Models Of Chemical Bonding. Chapter Nine. AP Chemistry

Transcription

1 Models Of Chemical Bonding Chapter Nine AP Chemistry

2 Q. Why do atoms bond? A. To lower the potential energy between positive and negative particles. Atoms like humans seek to become more stable.

3 Metals - Nonmetals Difference based upon their properties of 1. ENC - Z eff 2. Number of valence electrons 3. Atomic size 4. IE 5. EA

4 General comparison of metals and nonmetals.

5 Three Types of Bonds 1. Metal with nonmetal: electron transfer - ionic bonding - Group 1-2A with Group 7A -upper 6A - delocalized electrons

6 2. Nonmetal with nonmetal: electron sharing - covalent bonding - electrons are localized

7 Octet rule when atoms lose, gain or share electrons they try to get 8 electrons (four electron pairs). Nearly all main group monatomic ions have a filled outer shell of 2 or 8 electrons.

8 3. Metal with metal: electron pooling - metallic bonding - delocalized electrons

9 Identify the Type of Bond ClO 2, MgCl 2, O 2, Fe, CaO C I C M I

10 Electron Dot (Lewis) Diagrams Primary concern is for s and p electrons in the outer most energy level. These electrons are the ones involved in chemical reactions. Outer level s and p electrons are known as valence electrons.

11 Drawing an Electron Dot Diagram Step 1: Write the element symbol. It stands for the nucleus and inner electrons. Cl

12 Step 2: Determine which electrons are the valence electrons. Outer most s and p electrons. Cl 1s 2 2s 2 2p 6 3s 2 3p 5 Outer level is n = 3 and there are 7 valence electrons.

13 Step 3: Draw the long hand electron configuration for the valence electrons. 3s 3p

14 Note: There are 3 paired e - and 1 unpaired e -. Each side of the element symbol represents one of the 4 possible orbits. Show paired e - as pairs. Show unpaired e - as unpaired.

15 x x x x Cl x x x

16 For metals the number of unpaired dots represents the maximum number of electrons lost. For nonmetals the number of unpaired dots represents the number of electrons gained or paired, or the number of bonds that may be formed when making a covalent compound.

17 Draw Electron Dot Diagrams For: Bromine: (35 e - ) {Ar}4s 2 3d 10 4p 5

18 x x x x Br x x x Elements in the same group have the same electron dot diagrams.

19 Phosphorus: (15 e - ) {Ne}3s 2 3p 3 3s 3p

20 x x x P x x

21 Oxygen: (8 e - ) {He}2s 2 2p 4 2s 2p

22 x x x x O x x

23 Lewis electron-dot symbols for elements in Periods 2 and 3

24 Bonding electrons (Bond Pairs) vs nonbonding electrons (Lone Pairs) HCl H has one valence electron Cl has seven valence electrons. H Cl

25 Bond Pairs H Cl Lone Pairs

26 Lewis Electron Dot for Na and Cl

27 The Ionic Compound NaCl Another text s representation = Lone Pairs

28 What is the Lewis structure of PBr 3? Sample problems Br P Br Br P Br Br Br What is the Lewis structure of N 2 H 2? H N N H H N N H H N N H

29 What is the Lewis structure of CH 3 OH? Sample problems Sample problems What is the Lewis structure of NO 2-1? O C H H H H O C H H H H N O O N O O N O O N O O N O O

30 Failures of the Lewis Model A number of molecules with odd numbers of e - sexist (no octet), e.g. NO. An atom may not have enough e - sto complete its octet without having ridiculous formal charges, e.g. BF 3. A central atom may clearly have more than 8 e - s, e.g. SF 6. O 2 is paramagnetic!

