a molecule to attract shared electrons to itself. For a molecule HX, the relative electronegativities of the H and X atoms are determined by comparing the measured H–X bond energy with the “expected” H–X bond energy. On the periodic table, electronegativity generally increases across a period and decreases down a group. The range of electronegativity values is from 4.0 for fluorine (the most electronegative) to 0.7 for cesium and francium (the least electronegative). 8.2 Electronegativity

8. 8.2 Electronegativity Reality Check If lithium and fluorine react, which has more attraction for an electron? Why? In a bond between fluorine and iodine, which has more attraction for an electron? Why?

9. 8.2 Electronegativity The Relationship Between Electronegativity and Bond Type

10. 8.2 Electronegativity Reality Check Arrange the following bonds from most to least polar: a) N–FO–FC–F b)C–FN–OSi–F c)Cl–ClB–ClS–Cl a) C–F, N–F, O–F b) Si–F, C–F, N–O c) B–Cl, S–Cl, Cl–Cl

11. Property of a molecule whose charge distribution can be represented by a center of positive charge and a center of negative charge. Use an arrow to represent a dipole moment.  Point to the negative charge center with the tail of the arrow indicating the positive center of charge. 8.3 Bond Polarity and Dipole Moment Dipole Moment

12. 8.3 Bond Polarity and Dipole Moment Dipole Moment

13. 8.3 Bond Polarity and Dipole Moment No Net Dipole Moment (Dipoles Cancel)

14. Atoms in stable compounds usually have a noble gas electron configuration. When two nonmetals react to form a covalent bond, they share electrons in a way that completes the valence electron configurations of both atoms. When a nonmetal and a representative-group metal react to form a binary ionic compound, the ions form so that the valence electron configuration of the nonmetal achieves the electron configuration of the next noble gas atom. The valence orbitals of the metal are emptied. 8.4 Ions: Electron Configurations and Sizes Element and Ion Stability

15. A series of ions/atoms containing the same number of electrons. O 2-, F -, Ne, Na +, Mg 2+, and Al 3+ All of these ions contain 10 electrons The radii of these ions are not all the same because each ion has a different number of protons! 8.4 Ions: Electron Configurations and Sizes

16. Trends for:  Atomic size, ion radius, ionization energy, electronegativity Electron configurations Formula prediction for ionic compounds Covalent bond polarity ranking You must be able to describe the trend and explain its occurance (Why does atomic radius decrease across a period?). 8.4 Ions: Electron Configurations and Sizes Periodic Table Predictions

17. What are the factors that influence the stability and the structures of solid binary ionic compounds? How strongly the ions attract each other in the solid state is indicated by the lattice energy. – Atomic Size – Ionic charge Which compound, MgO or NaCl, has a greater lattice energy? Ionic charge always trumps atomic size. 8.5 Energy Effects in Binary Compounds

18. The change in energy that takes place when separated gaseous ions are packed together to form an ionic solid. k = proportionality constant Q 1 and Q 2 = charges on the ions r = shortest distance between the centers of the cations and anions 8.5 Energy Effects in Binary Compounds Lattice Energy

19. 1. Sublimation of the solid metal. M(s)  M(g) [endothermic] 2.Ionization of the metal atoms. M(g)  M + (g) + e  [endothermic] 3.Dissociation of the nonmetal. 1 /2X 2 (g)  X(g) [endothermic] 4. Formation of X  ions in the gas phase. X(g) + e   X  (g) [exothermic] 5. Formation of the solid MX. M + (g) + X  (g)  MX(s) [quite exothermic] 8.5 Energy Effects in Binary Compounds Formation of Species Using Ions

20. 8.5 Energy Effects in Binary Compounds Comparing Energy Changes

21. None of the bonds reach 100% ionic character even with compounds that have a large electronegativity difference (when tested in the gas phase). 8.6 Partial Ionic Character of Covalent Bonds

22. 8.6 Partial Ionic Character of Covalent Bonds Why is CsF less ionic than LiF?

