资源简介

(共125张PPT)
Chapter 7
Molecular Geometry and Bonding Theories
Molecular Geometry and Bonding Theories
7
7.1 Molecular Geometry
The VSEPR Model
Electron-Domain Geometry and Molecular Geometry
Deviation from Ideal Bond Angles
Geometry of Molecules with More Than One Central Atom
7.2 Molecular Geometry and Polarity
7.3 Valence Bond Theory
7.4 Hybridization of Atomic Orbitals
Hybridization of s and p Orbitals
Hybridization of s, p, and d Orbitals
7.5 Hybridization of Molecules Containing Multiple Bonds
7.6 Molecular Orbital Theory
Bonding and Antibonding Molecular Orbitals
σ Molecular Orbitals
Bond Order
π Molecular Orbitals
Molecular Orbital Diagrams
7.7 Bonding Theories and Descriptions
of Molecules with Delocalized Bonding
Molecular Geometry
Molecular shape can be predicted by using the valence-shell electron-pair repulsion (VSEPR) model.
A is the central atom surrounded by x B atoms.
x can have integer values of 2 to 6.
7.1
ABx
The VSEPR Model
The basis of the VSEPR model is that electrons repel each other.
Electrons are found in various domains.
2 double bonds
2 electron domains (on central atom)
1 single bond
1 double bond
1 lone pair
3 single bonds
1 lone pair
3 electron domains
(on central atom)
4 electron domains
(on central atom)
Lone pairs
Single bonds
Double bonds
Triple bonds
Electron-Domain Geometry and Molecular Geometry
The basis of the VSEPR model is that electrons repel each other.
Electrons will arrange themselves to be as far apart as possible.
Arrangements minimize repulsive interactions.
2 electron domains
Linear
3 electron domains
Trigonal planar
Electron-Domain Geometry and Molecular Geometry
5 electron domains
Trigonal bipyramidal
6 electron domains
Octahedral
4 electron domains
Tetrahedral
Electron-Domain Geometry and Molecular Geometry
The electron domain geometry is the arrangement of electron domains around the central atom.
The molecular geometry is the arrangement of bonded atoms.
In an ABx molecule, a bond angle is the angle between two adjacent A-B bonds.
Trigonal bipyramidal
Octahedral
Tetrahedral
Linear
Trigonal planar
180°
120°
109.5°
90°
120°
90°
Electron-Domain Geometry and Molecular Geometry
AB5 molecules contain two different bond angles between adjacent bonds.
Trigonal bipyramidal
Axial positions; perpendicular to the trigonal plane
Equatorial positions; three bonds arranged in a trigonal plane.
120°
90°
Electron-Domain Geometry and Molecular Geometry
When the central atom in an ABx molecule bears one or more lone pairs, the electron-domain geometry and the molecular geometry are no longer the same.
O
O
O




=



Electron-Domain Geometry and Molecular Geometry
Electron-Domain Geometry and Molecular Geometry
Electron-Domain Geometry and Molecular Geometry
The steps to determine the electron-domain and molecular geometries are as follows:
Step 1: Draw the Lewis structure of the molecule or polyatomic ion.
Step 2: Count the number of electron domains on the central atom.
Step 3: Determine the electron-domain geometry by applying the VSEPR model.
Step 4: Determine the molecular geometry by considering the positions of the atoms only.
Determine the shapes of (a) SO3 and (b) ICl4-.
Worked Example 7.1
Strategy Use Lewis structures and the VSEPR model to determine first the electron-domain geometry and then the molecular geometry (shape).
(a) The Lewis structure of SO3 is:
There are three electron domains on the central atom: one double bond and two single bonds.
(b) The Lewis structure of ICl4- is:
There are six electron domains on the central atom in ICl4-: four single bonds
and two lone pairs.
Worked Example 7.1 (cont.)
Solution
(a) According to the VSEPR model, three electron domains will be arranged in a trigonal plane. Since there are no lone pairs on the central atom in SO3, the molecular geometry is the same as the electron-domain geometry. Therefore, the shape of SO3 is trigonal planar.
(b) Six electron domains will be arranged in an octahedron. Two lone pairs on an octahedron will be located on opposite sides of the central atom, making the shape of ICl4- square planar.
Think About It Compare these results with the information in Figure 7.2 and Table 7.2. Make sure that you can draw Lewis structures correctly. Without a correct Lewis structure, you will be unable to determine the shape of a molecule.
