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Chapter 11
Theories of Covalent Bonding
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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
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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.
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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.
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Figure 11.1 Orbital overlap and spin pairing in three diatomic molecules.
Hydrogen, H2
Hydrogen fluoride, HF
Fluorine, F2
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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
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Figure 11.2 The sp hybrid orbitals in gaseous BeCl2.
atomic orbitals
hybrid orbitals
orbital box diagrams
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Figure 11.2 (continued)
The sp hybrid orbitals in gaseous BeCl2.
orbital box diagrams with orbital contours
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Figure 11.3 The sp2 hybrid orbitals in BF3.
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Figure 11.4 The sp3 hybrid orbitals in CH4.
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Figure 11.5 The sp3 hybrid orbitals in NH3.
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Figure 11.5 (continued)
The sp3 hybrid orbitals in H2O.
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Figure 11.6 The sp3d hybrid orbitals in PCl5.
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Figure 11.7 The sp3d2 hybrid orbitals in SF6.
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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
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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.
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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
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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
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Figure 11.10 The and bonds in ethylene (C2H4).
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Figure 11.11 The and bonds in acetylene (C2H2).
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Figure 11.12 Electron density and bond order.
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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
spsp33 hybridized hybridized
CC
C
O
H
H
HHH
H
spsp22 hybridized hybridized
bondsbond
CC
C
O
sp3
sp3
sp3
sp3
sp3
sp3
sp3
sp3
sp2 sp2
sp2
sp2
sp2sp2
H
HH
HH
H
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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).
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Amplitudes of wave functions added
Figure 11.13
An analogy between light waves and atomic wave functions.
Amplitudes of wave functions
subtracted.
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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.
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Figure 11.15 The MO diagram for H2.
En
erg
y
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.
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Figure 11.16 MO diagram for He2+ and He2.
En
erg
y
MO of He+
*1s
1s
AO of He+
1s
MO of He2
AO of He
1s
AO of He
1s
*1s
1s
En
erg
y
He2+ bond order = 1/2 He2 bond order = 0
AO of He
1s
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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
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*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
ner
gy
Li2Be2
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Figure 11.18Contours and energies of s and p MOs through
combinations of 2p atomic orbitals.
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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
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Figure 11.20
MO occupancy and molecular properties for B2 through Ne2
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Figure 11.21
The paramagnetic properties of O2
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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.
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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