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AP CHEMISTRY CHAPTER 8 BONDING. bond energy- energy required to break a chemical bond -We can measure bond energy to determine strength of interaction.

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Presentation on theme: "AP CHEMISTRY CHAPTER 8 BONDING. bond energy- energy required to break a chemical bond -We can measure bond energy to determine strength of interaction."— Presentation transcript:

1 AP CHEMISTRY CHAPTER 8 BONDING

2 bond energy- energy required to break a chemical bond -We can measure bond energy to determine strength of interaction

3 ionic compound- a metal reacts with a nonmetal Ionic bonds form when an atom that loses electrons easily reacts with an atom that has a high affinity for electrons. The charged ions are held together by their mutual attraction (Coulombic attraction). Ionic bonds form because the ion pair has lower energy than the separated ions. All bonds form in order to reach a lower energy level.

4 Bond length- the distance where the energy is at a minimum. We have a balance among proton-proton repulsion, electron-electron repulsion, and proton- electron attraction. In H 2, the two e − will usually be found between the two H atoms because they are spontaneously attracted to both protons. Therefore, electrons are shared by both nuclei. This is called covalent bonding.

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6 Polar covalent bonds occur when electrons are not shared equally. One end of the molecule may have a partial charge. This is called a dipole.  +  H F H H  +  - O  -

7 Electronegativity- the ability of an atom in a molecule to attract shared electrons to itself. -determined by comparing the measured bond energy and the expected bond energy. Expected H—X = H—H bond energy + X—X bond energy bond energy 2 Electronegativity difference = (Actual H—X bond energy) – (expected H—X bond energy) If X has a greater electronegativity than H, the e − ’s are closer to X and the molecule is polar. If the electronegativities are the same, the molecule is nonpolar.

8 Periodic Trends- Electronegativity generally increases across a period and decreases down a group. It ranges from 0.79 for cesium to 4.0 for fluorine.

9  +  − H—F polar H—H nonpolar has dipole moment O  +  + H H O S Obent, polar O  − has dipole moment planar no dipole moment CH 4 tetrahedral NH 3 trigonal pyramidal no dipole moment has dipole moment

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13 Electron Configurations: Stable compounds usually have atoms with noble gas electron configurations. Two nonmetals react to form a covalent bond by sharing electrons to gain valence electron configurations.

14 When a nonmetal and a group A metal react to form a binary ionic compound, the ions form so that the valence electron configuration of the nonmetal is completed and the valence orbitals of the metal are emptied to give both noble gas configurations.

15 Ions form to get noble gas configurations. -exceptions in Group A metals: Sn 2+ & Sn 4+ Pb 2+ &Pb 4+ Bi 3+ & Bi 5+ Tl + & Tl 3+ Metals with d electrons will lose their highest numerical energy level electrons before losing their inner d electrons.

16 Size of Ions Positive ions (cations) are smaller than their parent atoms since they are losing electrons. (More protons than electrons=greater nuclear pull) Negative ions (anions) are larger than their parent atoms since they are gaining electrons. (Fewer protons than electrons= lower nuclear pull) Think: Monster Ants & miniature cats

17 Ion size increases going down a group.

18 Isoelectronic ions –ions containing the same number of electrons O 2−, F −, Na +, Mg 2+, Al 3+ all have the Ne configuration. They are isoelectronic. *** For an isoelectronic series, size decreases as Z increases.

19 Lattice energy- the change in energy that takes place when separated gaseous ions are packed together to form an ionic solid. Na + (g) + Cl − (g)  NaCl(s) If exothermic, the sign will be negative and the ionic solid will be the stable form. We can use a variety of steps to determine the heat of formation of an ionic solid from its elements. This is called the Born-Haber cycle. See examples on pages 366 & 368.

20 Lattice energy can be calculated using the following: where k is a proportionality constant that depends on the structure of the solid and the electron configuration of the ions. Q 1 & Q 2 are the charges on the ions. r is the distance between the center of the cation and the anion. Since the ions will have opposite charges, lattice energy will be negative (exothermic). The attractive force between a pair of oppositely charged ions increases with increased charge on the ions or with decreased ionic sizes.

