Chapter 11 Theories of Covalent Bonding
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1 Chapter 11 Theories of Covalent Bonding 11-1
2 Theories of Covalent Bonding 11.1 Valence Bond (VB) Theory and Orbital Hybridization 11.2 Modes of Orbital Overlap and the Types of Covalent Bonds 11.3 Molecular Orbital (MO) Theory and Electron Delocalization 11-2
3 Valence Bond (VB) Theory The basic principle of VB theory: A covalent bond forms when the orbitals of two atoms overlap and a pair of electrons occupy the overlap region. The space formed by the overlapping orbitals can accommodate a maximum of two electrons and these electrons must have opposite (paired) spins. The greater the orbital overlap, the stronger the bond. Extent of orbital overlap depends on orbital shape and direction. 11-3
4 Figure 11.1 Orbital overlap and spin pairing in H 2. A covalent bond results from the overlap of orbitals from two atoms. The shared space is occupied by two electrons, which have opposite spins. 11-4
5 Figure 11.2 Orbital orientation and maximum overlap. Hydrogen fluoride, HF. Fluorine, F 2. The greater the extent of orbital overlap, the stronger the bond. 11-5
6 VB Theory and Orbital Hybridization The orbitals that form when bonding occurs are different from the atomic orbitals in the isolated atoms. If no change occurred, we could not account for the molecular shapes that are observed. Atomic orbitals mix or hybridize when bonding occurs to form hybrid orbitals. The spatial orientation of these hybrid orbitals correspond with observed molecular shapes. 11-6
7 Features of Hybrid Orbitals The number of hybrid orbitals formed equals the number of atomic orbitals mixed. The type of hybrid orbitals formed varies with the types of atomic orbitals mixed. The shape and orientation of a hybrid orbital maximizes overlap with the other atom in the bond. 11-7
8 Figure 11.3 Formation and orientation of sp hybrid orbitals and the bonding in BeCl 2. atomic orbitals hybrid orbitals One 2s and one 2p atomic orbital mix to form two sp hybrid orbitals. orbital box diagrams 11-8
9 Figure 11.3 continued box diagram with orbital contours 11-9 Overlap of Be and Cl orbitals to form BeCl 2.
10 Figure 11.4 The sp 2 hybrid orbitals in BF 3. Mixing one s and two p orbitals gives three sp 2 hybrid orbitals. The third 2p orbital remains unhybridized
11 Figure 11.4 continued The three sp 2 orbitals point to the corners of an equilateral triangle, their axes 120 apart. Each half-filled sp 2 orbital overlaps with the half-filled 2p orbital of a F atom
12 Figure 11.5 The sp 3 hybrid orbitals in CH 4. The four sp 3 orbitals adopt a tetrahedral shape
13 Figure 11.6 The sp 3 hybrid orbitals in NH 3. The N lone pair occupies an sp 3 hybrid orbital, giving a trigonal pyramidal shape
14 Figure 11.6 continued The sp 3 hybrid orbitals in H 2 O. The O lone pairs occupy sp 3 hybrid orbitals, giving a bent shape
15 Figure 11.7 The sp 3 d hybrid orbitals in PCl 5. The formation of more than four bonding orbitals requires d orbital involvement in hybridization
16 11-16 Figure 11.8 The sp 3 d 2 hybrid orbitals in SF 6.
