2. Chemical Bonding Chapter 8 Sections 1-8

3. A chemical bond is: a strong electrostatic force of attraction between atoms in a molecule or compound. Bonding between atoms occurs because it creates a more stable arrangement for the atoms.

4. Lewis Symbols – Dot Diagrams Convenient way to show the valence electrons

5. Three types of bonding Metallic bonding – results from the attraction between metal atoms and the surrounding sea of electrons Ionic bonding – results from the electrical attraction between positive and negative ions. Covalent bonding – results from the sharing of electron pairs between two atoms

6. Ionic Bonding Many atoms transfer electrons and other atoms accept electrons, creating cations (positive metal ions) and anions (negative nonmetal ions). The resulting ions are attracted to each other by electrostatic force.

7. Ionic Bonding The ions closely pack together in a crystal lattice. This arrangement maximizes the attractive forces among cations and anions while minimizing repulsive forces.

8. Because force is proportional to the charge on each ion, larger charges lead to stronger interactions. Because force is inversely proportional to the square of the distance between the centers of the ions, smaller ions lead to stronger interactions.

9. Ionic bonding between Na and Cl

11. Covalent bonding In many cases electrons do not completely transfer from one atom to another. The electrons between atoms are shared.

12. Covalent bonding between H 2 Hydrogen’s electron configuration is 1s 1 Because both H atoms need 1 more electron to become isoelectronic with He, it is unlikely that either will give up an electron.

13. Covalent bonding between H 2 ↑ 1s ↓ They share the two electrons. H· + H ·  H : H

14. Types of Covalent Bonds When electrons are shared equally the bond is called a NONPOLAR covalent bond. (i.e. H 2 ) Sometimes the electrons between two atoms are NOT shared equally. The bond created is called a POLAR covalent bond..... H· + ·Cl:  H:Cl:....

15. Polar Covalent Bonding An example of this would be HCl. HCl molecule Hydrogen atom [Ne] ↑↓ ↑↓ ↑↓ ↓ 3s 3p ↑ 1s Chlorine atom

16. How to classify bond types Electronegativity – ability of an atom in a molecule to attract shared electrons to it Each element on the periodic table is assigned an electronegativity value (see page 353) that ranges from 0.7 to 4.0. The difference in the electronegativity determines the bonding type (ionic, polar covalent, or nonpolar covalent).

17. Electronegativity Values

18. If the electronegativity difference is: 1.7 and higherionic 0.3 to 1.7polar covalent 0.0 to 0.3nonpolar covalent

19. What if I get an electronegativity difference that is 0.3 or 1.7? These cut-off numbers are guidelines. It is a gradual change not stair-step.

20. Ionic Character As the electronegativity difference increases, the ionic character increases as well!

21. Practice Problems What type of bond will occur between iodine and the following elements: cesium, iron, and sulfur? Bonding between I and: Electronegativity difference Bond Type Cesium Iron Sulfur

22. Determine the type of bond between the following pairs. Bonding between Electronegativity difference Bond type Li & Cl S & O Ca & Br P & H Si & Cl S & Br

23. Other ways to determine bonding Electronegativity is not the only factor in determining bonding. Generally, bonds between a metal and nonmetal are ionic, and between two nonmetals the bonds are covalent. Examination of the properties of a compound is the best way to determine the type of bonding.

24. Ionic Bonding Ionic compounds are formed to maximize stability. Nonmetal - will gain electrons to become isoelectronic with nearest noble gas; called an anion Metal – will lose electrons to become isoelectronic with noble gas; called a cation

25. Transition Metals Zinc Electron configuration is 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 When it forms the +2 ion, it loses the 2 valence electrons in the 4s sublevel. Zn 2+ configuration is 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10

26. Practice Problems Write the electron configurations for the following ions. – Fe 2+ – S 2- – Mg 2+ Use electron configurations to explain why the most probable charge on the strontium ion is +2.