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22 The Structure of Lithium Fluoride

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24 Compounds with more than 50% ionic character are considered to be ionic (electronegativity diff. of about 1.7). There are probably no totally ionic bonds. Percent ionic character in binary compounds can be calculated. Percent ionic character increases with electronegativity difference.

25 The Relationship Between the Ionic Character of a Covalent Bond and the Electronegativity Difference of the Bonded Atoms

26 Three Possible Types of Bonds

27 Polyatomic ions are held together by covalent bonds. We call Na 2 SO 4 ionic even though it has 4 covalent bonds and 2 ionic bonds. Ionic compound- any solid that conducts an electrical current when melted or dissolved in water Salt- an ionic compound

28 A chemical bond is a model “invented” by scientists to explain stability of compounds. A bond really represents an amount of energy. The bonding model helps us understand and describe molecular structure. It is supported by much research data. However, some data suggests that electrons are delocalized. That is, they are not associated with a particular atom in a molecule.

29 Single bond- one pair of shared electrons Double bond- two pair of shared electrons Triple bond- three pair of shared electrons

30 These values may be slightly different from those in your text. Use the textbook values for your homework.

31 Looking at the chart on the previous slide, what is the relationship between bond length and bond energy? Is there a relationship between number of bonds and bond energy?

32 Bond energies and bond lengths are given in tables on page 374. We can use bond energies to calculate heats of reaction.  H =  D(bonds broken)-  D(bonds formed) 2H 2 + O 2  2H 2 O Ex.  H = [2(432) + 495] –[4(467)] = −509 kJ 2 H−H O=O 4H−O exothermic

33 Bonding Models: Molecular Orbital Model- Electrons occupy orbitals in a molecule in much the same way as they occupy orbitals in atoms. Electrons do not belong to any one atom. -very complex model

34 Localized electron model- molecules are composed if atoms that are bound together by sharing pairs of electrons using the atomic orbitals of the bound atoms traditional model

35 lone pair- pair of electrons localized on an atom (nonbonding) shared pair or bonding pair- electrons found in the space between atoms

36 Lewis structure -shows how the valence electrons are arranged among the atoms in the molecule

37 The most important requirement for the formation of a stable compound is that the atoms achieve noble gas configurations ionic [ Na ] + [Cl] − only valence electrons are included molecular H 2 O H – O - H

38 duet rule- hydrogen forms stable molecules when it shares two electrons H:H -filled valence shell Why does He not form bonds? Its valence orbitals are already filled. octet rule – most elements need 8 electrons to complete their valence shell Cl-Cl

39 Rules for writing Lewis structures 1. Add up the number of valence electrons from all atoms. 2. Use 2 electrons to form a bond between each pair of bound atoms. A dash represents a pair of shared electrons. 3. Arrange the remaining electrons to satisfy the duet rule for H and the octet rule for most others.

40 Ex. H 2 S # of valence electrons: = 8 H − S − H

41 Ex. CO 2 # of valence electrons = = 16 O – C – O This uses 20 electrons! O = C = O

42 NH 3 has 8 valence electrons H N− H H

43 HCN HCN has 10 valence electrons. H−C≡N

44 NO + NO + has −1 = 10 electrons N≡O

45 CO 3 2− Carbonate has = 24 valence electrons. O 2− C O O

46 Exceptions: Boron and beryllium tend to form compounds where the B or Be atom have fewer than 8 electrons around them. BF 3 = 24 valence electrons F B F F Common AP equation: NH 3 + BF 3  H 3 NBF 3

47 C, N, O, F always obey the octet rule.

48 Some elements in Period 3 and beyond exceed the octet rule. Ex. SF 6 S has 12 electrons around it 48 valence electrons F F F F S F F

49 d orbitals are used to accommodate the extra electrons. Elements in the 1st or 2nd period of the table can’t exceed the octet rule because there is no d sublevel. If the octet rule can be exceeded, the extra electrons are placed on the central atom.