17 Table 11.1 Composition and Orientation of Hybrid Orbitals
18 Figure 11.9 From molecular formula to hybrid orbitals. Molecular Formula Step 1 Figure 10.1 Lewis structure Step 2 Figure Molecular shape and e - group arrangement Step 3 Table 11.1 Hybrid orbitals 11-18
19 Sample Problem 11.1 Postulating Hybrid Orbitals in a Molecule PROBLEM: Use partial orbital diagrams to describe how mixing of the atomic orbitals of the central atom(s) leads to hybrid orbitals in each of the following: (a) Methanol, CH 3 OH (b) Sulfur tetrafluoride, SF 4 PLAN: We use the molecular formula to draw the Lewis structure and determine the electron-group arrangement around each central atom. We then postulate the type of hybrid orbitals required and write a partial orbital diagram. SOLUTION: (a) CH 3 OH The electron-group arrangement is tetrahedral around both the C and the O atom
20 Sample Problem p isolated C atom 2s C has four half-filled sp 3 orbitals. sp 3 hybridized C atom 2s 2p isolated O atom The O atom has two half-filled sp 3 orbitals and two filled with lone pairs. sp 3 hybridized O atom 11-20
21 Sample Problem 11.1 (a) SF 4 The electron-group arrangement is trigonal bipyramidal, so the central S atom is sp 3 d hybridized. 3s 3d 3p isolated S atom 3d sp 3 d hybridized S atom 11-21
22 Limitations of the Hybridization Model Hybridization is not always consistent with observed molecular shapes. This is particularly true for the bonding of larger atoms. The bond angle in H 2 S is closer to the angle between unhybridized p orbitals. d-orbitals do not hybridize effectively with s and p orbitals, which are much lower in energy and more stable
23 Types of Covalent Bonds A sigma (σ) bond is formed by end-to-end overlap of orbitals. All single bonds are σ bonds. A pi (π) bond is formed by sideways overlap of orbitals. A π bond is weaker than a σ bond because sideways overlap is less effective than end-to-end overlap. A double bond consists of one σ bond and one π bond 11-23
24 Figure The σ bonds in ethane (C 2 H 6 ). Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. both C are sp 3 hybridized σ bond formed by s-sp 3 overlap End-to-end sp 3 -sp 3 overlap to form a σ bond A σ bond is cylindrically symmetrical, with its highest electron density along the bond axis
25 Figure continued There is relatively even distribution of electron density over all σ bonds
26 Figure The σ and π bonds in ethylene (C 2 H 4 ). Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. unhybridized 2p orbitals A π bond has two regions of electron density
27 Figure The σ and π bonds in acetylene (C 2 H 2 ). Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Each C is sp hybridized and has two unhybridized p orbitals
28 Figure Electron density and bond order in ethane, ethylene, and acetylene. A double bond is less than twice as strong as a single bond, because a π bond is weaker than a σ bond. However, in terms of bond order, a single bond has BO = 1, a double bond has BO = 2, and a triple bond has BO =
29 Sample Problem 11.2 Describing the Types of Bonds in Molecules PROBLEM: Describe the types of bonds and orbitals in acetone, (CH 3 ) 2 CO. PLAN: We use the Lewis structures to determine the arrangement of groups and shape at each central atom. We postulate the hybrid orbitals, taking note of the multiple bonds present. SOLUTION: sp 2 sp 3 sp 2 sp
30 Sample Problem 11.2 The sp 3 hybridized C atoms form σ bonds using sp 3 hybrid orbitals. The sp 2 hybridized C and O atoms form σ bonds using sp 2 hybrid orbitals, and the π bond of the C=O double bond is formed using p orbitals. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. σ bonds π bond (shown with molecule rotated 90 )
31 Figure Restricted rotation around a π bond. cis-1,2-dichloroethylene trans-1,2-dichloroethylene 11-31
32 Molecular Orbital (MO) Theory The combination of orbitals to form bonds is viewed as the combination of wave functions. Atomic wave functions (AOs) combine to form molecular wave functions (MOs). Addition of AOs forms a bonding MO, which has a region of high electron density between the nuclei. Subtraction of AOs forms an antibonding MO, which has a node, or region of zero electron density, between the nuclei
