Chapter 11 Theories of Covalent Bonding

Theories of Covalent Bonding 11.1 Valence Bond (VB) Theory and Orbital Hybridization 11.2 The Mode of Orbital Overlap and the Types of Covalent Bonds 11.3 Molecular Orbital (MO)Theory and Electron Delocalization

The Central Themes of VB Theory Basic Principle A covalent bond forms when the orbitals of two atoms overlap and the overlap region, which is between the nuclei, is occupied by a pair of electrons. The two wave functions are in phase so the amplitude increases between the nuclei.

The Central Themes of VB Theory A set of overlapping orbitals has a maximum of two electrons that must have opposite spins. The greater the orbital overlap, the stronger (more stable) the bond. The valence atomic orbitals in a molecule are different from those in isolated atoms. There is a hybridization of atomic orbitals to form molecular orbitals.

Orbital overlap and spin pairing in three diatomic molecules. Figure 11.1 Hydrogen, H2 Hydrogen fluoride, HF Fluorine, F2

Hybrid Orbitals Key Points The number of hybrid orbitals obtained equals the number of atomic orbitals mixed. The type of hybrid orbitals obtained varies with the types of atomic orbitals mixed. Types of Hybrid Orbitals sp sp2 sp3 sp3d sp3d2

The sp hybrid orbitals in gaseous BeCl2. Figure 11.2 The sp hybrid orbitals in gaseous BeCl2. atomic orbitals hybrid orbitals orbital box diagrams

The sp hybrid orbitals in gaseous BeCl2. Figure 11.2 (continued) The sp hybrid orbitals in gaseous BeCl2. orbital box diagrams with orbital contours

The sp2 hybrid orbitals in BF3. Figure 11.3 The sp2 hybrid orbitals in BF3.

The sp3 hybrid orbitals in CH4. Figure 11.4 The sp3 hybrid orbitals in CH4.

The sp3 hybrid orbitals in NH3. Figure 11.5 The sp3 hybrid orbitals in NH3.

The sp3 hybrid orbitals in H2O. Figure 11.5 (continued) The sp3 hybrid orbitals in H2O.

The sp3d hybrid orbitals in PCl5. Figure 11.6 The sp3d hybrid orbitals in PCl5.

The sp3d2 hybrid orbitals in SF6. Figure 11.7 The sp3d2 hybrid orbitals in SF6.

Molecular shape and e- group arrangement Figure 11.8 The conceptual steps from molecular formula to the hybrid orbitals used in bonding. Figure 10.1 Step 1 Figure 10.12 Step 2 Step 3 Table 11.1 Molecular shape and e- group arrangement Molecular formula Lewis structure Hybrid orbitals

SAMPLE PROBLEM 11.1 Postulating Hybrid Orbitals in a Molecule PROBLEM: Use partial orbital diagrams to describe mixing of the atomic orbitals of the central atom leads to hybrid orbitals in each of the following: (a) Methanol, CH3OH (b) Sulfur tetrafluoride, SF4 PLAN: Use the Lewis structures to ascertain the arrangement of groups and shape of each molecule. Postulate the hybrid orbitals. Use partial orbital box diagrams to indicate the hybrid for the central atoms. SOLUTION: (a) CH3OH The groups around C are arranged as a tetrahedron. O also has a tetrahedral arrangement with 2 nonbonding e- pairs.

SAMPLE PROBLEM 11.1 Postulating Hybrid Orbitals in a Molecule continued single C atom single O atom hybridized C atom hybridized O atom (b) SF4 has a seesaw shape with 4 bonding and 1 nonbonding e- pairs. S atom hybridized S atom

relatively even distribution of electron density over all s bonds Figure 11.9 The s bonds in ethane(C2H6). both C are sp3 hybridized s-sp3 overlaps to s bonds sp3-sp3 overlap to form a s bond relatively even distribution of electron density over all s bonds

The s and p bonds in ethylene (C2H4). Figure 11.10 The s and p bonds in ethylene (C2H4).

The s and p bonds in acetylene (C2H2). Figure 11.11 The s and p bonds in acetylene (C2H2).

Electron density and bond order. Figure 11.12 Electron density and bond order.