23. Any compound that conducts an electric current when melted will be classified as ionic. Usually have a symmetrical crystal structure. Some ionic compounds will not dissolve in water, but those that do will conduct an electric current. 8.6 Partial Ionic Character of Covalent Bonds Ionic Compounds

24. 1.A model does not equal reality. 2.Models are oversimplifications, and are therefore often wrong. 3.Models become more complicated and are modified as they age. 4.We must understand the underlying assumptions in a model so that we don’t misuse it. 5.When a model is wrong, we often learn much more than when it is right. 8.7 The Chemical Covalent Bond: A Model Fundamental Properties of Models

25. To break bonds, energy must be added to the system (endothermic). To form bonds, energy is released (exothermic).  H =  n×D(bonds broken) –  n×D(bonds formed) D represents the bond energy per mole of bonds (always has a positive sign). 8.8 Covalent Bond Energies and Chemical Reactions Bond Energies

26. 8.8 Covalent Bond Energies and Chemical Reactions Predict  H for the following reaction: Given the following information: Bond Energy (kJ/mol) C–H 413 C–N 305 C–C 347 891  H = –42 kJ Reality Check

27. Assumption: A molecule is composed of atoms that are bound together by sharing pairs of electrons using the atomic orbitals of the bound atoms. Electron pairs are assumed to be localized on a particular atom or in the space between two atoms:  Lone pairs – pairs of electrons localized on an atom  Bonding pairs – pairs of electrons found in the space between the atoms 8.9 The Localized Electron Bond Model Localized Electron Model

28. 1.Description of valence electron arrangement (Lewis structure). 2.Prediction of geometry (VSEPR model). 3.Description of atomic orbital types used to share electrons or hold lone pairs. 8.9 The Localized Electron Bond Model Localized Electron Model

29. 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. 8.10 Lewis Structures

30. Hydrogen forms stable molecules where it shares two electrons. 8.10 Lewis Structures Elements form stable molecules when surrounded by eight electrons. Duet Rule Octet Rule

32. 1.Determine the central atom. 2.Determine total number of valence electrons. 3.Attach ligands to the central atom. 4.Put lone pairs on ligands. 5.Put extra valence electron on central atom as lone pairs. 6.If there are problems, there may be 2x or 3x bonds. 8.10 Lewis Structures Steps for Lewis Structures (Ionic/Covalent Compounds)

33. 8.10 Lewis Structures Reality Check Draw a Lewis structure for each of the following molecules: NH 3 CO 2 CCl 4

34. Boron tends to be stable with only 6 electrons. Elements in with valence electrons beyond the 3 rd energy level can utilize the d orbital and have more than 8 electrons. 8.11 Exceptions to the Octet Rule

35. 8.11 Exceptions to the Octet Rule Draw a Lewis structure for each of the following molecules: BF 3 PCl 5 SF 6 Reality Check

36. C, N, O, and F should always be assumed to obey the octet rule. B and Be often have fewer than 8 electrons around them in their compounds. Second-row elements never exceed the octet rule. Third-row and heavier elements often satisfy the octet rule but can exceed the octet rule by using their empty valence d orbitals. When writing the Lewis structure for a molecule, satisfy the octet rule for the atoms first. If electrons remain after the octet rule has been satisfied, then place them on the elements having available d orbitals (elements in Period 3 or beyond). 8.11 Exceptions to the Octet Rule

37. 8.12 Resonance Actual structure is an average of the resonance structures. Electrons are really delocalized – they can move around the entire molecule.

38. 8.12 Resonance Used to evaluate nonequivalent Lewis structures. Atoms in molecules try to achieve formal charges as close to zero as possible. Any negative formal charges are expected to reside on the most electronegative atoms. Formal charge = (# valence e – on free atom) – (1/2 bonding electrons) – (electrons in lone pairs). Assume:  Lone pair electrons belong entirely to the atom in question.  Shared electrons are divided equally between the two sharing atoms. Formal Charge

39. 8.12 Resonance Consider the Lewis structure for POCl 3. Assign the formal charge for each atom in the molecule. P: 5 – ½ (8) - 0 = +1 O: 6 – ½ (2) - 6 = –1 Cl: 7 – ½ (2) - 6 = 0 Reality Check

40. The sum of the formal charges of all atoms in a given molecule or ion must equal the overall charge on that species. If nonequivalent Lewis structures exist for a species, those with formal charges closest to zero and with any negative formal charges on the most electronegative atoms are considered to best describe the bonding in the molecule or ion. 8.12 Resonance Rules Governing Formal Charge

41. VSEPR: Valence Shell Electron-Pair Repulsion. The structure around a given atom is determined principally by minimizing electron pair repulsions. You will be able to determine the structure of simple ionic/covalent compounds base on the arrangement of their bonds and lone pair electron positions. You will need to memorize the structures names. 8.13 Molecular Structure: The VSEPR Model VSEPR Model