Deviation from Ideal Bond Angles
Some electron domains are better than others at repelling neighboring domains.
Lone pairs take up more space than bonded pairs of electrons.
Multiple bonds repel more strongly than single bonds.
Deviation from Ideal Bond Angles
Some electron domains are better than others at repelling neighboring domains.
Lone pairs take up more space than bonded pairs of electrons.
Multiple bonds repel more strongly than single bonds.
Geometry of Molecules with More Than One Central Atom
The geometry of more complex molecules can be determined by treating them as though they have multiple central atoms.
Central C atom
No. of electron domains: 4
Electron-domain geometry: tetrahedral
Molecular geometry: tetrahedral
Central O atom
No. of electron domains: 4
Electron-domain geometry: tetrahedral
Molecular geometry: bent
Worked Example 7.2
Strategy The leftmost C atom is surrounded by four electron domains: one C C bond and three C H bonds. The middle C atom is surrounded by three electron domains: one C C bond, one C O bond, and one C=O bond. The O atom is surrounded by four electron domains: one O C bond, one O H bond, and two
lone pairs.
Acetic acid, the substance that gives vinegar its characteristic smell and sour taste, is sometimes used in combination with corticosteroids to treat certain types of ear infections. Its Lewis structure is
Determine the molecular geometry about each of the central atoms, and determine the approximate value of each of the bond angles in the molecule. Which if any of the bond angles would you expect to be smaller than the ideal values
Worked Example 7.2 (cont.)
Solution The electron-domain geometry of the leftmost C is tetrahedral. Because all four electron domains are bonds, the molecular geometry of this part of the molecule is also tetrahedral. The electron-domain geometry of the middle C is trigonal planar. Again, because all the domains are bonds, the molecular geometry is also trigonal planar. The electron-domain geometry of the O atom is tetrahedral. Because two of the domains are lone pairs, the molecular geometry about the O atom is bent.
Worked Example 7.2 (cont.)
Solution (cont.) Bond angles are determined using electron-domain geometry. Therefore, the approximate bond angles about the leftmost C are 109.5°C, those about the middle C are 120°, and those about the O are 109.5°. The angle between the two single bonds on the middle carbon will be less than 120° because the double bond repels the single bonds more strongly than they repel each other. Likewise, the bond angle between the two bonds on the O will be less than 109.5° because the lone pairs on O repel the single bonds more strongly than they repel each other and push the two bonding pairs closer together. The angles are labeled as follows:
~109.5°
>120°
<120°
<109.5°
Think About It Compare these answers with the information in Figure 7.2 and Table 7.2
Molecular Geometry and Polarity
Molecular polarity is one of the most important consequences of molecular geometry.
A diatomic molecule is polar when the electronegativites of the two atoms are different.
7.2
H
F



H
F



δ+
δ
Molecular Geometry and Polarity
The polarity of a molecule made up of three or more atoms depends on:
(1) the polarity of the individual bonds
(2) the molecular geometry
The bonds in CO2 are polar but the molecule is nonpolar.
Carbon dioxide, CO2
Molecular Geometry and Polarity
The polarity of a molecule made up of three or more atoms depends on:
(1) the polarity of the individual bonds
(2) the molecular geometry
The bonds in H2O are polar and the molecule is polar.
Water, H2O
Molecular Geometry and Polarity
The polarity of a molecule made up of three or more atoms depends on:
(1) the polarity of the individual bonds
(2) the molecular geometry
The bonds in BF3 are polar but the molecule is nonpolar.
Boron trifluoride, BF3
Worked Example 7.3 (a)
Strategy Draw the Lewis structure, use the VSEPR model to determine its molecular geometry, and then determine whether the individual bond dipoles cancel.
(a) The Lewis structure of PCl5 is
With five identical electron domains around the central atom, the electron-domain and molecular geometries are trigonal bipyramidal. The equatorial bond dipoles will cancel one another, just as in the case of BF3, and the axial bond dipoles will also cancel each other.
Determine whether PCl5 is polar.
Solution PCl5 is nonpolar.
Worked Example 7.3 (b)
Strategy Draw the Lewis structure, use the VSEPR model to determine its molecular geometry, and then determine whether the individual bond dipoles cancel.
(b) The Lewis structure of H2CO is
The bond dipoles, although symmetrically distributed around the C atom, are not identical and therefore will not sum to zero.
Determine whether (b) H2CO (C double bonded to O) is polar.
Solution H2CO is polar.
