无机化学课件：07 Bonding Theory

1、Chemistry: Atoms FirstJulia Burdge & Jason OverbyChapter 7Molecular Geometry and Bonding Theories Molecular Geometry and Bonding Theories77.1Molecular GeometryThe VSEPR ModelElectron-Domain Geometry and Molecular GeometryDeviation from Ideal Bond AnglesGeometry of Molecules with More Than One Centra

2、l Atom7.2Molecular Geometry and Polarity7.3Valence Bond Theory7.4Hybridization of Atomic OrbitalsHybridization of s and p OrbitalsHybridization of s, p, and d Orbitals7.5Hybridization of Molecules Containing Multiple Bonds7.6Molecular Orbital TheoryBonding and Antibonding Molecular Orbitals Molecula

3、r OrbitalsBond Order Molecular OrbitalsMolecular Orbital Diagrams7.7Bonding Theories and Descriptions of Molecules with Delocalized BondingMolecular GeometryMolecular shape can be predicted by using the valence-shell electron-pair repulsion (VSEPR) model.A is the central atom surrounded by x B atoms

4、.x can have integer values of 2 to 6.7.1ABxThe VSEPR ModelThe basis of the VSEPR model is that electrons repel each other.Electrons are found in various domains.2 double bonds2 electron domains (on central atom)1 single bond1 double bond1 lone pair3 single bonds1 lone pair 3 electron domains(on cent

5、ral atom)4 electron domains(on central atom)Lone pairsSingle bondsDouble bondsTriple bondsElectron-Domain Geometry and Molecular GeometryThe 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 in

6、teractions.2 electron domainsLinear3 electron domainsTrigonal planarElectron-Domain Geometry and Molecular Geometry5 electron domainsTrigonal bipyramidal6 electron domainsOctahedral4 electron domainsTetrahedralElectron-Domain Geometry and Molecular GeometryThe electron domain geometry is the arrange

7、ment 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 bipyramidalOctahedralTetrahedralLinearTrigonal planar180120109.59012090Electron-Domain Geometry and Molecul

8、ar GeometryAB5 molecules contain two different bond angles between adjacent bonds.Trigonal bipyramidalAxial positions; perpendicular to the trigonal planeEquatorial positions; three bonds arranged in a trigonal plane.12090Electron-Domain Geometry and Molecular GeometryWhen the central atom in an ABx

9、 molecule bears one or more lone pairs, the electron-domain geometry and the molecular geometry are no longer the same.OOO=Electron-Domain Geometry and Molecular GeometryElectron-Domain Geometry and Molecular GeometryElectron-Domain Geometry and Molecular GeometryThe steps to determine the electron-

10、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 consid

11、ering the positions of the atoms only.Determine the shapes of (a) SO3 and (b) ICl4-. Worked Example 7.1Strategy 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

12、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 bondsand two lone pairs. Worked Example 7.1 (cont.)Solution (a) According to the VSEPR model, three electron domains will be

13、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

14、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

15、the shape of a molecule.Deviation from Ideal Bond AnglesSome 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 AnglesSome electron domain

16、s 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 AtomThe geometry of more complex molecules can be determined by treating them a

17、s though they have multiple central atoms.Central C atomNo. of electron domains:4Electron-domain geometry: tetrahedralMolecular geometry: tetrahedralCentral O atomNo. of electron domains:4Electron-domain geometry: tetrahedralMolecular geometry: bent Worked Example 7.2Strategy The leftmost C atom is

18、surrounded by four electron domains: one CC bond and three CH bonds. The middle C atom is surrounded by three electron domains: one CC bond, one CO bond, and one C=O bond. The O atom is surrounded by four electron domains: one OC bond, one OH bond, and twolone pairs.Acetic acid, the substance that g

19、ives vinegar its characteristic smell and sour taste, is sometimes used in combination with corticosteroids to treat certain types of ear infections. Its Lewis structure isDetermine the molecular geometry about each of the central atoms, and determine the approximate value of each of the bond angles

20、 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 i

21、s 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

22、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 109.5C, those about the middle C are 120, and those about the O are 109.5. The angle between the two single

23、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

24、each other and push the two bonding pairs closer together. The angles are labeled as follows:109.5120120109.5Think About It Compare these answers with the information in Figure 7.2 and Table 7.2Molecular Geometry and PolarityMolecular polarity is one of the most important consequences of molecular g

25、eometry.A diatomic molecule is polar when the electronegativites of the two atoms are different.7.2HFHF+Molecular Geometry and PolarityThe polarity of a molecule made up of three or more atoms depends on:(1)the polarity of the individual bonds(2)the molecular geometryThe bonds in CO2 are polar but t

26、he molecule is nonpolar.Carbon dioxide, CO2Molecular Geometry and PolarityThe polarity of a molecule made up of three or more atoms depends on:(1)the polarity of the individual bonds(2)the molecular geometryThe bonds in H2O are polar and the molecule is polar.Water, H2OMolecular Geometry and Polarit

27、yThe polarity of a molecule made up of three or more atoms depends on:(1)the polarity of the individual bonds(2)the molecular geometryThe 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 determi

28、ne its molecular geometry, and then determine whether the individual bond dipoles cancel.(a) The Lewis structure of PCl5 isWith five identical electron domains around the central atom, the electron-domain and molecular geometries are trigonal bipyramidal. The equatorial bond dipoles will cancel one

