Chapter 8 & 9 Concepts of Chemical. Bonding

Transcription

1 Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 8 & 9 Concepts of John D. Bookstaver St. Charles Community College St. Peters, MO 2006, Prentice Hall, Inc.

2 Bonds Three basic types of bonds: Ø Ionic Ø Covalent Ø Metallic

3 Ionic

4 Energetics of Ionic As we saw in the last chapter, it takes 495 kj/mol to remove electrons from sodium.

5 Energetics of Ionic We get 349 kj/mol back by giving electrons to chlorine.

6 Energetics of Ionic But these numbers don t explain why the reaction of sodium metal and chlorine gas to form sodium chloride is so exothermic!

7 Energetics of Ionic There must be a third piece to the puzzle. What is as yet unaccounted for is the electrostatic attraction between the newly formed sodium cation and chloride anion.

8 Lattice Energy This third piece of the puzzle is the lattice energy: The energy required to completely separate a mole of a solid ionic compound into its gaseous ions. The energy associated with electrostatic interactions is governed by Coulomb s law: E el = κ Q 1 Q 2 d

9 Coulomb s Law E el = κ Q 1 Q 2 d Where Q 1 Q 2 are the charges on the particles, d is the distance between their centers and κ is a constant, 8.99 x 10 9 J-m/C 2

10 Lattice Energy Lattice energy, then, increases with the. It also increases with.

11 Example Which substance would you predict to have the greatest lattice energy, AgCl, CuO, or CrN? Explain.

12 Energetics of Ionic By accounting for all three energies (ionization energy, electron affinity, and lattice energy), we can get a good idea of the energetics involved in such a process.

13 Energetics of Ionic These phenomena also helps explain the octet rule. Metals, for instance, tend to stop losing electrons once they attain a noble gas configuration because energy would be expended that cannot be overcome by lattice energies.

14 Covalent In these bonds atoms share electrons. There are several electrostatic interactions in these bonds:

15 Polar Covalent Bonds Although atoms often form compounds by sharing electrons, the electrons are not always shared equally. Fluorine pulls harder on the electrons it shares with hydrogen than hydrogen does. Therefore, the fluorine end of the molecule has more electron density than the hydrogen end.

16 Electronegativity: The ability of atoms in a molecule to attract electrons to itself. On the periodic chart, electronegativity increases as you go Ø from left to right across a row. Ø from the bottom to the top of a column. Ø With the most electronegative element being fluorine.

17 Polar Covalent Bonds When two atoms share electrons unequally, a bond dipole results. The dipole moment, µ, produced by two equal but opposite charges separated by a distance, r, is calculated: µ = Qr It is measured in debyes (D). 1 D = 3.34 x C-m

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19 Polar Covalent Bonds The greater the difference in electronegativity, the more polar is the bond.

20 Example The dipole moment of chlorine monofluoride, ClF (g), is 0.88D. The bond length of the molecule is 1.63Å. a. Which atom is expected to have a negative charge? b. What is the charge on that atom in e?

21 Example Arrange the following bonds in order of increasing polarity: S-Cl, S-Br, Se-Cl or Se-Br. Indicate in each case which atom has the partial negative charge. Which of the bonds above would be expected to be most soluble in water?

22 Lewis Structures Lewis structures are representations of molecules showing all electrons, bonding and nonbonding.

23 Writing Lewis Structures PCl (7) = Find the sum of valence electrons of all atoms in the polyatomic ion or molecule. Ø If it is an anion, add one electron for each negative charge. Ø If it is a cation, subtract one electron for each positive charge.

24 Writing Lewis Structures Keep track of the electrons: 2. The central atom is the least electronegative element that isn t hydrogen. Connect the outer atoms to it by single bonds = 20

25 Writing Lewis Structures 3. Fill the octets of the outer atoms. Keep track of the electrons: 26-6 = = 2

26 Writing Lewis Structures 4. Fill the octet of the central atom. Keep track of the electrons: 26-6 = = 2-2 = 0

27 Writing Lewis Structures 5. If you run out of electrons before the central atom has an octet form multiple bonds until it does.

28 Examples Draw the Lewis structures of CH 2 Cl 2, C 2 H 4, BrO 3-, NO + Recall, isomers are compounds that have the same structures, by different arrangements. Which of these four would be expected to have an isomer? What would its structure be?