Think About It Make sure that your Lewis structures are correct and that you count electron domains on the central atom carefully. This will give you the correct electron-domain and molecular geometries. Molecular polarity depends on both the individual bond dipoles and the molecular geometries.
Molecular Geometry and Polarity
Dipole moments can be used to distinguish between structural isomers.
trans-dichloroethylene
nonpolar
cis-dichloroethylene
polar
Valence Bond Theory
According to valence bond theory, atoms share electrons when atomic orbitals overlap.
A bond forms when single occupied atomic orbitals on two atoms overlap.
The two electrons shared in the region of orbital overlap must be of opposite spin.
Formation of a bond results in a lower potential energy for the system.
7.3
Valence Bond Theory
The H H bond in H2 forms when the singly occupied 1s orbitals of the two H atoms overlap:
Valence Bond Theory
The F F bond in F2 forms when the singly occupied 2p orbitals of the two F atoms overlap:
Valence Bond Theory
The H F bond in HF forms when the singly occupied 1s orbital on the H atom overlaps with the single occupied 2p orbital of the F atom:
Worked Example 7.4
Strategy The ground-state electron configuration of Se is [Ar]4s23d104p4. Its orbital diagram (showing only the 4p orbitals) is
Hydrogen selenide (H2Se) is a foul-smelling gas that can cause eye and respiratory tract inflammation. The H Se H bond angle in H2Se is approximately 92°. Use valence bond theory to describe the bonding in this molecule.
Solution Two of the 4p orbitals are singly occupied and therefore available for bonding. The bonds in H2Se form as the result of the overlap of a hydrogen 1s orbital with each of these orbitals on the Se atom.
Think About It Because the 4p orbitals on the Se atom are all mutually perpendicular, we should expect the angles between bonds formed by their overlap to be approximately 90°.
Hybridization of Atomic Orbitals
Hybridization or mixing of atomic orbitals can account for observed bond angles in molecules that could not be described by the direct overlap of atomic orbitals.
7.4
Hybridization of Atomic Orbitals
BeCl2
Lewis theory and VSEPR theory predict Cl Be Cl bond angle of 180°
A ground state beryllium atom can not form two bonds; there are no unpaired electrons.
2 electron domains
Linear molecular geometry
Hybridization of Atomic Orbitals
BeCl2
An excited state configuration for Be has two unpaired electrons and can form to bonds.
The two bonds formed would not be equivalent.
Hybridization of s and p Orbitals
BeCl2
Experimentally the bond in BeCl2 bonds are identical in length and strength.
Mixing of one s orbital and one p orbital to yield two sp orbitals.
Hybridization of s and p Orbitals
BeCl2
The 2s orbital and one of the 2p orbitals on Be combine to form two sp hybrid orbitals.
Like any two electron domains, the hybrid orbitals on Be are 180° apart.
Hybridization of s and p Orbitals
BeCl2
The hybrid orbitals on Be each overlap with a singly occupied 3p orbital on a Cl atom.
The energy required to form an excited state Be atom is more than compensated for by the energy given off when a bond forms.
Hybridization: the mixing of nonequivalent atomic orbitals in an atom (usually a central atom) to generate a set of hypothetical equivalent bonding orbitals, called hybrid orbitals,
y +
y -
Hybridization of s and p Orbitals
Determine the number and type of hybrid orbitals necessary to rationalize the bonding in BF3
Step 1: Draw the Lewis structure:
Hybridization of s and p Orbitals
Determine the number and type of hybrid orbitals necessary to rationalize the bonding in BF3
Step 2: Count the number of electron domains on the central atom. This is the number of hybrid orbitals necessary to account for the molecule’s geometry. (This is also the number of atomic orbitals that must undergo hybridization.)
Three electron domains
Three hybrid orbitals required
Hybridization of s and p Orbitals
Determine the number and type of hybrid orbitals necessary to rationalize the bonding in BF3
Step 3: Draw the ground-state orbital diagram for the central atom.
Step 4: Maximize the number of unpaired valence electrons by promotion.
Hybridization of s and p Orbitals
Determine the number and type of hybrid orbitals necessary to rationalize the bonding in BF3
Step 5: Combine the necessary number of atomic orbitals to generate the required number of hybrid orbitals.
Step 6: Place electrons in the hybrid orbitals, putting one electron in each orbital before pairing any electrons.
Hybridization of s and p Orbitals
Mixing of one s orbital and two p orbitals to yield three sp2 orbitals.
