Chapter 10 Molecular Geometry and Chemical Bonding Theory

Chapter 10 Molecular Geometry and Chemical Bonding Theory 10 | 1

In this chapter, we discuss how to explain the geometries of molecules in terms of their electronic structures. We also explore two theories of chemical bonding: valence bond theory and molecular orbital theory. 10 | 2

Molecular geometry is the general shape of a molecule, as determined by the relative positions of the atomic nuclei. 10 | 3

The valence-shell electron-pair repulsion (VSEPR) model predicts the shapes of molecules and ions by assuming that the valence-shell electron pairs are arranged about each atom so that electron pairs are kept as far away from one another as possible, thereby minimizing electron pair repulsions. The diagram on the next slide illustrates this. 10 | 4

Two electron pairs are 180° apart ( a linear arrangement). Three electron pairs are 120° apart in one plane (a trigonal planar arrangement). Four electron pairs are 109. 5° apart in three dimensions (a tetrahedral arrangment). 10 | 5

Five electron pairs are arranged with three pairs in a plane 120° apart and two pairs at 90°to the plane and 180° to each other (a trigonal bipyramidal arrangement). Six electron pairs are 90° apart (an octahedral arrangement). This is illustrated on the next slide. 10 | 6

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These arrangements are illustrated below with balloons and models of molecules for each. 10 | 8

To describe the molecular geometry, we describe the relative positions of the atoms, not the lone pairs. The direction in space of the bonding pairs gives the molecular geometry. 10 | 9

The diagrams below illustrate molecular geometry and the impact of lone pairs on it for linear and trigonal planar electron-pair arrangements. 10 | 10

Molecular geometries with a tetrahedral electronpair arrangement are illustrated below. 10 | 11

Molecular geometries for the trigonal bipyramidal electron-pair arrangement are shown on the next slide. 10 | 12

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Molecular geometries for the octahedral electronpair arrangement are shown below. 10 | 14

The VSEPR model considers a double or triple bond as though it were one lone pair. When resonance structures are required for the electron-dot diagram, you may choose any one to determine the electron-pair arrangement and the molecular geometry. 10 | 15

Predicting Molecular Geometry Using VSEPR 1. Write the electron-dot formula from the formula. 2. Based on the electron-dot formula, determine the number of electron pairs around the central atom (including bonding and nonbonding pairs). 3. Determine the arrangement of the electron pairs about the central atom (Figure 10. 3). 4. Obtain the molecular geometry from the directions of the bonding pairs for this arrangement (Figure 10. 4). 10 | 16

? Use the VSEPR model to predict the geometries of the following molecules: a. As. F 3 b. PH 4+ c. BCl 3 10 | 17

As. F 3 has 1(5) + 3(7) = 26 valence electrons; As is the central atom. The electron-dot formula is There are four regions of electrons around As: three bonds and one lone pair. The electron regions are arranged tetrahedrally. One of these regions is a lone pair, so the molecular geometry is trigonal pyramidal. 10 | 18

PH 4+ has 1(5) + 4(1) – 1 = 9 valence electrons; P is the central atom. The electron-dot formula is There are four regions of electrons around P: four bonding electron pairs. The electron-pairs arrangement is tetrahedral. All regions are bonding, so the molecular geometry is tetrahedral. 10 | 19

BCl 3 has 1(3) + 3(7) = 24 valence electrons; B is the central atom. The electron-dot formula is There are three regions of electrons around B; all are bonding. The electron-pair arrangement is trigonal planar. All of these regions are bonding, so the molecular geometry is trigonal planar. 10 | 20

The electron-pair arrangement is tetrahedral. Any three pairs are arranged as a trigonal pyramid. When one pair of the four is a lone pair, the geometry is trigonal pyramidal. 10 | 21

? Using the VSEPR model, predict the geometry of the following species: a. ICl 3 b. ICl 4 - 10 | 22

ICl 3 has 1(7) + 3(7) = 28 valence electrons. I is the central atom. The electron-dot formula is There are five regions: three bonding and two lone pairs. The electron-pair arrangement is trigonal bipyramidal. The geometry is T-shaped. 10 | 23

ICl 4 - has 1(7) + 4(7) + 1 = 36 valence electrons. I is the central element. The electron-dot formula is There are six regions around I: four bonding and two lone pairs. The electron-pair arrangement is octahedral. The geometry is square planar. 10 | 24

Predicting Bond Angles The angles 180°, 120°, 109. 5°, and so on are the bond angles when the central atom has no lone pair and all bonds are with the same other atom. When this is not the case, the bond angles deviate from these values in sometimes predictable ways. Because a lone pair tends to require more space than a bonding pair, it tends to reduce the bond angles. 10 | 25

The impact of lone pair(s) on bond angle for tetrahedral electron-pair arrangements has been experimentally determined. 10 | 26

Multiple bonds require more space than single bonds and, therefore, constrict the bond angle. This situation is illustrated below, again with experimentally determined bond angles. 10 | 27

Dipole Moment A quantitative measure of the degree of charge separation in a molecule. 10 | 28