27. Does the size go up or down when losing an electron to form a cation? Size of Ions

28. CATIONS are SMALLER than the atoms from which they come. CATIONS are SMALLER than the atoms from which they come. The electron/proton attraction has gone UP and so size DECREASES. The electron/proton attraction has gone UP and so size DECREASES. Li,152 pm 3e and 3p Li +, 78 pm 2e and 3 p + Forming a cation. Size of Ions

29. Does the size go up or down when gaining an electron to form an anion? Size of Ions

30. ANIONS are LARGER than the atoms from which they come. ANIONS are LARGER than the atoms from which they come. The electron/proton attraction has gone DOWN and so size INCREASES. The electron/proton attraction has gone DOWN and so size INCREASES. Trends in ion sizes are the same as atom sizes. Trends in ion sizes are the same as atom sizes. Forming an anion. F, 71 pm 9e and 9p F -, 133 pm 10 e and 9 p - Size of Ions

31. Trends in Ion Sizes Figure 8.13

32. Which is Bigger? Cl or Cl - ? K + or K ? Ca or Ca +2 ? I - or Br - ?

33. Which is Bigger? Cl or Cl - ? Cl - K + or K ?K Ca or Ca +2 ? Ca I - or Br - ? I -

34. Lattice Energy Effects 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.

35. Lattice Energy To determine which compound will have the highest lattice energy, take into consideration the following: – The size of the ions in the compound The smaller the size, the greater the lattice energy – The charge of the ions in the compound The greater the charge, the greater the lattice energy

36. Calculating ∆H f We can take advantage of the fact the energy is a state function and break the reaction into steps, the sum of which is the overall reaction. Let’s do #41 Na(s) + ½ Cl 2 (g)  NaCl(s) Given the following: Lattice energy-786 kJ/mol Ionization energy for Na495 kJ/mol Electron affinity for Cl-349 kJ/mol Bond energy of Cl 2 239 kJ/mol Enthalpy sublimation for Na109 kJ/mol

37. Process Step 1: Sublimation of Na Na(s)  Na(g)109 kJ/mol Step 2: Ionization of Na Na (g)  Na + (g) + e - 495 kJ/mol Step 3: Dissociation of Cl 2 ½ Cl 2 (g)  Cl(g)119.5 kJ/mol Step 4: Formation of Cl - (Electron Affinity) Cl (g) + e -  Cl - (g)-349 kJ/mol Step 5: Formation of NaCl Na + (g) + Cl - (g)  NaCl(s)-786 kJ/mol Na(s) + ½ Cl 2 (g)  NaCl(s)-411.5 kJ/mol

38. Drawing Lewis Structures Valence Electron Review Valence electrons are in outermost level You can use periodic table or electron configuration to determine valence electrons Example: Phosphorus – Located in Group 15 or 5A – Electron configuration is 1s 2 2s 2 2p 6 3s 2 3p 3 – Contains 5 valence electrons Complete Exercise 1 on worksheet

39. Drawing Lewis Structures Octet Rule Most useful rule for creating Lewis structures Every atom usually has 8 valence electrons Exception: hydrogen is good with 2 (like He) Lines are used to link atoms together (same as using 2 dots) Same as

40. Steps to Drawing Lewis Structures 1.Count valence electrons. 2.Connect atoms together with bonds. In molecules with a single atom of one element and several atoms of another element, the single atom is generally in the center with the other atoms attached to it. 3.Add electrons around outside of atoms to give each atom 8 electrons (or 2 in the case of hydrogen). 4.Count electrons used. This number must be the same as valence electrons.

41. PCl 3 Bonding Pairs Lone Pairs (a.k.a. nonbonding electrons) 5+(3*7)=26 e - Complete Exercise 2.

42. Helpful Hints Carbon atoms form 4 bonds. Nitrogen atoms form 3 bonds. Oxygen atoms form 2 bonds. Hydrogen atoms form 1 bond. Fluorine atoms form 1 bond. Other halogens (Cl, Br, and I) frequently form 1 bond (but not always).

43. Determining the Central Atom In a molecule, the atom that typically forms the greatest number of bonds is in the center, with other atoms attached to it. Example: CH 3 Cl – Carbon forms 4 bonds – Hydrogen forms 1 bond – Chlorine forms 1 bond SO CARBON IS IN THE MIDDLE WITH HYDROGEN AND CHLORINE AROUND IT! Don’t forget electrons on chlorine to make 8! Complete exercise 3.