31 What is the Lewis structure of O 2? O O O + O O O O O paramagnetic: unpaired e - s 31

32 Depicting Ion Formation Depict the formation of Na + and O 2- ions from the atoms, and determine the formula of the compound. O Na 3s 2s 3p 2p O 2-2s 2p Na 3s 3p Na. Na.. + : O:. 2 Na + 2Na + + : O 2- : : :

33 Three ways to represent the formation of Li + and F - through electron transfer. Electron configurations Li 1s 2 2s 1 + F 1s 2 2s 2 2p 5 Li + 1s 2 + F - 1s 2 2s 2 2p 6 Li Orbital diagrams 1s 2s 2p Li + 1s 2s 2p + F + F - 1s 2s 2p 1s 2s 2p Lewis electron-dot symbols. Li. + : F : Li+ + : F : - : : :

34 Energy Considerations and Lattice Energy (Born-Haber Cycle) Electron transfer is an energy adsorbing process. It is the energy released when the ions come together that provides the motivation for the reaction to occur.

35 Usually several steps in the formation of a bond. 1. atomization (gasification) (s g) 2. molecules to atoms (dissociation) 3. IE ( H ) IE 4. EA ( H ) EA 5. Lattice Energy H symbols Lattice Energy is hard to determine.

36 Adding the energy of these steps gives the energy associated with the formation of one mole of the product. H o 1 + Ho 2 + Ho 3 + Ho 4 + Ho 5 = H o Total H o Total = Ho f (Heat of Formation) Note: Another name for heat is ENTHALPY

37 Steps 1-3 are endothermic (taking energy in) and therefore have + H o values, while steps 4 and 5 are exothermic (releasing energy) and therefore have - H o values.

38 The Lattice Energy (Step 5) is highly exothermic (negative) and dominates the multiple step reaction. If this were not so, the reaction would be endothermic and therefore probably not take place.

39 Lattice Energies Calculated through the Born-Haber Cycle Step 1) Li (S) Li (g) H o step 1 = 161 kj Step 2) Converting F 2 into 2 F atoms 1/2 F 2 (g) F (g) H0 step 2 = 1/2 Bond Energy (BE) of F 2 = 1/2 ( 159 kj) = 79.5 kj Step 3) Removing the 2s electron for Li to Li + Li (g) Li + (g) + e - H 0 step 3 = IE 1 = 520 kj Step 4) Adding an electron to F to form F - F (g) + e - F - (g) H 0 step 4 = EA = kj Step 5) Formation of the crystalline solid from the gaseous ions Li + (g) + F - (g) LiF (s) H 0 step 5 = H 0 LiF (Lattice Energy) We know the Energy change for the formation Reaction: Li (s) + 1/2 F 2 (g) LiF (s) H 0 overall = H 0 f = -617 kj

40 The Born-Haber cycle for lithium fluoride 40

41 We can calculate the Lattice Energy ( step 5 ) from Hess s Law: H 0 f = -617 kj = H 0 step 1 + H 0 step 2 + H 0 step 3 + H 0 step 4 + H 0 LiF H 0 LiF = H f - [ H 0 step 1 + H 0 step 2 + H 0 step 3 + H 0 step 4 ] H 0 LiF = -617 kj/mole - [ 161 kj kj/mole kj/mole + ( -328 kj/mole)] H 0 LiF = kj/mole

42 Born-Haber example of lithium fluoride 1) Atomization enthalpy of lithium 2) Ionization enthalpy of lithium 3) Atomization enthalpy of fluorine 4) Electron affinity of fluorine 5) Lattice enthalpy 42

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44 Lattice Energies for MgO Mg (s) Mg (g) H 0 atom = 148 kj/mol Mg (g) Mg 2+ (g) + 2e - H 0 = IE 1 + IE 2 = 738 kj kj H 0 = 2188 kj 1/2 O 2 (g) O (g) H 0 = 1/2 bond Energy of O 2 H 0 = 1/2 x 498 kj/mol = 249 kj O (g) + e - O - (g) H 0 = EA 1 = -141 kj O - (g) + e - O 2- (g) H 0 = EA 2 = 878 kj O (g) + 2 e - O 2- (g) H 0 = EA 1 + EA 2 = 737 kj With all of these endothermic steps, it is the enormous lattice energy ( H 0 MgO = kj/mole ) that more than compensates for the endothermic steps to assure that MgO is formed every time that Mg metal is burned in Air. [ H 0 f of MgO (s) = -601 kj/mole]