Measurements are based on the fact that polar molecules are oriented by an electric field. This orientation affects the capacitance of the charged plates that create the electric field. 10 | 29

In part A, there is no electric field; molecules are oriented randomly. In part B, there is an electric field; molecules align themselves against the field. 10 | 30

A polar bond is characterized by separation of electrical charge. Polar molecules, therefore, have nonzero dipole moments. For HCl, we can represent the charge separation using d+ and d- to indicate partial charges. Because Cl is more electronegative than H, it has the dcharge, while H has the d+ charge. 10 | 31

The figure below shows the orbitals involved in HCl bond: the H 1 s and the Cl 3 p. 10 | 32

To determine whether a molecule is polar, we need to determine the electron-dot formula and the molecular geometry. We then use vectors to represent the charge separation. They begin at d+ atoms and go to d- atoms. Vectors have both magnitude and direction. We then sum the vectors. If the sum of the vectors is zero, the dipole moment is zero. If there is a net vector, the molecule is polar. 10 | 33

To illustrate this process, we use arrows with a + on one end of the arrow. We’ll look at CO 2 and H 2 O. CO 2 is linear, and H 2 O is bent. The vectors add to zero (cancel) for CO 2. Its dipole moment is zero. For H 2 O, a net vector points up. Water has a dipole moment. 10 | 34

The relationship between molecular geometry and dipole moment is summarized in Table 10. 1. 10 | 35

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Polar molecules experience attractive forces between molecules; in response, they orient themselves in a d+ to d- manner. This has an impact on molecular properties such as boiling point. The attractive forces due to the polarity lead the molecule to have a higher boiling point. 10 | 37

We can see this illustrated with two compounds: cis-1, 2 -dichloroethene trans-1, 2 -dichloroethene The net polarity is down; this is a polar molecule. There is no net polarity; this is a nonpolar molecule. Boiling point 60°C. Boiling point 48°C. 10 | 38

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The formula AX 3 can have the following molecular geometries and dipole moments: Trigonal planar (zero) Trigonal pyramidal (can be nonzero) T-shaped (can be nonzero) Molecule Y is likely to be trigonal planar, but might be trigonal pyramidal or T-shaped. Molecule Z must be either trigonal pyramidal or Tshaped. 10 | 40

? Which of the following molecules would be expected to have a zero dipole moment? a. Ge. F 4 b. SF 2 c. Xe. F 2 d. As. F 3 10 | 41

Ge. F 4: 1(4) + 4(7) = 32 valence electrons. Ge is the central atom. 8 electrons are bonding; 24 are nonbonding. Tetrahedral molecular geometry. Ge. F 4 is nonpolar and has a zero dipole moment. 10 | 42

SF 2: 1(6) + 2(7) = 20 valence electrons. S is the central atom. 4 electrons are bonding; 16 are nonbonding. Bent molecular geometry. SF 2 is polar and has a nonzero dipole moment. 10 | 43

Xe. F 2: 1(8) + 2(7) = 22 valence electrons. Xe is the central atom. 4 electrons are bonding; 18 are nonbonding. Linear molecular geometry. Xe. F 2 is nonpolar and has a zero dipole moment. 10 | 44

As. F 3: 1(5) + 3(7) = 26 valence electrons. As is the central atom. 6 electrons are bonding; 20 are nonbonding. Trigonal pyramidal molecular geometry. As. F 3 is polar and has a nonzero dipole moment. 10 | 45

Which of the following molecules would be expected to have a zero dipole moment? a. Ge. F 4 tetrahedral molecular geometry zero dipole moment b. SF 2 bent molecular geometry nonzero dipole moment c. Xe. F 2 linear molecular geometry zero dipole moment d. As. F 3 trigonal pyramidal molecular geometry nonzero dipole moment 10 | 46

Valence bond theory is an approximate theory put forth to explain the electron pair or covalent bond by quantum mechanics. 10 | 47

A bond forms when • An orbital on one atom comes to occupy a portion of the same region of space as an orbital on the other atom. The two orbitals are said to overlap. • The total number of electrons in both orbitals is no more than two. 10 | 48

The greater the orbital overlap, the stronger the bond. Orbitals (except s orbitals) bond in the direction in which they protrude or point, so as to obtain maximum overlap. 10 | 49

Hybrid orbitals are orbitals used to describe the bonding that is obtained by taking combinations of the atomic orbitals of the isolated atoms. The number of hybrid orbitals formed always equals the number of atomic orbitals used. 10 | 50

Hybrid orbitals are named by using the atomic orbitals that combined: • one s orbital + one p orbital gives two sp orbitals • one s orbital + two p orbitals gives three sp 2 orbitals • one s orbital + three p orbitals gives four sp 3 orbitals • one s orbital + three p orbitals + one d orbital gives five sp 3 d orbitals • one s orbital + three p orbitals + two d orbitals gives six sp 3 d 2 orbitals 10 | 51

Hybrid orbitals have definite directional characteristics, as described in Table 10. 2. 10 | 52