SAMPLE PROBLEM 11.2 Describing the Bond in Molecules PROBLEM: Describe the types of bonds and orbitals in acetone, (CH3)2CO. PLAN: Use the Lewis structures to ascertain the arrangement of groups and shape at each central atom. Postulate the hybrid orbitals taking note of the multiple bonds and their orbital overlaps. SOLUTION: sp3 hybridized sp2 hybridized bond bonds

The Central Themes of MO Theory A molecule is viewed on a quantum mechanical level as a collection of nuclei surrounded by delocalized molecular orbitals. Atomic wave functions are summed to obtain molecular wave functions. If wave functions reinforce each other, a bonding MO is formed (region of high electron density exists between the nuclei). If wave functions cancel each other, an antibonding MO is formed (a node of zero electron density occurs between the nuclei).

Amplitudes of wave functions subtracted. Figure 11.13 An analogy between light waves and atomic wave functions. Amplitudes of wave functions subtracted. Amplitudes of wave functions added

Figure 11.14 Contours and energies of the bonding and antibonding molecular orbitals (MOs) in H2. The bonding MO is lower in energy and the antibonding MO is higher in energy than the AOs that combined to form them.

The MO diagram for H2. Energy AO of H AO of H Figure 11.15 The MO diagram for H2. Filling molecular orbitals with electrons follows the same concept as filling atomic orbitals. s*1s Energy AO of H 1s AO of H 1s H2 bond order = 1/2(2-0) = 1 s1s MO of H2

MO diagram for He2+ and He2. Figure 11.16 MO diagram for He2+ and He2. Energy s*1s s1s s*1s AO of He 1s AO of He+ 1s AO of He 1s AO of He 1s Energy s1s MO of He+ MO of He2 He2 bond order = 0 He2+ bond order = 1/2

SAMPLE PROBLEM 11.3 Predicting Stability of Species Using MO Diagrams PROBLEM: Use MO diagrams to predict whether H2+ and H2- exist. Determine their bond orders and electron configurations. PLAN: Use H2 as a model and accommodate the number of electrons in bonding and antibonding orbitals. Find the bond order. SOLUTION: bond order = 1/2(1-0) = 1/2 bond order = 1/2(2-1) = 1/2 s s s s H2+ does exist H2- does exist 1s AO of H AO of H- configuration is (s1s)1 1s AO of H 1s AO of H configuration is (s1s)2(s2s)1 MO of H2- MO of H2+

Bonding in s-block homonuclear diatomic molecules. Figure 11.17 Bonding in s-block homonuclear diatomic molecules. s*2s s2s s*2s s2s Energy 2s 2s 2s Be2 Li2 Be2 bond order = 0 Li2 bond order = 1

Contours and energies of s and p MOs through combinations of 2p atomic orbitals. Figure 11.18

Relative MO energy levels for Period 2 homonuclear diatomic molecules. Figure 11.19 Relative MO energy levels for Period 2 homonuclear diatomic molecules. without 2s-2p mixing with 2s-2p mixing MO energy levels for O2, F2, and Ne2 MO energy levels for B2, C2, and N2

MO occupancy and molecular properties for B2 through Ne2 Figure 11.20 MO occupancy and molecular properties for B2 through Ne2

The paramagnetic properties of O2 Figure 11.21 The paramagnetic properties of O2

SAMPLE PROBLEM 11.4 Using MO Theory to Explain Bond Properties PROBLEM: As the following data show, removing an electron from N2 forms an ion with a weaker, longer bond than in the parent molecules, whereas the ion formed from O2 has a stronger, shorter bond: Bond energy (kJ/mol) Bond length (pm) N2 N2+ O2 O2+ 945 110 498 841 623 112 121 Explain these facts with diagrams that show the sequence and occupancy of MOs. PLAN: Find the number of valence electrons for each species, draw the MO diagrams, calculate bond orders, and then compare the results. SOLUTION: N2 has 10 valence electrons, so N2+ has 9. O2 has 12 valence electrons, so O2+ has 11.

SAMPLE PROBLEM 11.4 Using MO Theory to Explain Bond Properties continued N2 N2+ O2 O2+ 2p s2s s2s 2p 2p 2p 2p antibonding e- lost bonding e- lost 2p 2p 2p s2s s2s bond orders 1/2(8-2)=3 1/2(7-2)=2.5 1/2(8-4)=2 1/2(8-3)=2.5