45 Periodic Trends for Lattice Energy Coulomb s Law Electrostatic Force α Charge A x Charge B Distance 2

46 Energy = F x D Therefore: Electrostatic Energy α Charge A x Charge B Distance

47 Electrostatic Energy α Cation Charge x Anion Charge (Cation radius + anion radius) α H 0 lattice

48 Trends in Lattice Energy

49 Small ions produce lattice energies that are greater than large ions of the same charge! Ions with greater charge will produce greater lattice energy than those of lower charge. E α charge and E α 1/ size

50 EXAMPLE MgO has ions about the same size as LiF.but the lattice energy is approximately 4 times as great as LiF because the charge on each ion is +2 and 2 respectively.

51 Which has greater lattice energy? (a) CaS or BaS Smaller (b) NaCl or NaF Smaller

52 Properties of Ionic Compounds Ionic Solids are: Brittle because of the tremendous lattice energy of ionic solids, they require tremendous forces to separate them. Hence they will shatter rather than deform (dent).

53 Electrostatic forces and the reason ionic compounds crack.

54 Do not conduct electricity (same reason as above) unless the ions are made mobile by fusion (melting) or in a water solution.

55 Electrical Conductance and Ion Mobility Solid Molten Dissolved in water

56 Melting and Boiling Points of Some Ionic Compounds Compound MP ( 0 C) BP ( 0 C) CsBr NaI MgCl 2 KBr CaCl 2 NaCl LiF KF MgO > Strong attractions = high MP and BP.

57 Vaporizing an Ionic Compound. Ionic attraction is so strong that the entire ion breaks away when vaporizing.

58 Covalent Bonding Sharing electrons is the way that most atoms interact chemically. Each nucleus attracts the other atom's electrons.

59 At some point, the maximum attraction is reached. Any distance closer would mean greater or repulsion. When optimization occurs a covalent bond is formed. Full outer shell or octet arrangement is achieved. Shared pair of electrons or bonded pair between the nuclei.

60 Covalent Bond Formation in H 2.

61 One shared pair = single bond = bond order = 1 H x x H

62 Two shared pair = double bond = bond order = 2 xx O xx xx x x x x xx O

63 Three shared pair = triple bond = bond order = 3 x x N x x x xx x N x x

64 Properties of Covalent Bonds A. Bond energy B. Bond length Properties depend upon: nucleus electron attraction and nucleus-nucleus and electronelectron repulsions.

65 The attractive and repulsive forces in covalent bonding.

66 Bond Energy Bond energy (bond enthalpy - BE) is defined as the amount of energy required to separate 1 mole of gaseous atoms. AB (g) + energy A (g) + B (g) H reactant bonds broken = + value (endothermic)

67 A (g) + B (g) AB (g) + energy H product bonds formed = - value (exothermic) H reactant bonds broken = H product bonds formed

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69 Bonds Length Bond length is equal to the distance between nuclei of the bonded atoms. Bigger atoms = greater bond length.

70 Bond Length and Covalent Radius. Internuclear distance (bond length) Covalent radius

71 Internuclear distance (bond length) Covalent radius Internuclear distance (bond length) Covalent radius Internuclear distance (bond length) Covalent radius

72 Greater bond order = shorter bond length = greater bond energy. Triple bonds are stronger than double which are stronger than single! Triple bonds are shorter than double which are shorter than single.