Hybridization of s and p Orbitals
Hybrid orbitals on boron overlap with 2p orbitals on fluorine.
Hybridization of s and p Orbitals
Determine the number and type of hybrid orbitals necessary to rationalize the bonding in CH4.
Step 1: Draw the Lewis structure:
Hybridization of s and p Orbitals
Step 2: The number of electron domains on the central atom is the number of hybrid orbitals necessary to account for the molecule’s geometry.
Four electron domains
Four hybrid orbitals required
Determine the number and type of hybrid orbitals necessary to rationalize the bonding in CH4.
Hybridization of s and p Orbitals
Step 3: Draw the ground-state orbital diagram for the central atom.
Step 4: Maximize the number of unpaired electrons by promotion.
Determine the number and type of hybrid orbitals necessary to rationalize the bonding in CH4.
Hybridization of s and p Orbitals
Step 5: Combine the necessary number of atomic orbitals to generate the required number of hybrid orbitals.
Step 6: Place electrons in the hybrid orbitals, putting one electron in each orbital before pairing any electrons.
Determine the number and type of hybrid orbitals necessary to rationalize the bonding in CH4.
Hybridization of s and p Orbitals
Mixing of one s orbital and three p orbitals to yield four sp3 orbitals
Hybridization of s and p Orbitals
Hybrid orbitals on carbon overlap with 1s orbitals on hydrogen.
Hybridization of s, p and d Orbitals
Determine the number and type of hybrid orbitals necessary to rationalize the bonding in PCl5.
Step 1: Draw the Lewis structure:
Hybridization of s, p and d Orbitals
Step 2: The number of electron domains on the central atom is the number of hybrid orbitals necessary to account for the molecule’s geometry.
Five electron domains
Five hybrid orbitals required
Determine the number and type of hybrid orbitals necessary to rationalize the bonding in PCl5.
Hybridization of s, p and d Orbitals
Step 3: Draw the ground-state orbital diagram for the central atom.
Step 4: Maximize the number of unpaired electrons by promotion.
Determine the number and type of hybrid orbitals necessary to rationalize the bonding in PCl5.
Hybridization of s, p and d Orbitals
Step 5: Combine the necessary number of atomic orbitals to generate the required number of hybrid orbitals.
Step 6: Place electrons in the hybrid orbitals, putting one electron in each orbital before pairing any electrons.
Determine the number and type of hybrid orbitals necessary to rationalize the bonding in PCl5.
Hybridization of s, p and d Orbitals
Hybrid orbitals on phosphorus overlap with 3p orbitals on chlorine.
Hybridization of s, p and d Orbitals
Worked Example 7.5
Strategy Starting with the Lewis structure, determine the number and type of hybrid orbitals necessary to rationalize the bonding in NH3.
The ground-state electron configuration of the N atom is [He]2s22p3. Its valence orbital diagram is
Ammonia (NH3) is a trigonal pyramidal molecule with H N H bond angles of about 107°. Describe the formation of three equivalent N H bonds, and explain the angle between them.
Worked Example 7.5 (cont.)
Solution Although the N atom has the three unpaired electrons needed to form three N H bonds, we would expect bond angles of ~90° (not 107°) to form from the overlap of the three mutually perpendicular 2p orbitals. Hybridization, therefore, is necessary to explain the bonding in NH3. Although we often need to promote an electron to maximize the number of unpaired electrons, no promotion is necessary for the nitrogen in NH3. We already have the three unpaired electrons necessary, and the promotion of an electron from the 2s orbital to one of the 2p orbitals would not result in any additional unpaired electrons. Furthermore, there are no empty d orbitals in the second shell.
Worked Example 7.5 (cont.)
Solution According to the Lewis structure, there are four electron domains on the central atom (three bonds and a lone pair of electrons). Four electron domains on the central atom require four hybrid orbitals, and four hybrid orbitals require the hybridization of four atomic orbitals: one s and three p. This corresponds to sp3 hybridization. Because the atomic orbitals involved in the hybridization contain a total of five electrons, we place five electrons in the resulting hybrid orbitals. This means that one of the hybrid orbitals will contain a lone pair of electrons:
Worked Example 7.5 (cont.)
Solution
Each N H bond is formed by overlap between an sp3 hybrid orbital on the N atom and the 1s atomic orbital on an H atom. Because there are four electron domains on the central atom, we expect them to be arranged in a tetrahedron. In addition, because one of the electron domains is a lone pair, we expect the H N H bond angles to be slightly smaller than the ideal tetrahedral bond angle
of 109.5°.
