2. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Bonding: General Concepts Chapter 8

3. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 2 Overview Types of chemical bonds, Electronegativity, Bond polarity and Dipole Moments. The Ions: electron configurations, size, formula, lattice energy calculations. Covalent bonds: model, bond energies, chemical reactions. Lewis structure, exceptions to the octet rule, resonance. Molecular structure models from Valence Shell Electron Pair Model “VSEPR” for single and multiple bonds.

4. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 3 Bonds Forces that hold groups of atoms together and make them function as a unit. NaCl – attraction is electrostatic since Na + and Cl - are the Stable forms for these elements. This is an example of “Ionic Bonding”

5. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 4 Bond Energy 4 It is the energy required to break a bond. 4 It gives us information about the strength of a bonding interaction, as well as radius.

6. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 5 Bond Length The distance where the system energy is a minimum.

7. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 6 Figure 8.1: (a) The interaction of two hydrogen atoms. (b) Energy profile as a function of the distance between the nuclei of the hydrogen atoms.

8. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 7 Change in electron density as two hydrogen atoms approach each other.

9. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 8 Covalent Bond  No electron transfer  Electrons are shared between two atoms, positioned between the two nuclei  Example: H 2, O 2, H 2 O, CO 2, etc.

10. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 9 Ionic Bonds 4 Formed from electrostatic attractions of closely packed, oppositely charged ions. 4 Formed when an atom that easily loses electrons reacts with one that has a high electron affinity.

11. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 10 Li + F Li + F - The Ionic Bond 1s 2 2s 1 1s 2 2s 2 2p 5 1s 2 1s 2 2s 2 2p 6 [He][Ne] Li Li + + e - e - + FF - F - Li + + Li + F -

12. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 11 Cation is always smaller than atom from which it is formed. Anion is always larger than atom from which it is formed. Ionic Compound

13. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 12 Ionic Bonds and Coulomb’s Law Q 1 and Q 2 = numerical ion charges r = distance between ion centers (in nm) 4 4 Attractive forces are (-), repulsive are (+).

14. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 13 Covalent or Ionic?  Covalent and ionic are simply extreme cases.  Most molecules share electrons but “Unequally” due to the difference in electronegativity and electron affinity.  This will give rise to a dipole moment and the molecule becomes polar.

15. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 14 Electronegativity  The ability of an atom in a molecule to attract shared electrons to itself.  Linus Pauling simple model Δ= (H  X) actual  (H  X) expected (H-H) + (X-X) 2 (H-X) experimental # (H-X) expected = If  = 0 => no polarity

16. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 15

17. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 16 Covalent share e - Polar Covalent partial transfer of e - Ionic transfer e - Increasing difference in electronegativity Classification of bonds by difference in electronegativity DifferenceBond Type 0 to 0.1Covalent  2 Ionic 0 < and <2 Polar Covalent

18. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 17

19. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 18 Classify the following bonds as ionic, polar covalent, or covalent: The bond in CsCl; the bond in H 2 S; and the NN bond in H 2 NNH 2. Cs – 0.7Cl – 3.03.0 – 0.7 = 2.3Ionic H – 2.1S – 2.52.5 – 2.1 = 0.4Polar Covalent N – 3.0 3.0 – 3.0 = 0Covalent 9.5

20. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 19 H F F H Polar covalent bond or polar bond is a covalent bond with greater electron density around one of the two atoms. The molecule is called “Dipolar”. electron rich region electron poor region e - riche - poor ++ -- Dipole Moment

21. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 20 (a) (a)When no electric field is present, the molecules are randomly oriented. Figure 8.2: The effect of an electric field on hydrogen fluoride molecules. (b) When the field is turned on, the molecules tend to line up with their negative ends toward the positive pole and their positive ends toward the negative pole.

22. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 21

23. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 22 May exhibit dipole moment depending on their structure i.e. arrangement in space Polyatomic Molecules

24. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 23 Comparison of Ionic and Covalent Compounds

25. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 24 Figure 8.4: (a) The charge distribution in the water molecule. (b) The water molecule in an electric field. V-Shape

26. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 25 Figure 8.5: (a) The structure and charge distribution of the ammonia molecule. The polarity of the N—H bonds occurs because nitrogen has a greater electronegativity than hydrogen. (b) The dipole moment of the ammonia molecule oriented in an electric field. Look for a “NET DIPOLE”

27. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 26 Look for a “NET DIPOLE” N HHHHHHHHHHHHH Trigonal Pyramidal Structure

28. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 27

29. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 28 The carbon dioxide molecule CO 2 : The opposed bond polarities cancel out, and the carbon dioxide has no dipole moment: Non-polar molecule Note: The C-O bond is polar, but the net dipole is Zero Example: SO 3, CCl 4, etc.

30. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 29

31. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 30 Which of the following molecules have a dipole moment? H 2 O, CO 2, SO 2, and CH 4 O H H dipole moment polar molecule S O O CO O no dipole moment nonpolar molecule dipole moment polar molecule C H H HH no dipole moment nonpolar molecule

32. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 31 Compounds Two nonmetals react: They share electrons to achieve NGEC (Noble Gas Electron Configurations) and form covalent bonds. A nonmetal and a representative group metal react (ionic compound): The valence orbitals of the metal are emptied (cation) to achieve NGEC. The valence electron configuration of the nonmetal (anion) achieves NGEC.

33. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 32 Ions  Ionic compounds are always electrically neutral e.g. they have the same amount of +ve and –ve charges.  Common ions have noble gas configurations.

34. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 33 Electron Configurations of Cations and Anions Na [Ne]3s 1 Na + [Ne] Ca [Ar]4s 2 Ca 2+ [Ar] Al [Ne]3s 2 3p 1 Al 3+ [Ne] Atoms lose electrons so that cation has a noble-gas outer electron configuration. H 1s 1 H - 1s 2 or [He] F 1s 2 2s 2 2p 5 F - 1s 2 2s 2 2p 6 or [Ne] O 1s 2 2s 2 2p 4 O 2- 1s 2 2s 2 2p 6 or [Ne] N 1s 2 2s 2 2p 3 N 3- 1s 2 2s 2 2p 6 or [Ne] Atoms gain electrons so that anion has a noble-gas outer electron configuration. Of Representative Elements

35. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 34 ns 1 ns 2 ns 2 np 1 ns 2 np 2 ns 2 np 3 ns 2 np 4 ns 2 np 5 ns 2 np 6 d1d1 d5d5 d 10 4f 5f Ground State Electron Configurations of the Elements

36. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 35 +1+2+3 -2-3 Cations and Anions Of Representative Elements

37. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 36 Notes on Ions  Hydrogen may form H + or H -  Tin forms Sn 2+ and Sn 4+.  Transition metals exhibit a more complicated behavior.

38. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 37 Example of Ionic Compounds  MgO magnesium oxide is formed of Mg 2+ and O 2-.  CaO formed from Ca 2+ and O 2-.  Al 2 O 3 is formed of 2Al 3+ and 3O 2-.

39. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 38 Figure 8.7: Sizes of ions related to positions of the elements on the periodic table.

40. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 39

41. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 40 PERIODIC TRENDS OF ELEMENTS

42. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 41 Isoelectronic Ions Contain the the same number of electrons 8 O1s 2 2s 2 2p 4 O 2- 1s 2 2s 2 2p 6 9 F1s 2 2s 2 2p 5 F - 1s 2 2s 2 2p 6 11 Na1s 2 2s 2 2p 6 3s 1 Na + 1s 2 2s 2 2p 6 12 Mg 1s 2 2s 2 2p 6 3s 2 Mg 2+ 1s 2 2s 2 2p 6 13 Al 1s 2 2s 2 2p 6 3s 2 3p 1 Al 3+ 1s 2 2s 2 2p 6 Which one you expect to have the smallest radius? And why?

43. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 42 Radii of Isoelectronic Ions O 2  > F  > Na + > Mg 2+ > Al 3+ largest smallest 13 protons vs. 10 electrons

44. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 43 Example Choose the largest ion in each of the following groups: 1.Li +, Na +, K +, Rb +, Cs + 2.Ba 2+, Cs +, I -, Te 2-

45. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 44 To Reiterate Cation is smaller than parent molecule. Anion is larger than parent molecule. Size increase down in a group (+ve or –ve). The larger the mass number the smaller the size for isoelectronic cations and anions.

46. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 45 Electron Configurations of Cations of Transition Metals When a cation is formed from an atom of a transition metal, electrons are always removed first from the ns orbital and then from the (n – 1)d orbitals. Fe: [Ar]4s 2 3d 6 Fe 2+ : [Ar]4s 0 3d 6 or [Ar]3d 6 Fe 3+ : [Ar]4s 0 3d 5 or [Ar]3d 5 Mn: [Ar]4s 2 3d 5 Mn 2+ : [Ar]4s 0 3d 5 or [Ar]3d 5

47. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 46 Why Compounds Exist? The driving force behind the naturally occurring compounds (such as NaCl, H 2 O, etc.) is to yield a stable lower energy form. A stable form is a an arrangement of atoms, held together by bonding that prevent decomposition. This bonding energy when ions condense from gas phase into ionic solid is called Lattice Energy.

48. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 47 Lattice Energy The change in energy when separated gaseous ions are packed together to form an ionic solid. M + (g) + X  (g)  MX(s) Lattice energy is negative (exothermic) from the point of view of the system.

49. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 48 Formation of an Ionic Solid 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 (LATTICE) MX M + (g) + X  (g)  MX(s) [quite exothermic] Lattice Energy

50. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 49 The energy changes involved in the formation of solid lithium fluoride from its elements. From Gas to Solid LatticeEnergy

51. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 50 The structure of lithium fluoride. Called also the NaCl structure where each ion is surrounded by 6 of the other ions Applicable for all alkali- metals/halogen except the Cesium salts.

52. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 51 Q 1, Q 2 = charges on the ions (Na +, Cl -, Ca ++, O -- ) r = shortest distance between centers of the cations and anions (Ionic radii increases from top to bottom in a group) The magnitude of Q’s and sizes of ions (ionic radii) will determine how strong is the lattice.

53. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 52 Comparison of the energy changes involved in the formation of solid sodium fluoride and solid magnesium oxide. Note all ions are isoelectronic

54. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 53 Lattice energy (E) increases as Q increases and/or as r decreases. Compd lattice energy MgF 2 MgO LiF LiCl 2957 3938 1036 853 Q= +2,-1 Q= +2,-2 r F < r Cl Electrostatic (Lattice) Energy

55. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 54 Partial Ionic Character of the Covalent Bond Introduce the concept of percent ionic character in a polar covalent bond % ionic character = Measured dipole of X-Y Calculated dipole of X + Y - x 100

56. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 55 Experiments showed that none of the studied systems have 100% ionic character despite large differences in electronegativity. The reason is “Ion Pairing” during motion, ions approach closely and for a short period of time their charges will be neutralized before of time their charges will be neutralized before moving away from each other. Ion Pairing

57. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 56 Molten NaCl conducts an electric current, indicating the presence of mobile Na + and Cl - ions. The more mobile the ions the Stronger the current and the brighter the lamp!

58. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 57 The relationship between the ionic character of a covalent bond and the electronegativity difference of the bonded atoms.>50%Ionic <50%Non-Ionic Note: These are “Molten” salts and not “Aqueous” salts! and not “Aqueous” salts!

59. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 58 A Model for Covalent Bond Models are attempts to explain how nature operates on the microscopic level based on experiences in the macroscopic world.

60. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 59 Fundamental Properties of Models 4 A model does not equal reality. 4 Models are oversimplifications, and are therefore often wrong. 4 Models become more complicated as they age. 4 We must understand the underlying assumptions in a model so that we don’t misuse it.

61. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 60 Generally, bonds occur when collections of atoms are more stable (lower in energy) than the separate atoms e.g. CH 4 is 1652 KJ lower in energy than 1 mole of C and 4 mole of H. Stability can be determined in terms if model called “Chemical Bond” C-H bond energy == 413 KJ/mol 1652 KJ/mol 4

62. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 61 The enthalpy change required to break a particular bond in one mole of gaseous molecules is the bond energy. H 2 (g) H (g) +  H 0 = 436.4 kJ Cl 2 (g) Cl (g) +  H 0 = 242.7 kJ HCl (g) H (g) +Cl (g)  H 0 = 431.9 kJ O 2 (g) O (g) +  H 0 = 498.7 kJ OO N 2 (g) N (g) +  H 0 = 941.4 kJ N N Bond Energy Bond Energies Single bond < Double bond < Triple bond

63. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 62 Using this model, one can determine other bond energies. CH 3 Cl is composed of 3 C-H bond and 1 C-Cl bond C-H bond is known C-H bond is known 413 KJ/mol C-Cl bond energy = 1572 – 3(413) = 339 KJ/mol

64. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 63 Average bond energy in polyatomic molecules H 2 O (g) H (g) +OH (g)  H 0 = 502 kJ OH (g) H (g) +O (g)  H 0 = 427 kJ Average OH bond energy = 502 + 427 2 = 464 kJ

65. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 64

66. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 65 Note : The shorter the bond the higher the bond energy Bond Energy and Bond Length

67. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 66 Bond Energies and Enthalpy of Reaction Bond breaking requires energy (endothermic). Bond formation releases energy (exothermic).  H =  D( bonds broken )   D( bonds formed ) Energy required Energy released Reactants Products

68. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 67 Use bond energies to calculate the enthalpy change for: H 2 (g) + F 2 (g) 2HF (g)  H 0 =  BE(reactants) –  BE(products) Type of bonds broken Number of bonds broken Bond energy (kJ/mol) Energy change (kJ) HH1436.4 FF 1156.9 Type of bonds formed Number of bonds formed Bond energy (kJ/mol) Energy change (kJ) HF2568.21136.4  H 0 = 436.4 + 156.9 – 2 x 568.2 = -543.1 kJ 9.10

69. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 68 Localized Electron Model A molecule is composed of atoms that are bound together by sharing pairs of electrons using the atomic orbitals of the bound atoms.

70. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 69 Localized Electron Model 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.

71. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 70 Lewis Structure 4 Shows how valence electrons are arranged among atoms in a molecule. 4 Reflects central idea that stability of a compound relates to noble gas electron configuration.

72. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 71 1. 1.Count valence electrons of all atoms and sum up. 2. 2.In ions, add 1 for each negative charge and subtract 1 for each positive charge from the total valence electrons. 3. 3.Draw skeletal structure of compound showing what atoms are bonded to each other. Put least electronegative element in the center. 4. 4.Complete an octet for all terminal atoms first except hydrogen which has duet. 5. 5.If structure contains excess of electrons, put theses electrons on central atom and also try to form double and triple bonds as needed for octet. Drawing Lewis Structures

73. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 72 Write the Lewis structure of nitrogen trifluoride (NF 3 ). Step 1 – N is less electronegative than F, put N in center FNF F Step 2 – Count valence electrons N - 5 (2s 2 2p 3 ) and F - 7 (2s 2 2p 5 ) 5 + (3 x 7) = 26 valence electrons Step 3 – Draw single bonds between N and F atoms and complete octets on N and F atoms. Step 4 - Check, are # of e - in structure equal to number of valence e - ? 3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons Actual in 3D

74. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 73 F F F F Lewis structure of F 2 lone pairs single covalent bond

75. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 74 8e - H H O ++ O HH O HHor 2e - Lewis structure of water Double bond – two atoms share two pairs of electrons single covalent bonds O C O or O C O 8e - double bonds Triple bond – two atoms share three pairs of electrons N N 8e - N N triple bond or

76. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 75 Write the Lewis structure of the carbonate ion (CO 3 2- ). Step 1 – C is less electronegative than O, put C in center OCO O Step 2 – Count valence electrons C - 4 (2s 2 2p 2 ) and O - 6 (2s 2 2p 4 ) -2 charge – 2e - 4 + (3 x 6) + 2 = 24 valence electrons Step 3 – Draw single bonds between C and O atoms and complete octet on C and O atoms. Step 4 - Check, are # of e - in structure equal to number of valence e - ? 3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons Step 5 - Too many electrons, form double bond and re-check # of e - 2 single bonds (2x2) = 4 1 double bond = 4 8 lone pairs (8x2) = 16 Total = 24

77. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 76 Carbon Dioxide CO 2 or O CO.......... No octet yet Octet O CO........ BUT Do not Break Symmetry O C O........

78. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 77 Exceptions to the Octet Rule The Incomplete Octet HHBe Be – 2e - 2H – 2x1e - 4e - BeH 2 BF 3 B – 3e - 3F – 3x7e - 24e - FBF F 3 single bonds (3x2) = 6 9 lone pairs (9x2) = 18 Total = 24

79. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 78 Exceptions to the Octet Rule Odd-Electron Molecules N – 5e - O – 6e - 11e - NO N O The Expanded Octet (central atom with principal quantum number n > 2) SF 6 S – 6e - 6F – 42e - 48e - S F F F F F F 6 single bonds (6x2) = 12 18 lone pairs (18x2) = 36 Total = 48

80. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 79 Comments About the Octet Rule 4 2nd row elements C, N, O, F observe the octet rule. 4 2nd row elements B and Be often have fewer than 8 electrons around themselves - they are very reactive (BF 3, BeH 2, etc.. ). 4 3rd row and heavier elements can exceed the octet rule using empty valence d orbitals (SF 6, PCl 5, etc.). 4 When writing Lewis structures, satisfy octets first, then place electrons around elements having available d orbitals.

81. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 80 Note: For molecules that has several Third row (or higher) elements the extra electrons should be placed on the central atom. I3-I3-I3-I3- IIIIIIIIIIII....................

82. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 81 Resonance Occurs when more than one valid Lewis structure can be written for a particular molecule. C=C<C C< C-C …. These are resonance structures. The actual structure is an average of the resonance structures. The value of the resonance bond is in between a single bond and double bond.

83. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 82 Resonance Structure of Ozone: O 3 OOO + - OOO + - OCO O -- OCO O - - OCO O - - What are the resonance structures of the carbonate (CO 3 2 -) ion?

84. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 83 How the charges in the previous examples were depicted? Calculate the Formal charge on each atom.

85. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 84 An atom’s formal charge is the difference between the number of valence electrons in an isolated atom and the number of electrons assigned to that atom in a Lewis structure. formal charge on an atom in a Lewis structure = 1 2 total number of bonding electrons () total number of valence electrons in the free atom - total number of nonbonding electrons - The sum of the formal charges of the atoms in a molecule or ion must equal the charge on the molecule or ion. Formal Charge

86. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 85 Two possible skeletal structures of formaldehyde (CH 2 O) HCOH H CO H Calculate and minimize Formal Charges

87. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 86 HCOH C – 4 e - O – 6 e - 2H – 2x1 e - 12 e - 2 single bonds (2x2) = 4 1 double bond = 4 2 lone pairs (2x2) = 4 Total = 12 formal charge on C = 4 -2 -2 -½ x 6 = -1 formal charge on O = 6 -2 -2 -½ x 6 = +1 formal charge on an atom in a Lewis structure = 1 2 total number of bonding electrons () total number of valence electrons in the free atom - total number of nonbonding electrons - +1

88. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 87 C – 4 e - O – 6 e - 2H – 2x1 e - 12 e - 2 single bonds (2x2) = 4 1 double bond = 4 2 lone pairs (2x2) = 4 Total = 12 H CO H formal charge on C = 4 -0 -0 -½ x 8 = 0 formal charge on O = 6 -4 -4 -½ x 4 = 0 formal charge on an atom in a Lewis structure = 1 2 total number of bonding electrons () total number of valence electrons in the free atom - total number of nonbonding electrons - 00

89. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 88 Formal Charge and Lewis Structures 1. 1.For neutral molecules, a Lewis structure in which there are NO formal charges is preferable to one in which formal charges are present. 2. 2.Lewis structures with large formal charges are less plausible (acceptable) than those with small formal charges. 3. 3.Among Lewis structures having similar distributions of formal charges, the most plausible structure is the one in which negative formal charges are placed on the more electronegative atoms. Which is the most likely Lewis structure for CH 2 O? HCOH +1 H CO H 00

90. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 89 Formal Charge Not as good Better Charges are minimized

91. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 90 VSEPR Model The structure around a given atom is determined principally by minimizing electron pair repulsions.

92. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 91 Predicting a VSEPR Structure 1.Draw Lewis structure. 2.Put pairs as far apart as possible. 3.Determine positions of atoms from the way electron pairs are shared. 4.Determine the name of molecular structure from positions of the atoms.

93. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 92

94. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 93 Valence shell electron pair repulsion (VSEPR) model: Predict the geometry of the molecule from the electrostatic repulsions between the electron (bonding and nonbonding) pairs. AB 2 20 Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry linear B B

95. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 94 Cl Be 2 atoms bonded to central atom 0 lone pairs on central atom

96. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 95 AB 2 20linear Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry VSEPR AB 3 30 trigonal planar

97. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 96

98. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 97 AB 2 20linear Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry VSEPR AB 3 30 trigonal planar AB 4 40 tetrahedral

99. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 98

100. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 99

101. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 100 AB 2 20linear Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry VSEPR AB 3 30 trigonal planar AB 4 40 tetrahedral AB 5 50 trigonal bipyramidal trigonal bipyramidal

102. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 101

103. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 102 AB 2 20linear Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry VSEPR AB 3 30 trigonal planar AB 4 40 tetrahedral AB 5 50 trigonal bipyramidal trigonal bipyramidal AB 6 60 octahedral

104. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 103

105. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 104 Molecular structure of PCl 6 - Octahedral Geometry

106. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 105 Octahedral electron arrangement for Xe

107. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 106 Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry VSEPR AB 3 30 trigonal planar AB 2 E21 trigonal planar bent

108. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 107 Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry VSEPR AB 3 E31 AB 4 40 tetrahedral trigonal pyramidal

109. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 108 Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry VSEPR AB 4 40 tetrahedral AB 3 E31tetrahedral trigonal pyramidal AB 2 E 2 22 tetrahedralbent H O H V-Shape

110. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 109

111. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 110 bonding-pair vs. bonding pair repulsion lone-pair vs. lone pair repulsion lone-pair vs. bonding pair repulsion <<

112. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 111 Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry VSEPR AB 5 50 trigonal bipyramidal trigonal bipyramidal AB 4 E41 trigonal bipyramidal distorted tetrahedronSee-Saw

113. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 112 Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry VSEPR AB 5 50 trigonal bipyramidal trigonal bipyramidal AB 4 E41 trigonal bipyramidal distorted tetrahedron AB 3 E 2 32 trigonal bipyramidal T-shaped Cl F F F

114. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 113 Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry VSEPR AB 5 50 trigonal bipyramidal trigonal bipyramidal AB 4 E41 trigonal bipyramidal distorted tetrahedron AB 3 E 2 32 trigonal bipyramidal T-shaped AB 2 E 3 23 trigonal bipyramidal linear I I I

115. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 114 Three possible arrangements of the electron pairs in the I 3 - ion. Least Lone pair repulsions

116. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 115 Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry VSEPR AB 6 60 octahedral AB 5 E51 octahedral square pyramidal Br FF FF F

117. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 116 Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry VSEPR AB 6 60 octahedral AB 5 E51 octahedral square pyramidal AB 4 E 2 42 octahedral square planar Xe FF FF

118. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 117 Possible electron-pair arrangements for XeF 4.

119. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 118 Predicting Molecular Geometry 1. 1.Draw Lewis structure for molecule. 2. 2.Count number of lone pairs on the central atom and number of atoms bonded to the central atom. 3. 3.Use VSEPR to predict the geometry of the molecule. What are the molecular geometries of SO 2 and SF 4 ? SO O AB 2 E bent S F F F F AB 4 E See-saw / distorted tetrahedron

120. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 119 POLARITY OF COMPOUNDS ALL MOLECULES WITHOUT LONE PAIRS WILL BE NON-POLAR AB 2, AB 3, AB 4, AB 5, AB 6

121. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 120 The molecular structure of methanol CH 3 OH. (a) The arrangement of electron pairs and atoms around the carbon atom. (b) The arrangement of bonding and lone pairs around the oxygen atom. (c) The molecular structure.