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74 The Relation of Bond Order, Bond Length and Bond Energy Bond Bond Order Average Bond Average Bond Length (pm) Energy (kj/mol) C O C O C O C C C C C C N N N N N N

75 Comparing Bond Length and Bond Strength Using the periodic table, rank the bonds in each set in order of decreasing bond length and bond strength: (a) S - F, S - Br, S - Cl (b) C = O, C - O, C O (a) The bond order is one for all and sulfur is bonded to halogens; bond length should increase and bond strength should decrease with increasing atomic radius. (b) The same two atoms are bonded but the bond order changes; bond length decreases as bond order increases while bond strength increases as bond order increases. (a) Atomic size increases going down a group. Bond length: S - Br > S - Cl > S - F Bond strength: S - F > S - Cl > S - Br (b) Using bond orders we get: Bond length: C - O > C = O > C O Bond strength: C O > C = O > C - O

76 Properties of Various Covalent Compounds 1. Molecular covalent cpds. have strong internal covalent bonds not easily broken. but the attractions between (intermolecular forces) the molecules are weak...these are not covalent bonds.

77 hence the MP and BP are very low. (H 2, O 2, Pentane - C 5 H 12 )

78 hence the MP and BP are very low. (H 2, O 2, Pentane - C 5 H 12 ) Strong covalent bonding forces within molecules Weak intermolecular forces between molecules

79 2. Network covalent compounds. No separate molecules Entire structure made of covalent bonds... therefore strong forces throughout. hence the MP and BP are very high. (Diamond)

80 Covalent Bonds of Network Covalent Solids

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82 Covalent cpds. are poor conductors of electricity - even when melted or dissolved in water - because their electrons are localized. good insulators.

83 Electronegativity and Polarity Electronegativity (EN) the ability of a bonded atom to attract a shared pair of electrons. Linus Pauling developed an EN scale for the elements.

84 Bond energy of H = 432 kj/mole 2 and F = 159 kj/mole should 2 have average of 296 kj/mole for the HF molecule bond energy. However the real bond energy is 565 kj/mole. Why?

85 Linus said the F must attract the shared pair of electrons more than the H setting up a partially negative area on the F and a slightly + area on the H. The result would be an electrostatic attraction creating more pull between the two atoms and therefore a stronger HF bond (565 kj/mole)

86 Electronegativity Trends Electronegativity (EN) increases across a period. The EN is inversely proportional to the atomic size.

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88 The Periodic Table of the Elements He Ne Ar Kr Electronegativity Ce Pr Nd Pm Th Pa U Np No Lr 2.2 Yb Lu Xe Rn

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90 Electronegativity and Atomic Size.

91 EN and Oxidation Number 1. The more EN atom in a bond is assigned all the shared electrons, the less EN atom is assigned none. 2. Each atom in a bond is assigned all of its unshared electrons.

92 The oxidation number is determined by: O.N. = # of valence e - - (# of shared e - + # of unshared e - ) HCl EN Cl > EN H Cl has 7 valence e - but is assigned 8 (2 shared + 6 unshared). Hence, O.N. = 7 8 = -1.

93 H has 1 valence e - and is assigned none. Hence its O.N. = 1 0 = +1. H 2 O For O: O.N. = 6 (4 + 4) = -2 For H: O.N. = 1- (0) = +1

94 Polarity Unequal sharing of e - results in a bond with a partial negative charge directed toward the element with greater EN and a partial positive charge directed toward the element with the lower EN. The unequal shared pair is called the POLAR Bond.

95 Polar covalent bonds are represented by a polar arrow that points to the negative pole. Polar arrow = xx H F xx x x

96 Another way to show bond polarity (Greek symbols) The bonding electrons spend more time on F than on H.

97 The bond in H-H and F-F, for example. are called nonpolar covalent since the atoms are identical. The EN of each atom is the same so the electrons are shared equally. There is no polarity.