44. Multiple Bonds It is possible for more than one pair of electrons to be shared between two atoms (multiple bonds): One shared pair of electrons = single bond (e.g. H 2 ); Two shared pairs of electrons = double bond (e.g. O 2 ); Three shared pairs of electrons = triple bond (e.g. N 2 ). Generally, bond distances shorten with multiple bonding. Covalent Bonding Octet in each case

46. Resonance Occurs when more than one valid Lewis structure can be written for a particular molecule.

47. Odd Number of Electrons… NO Number of valence electrons = 11 NO 2 Number of valence electrons = 17 Resonance occurs when more than one valid Lewis structure can be written for a particular molecule (i.e. rearrange electrons) Molecules and atoms which are neutral (contain no formal charge) and with an unpaired electron are called Radicals

48. Beyond the Octet Elements in the 3 rd period or higher can have more than an octet if needed. Atoms of these elements have valence d orbitals, which allow them to accommodate more than eight electrons.

49. More than an Octet… PCl 5 Elements from the 3rd period and beyond, have ns, np and unfilled nd orbitals which can be used in bonding P : (Ne) 3s 2 3p 3 3d 0 Number of valence electrons = 5 + (5 x 7) = 40 SF 4 S : (Ne) 3s 2 3p 4 3d 0 Number of valence electrons = 6 + (4 x 7) = 34 The Larger the central atom, the more atoms you can bond to it – usually small atoms such as F, Cl and O allow central atoms such as P and S to expand their valency.

50. Formal Charge Difference between the # of valence electrons in the free atom and the # of electrons assigned to that atom in the Lewis structure. FC = formal charge; G.N. = Group Number #BE = bonding electrons; #LPE = lone pair electrons If Step 4 leads to a positive formal charge on an inner atom beyond the second row, shift electrons to make double or triple bonds to minimize formal charge, even if this gives an inner atom with more than an octet of electrons.

51. Formal Charge Not as good Better

52. Molecular Shapes Lewis structures give atomic connectivity: they tell us which atoms are physically connected together. They do not tell us the shape. The shape of a molecule is determined by its bond angles. Consider CCl 4 : experimentally we find all Cl- C-Cl bond angles are 109.5 . Therefore, the molecule cannot be planar. All Cl atoms are located at the vertices of a tetrahedron with the C at its center.

53. Molecular Shape of CCl 4

54. VSEPR Theory  In order to predict molecular shape, we assume the valence electrons repel each other. Therefore, the molecule adopts whichever 3D geometry minimized this repulsion.  We call this process Valence Shell Electron Pair Repulsion (VSEPR) theory.

55. Why is VSEPR Theory Important? Gives a specific shape due to the number of bonded and non-bonded electron pairs in a molecule Tells us the actual 3-D structure of a molecule In bonding, electron pairs want to be as far away from each other as possible.

56. VSEPR and Resulting Geometries

57. How does VSEPR THEORY work? We can use VSEPR theory using 4 steps. 1.Draw the Lewis Structure for the molecule. Example: SiF 4 F F-Si-F F

58. How does VSEPR THEORY work? We can use VSEPR theory using 4 steps 1.Draw the Lewis Structure for the molecule. 2.Tally the number of bonding pairs and lone (non-bonding) pairs on the center atom. F F-Si-F F Bonding pairs: 4 Lone pairs on central atom: 0

59. How does VSEPR THEORY work? We can use VSEPR theory using 4 steps 1.Draw the Lewis Structure for the molecule 2.Tally the number of bonding pairs and lone pairs on the center atom. 3.Arrange the rest of the atoms so that they are as far away from each other as possible. Si F F F F

60. How does VSEPR THEORY work? We can use VSEPR theory using 4 steps 1.Draw the Lewis Structure for the molecule 2.Tally the number of bonding pairs and lone pairs on the center atom. 3.Arrange the rest of the atoms so that they are as far away from each other as possible 4.Give the type of geometry the molecule has: Tetrahedral

61. Another Example: To determine the electron pair geometry: 1) draw the Lewis structure; 2) count the total number of electron pairs around the central atom. 3) arrange the electron pairs in one of the geometries to minimize e  -e  repulsion. 4) multiple bonds count as one bonding pair for VSEPR

62. The VSEPR Model Predicting Molecular Geometries

63. The VSEPR Model Predicting Molecular Geometries

64. The VSEPR Model Difference between geometry and shape Geometry: We determine the geometry only looking at electrons. All the atoms that obey the octet rule have the same tetrahedral-like geometry. Shape: We name the shape by the positions of atoms. We ignore lone pairs in the shape.