Solution This analysis agrees with the experimentally observed geometry and bond angles of 107° in NH3.
Hybridization in Molecules Containing Multiple Bonds
Valence bond theory and hybridization can be used to describe the bonding in molecules containing double and triple bonds.
7.5
Each carbon has three electron domains:
2 single bonds
1 double bond
ethylene (C2H4)
Expect sp2 hybridization
Hybridization in Molecules Containing Multiple Bonds
Maximize unpaired electrons on carbon by promotion:
Hybridize the required number of atomic orbitals (one for each electron domain on carbon)
Hybridization in Molecules Containing Multiple Bonds
Three equivalent sp2 hybrid orbitals explain three bonds around carbon
Hybridization scheme for the carbon atoms in ethylene:
One unhybridized atomic 2p orbital gives rise to multiple bonds
ethylene (C2H4)
Hybridization in Molecules Containing Multiple Bonds
A sigma (σ) bond forms when sp2 hybrid orbitals on the C atoms overlap.
In a sigma bond, the shared electron density lies directly along the internuclear axis.
Hybridization in Molecules Containing Multiple Bonds
The ethylene molecule contains five sigma bonds:
1 between the two carbon atoms (sp2 and sp2 overlap)
4 between the C and H atoms (sp2 and 1s overlap)
Hybridization in Molecules Containing Multiple Bonds
The remaining unhybridized p orbital is perpendicular to the plane in which the atoms of the molecule lie.
The unhybridized p orbitals overlap in a sideways fashion to form a pi (π) bond.
Worked Example 7.6
Strategy Remember, a single bond is composed of a sigma bond, whereas a double bond is usually composed of one sigma bond and one pi bond.
Thalidomide (C13H10N2O4) is a sedative and antiemetic that was widely prescribed during the 1950s, although not in the United States, for pregnant women suffering from morning sickness. Its use was largely discontinued when it was determined to be responsible for thousands of devastating birth defects. Determine the number of carbon-carbon sigma bonds and the total number of pi bonds in thalidomide.
Solution Thalidomide contains 12 carbon-carbon sigma bonds and a total of seven pi bonds (three in carbon-carbon double bonds and four in carbon-oxygen double bonds.
Think About It The Lewis structure given to thalidomide is one of two possible resonance structures. Draw the other resonance structure, and count sigma and pi bonds again. Make sure you get the same answer.
Hybridization in Molecules Containing Multiple Bonds
Sigma bonds exhibit free rotation around the bond axis.
All three Lewis structures represent the same molecule.
Hybridization in Molecules Containing Multiple Bonds
Pi bonds restrict free rotation around the bond axis.
There are two isomers of 1,2-dichloroethylene
cis-1,2-dichloroethylene
trans-1,2-dichloroethylene
Hybridization in Molecules Containing Multiple Bonds
The acetylene molecule is linear with sp hybridized carbons.
Promotion of an electron maximizes the number of unpaired electrons:
acetylene (C2H2)
Hybridization in Molecules Containing Multiple Bonds
The acetylene molecule is linear with sp hybridized carbons.
acetylene (C2H2)
The 2s orbital and one of the 2p orbitals then mix to form two sp hybrid orbitals:
3 sigma bonds:
1 between the two carbon atoms (sp and sp)
2 between the C and H atoms (sp and 1s)
2 pi bonds
2 between the two carbon atoms (2p and 2p)
Hybridization in Molecules Containing Multiple Bonds
The acetylene molecule is linear with sp hybridized carbons.
acetylene (C2H2)
Two equivalent sp hybrid orbitals give rise to 2 sigma bonds
Two unhybridized atomic 2p orbitals gives rise to 2 pi bonds
Hybridization in Molecules Containing Multiple Bonds
Formation of the C C sigma bond in acetylene:
acetylene (C2H2)
3 sigma bonds:
1 between the two carbon atoms (sp and sp)
2 between the C and H atoms (sp and 1s)
Hybridization in Molecules Containing Multiple Bonds
Formation of the C C pi bond in acetylene:
acetylene (C2H2)
2 pi bonds
2 between the two carbon atoms (2p and 2p)
Worked Example 7.7
Strategy The Lewis structure of formaldehyde is
The C and O atoms each have three electron domains around them. [Carbon has two single bonds (C H) and a double bond (C=O); oxygen has a double bond (O=C) and two lone pairs.]