98 Determining Bond Polarity from Electronegativity Values (a) Indicate the polarity of the following bonds with a polarity arrow: O - H, O - Cl, C - N, P - N, N - S, C - Br, As - S (b) Rank those bonds in order of increasing polarity. a) the EN of O = 3.5 and of H = 2.1: O - H the EN of O = 3.5 and of Cl = 3.0: O - Cl the EN of C = 2.5 and of P = 2.1: C - P the EN of P = 2.1 and of N = 3.0: P - N the EN of N = 3.0 and of S = 2.1: N - S the EN of C = 2.5 and of Br = 2.8: C - Br the EN of As = 2.0 and of O = 3.5: As - O b) C - Br < C - P < O - Cl < P - N < N - S < O - H < As - O 0.3 < 0.4 < 0.5 < 0.9 < 0.9 < 1.4 < 1.5

99 Determining Bond Polarity from EN Values (a) Use a polar arrow to indicate the polarity of each bond: N-H, F-N, I-Cl. (b) Rank the following bonds in order of increasing polarity: H-N, H-O, H-C. (a) The EN of N = 3.0, H = 2.1; F = 4.0; I = 2.5, Cl = 3.0 N - H F - N I - Cl (b) The order of increasing EN is C < N < O; all have an EN larger than that of H. H-C < H-N < H-O

100 Boundary ranges for classifying ionic character of chemical bonds. EN

101 Therefore: Cl-Cl = = 0.0 = (Nonpolar) H-O = = 1.4 = (Polar) Na-Cl = = 2.1 = (Ionic)

102 Another way to look at electronegativity difference If difference is 0.0 then the cpd has 0 % ionic character If the difference is 1.7 it has 50% ionic character If the difference is 3.3 it has 100% ionic character.

103 Percent Ionic Character of Electronegativity Difference ( EN).

104

105 The symmetry of molecules plays an important part in determining the final outcome of their bond type. CCl 4 has four slightly polar bonds and yet is a nonpolar molecule because the molecule is symmetrical.

106 What about these molecules?

107 The Continuum of Bonding Across a Period NaCl MgCl AlCl SiCl PCl S Cl Cl Highly ionic Polar NP covalent

108 Properties of the Period 3 Chlorides

109 Metallic Bonding What holds atoms together in a piece of metal? Electron Sea Model of Metallic Bonding: All metal atoms in the sample contribute their valence electrons to form a sea of electrons that is delocalized throughout the substance.

110 Metal atoms and their core electrons are submerged in this sea of electrons in an orderly way.

111 Atoms are not held in place as tightly/rigidly as they are in ionic substances. Atoms are not linked to another atom either like they are in covalent substances. Atoms are held together because of the mutual attraction of the metal nuclei for the mobile electrons.

112 Metal Characteristics: 1. Most elements are metals. 2. Most metals are solids. Except Hg which is a liquid. 3. Color: Most are silver-gray Au is yellow (gold) Cu is reddish brown

113 4. reflects light when polished. 5. Malleable: can be hammered into shapes. 6. Ductile: can be stretched into wires. 7. conducts electrical current very well in both L and S phase.

114 8. conducts heat well in both L and S phase. 9. most moderate MP and high BP. The Unusually Low Melting Point of Gallium

115 Melting and Boiling Points of Some Metals Element MP( 0 C) BP( 0 C) Lithium (Li) Tin (Sn) Aluminum (Al) Barium (Ba) Silver (Ag) Copper (Cu) Uranium (U)

116 Mobile cations slide by each other.

117 Properties of Metals Explained Melting only requires moderate temperatures because the attractions between the mobile cations and the electrons need not be broken. Boiling requires each cation and its electrons to break away from the others. Hence, a very high BP.

118 Trends in MP Alkaline Earth Metals have a higher MP than the Alkali Metals. Alkaline Earth Metals have 2 valence electrons and form +2 cations.

119 Greater attraction between these cations and double valence electrons provides a stronger metallic bond than occurs in the Alkali Metals. Metals are good conductors of electricity because of the mobile electrons.

120 Metals are good conductors of heat because of the delocalized electrons. The electrons disperse the heat from your hand more quickly than localized electron pairs in covalently bonded substances.

121 Melting points of the Group 1A(1) and Group 2A(2) elements.

122 The End