42. 8.13 Molecular Structure: The VSEPR Model 1.Draw the Lewis structure for the molecule. 2.Count the electron pairs and arrange them in the way that minimizes repulsion (put the pairs as far apart as possible. 3.Determine the positions of the atoms from the way electron pairs are shared (how electrons are shared between the central atom and surrounding atoms). 4.Determine the name of the molecular structure from positions of the atoms. Steps to Apply the VSEPR Model

43. 8.13 Molecular Structure: The VSEPR Model Determine the shape for each of the following molecules, and include bond angles: HCN PH 3 SF 4 HCN – linear, 180 o SF 4 – see saw, 90 o, 120 o PH 3 – trigonal pyramid, 109.5 o (107 o ) Reality Check

44. 8.13 Molecular Structure: The VSEPR Model Determine the shape for each of the following molecules, and include bond angles: O 3 KrF 4 O 3 – bent, 120 o KrF 4 – square planar, 90 o, 180 o Reality Check

45. 8.13 Molecular Structure: The VSEPR Model True or false: A molecule that has polar bonds will always be polar. -If true, explain why. -If false, provide a counter-example. Reality Check

46. 8.13 Molecular Structure: The VSEPR Model True or false: Lone pairs make a molecule polar. -If true, explain why. -If false, provide a counter-example. Think about XeF 4... Reality Check

47. 8.13 Molecular Structure: The VSEPR Model

48. 8.13 Molecular Structure: The VSEPR Model

49. 8.13 Molecular Structure: The VSEPR Model

50. 8.13 Molecular Structure: The VSEPR Model

51. 9.1 Hybridization and the Localized Electron Model Hybridization helps explain why molecules are in a given shape. Hybridization involves combining traditional electron orbitals (s, p, d, and f) to make equal energy orbitals. Let’s look at methane (CH 4 )

52. Carbon’s valence electrons are s 2 p 2. – s orbitals are sphere shaped and can only bond at 180 o – p orbitals are 90 o apart and can only bond at angles that are multiples of 90 o (0 o – 360 o ) All four bond angles in CH 4, which is tetrahedral, are observed at 109.5 o. So the electron orbitals much be changed in some way to have such a bond angle. This change is called hybridization. 9.1 Hybridization and the Localized Electron Model

53. 9.1 Hybridization and the Localized Electron Model Combination of one s and three p orbitals. Whenever a set of equivalent tetrahedral atomic orbitals is required by an atom, the localized electron model assumes that the atom adopts a set of sp 3 orbitals; the atom becomes sp 3 hybridized. sp 3 Hybridization

54. 9.1 Hybridization and the Localized Electron Model

55. 9.1 Hybridization and the Localized Electron Model Tetrahedral with Four sp 3 Orbitals

56. 9.1 Hybridization and the Localized Electron Model Draw the Lewis structure for C 2 H 4 (ethylene)?  What is the shape of an ethylene molecule? trigonal planar around each carbon atom  What are the approximate bond angles around the carbon atoms? 120 o Why can’t sp 3 hybridization account for the ethylene molecule? Reality Check

57. Combination of one s and two p orbitals. Gives a trigonal planar arrangement of atomic orbitals. One p orbital is not used.  Oriented perpendicular to the plane of the sp 2 orbitals. 9.1 Hybridization and the Localized Electron Model sp 2 Hybridization and Energy Diagram

58. Sigma bonds are an electron pair is shared in an area centered on a line running between the atoms (single bonds). Pi bonds form double and triple bonds by sharing electron pair(s) in the space above and below the σ bond. Uses the unhybridized p orbitals. 9.1 Hybridization and the Localized Electron Model Sigma (σ) and Pi (π) Bonds

59. 9.1 Hybridization and the Localized Electron Model Combination of one s and one p orbital. Gives a linear arrangement of atomic orbitals. Two p orbitals are not used.  Needed to form the  bonds. sp Hybridization and Energy Diagram

60. 9.1 Hybridization and the Localized Electron Model The Orbitals for CO 2

61. 9.1 Hybridization and the Localized Electron Model Combination of one d, one s, and three p orbitals. Gives a trigonal bipyramidal arrangement of five equivalent hybrid orbitals. sp 3 d Hybridization and Energy Diagram

62. 9.1 Hybridization and the Localized Electron Model Combination of two d, one s, and three p orbitals. Gives an octahedral arrangement of six equivalent hybrid orbitals. sp 3 d 2 Hybridization and Energy Diagram

63. 8.13 Molecular Structure: The VSEPR Model