In addition to its use in aqueous solution as a preservative for laboratory specimens, formaldehyde gas is used as an antibacterial fumigant. Use hybridization to explain the bonding in formaldehyde (CH2O).
Worked Example 7.7 (cont.)
Solution Three electron domains correspond to sp2 hybridization. For carbon, promotion of an electron from the 2s orbital to the empty 2p orbital is necessary to maximize the number of unpaired electrons. For oxygen, no promotion is necessary. Each undergoes hybridization to produce sp2 hybrid orbitals; and each is left with a singly occupied, unhybridized p orbital.
Worked Example 7.7 (cont.)
Solution
A sigma bond is formed between the C and O atoms by the overlap of one of the sp2 hybrid orbitals from each of them. Two more sigma bonds form between the C atom and the H atoms by the overlap of carbon’s remaining sp2 hybrid orbitals with the 1s orbital on each H atom. Finally, the remaining p orbitals on C and O overlap to form a pi bond.
Think About It Our analysis describes the formation of both a sigma bond and a pi bond between the C and O atoms. This corresponds correctly to the double bond predicted by the Lewis structure.
Molecular Orbital Theory
Lewis structures and valence bond theory fail to predict some important properties of molecules.
Paramagnetism is a result of a molecule’s electron configuration.
Species that contain one or more unpaired electrons are paramagnetic.
Paramagnetic species are attracted to magnet fields.
7.6
The Lewis structure for O2 shows no unpaired electrons.
O2 exhibits paramagnetism.
Molecular Orbital Theory
Lewis structures and valence bond theory fail to predict some important properties of molecules.
Species that contain paired electrons are diamagnetic.
Diamagnetic species are weakly repelled by magnetic fields.
Nitrogen, N2, is diamagnetic.
Molecular Orbital Theory
Another bonding theory is needed to describe the paramagnetism of O2.
In molecular orbital theory, the atomic orbitals combine to form new orbitals that are the “property” of the entire molecule.
The new orbitals are called molecular orbitals.
Molecular orbitals have characteristics similar to atomic and hybrid orbitals:
specific shapes
specific energies
accommodate a maximum of 2 electrons each
electron filling follows the Pauli exclusion principle
the number of molecular orbitals obtained equals the number of orbitals combined
Bonding and Antibonding Molecular Orbitals
H2 is the simplest homonuclear diatomic molecule.
Valence bond theory:
H2 forms when from the overlap of the 1s orbitals.
Molecular orbital theory:
H2 forms when the 1s orbitals combine to give molecular orbitals.
Molecular orbitals result from the constructive and destructive combination of atomic orbitals.
Molecular Orbital Theory (MO)
Atomic orbitals combine to form new molecular orbitals which are spread out over the entire molecule. Electrons are in orbitals that belong to the molecule as a whole.
Molecular orbitals (wave functions) result from adding and/or subtracting atomic orbitals (wave functions).
H2+: The Simplest Molecule
Electron-nuclear
attraction
internuclear
repulsion
Born-Oppenheimer approximation
set of wavefunctions, y(x,y,z;R), describing the quantum state of the electron called molecular orbitals
Linear Combinations of Atomic Orbitals LCAO
which implies
because cA=cB or cA=-cB, there are only two possible molecular orbitals. Orbitals are conserved.
lower energy
higher energy
s (sigma)
Energy as a Function of Internuclear Separation
higher energy
Pictorial representation.
H + H H2
1s1 1s1 1s2
bonding orbital
antibonding orbital
1s 1s
H atomic orbitals
H2 molecular orbitals
sum
difference
H2: two electrons
Molecular Orbital Theory
Constructive combination of the two 1s orbitals gives rise to a molecular orbital that lies along the internuclear axis.
Constructive combination increases the electron density between the two nuclei.
This molecular orbital is referred to as a bonding molecular orbital.
Bonding and Antibonding Molecular Orbitals
Destructive combination of the two 1s orbitals gives rise to a molecular orbital that lies along the internuclear axis, but does not lie between the two nuclei.
Electron density in this molecular orbital pulls the two nuclei in opposite directions.
This molecular orbital is referred to as an antibonding molecular orbital.
Bonding and Antibonding Molecular Orbitals
Molecular orbitals that lie along the internuclear axis are referred to as σ molecular orbitals.
Examples:
σ1s bonding molecular orbital from the combination of two 1s orbitals
σ*1s antibonding molecular orbital from the combination of two 1s orbitals
The asterisk distinguishes an antibonding molecular orbital from a bonding orbital.
