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Chapter Nine: COVALENT BONDING: ORBITALS. Assignment 1-85 題中每 5 題裡任選 1-2 題 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide.

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Presentation on theme: "Chapter Nine: COVALENT BONDING: ORBITALS. Assignment 1-85 題中每 5 題裡任選 1-2 題 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide."— Presentation transcript:

1 Chapter Nine: COVALENT BONDING: ORBITALS

2 Assignment 1-85 題中每 5 題裡任選 1-2 題 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 2

3 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 3 Exercise Draw the Lewis structure for methane, CH 4. What is the shape of a methane molecule? What are the bond angles? 9.1

4 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 4 Concept Check What is the valence electron configuration of a carbon atom? Why can’t the bonding orbitals for methane be formed by an overlap of atomic orbitals? 9.1

5 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 5 Bonding in Methane Assume that the carbon atom has four equivalent atomic orbitals, arranged tetrahedrally. 9.1

6 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 6 Hybridization Mixing of the native atomic orbitals to form special orbitals for bonding. 9.1

7 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 7 sp 3 Hybridization Combination of one s and three p orbitals. Whenever a set of equivalent tetrahedral atomic orbitals is required by an atom, the localized electron model assumes that the atom adopts a set of sp 3 orbitals; the atom becomes sp 3 hybridized. 9.1

8 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 8 An Energy-Level Diagram Showing the Formation of Four sp 3 Orbitals 9.1

9 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 9 The Formation of sp 3 Hybrid Orbitals 9.1

10 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 10 Tetrahedral Set of Four sp 3 Orbitals 9.1

11 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 11 Exercise Draw the Lewis structure for O 2. What is the shape of an oxygen molecule? What is the approximate angle between lone pairs of electrons on each of the oxygen atoms? 9.1

12 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 12 Concept Check Why can’t sp 3 hybridization account for the oxygen molecule? 9.1

13 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 13 sp 2 Hybridization Combination of one s and two p orbitals. Gives a trigonal planar arrangement of atomic orbitals. One p orbital is not used. –Oriented perpendicular to the plane of the sp 2 orbitals 9.1

14 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 14 Sigma (σ) Bond Electron pair is shared in an area centered on a line running between the atoms. 9.1

15 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 15 Pi (π) Bond Forms double and triple bonds by sharing electron pair(s) in the space above and below the σ bond. Uses the unhybridized p orbitals. 9.1

16 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 16 An Orbital Energy-Level Diagram for sp 2 Hybridization 9.1

17 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 17 The Hybridization of the s, p x, and p y Atomic Orbitals 9.1

18 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 18 Formation of C=C Double Bond in Ethylene 9.1

19 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 19 Exercise Draw the Lewis structure for CO 2. What is the shape of a carbon dioxide molecule? What are the bond angles? 9.1

20 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 20 sp Hybridization Combination of one s and one p orbital. Gives a linear arrangement of atomic orbitals. Two p orbitals are not used. –Needed to form the π bonds 9.1

21 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 21 The Orbital Energy-Level Diagram for the Formation of sp Hybrid Orbitals on Carbon 9.1

22 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 22 When One s Orbital and One p Orbital are Hybridized, a Set of Two sp Orbitals Oriented at 180 Degrees Results 9.1

23 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 23 The Orbitals for CO 2 9.1

24 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 24 Exercise Draw the Lewis structure for PCl 5. What is the shape of a phosphorus pentachloride molecule? What are the bond angles? 9.1

25 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 25 dsp 3 Hybridization Combination of one d, one s, and three p orbitals. Gives a trigonal bipyramidal arrangement of five equivalent hybrid orbitals. 9.1

26 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 26 The Orbitals Used to Form the Bonds in PCl 5 9.1

27 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 27 Exercise Draw the Lewis structure for XeF 4. What is the shape of a xenon tetrafluoride molecule? What are the bond angles? 9.1

28 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 28 d 2 sp 3 Hybridization Combination of two d, one s, and three p orbitals. Gives an octahedral arrangement of six equivalent hybrid orbitals. 9.1

29 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 29 How is the Xenon Atom in XeF 4 Hybridized? 9.1

30 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 30 Concept Check Draw the Lewis structure for HCN. Which hybrid orbitals are used? Draw HCN: –Showing all bonds between atoms –Labeling each bond as  or  9.1

31 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 31 Concept Check Determine the bond angle and expected hybridization of the central atom for each of the following molecules: NH 3 SO 2 KrF 2 CO 2 ICl 5 9.1

32 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 32 The Localized Electron Model Draw the Lewis structure(s). Determine the arrangement of electron pairs (VSEPR model). Specify the necessary hybrid orbitals. 9.1

33 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 33 Molecular Orbital Model The electron probability of both molecular orbitals is centered along the line passing through the two nuclei. –Sigma (σ) molecular orbitals In the molecule only the molecular orbitals are available for occupation by electrons. 9.2

34 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 34 Molecular Orbital Model ( continued ) MO 1 is lower in energy than the s orbitals of free atoms, while MO 2 is higher in energy than the s orbitals. –Bonding molecular orbital – lower in energy –Antibonding molecular orbital – higher in energy 9.2

35 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 35 Molecular Orbital Model ( continued ) The molecular orbital model produces electron distributions and energies that agree with our basic ideas of bonding. The labels on molecular orbitals indicate their symmetry (shape), the parent atomic orbitals, and whether they are bonding or antibonding. 9.2

36 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 36 Molecular Orbital Model ( continued ) Molecular electron configurations can be written similar to atomic electron configurations. Each molecular orbital can hold 2 electrons with opposite spins. Orbitals are conserved. 9.2

37 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 37 Bonding in H 2 9.2

38 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 38 Sigma Bonding and Antibonding Orbitals 9.2

39 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 39 Bond Order Larger bond order means greater bond strength. B.O. = (# of bonding e - – # of antibonding e - )/2 9.2

40 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 40 Homonuclear Diatomic Molecules Composed of 2 identical atoms. Only the valence orbitals of the atoms contribute significantly to the molecular orbitals of a particular molecule. 9.3

41 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 41 Pi Bonding and Antibonding Orbitals 9.3

42 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 42 Magnetic Properties of Liquid Nitrogen and Oxygen

43 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 43 Paramagnetism Paramagnetism – substance is attracted into the inducing magnetic field. –Unpaired electrons (O 2 ) Diamagnetism – substance is repelled from the inducing magnetic field. –Paired electrons (N 2 ) 9.3

44 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 44 Apparatus Used to Measure the Paramagnetism of a Sample

45 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 45 Molecular Orbital Summary of Second Row Diatomic Molecules 9.3

46 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 46 Heteronuclear Diatomic Molecules Composed of 2 different atoms. 9.4

47 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 47 Orbital Energy-Level Diagram for the HF Molecule 9.4

48 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 48 The Electron Probability Distribution in the Bonding Molecular Orbital of the HF Molecule 9.4

49 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 49 The Sigma System for Benzene 9.5

50 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 50 The Pi System for Benzene 9.5

51 Copyright © Houghton Mifflin Company. All rights reserved.Chapter 9 | Slide 51 Pi Bonding in the Nitrate Ion 9.5


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