29 Writing Lewis Structures Then assign formal charges. Ø For each atom, count the electrons in lone pairs and half the electrons it shares with other atoms. Ø Subtract that from the number of valence electrons for that atom: The difference is its formal charge.

30 Writing Lewis Structures The best Lewis structure Ø is the one with the fewest charges. Ø puts a negative charge on the most electronegative atom.

31 Example There are three possible structures for NCO -. Draw each of these structures and indicate the preferred one. Why is this one preferred?

32 Resonance This is the Lewis structure we would draw for ozone, O

33 Resonance But this is at odds with the true, observed structure of ozone, in which Ø both O O bonds are the same length. Ø both outer oxygens have a charge of -1/2.

34 Resonance One Lewis structure cannot accurately depict a molecule such as ozone. We use multiple structures, resonance structures, to describe the molecule.

35 Resonance Just as green is a synthesis of blue and yellow ozone is a synthesis of these two resonance structures.

36 Resonance In truth, the electrons that form the second C O bond in the double bonds below do not always sit between that C and that O, but rather can move among the two oxygens and the carbon. They are not localized, but rather are delocalized.

37 Resonance The organic compound benzene, C 6 H 6, has two resonance structures. It is commonly depicted as a hexagon with a circle inside to signify the delocalized electrons in the ring.

38 Examples Draw the resonance structures of the carbonate anion.

39 Exceptions to the Octet Rule There are three types of ions or molecules that do not follow the octet rule: Ø Ions or molecules with an odd number of electrons. Ø Ions or molecules with less than an octet. Ø Ions or molecules with more than eight valence electrons (an expanded octet).

40 Odd Number of Electrons Though relatively rare and usually quite unstable and reactive, there are ions and molecules with an odd number of electrons. (i.e.) Chlorine dioxide was the first oxide of chlorine discovered in 1822 and was recently used to kill Anthrax spores released in the U.S. Senate building in October 2001 due to its high reactivity.

41 Fewer Than Eight Electrons Consider BF 3 : Ø Giving boron a filled octet places a negative charge on the boron and a positive charge on fluorine. Ø This would not be an accurate picture of the distribution of electrons in BF 3.

42 Fewer Than Eight Electrons Therefore, structures that put a double bond between boron and fluorine are much less important than the one that leaves boron with only 6 valence electrons.

43 Fewer Than Eight Electrons The lesson is: If filling the octet of the central atom results in a negative charge on the central atom and a positive charge on the more electronegative outer atom, don t fill the octet of the central atom.

44 More Than Eight Electrons The only way PCl 5 can exist is if phosphorus has 10 electrons around it. It is allowed to expand the octet of atoms on the 3rd row or below. Ø Presumably d orbitals in these atoms participate in bonding.

45 More Than Eight Electrons Even though we can draw a Lewis structure for the phosphate ion that has only 8 electrons around the central phosphorus, the better structure puts a double bond between the phosphorus and one of the oxygens.

46 More Than Eight Electrons This eliminates the charge on the phosphorus and the charge on one of the oxygens. The lesson is: When the central atom is on the 3rd row or below and expanding its octet eliminates some formal charges, do so.

47 Summary C, N, O and F always obey the octet rule B, Be and Al are often satisfied with less than an octet Second row elements never exceed the octet rule Third row and beyond can use valence shell expansion to exceed the octet.

48 Molecular Shapes The shape of a molecule plays an important role in its reactivity. By noting the number of bonding and nonbonding electron pairs we can easily predict the shape of the molecule.

49 What Determines the Shape of a Molecule? Simply put, electron pairs, whether they be bonding or nonbonding, repel each other. By assuming the electron pairs are placed as far as possible from each other, we can predict the shape of the molecule.

50 Electron Domains This molecule has four electron domains. We can refer to the electron pairs as electron domains. In a double or triple bond, all electrons shared between those two atoms are on the same side of the central atom; therefore, they count as one electron domain.

51 Valence Shell Electron Pair Repulsion Theory (VSEPR) The best arrangement of a given number of electron domains is the one that minimizes the repulsions among them. See the summary chart

52 Lewis Structures predict the two dimensional arrangement of electrons in a molecule. VSEPR theory allows us to extend the Lewis structure of a molecule to three dimensional space. Neither of these models allows us to understand the actual formation of the covalent bond.

53 Binary Covalent Bonds Occur when two adjacent orbitals overlap. Examples: H 2, HF, F 2 Optimal bonding occurs when there is an equilibrium between bond length and repelling nuclei.