σ Molecular Orbitals
Molecular orbitals have specific energies.
Electrons in bonding molecular orbitals stabilize the molecule and are lower in energy than the isolated atomic orbitals.
Electrons in antibonding molecular orbitals destabilize the molecule and are higher in energy than the isolated atomic orbitals.
Bond Order
The bond order indicates how stable a molecule is.
The higher the bond order, the more stable a molecule is.
Bond Order
H2
According to molecular orbital theory, H2 is a stable molecule.
Bond Order
He2
According to molecular orbital theory, He2 is NOT a stable molecule and does not exist.
π Molecular Orbitals
p atomic orbitals also form molecular orbitals by both constructive and destructive combination.
The orientations of px, py, and pz give rise to two different types of molecular orbitals:
σ molecular orbitals – electron density along the internuclear axis
px orbitals point towards each other
bonding and antibonding σ molecular orbitals
π Molecular Orbitals
p atomic orbitals also form molecular orbitals by both constructive and destructive combination.
π molecular orbitals – electron density above and below the internuclear axis
Molecular Orbital Theory
Molecular orbitals resulting from the combination of p atomic orbitals are higher than those resulting from the combination of s atomic orbitals.
This order of orbital energies assumes no mixing of s and p orbitals.
s orbitals only interact with s orbitals
p orbitals only interact with p orbitals
This is found in O2, Fe2, Ne2.
Molecular Orbital Theory
Molecular orbitals resulting from the combination of p atomic orbitals are higher than those resulting from the combination of s atomic orbitals.
This order of orbital energies assumes some mixing of s and p orbitals.
This arrangement of orbitals is found in Li2, B2, C2 and N2.
The number of molecular orbitals (MOs) formed is always equal to the number of atomic orbitals combined.
The more stable the bonding MO, the less stable the corresponding antibonding MO.
The filling of MOs proceeds from low to high energies.
Each MO can accommodate up to two electrons.
Use Hund’s rule when adding electrons to MOs of the same energy.
The number of electrons in the MOs is equal to the sum of all the electrons on the bonding atoms.
Molecular Orbital (MO) Configurations
Molecular Orbital Diagrams
Molecular orbital diagrams for second-period homonuclear diatomic molecules.
Worked Example 7.8
Strategy Start with the molecular orbital diagram for O2 and add an electron to the lowest-energy molecular orbital available.
The superoxide ion (O2-) has been implicated in a number of degenerative conditions, including aging and Alzheimer’s disease. Using molecular orbital theory, determine whether (O2-) is paramagnetic or diamagnetic, and then calculate its bond order.
Solution In this case, either of the two singly occupied π*2p orbitals can accommodate an additional electron. This gives a molecular orbital diagram in which there is one unpaired electron, making (O2-) paramagnetic. The new diagram has six electrons in bonding molecular orbitals and three in antibonding molecular orbitals. We can ignore the electrons in the σ2s and σ*2s orbitals because their contributions to the bond order cancel each other. The bond order is (6 – 3)/2 = 1.5.
Think About It Experiments confirm that the superoxide ion is paramagnetic. Also, any time we add one or more electrons to an antibonding molecular orbital, as we did in this problem, we should expect the bond order to decrease. Electrons in antibonding orbitals cause a bond to be less stable.
Heteronuclear Diatomic Molecules of First- and Second-Period Elements
HF
CO
C
O
受到四个原子轨道的影响
MO Theory for Heteronuclear Diatomics
MO’s will no longer contain equal contributions from each AO.
AO’s interact if symmetries are compatible.
AO’s interact if energies are close.
No interaction will occur if energies are too far apart. A nonbonding orbital will form.
YY makes a greater contribution to the YMO
YX makes a greater contribution to the Y*MO
Example HF
The F (2s) is much lower in energy than the H (1s) so they do not mix.
The F (2s) orbital makes a non-bonding MO.
We certainly don’t have to worry about the F (1s) because is MUCH lower in energy.
The H (1s) and F (2p)’s are close in energy and do interact.
The 2px and 2py don’t have the appropriate symmetry though and therefore form nonbonding MO’S
Only the 2pz and 1s mix.