33 Figure An analogy between light waves and atomic wave functions. Amplitudes of wave functions added Amplitudes of wave functions subtracted
34 Figure Contours and energies of H 2 bonding and antibonding MOs. The bonding MO is lower in energy and the antibonding MO is higher in energy than the AOs that combined to form them
35 Molecular Orbital Diagrams An MO diagram, just like an atomic orbital diagram, shows the relative energy and number of electrons in each MO. The MO diagram also shows the AOs from which each MO is formed. Bond order is calculated as follows: ½[(# of e - in bonding MO) (# of e - in antibonding MO)] 11-35
36 Figure MO diagram for H 2. H 2 bond order = ½ (2 0) =
37 Electrons in Molecular Orbitals Electrons are placed in MOs just as they are in AOs. MOs are filled in order of increasing energy. An MO can hold a maximum of 2 e - with opposite spins. Orbitals of equal energy are half-filled, with spins parallel, before pairing spins. A molecular electron configuration shows the type of MO and the number of e - each contains. For H 2 the configuration is (σ 1s )
38 Figure MO diagram for He 2+ and He 2. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. He 2+ bond order = ½ He 2 bond order = 0 (σ 1s ) 2 * (σ 1s ) 1 (σ 1s ) 2 * (σ 1s )
39 Sample Problem 11.3 Predicting Stability of Species Using MO Diagrams PROBLEM: Use MO diagrams to find bond orders and predict whether H 2+ and H 2 exist. If either exists, write its electron configuration. PLAN: SOLUTION: Since the 1s AOs form the MOs, the MO diagrams are similar to the one for H 2. We find the number of electrons in each species and distribute them one at a time to the MOs following the rules for orbital filling. We calculate the bond order and predict stability. H 2+ has one electron to place in its MOs while H 2- has three electrons to place
40 Sample Problem 11.3 For H 2+, the bond order is ½(1 0) = ½; so we predict that H 2+ exists. The configuration is (σ 1s ) 1. For H 2-, the bond order is ½(2 1) = ½; so we predict that H 2- exists. * 1s The configuration is (σ 1s ) 2 (σ )
41 Figure Bonding in s-block homonuclear diatomic molecules. Li 2 Li 2 bond order = 1 Be 2 Be 2 bond order =
42 Figure Shapes and energies of σ and π MOs from combinations of 2p atomic orbitals
43 Figure Relative MO energy levels for Period 2 homonuclear diatomic molecules. without 2s-2p mixing with 2s-2p mixing MO energy levels for O 2, F 2, and Ne 2 MO energy levels for B 2, C 2, and N
44 Figure MO occupancy and molecular properties for B 2 through Ne
45 Figure The paramagnetic properties of O
46 Sample Problem 11.4 Using MO Theory to Explain Bond Properties PROBLEM: Explain the following data with diagrams showing the occupancy of MOs: N 2 N + 2 O 2 O + 2 Bond energy (kj/mol) Bond length (pm) PLAN: The data show that removing an electron from each parent molecule has opposite effects: N 2+ has a weaker longer bond than N 2, but O 2+ has a stronger, shorter bond than O 2. We determine the valence electrons in each species, draw the sequence of MO energy levels (showing orbital mixing in N 2 but not in O 2 ), and fill them with electrons. We then calculate bond orders, which relate directly to bond energy and inversely to bond length
47 Sample Problem 11.4 SOLUTION: N 2 N 2 + O 2 O 2 + σ * 2p σ * 2p π * 2p π * 2p σ 2p π 2p π 2p σ 2p σ * 2s σ * 2s σ 2s σ 2s 11-47
48 Sample Problem 11.4 Calculating bond orders: For N 2 ½(8 2) = 3 For N 2+ ½(7 2) = 2.5 N + 2 has a longer, weaker bond than N 2 because to form N 2+, a bonding electron is removed and the bond order decreases. For O 2 ½(8 4) = 2 For O 2+ ½(8 3) = 2.5 O + 2 has a shorter, stronger bond than O 2 because to form O 2+, an antibonding electron is removed and the bond order increases
49 Figure 11.24A The MO diagram for HF
50 Figure 11.24B The MO diagram for NO
51 Figure The lowest energy π-bonding MOs in benzene and ozone
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