54 More to the Story Beyond Binary Compounds The covalent bonding in polyatomic molecules is more complex. It can be explained with Linus Pauling s hybrid orbitals Let s consider BeF 2

55 Hybrid Orbitals Consider beryllium: Ø In its ground electronic state, it would not be able to form bonds because it has no singly-occupied orbitals.

56 Hybrid Orbitals But if it absorbs the small amount of energy needed to promote an electron from the 2s to the 2p orbital, it can form two bonds.

57 Hybrid Orbitals Mixing the s and p orbitals yields two degenerate orbitals that are hybrids of the two orbitals. Ø These sp hybrid orbitals have two lobes like a p orbital. Ø One of the lobes is larger and more rounded as is the s orbital.

58 Hybrid Orbitals These two degenerate orbitals would align themselves 180 from each other. This is consistent with the observed geometry of beryllium compounds: linear.

59 Hybrid Orbitals With hybrid orbitals the orbital diagram for beryllium would look like this. The sp orbitals are higher in energy than the 1s orbital but lower than the 2p.

60 Hybrid Orbitals Using a similar model for boron leads to

61 Hybrid Orbitals three degenerate sp 2 orbitals.

62 Hybrid Orbitals With carbon we get

63 Hybrid Orbitals four degenerate sp 3 orbitals.

64 Hybrid Orbitals For geometries involving expanded octets on the central atom, we must use d orbitals in our hybrids.

65 Hybrid Orbitals This leads to five degenerate sp 3 d orbitals or six degenerate sp 3 d 2 orbitals.

66 Hybrid Orbitals Once you know the electron-domain geometry, you know the hybridization state of the atom.

67 Predicting Hybrid Orbitals Summary Draw the Lewis structure Determine the electron-domain geometry Specify the hybrid orbitals needed to accommodate the electron pairs based of their arrangement.

68 Covalent Bond Strength Most simply, the strength of a bond is measured by determining how much energy is required to break the bond. This is the bond enthalpy. The bond enthalpy for a Cl Cl bond, D(Cl Cl), is measured to be 242 kj/mol.

69 Average Bond Enthalpies This table lists the average bond enthalpies for many different types of bonds. Average bond enthalpies are positive, because bond breaking is an endothermic process.

70 Average Bond Enthalpies NOTE: These are average bond enthalpies, not absolute bond enthalpies; the C H bonds in methane, CH 4, will be a bit different than the C H bond in chloroform, CHCl 3.

71 Enthalpies of Reaction Yet another way to estimate ΔH for a reaction is to compare the bond enthalpies of bonds broken to the bond enthalpies of the new bonds formed. In other words, ΔH rxn = Σ(bond enthalpies of bonds broken) - Σ(bond enthalpies of bonds formed)

72 Enthalpies of Reaction CH 4 (g) + Cl 2 (g) CH 3 Cl(g) + HCl(g) In this example, one C H bond and one Cl Cl bond are broken; one C Cl and one H Cl bond are formed.

73 Enthalpies of Reaction So, ΔH rxn = [D(C H) + D(Cl Cl) - [D(C Cl) + D(H Cl) = [(413 kj) + (242 kj)] - [(328 kj) + (431 kj)] = (655 kj) - (759 kj) = -104 kj

74 Example Use the bond enthalpies on page 301 to calculate the heat of combustion of methane gas with O 2 to produce water vapor and carbon dioxide gas.

75 Bond Enthalpy and Bond Length We can also measure an average bond length for different bond types. As the number of bonds between two atoms increases, the bond length decreases.

76 Sample Integrative Exercise Phosgene, a substance used in poisonous gas warfare in World War I, is so named because it was first prepared by the action of sunlight on a mixture of carbon monoxide and chlorine gases. Its name comes from the Greek words phos (light) and genes (born of). Phosgene has the following elemental composition: 12.14% C, 16.17% O, and 71.69% Cl by mass. Its molar mass is 98.9 g/mol. (a) Determine the molecular formula of this compound. (b) Draw three Lewis structures for the molecule that satisfy the octet rule for each atom. (The Cl and O atoms bond to C.) (c) Using formal charges, determine which Lewis structure is the most important one. (d) Using average bond enthalpies, estimate ΔH for the formation of gaseous phosgene from CO(g) and Cl2(g).