65. The VSEPR Model Predicting Shape Shape

66. The VSEPR Model Predicting Shape Shape

67. The VSEPR Model The Effect of Nonbonding Electrons and Multiple Bonds on Bond Angles By experiment, the H-X-H bond angle decreases on moving from C to N to O: Since electrons in a bond are attracted by two nuclei, they do not repel as much as lone pairs. Therefore, the bond angle decreases as the number of lone pairs increase.

68. The VSEPR Model The Effect of Nonbonding Electrons and Multiple Bonds on Bond Angles Similarly, electrons in multiple bonds repel more than electrons in single bonds.

69. The VSEPR Model Molecules with Expanded Valence Shells Atoms that have expanded octets have AB 5 (trigonal bipyramidal) or AB 6 (octahedral) electron pair geometries. Examples: PF 5 trigonal bipyramidalSCl 6 octahedral

70. The VSEPR Model Molecules with Expanded Valence Shells

71. The VSEPR Model Molecules with Expanded Valence Shells

72. The VSEPR Model Molecules with More than One Central Atom In acetic acid, CH 3 COOH, there are three central atoms. We assign the geometry about each central atom separately.

73. Hybrid Orbitals In bonding, s and p orbitals are used in bonding. It is easy to tell which ones are used by looking at our molecule. For example, CH 4. Looking again at the Lewis structure, we see that there are 4 bonds. We call this sp 3 hybridized.

74. Hybrid Orbitals Regions of electron density-EACH BOND AND LONE PAIR OF ELECTRONS ON THE CENTRAL ATOM IS KNOWN AS A REGION OF ELECTRON DENSITY. 2 regions of electron density-sp hybridized 3 regions of electron density-sp 2 hybridized 4 regions of electron density-sp 3 hybridized

75. Hybridization sp Hybrid Orbitals The two lobes of an sp hybrid orbital are 180  apart.

76. Hybrid Orbitals sp 2 Hybrid Orbitals Important: when we mix n atomic orbitals we must get n hybrid orbitals. sp 2 hybrid orbitals are formed with one s and two p orbitals. (Therefore, there is one unhybridized p orbital remaining.) The large lobes of sp 2 hybrids lie in a trigonal plane. All molecules with trigonal planar electron pair geometries have sp 2 orbitals on the central atom.

77. Hybridization

78. sp 3 Hybrid Orbitals sp 3 Hybrid orbitals are formed from one s and three p orbitals. Therefore, there are four large lobes. Each lobe points towards the vertex of a tetrahedron. The angle between the large lobes is 109.5  All molecules with tetrahedral electron pair geometries are sp 3 hybridized.

79. Hybridization

80. Hybrid Orbitals

81. Summary To assign hybridization: 1.Draw a Lewis structure. 2.Assign the geometry using VSEPR theory. 3.Use the geometry to determine the hybridization. 4.Name the shape by the positions of the atoms.

82. Hybridization and Multiple Bonds Multiple bonds overlap differently and are called  -bonds and  -bonds All single bonds are  Double bonds contain 1  and 1  bond Triple bonds contain 1  and 2  bonds

83. Bond Energy

84. Covalent Bonding & Orbital Overlap As two nuclei approach each other their atomic orbitals overlap. As the amount of overlap increases, the energy of the interaction decreases. At some distance the minimum energy is reached. The minimum energy corresponds to the bonding distance (or bond length).

85. Covalent Bonding & Orbital Overlap As the two atoms get closer, their nuclei begin to repel and the energy increases. At the bonding distance, the attractive forces between nuclei and electrons just balance the repulsive forces (nucleus-nucleus, electron-electron).

86. Bond Energies Bond breaking requires energy (endothermic). Bond formation releases energy (exothermic).  H =  D( bonds broken )   D( bonds formed ) energy required energy released

87. Bond Energies