Energy
MO of HF
AO of H
1s
s
2px
2py
s
AO of F
2p
The MO diagram for HF
Energy
The MO diagram for NO
MO of NO
2s
AO of N
2p
s* s
s s
2s
AO of O
2p
p
s p
* p
s* s
possible Lewis structures
AO(2)
AO(1)
Molecular Orbital Theory
AO(2)
AO(1)
AO(2)
AO(1)
AO(2)
AO(1)
Most covalent
Polar Covalent
Ionic
Remember that the closer to AO’s of appropriate symmetry are in energy, the more they interact with one another and the more stable the bonding MO that will be formed. This means that as the difference in electronegativity between two atoms increases, the stabilization provided by covalent bonding decreases (and the polarity of the bond increases). If the difference in energy of the orbitals is sufficiently large, then covalent bonding will not stabilize the interaction of the atoms. In that situation, the less electronegative atom will lose an electron to the more electronegative atom and two ions will be formed.
MO Theory and Polyatomic Molecules
MO diagrams are complicated for polyatomics.
For example, CO2 requires drawing a diagram with four sets of orbitals (3 AO’s and 1 MO).
To simplify the problem we use the ligand group orbital approach (LGO)
Ligand Group Orbital (LGO) Approach to MO’s
Consider XH2, a linear molecule, oriented along the z-axis. Let X have 2s and 2p orbitals.
The 1s orbitals on H have two possible phases.
Take the two H’s as a group to make LGO’s.
Draw MO diagram.
The MO’s will look like:
NOTE: The s-bonding character in orbitals Y1 and Y2 is spread over all three atoms indicating the bond character is delocalized.
A bent triatomic H2O
How do we know which orbitals interact
Answer: Only orbitals with the same symmetry label can interact.
MO diagram for H2O
Bonding Theories and Descriptions of Molecules with Delocalized Bonding
Lewis Theory
Strength:
qualitative prediction of bond strength and bond length
Weakness:
two dimensional model, real molecules are three dimensional
fails to explain why bonds form
7.7
Valence-Shell Electron-Pair Repulsion Model
Strength:
predict the shape of many molecules and polyatomic ions
Weakness:
fails to explain why bonds form (based on Lewis theory)
Bonding Theories and Descriptions of Molecules with Delocalized Bonding
Valence Bond Theory
Strength:
covalent bonds form when atomic orbitals overlap
Weakness:
fails to explain the bonding in many molecules
Hybridization of Atomic Orbitals
Strength:
an extension of valence bond theory. Using hybrid orbitals it is possible to explain the bonding and geometry of more molecules
Weakness:
fails to predict some important properties, such as magnetism
Bonding Theories and Descriptions of Molecules with Delocalized Bonding
Molecular Orbital Theory
Strength:
accurately predict the magnetic and other properties of molecules
Weakness:
complex
Bonding Theories and Descriptions of Molecules with Delocalized Bonding
Some molecules are best described using a combination of models.
Benzene, C6H6, is represented with two resonance structures:
The π bonds in benzene are delocalized -- spread out over the entire molecule.
Bonding Theories and Descriptions of Molecules with Delocalized Bonding
The π bonds in benzene are delocalized -- spread out over the entire molecule.
Worked Example 7.9
Strategy The Lewis structure of the carbonate ion shows three electron domains around the central C atom, so the carbon must be sp2 hybridized.
It takes three resonance structures to represent the carbonate ion CO32-:
None of the three, though, is a completely accurate description. As with benzene, the bonds that are shown in the Lewis structure as one double and two single are actually three equivalent bonds. Use a combination of valence bond theory and molecular orbital theory to explain the bonding in CO32-.
Worked Example 7.9 (cont.)
Solution Each of the sp2 hybrid orbitals on the C atom overlaps with a singly occupied p orbital on an O atom, forming three σ bonds. Each O atom has an additional, singly occupied p orbital perpendicular to the one involved in σ bonding. The unhybridized p orbital on C overlaps with the p orbitals on O to form π bonds, which have electron densities above and below the plane of the molecule. Because the species can be represented by resonance structures, we know that the π bonds are delocalized.
Think About It Although the Lewis structure of CO32- shows three electron domains on one of the O atoms, we generally do not treat terminal atoms (those with single bonds to only one other atom) as though they are hybridized because it is unnecessary to do so.
Chapter Summary: Key Points
7
Molecular Geometry
The VSEPR Model
Bond Angles
Molecular Polarity
Valence Bond Theory
Hybridization
Multiple Bonding
Molecular Orbital Theory
Bonding Molecular Orbitals
Antibonding Molecular Orbitals
Bond Order
Molecular Orbital Diagrams
Delocalized Bonding

展开更多......

收起↑