PowerPoint · PPT file · Web view · 2013-06-25There is a hybridization of atomic orbitals to form ... Times Arial Symbol Blank PowerPoint Presentation PowerPoint Presentation PowerPoint

11-1

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Chapter 11

Theories of Covalent Bonding

11-2


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

11-3


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 increasesbetween the nuclei.

11-4


The Central Themes of VB Theory

Themes

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 molecularorbitals.

11-5


Figure 11.1 Orbital overlap and spin pairing in three diatomic molecules.

Hydrogen, H2

Hydrogen fluoride, HF

Fluorine, F2

11-6


Hybrid Orbitals

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.

Key Points

sp sp2 sp3 sp3d sp3d2

Types of Hybrid Orbitals

11-7


Figure 11.2 The sp hybrid orbitals in gaseous BeCl2.

atomic orbitals

hybrid orbitals

orbital box diagrams

11-8


Figure 11.2 (continued)

The sp hybrid orbitals in gaseous BeCl2.

orbital box diagrams with orbital contours

11-9


Figure 11.3 The sp2 hybrid orbitals in BF3.

11-10


Figure 11.4 The sp3 hybrid orbitals in CH4.

11-11


Figure 11.5 The sp3 hybrid orbitals in NH3.

11-12


Figure 11.5 (continued)

The sp3 hybrid orbitals in H2O.

11-13


Figure 11.6 The sp3d hybrid orbitals in PCl5.

11-14


Figure 11.7 The sp3d2 hybrid orbitals in SF6.

11-15


11-16


Figure 11.8

The conceptual steps from molecular formula to the hybrid orbitals used in bonding.

Molecular formula

Lewis structure

Molecular shape and e- group arrangement

Hybrid orbitals

Figure 10.1

Step 1

Figure 10.12

Step 2 Step 3

Table 11.1

11-17


SAMPLE PROBLEM 11.1 Postulating Hybrid Orbitals in a Molecule

SOLUTION:

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:

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.

(a) Methanol, CH3OH (b) Sulfur tetrafluoride, SF4

(a) (a) CH3OH H

CH H

OH

The groups around C are arranged as a tetrahedron.

O also has a tetrahedral arrangement with 2 nonbonding e- pairs.

11-18


SFF

F

F

SAMPLE PROBLEM 11.1 Postulating Hybrid Orbitals in a Molecule

continued

2p

2s single C atomsingle C atom

sp3

hybridized hybridized C atomC atom

2p

2s single O atomsingle O atom

sp3

hybridized hybridized O atomO atom

(b) SF4 has a seesaw shape with 4 bonding and 1 nonbonding e- pairs.

3p

3s

3d

S atomS atomsp3d

3d

hybridized hybridized S atomS atom

11-19


Figure 11.9 The bonds in ethane(C2H6).

both C are sp3 hybridizeds-sp3 overlaps to bonds

sp3-sp3 overlap to form a bondrelatively even

distribution of electron density over all bonds

11-20


Figure 11.10 The and bonds in ethylene (C2H4).

11-21


Figure 11.11 The and bonds in acetylene (C2H2).

11-22


Figure 11.12 Electron density and bond order.

11-23


SAMPLE PROBLEM 11.2 Describing the Bond in Molecules

SOLUTION:

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.

H3C

C

CH3

O

spsp33 hybridized hybridized


CC

C

O

H

H

HHH

H


bondsbond

CC

C

O

sp3

sp3

sp3

sp3sp3

sp3

sp3

sp3

sp2 sp2sp2

sp2

sp2sp2

H

HH

HH

H

11-24


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).

11-25


Amplitudes of wave functions added

Figure 11.13

An analogy between light waves and atomic wave functions.

Amplitudes of wave functions

subtracted.

11-26


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.

11-27


Figure 11.15 The MO diagram for H2.

Ene

rgy

MO of H2

*1s

1s

AO of H

1s

AO of H

1s

H2 bond order = 1/2(2-0) = 1

Filling molecular orbitals with electrons follows the same concept as filling atomic orbitals.

11-28


Figure 11.16 MO diagram for He2+ and He2.

Ene

rgy

MO of He+

*1s

1s

AO of He+

1s

MO of He2

AO of He

1s

AO of He

1s

*1s

1s

Ene

rgy

He2+ bond order = 1/2 He2 bond order = 0

AO of He

1s

11-29


SAMPLE PROBLEM 11.3 Predicting Stability of Species Using MO Diagrams

SOLUTION:

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.

1s1s

AO of HAO of H

1s1s

AO of HAO of H

MO of HMO of H22++

bond order = 1/2(1-0) = 1/2

HH22++ does exist does exist

MO of HMO of H22--

bond order = 1/2(2-1) = 1/2

H2- does exist

1s1s 1s1s

AO of HAO of H AO of HAO of H--

configuration is (1s)1configuration is (1s)2(2s)1

11-30


*2s

2s

2s2s

Figure 11.17

2s 2s

*2s

2s

Li2 bond order = 1 Be2 bond order = 0

Bonding in s-block homonuclear diatomic molecules.E

nerg

y

Li2Be2

11-31


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

11-32


Figure 11.19 Relative MO energy levels for Period 2 homonuclear diatomic molecules.

MO energy levels for O2, F2, and Ne2

MO energy levels for B2, C2, and N2

without 2s-2p mixing

with 2s-2p mixing

11-33


Figure 11.20

MO occupancy and molecular properties for B2 through Ne2

11-34


Figure 11.21

The paramagnetic properties of O2

11-35


SAMPLE PROBLEM 11.4 Using MO Theory to Explain Bond Properties

SOLUTION:

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:

PLAN: Find the number of valence electrons for each species, draw the MO diagrams, calculate bond orders, and then compare the results.

Explain these facts with diagrams that show the sequence and occupancy of MOs.Explain these facts with diagrams that show the sequence and occupancy of MOs.

Bond energy (kJ/mol)Bond energy (kJ/mol)

Bond length (pm)Bond length (pm)

NN22 NN22++ OO22 OO22

++

945945

110110

498498841841 623623

112112121121112112

N2 has 10 valence electrons, so N2+ has 9.

O2 has 12 valence electrons, so O2+ has 11.

11-36


SAMPLE PROBLEM 11.4 Using MO Theory to Explain Bond Properties

continued

2s

2s

2p

2p

2p

2p

N2 N2+ O2 O2

+

bond orders

1/2(8-2)=3 1/2(7-2)=2.5 1/2(8-4)=2 1/2(8-3)=2.5

2s

2s

2p

2p

2p

2p

bonding e- lost

antibonding e- lost

Documents

PowerPoint · PPT file · Web view · 2013-06-25There is a hybridization of atomic orbitals to form ... Times Arial Symbol Blank PowerPoint Presentation PowerPoint Presentation PowerPoint