29、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 indi

30、vidual bond dipoles cancel.(b) The Lewis structure of H2CO isThe 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 yo

31、ur 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 PolarityDipol

32、e moments can be used to distinguish between structural isomers.trans-dichloroethylenenonpolarcis-dichloroethylenepolarValence Bond TheoryAccording to valence bond theory, atoms share electrons when atomic orbitals overlap.A bond forms when single occupied atomic orbitals on two atoms overlap.The tw

33、o 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.7.3Valence Bond TheoryThe HH bond in H2 forms when the singly occupied 1s orbitals of the two H atoms overlap:Valence Bond TheoryThe FF bond in F2 forms

34、 when the singly occupied 2p orbitals of the two F atoms overlap:Valence Bond TheoryThe HF 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 7.4Strategy The ground-state electron configuration of Se is Ar4s23

35、d104p4. Its orbital diagram (showing only the 4p orbitals) isHydrogen selenide (H2Se) is a foul-smelling gas that can cause eye and respiratory tract inflammation. The HSeH bond angle in H2Se is approximately 92. Use valence bond theory to describe the bonding in this molecule.Solution Two of the 4p

36、 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

37、 between bonds formed by their overlap to be approximately 90.Hybridization of Atomic OrbitalsHybridization 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.7.4Hybridization of Atomic OrbitalsBeCl2Lewi

38、s theory and VSEPR theory predict ClBeCl bond angle of 180A ground state beryllium atom can not form two bonds; there are no unpaired electrons.2 electron domainsLinear molecular geometryHybridization of Atomic OrbitalsBeCl2An excited state configuration for Be has two unpaired electrons and can for

39、m to bonds. The two bonds formed would not be equivalent.Hybridization of s and p OrbitalsBeCl2Experimentally 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 OrbitalsBeCl2The 2s orbital and one o

40、f 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 OrbitalsBeCl2The hybrid orbitals on Be each overlap with a singly occupied 3p orbital on a Cl atom.The energy required to form an excited sta

41、te 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 Orbita

42、lsDetermine the number and type of hybrid orbitals necessary to rationalize the bonding in BF3Step 1: Draw the Lewis structure:Hybridization of s and p OrbitalsDetermine the number and type of hybrid orbitals necessary to rationalize the bonding in BF3Step 2: Count the number of electron domains on

43、the central atom. This is the number of hybrid orbitals necessary to account for the molecules geometry. (This is also the number of atomic orbitals that must undergo hybridization.) Three electron domainsThree hybrid orbitals requiredHybridization of s and p OrbitalsDetermine the number and type of

44、 hybrid orbitals necessary to rationalize the bonding in BF3Step 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 OrbitalsDetermine the number and type of hybrid orbitals necessary to rationa

45、lize the bonding in BF3Step 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 OrbitalsMixing of one s orbital

46、 and two p orbitals to yield three sp2 orbitals.Hybridization of s and p OrbitalsHybrid orbitals on boron overlap with 2p orbitals on fluorine.Hybridization of s and p OrbitalsDetermine the number and type of hybrid orbitals necessary to rationalize the bonding in CH4.Step 1: Draw the Lewis structur

47、e:Hybridization of s and p OrbitalsStep 2: The number of electron domains on the central atom is the number of hybrid orbitals necessary to account for the molecules geometry. Four electron domainsFour hybrid orbitals requiredDetermine the number and type of hybrid orbitals necessary to rationalize

48、the bonding in CH4.Hybridization of s and p OrbitalsStep 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

49、OrbitalsStep 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 rationali

50、ze the bonding in CH4.Hybridization of s and p OrbitalsMixing of one s orbital and three p orbitals to yield four sp3 orbitalsHybridization of s and p OrbitalsHybrid orbitals on carbon overlap with 1s orbitals on hydrogen.Hybridization of s, p and d OrbitalsDetermine the number and type of hybrid or

51、bitals necessary to rationalize the bonding in PCl5.Step 1: Draw the Lewis structure:Hybridization of s, p and d OrbitalsStep 2: The number of electron domains on the central atom is the number of hybrid orbitals necessary to account for the molecules geometry.Five electron domainsFive hybrid orbita

52、ls requiredDetermine the number and type of hybrid orbitals necessary to rationalize the bonding in PCl5.Hybridization of s, p and d OrbitalsStep 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

53、 of hybrid orbitals necessary to rationalize the bonding in PCl5.Hybridization of s, p and d OrbitalsStep 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

54、pairing any electrons.Determine the number and type of hybrid orbitals necessary to rationalize the bonding in PCl5.Hybridization of s, p and d OrbitalsHybrid orbitals on phosphorus overlap with 3p orbitals on chlorine.Hybridization of s, p and d Orbitals Worked Example 7.5Strategy Starting with the

55、 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 He2s22p3. Its valence orbital diagram isAmmonia (NH3) is a trigonal pyramidal molecule with HNH bond angles of about 107. Describe t

56、he formation of three equivalent NH 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 NH bonds, we would expect bond angles of 90 (not 107) to form from the overlap of the three mutually perpendicul

57、ar 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 promoti

58、on 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 a

59、tom (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 hy

60、bridization 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.)SolutionEach NH bond is formed by overlap between an sp3 hybrid orbital on the N atom a