50 Examples of exceptions Ex. I 3 −, ClF 3, RnCl 2 I - I - I F F - Cl - F Cl - Rn - Cl

51 Resonance- -occurs when more than one valid Lewis structure can be written for a particular molecule actual structure is an average of all resonance structures -this concept is needed to fit the localized electron model (electrons are really delocalized)

52 Ex. Benzene, C 6 H 6 All bond lengths and angles are the same.

53 Ex. SO 3 All 3 structures are equivalent. The bonds can be thought of as 1 1/3 bonds.

54 Formal Charge -used to determine the most accurate Lewis structure -is the difference between the # of valence electrons on the free atom and the # of valence electrons assigned to the atom in the molecule

55 - atoms try to achieve formal charges as close to zero as possible -any negative formal charges are expected to reside on the most electronegative atoms -Sum of the formal charges must equal the overall charge on the molecule (zero) or ion.

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57 Ex. SO 4 2− O 2− O 2− O S O O S O O O Formal charge only needs to be considered on the AP test if it is specifically asked for.

58 VSEPR-Valence Shell Electron Pair Repulsion -allows us to use electron dot structures to determine molecular shapes -the structure around a given atom is determined primarily by minimizing electron repulsions -bonding and nonbonding pairs of electrons around an atom position themselves as far apart as possible

59 Steps: 1. Draw Lewis structure 2. Count effective electron pairs on central atom (double and triple bonds count as one) 3. Arrange the electron pairs as far apart as possible

60 Shapes AX 2 (A represents central atom, X represents attached atom, E represents unshared electron pair) X – A – X linear 180 o bond angle O=C=O Cl – Be – Cl

61 AX 3 Shape is trigonal planar X X A 120 o bond angle F F X BF 3 B F Any resonance SO 3 structure can be used to O− S = O determine shape. O

62 AX 2 E Shape is bent Bond angle is < 120 o X X A E Ex. SnCl 2 Cl Cl Sn

63 AX4 Shape is tetrahedral Bond angle is o X Ex. CH 4 H X A X H C H X H

64 Figure 8.14 The Molecular Structure of Methane

65 AX 3 E Shape is trigonal pyramidal Bond angle is < o Ex. NH 3 H - N- H H

66 Figure 8.15 The Molecular Structure of NH 3

67 AX 2 E 2 Shape is bent Bond angle is < o Unshared electron pairs repel more than shared pair. Lone pairs require more space than share pairs. E Ex. H 2 O X A X E H – O − H

68 Figure 8.16 The Molecular Structure of H 2 O

69 Figure 8.17 The Bond Angles in the CH 4, NH 3, and H 2 O Molecules

70 AX 5 Shape is trigonal bipyramidal Bond angles are 120 o (equatorial) and 90 o (axial) X X A X X X Ex. PCl 5 Cl Cl P Cl Cl Cl

71 AX 4 E Shape is see-saw Bond angles are <90 o and <120 o X E A X X X Ex. SF 4 34 electrons F S F F F

72 Figure 8.20 Three Possible Arrangements of the Electron Pairs in the I 3 − Ion

73 AX 3 E 2 Shape is T-shaped Bond angle is <90 o X E A X E X Ex. ClF 3 F Cl F F

74 AX 2 E 3 shape is linear bond angle is 180 o X E A E E X Ex. XeF 2 F Xe F

75 Figure 8.19 Possible Electron Pair Arrangements for XeF 4

76 AX 6 shape is octahedral bond angle is 90 o X X X A X X X Ex. SF 6 F F F S F F F

77 AX 5 E Shape is square pyramidal Bond angle is <90 o X X X A X X E Ex. BrF 5 F F F Br F F

78 AX 4 E 2 Shape is square planar. Bond angle is 90 o. E X X A X X E Youtube VSEPR annimation VSEPR OKState


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