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To obtain the bonding description about any atom in a molecule: 1. Write the Lewis electron-dot formula. 2. Use VSEPR to determine the electron arrangement about the atom. 3. From the arrangement, deduce the hybrid orbitals. 4. Assign the valence electrons to the hybrid orbitals one at a time, pairing only when necessary. 5. Form bonds by overlapping singly occupied hybrid orbitals with singly occupied orbitals of another atom. 10 | 54

Let’s look at the methane molecule, CH 4. Simply using the atomic orbital diagram, it is difficult to explain its four identical C—H bonds. The valence bond theory allows us to explain this in two steps: promotion and hybridization. 10 | 55

First, the paired 2 s electron is promoted to the unfilled orbital. Now each orbital has one electron. Second, these orbitals are hybridized, giving four sp 3 hybrid orbitals. 10 | 56

? Use valence bond theory to describe the bonding about an N atom in N 2 F 4. The Lewis electron-dot structure shows three bonds and one lone pair around each N atom. They have a tetrahedral arrangement. A tetrahedral arrangement has sp 3 hybrid orbitals. 10 | 57

The orbital diagram of the ground-state N atom is 1 s 2 s 2 p The sp 3 hybridized N atom is sp 3 1 s Consider one N in N 2 F 4: the two N—F bonds are formed by the overlap of a half-filled sp 3 orbital with a half-filled 2 p orbital on F. The N—N bond forms from the overlap of a half-filled sp 3 orbital on each. The lone pair occupies one sp 3 orbital. 10 | 58

? Use valence bond theory to describe the bonding in the Cl. F 2 - ion. The valence orbital diagram for the Cl- ion is 3 s 3 p 3 d After the promotion to get two half-filled orbitals, the orbital diagram is 3 s 3 p 3 d 10 | 59

The sp 3 d hybridized orbital diagram is sp 3 d 3 d Two Cl—F bonds are formed from the overlap of two half-filled sp 3 d orbitals with half-filled 2 p orbitals on the F atom. These use the axial positions of the trigonal bipyramid. Three lone pairs occupy three sp 3 d orbitals. These are in the equatorial position of the trigonal bipyramid. 10 | 60

One hybrid orbital is required for each bond (whether a single or a multiple bond) and for each lone pair. Multiple bonding involves the overlap of one hybrid orbital and one (for a double bond) or two (for a triple bond) nonhybridized p orbitals. 10 | 61

To describe a multiple bond, we need to distinguish between two kinds of bonds. A s bond (sigma) has a cylindrical shape about the bond axis. It is formed either when two s orbitals overlap or with directional orbitals (p or hybrid), when they overlap along their axis. A p bond (pi) has an electron distribution above and below the bond axis. It is formed by the sideways overlap of two parallel p orbitals. This overlap occurs when two parallel half-filled p orbitals are available after s bonds have formed. 10 | 62

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Figure A illustrates the s bonds in C 2 H 4. The top of Figure B shows the p orbital on each carbon at 90° to each other, with no overlap. The bottom of Figure B shows parallel p orbitals overlapping to form a p bond. 10 | 64

In acetylene, C 2 H 2, each C has two s bonds and two p bonds. The s bonds form using the sp hybrid orbital on C. This is shown in part A. The two p bonds form from the overlap of two sets of parallel p orbitals. This is illustrated in Figure B. 10 | 65

The description of a p bond helps to explain the cistrans isomers of 1, 2 -dichloroethene. The overlap of the parallel p orbitals restricts the rotation around the C=C bond. This fixes the geometric positions of Cl: either on the same side (cis) or on different sides (trans) of the C=C bond. cis trans 10 | 66

One single bond and one triple bond requires two hybrid orbitals and two sets of two parallel p orbitals. That requires sp hybridization. 10 | 67

? Describe the bonding about the C atom in formaldehyde, CH 2 O, using valence bond theory. The electron arrangement is trigonal pyramidal using sp 2 hybrid orbitals. The ground-state orbital diagram for C is 1 s 2 s 2 p 10 | 68

After promotion, the orbital diagram is 1 s 2 s 2 p After hybridization, the orbital diagram is 1 s sp 2 2 p 10 | 69

The C—H s bonds are formed from the overlap of two C sp 2 hybrid orbitals with the 1 s orbital on the H atoms. The C—O s bond is formed from the overlap of one sp 2 hybrid orbital and one O half-filled p orbital. The C—O p bond is formed from the sideways overlap of the C 2 p orbital and an O 2 p orbital. 10 | 70

As atoms approach one another, their atomic orbitals overlap and form molecular orbitals. Molecular orbitals concentrated in regions between nuclei are called bonding orbitals. They are obtained by adding atomic orbitals. Molecular orbitals having zero values in regions between nuclei (and are in other regions) are called antibonding orbitals. They are obtained by subtracting atomic orbitals. 10 | 71

The figure on the next slide illustrates the combination of two 1 s orbitals. The top shows the formation of the bonding molecular orbital; the bottom shows the formation of the antibonding molecular orbital. 10 | 72

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Once a molecular orbital is formed, it can be occupied in the same way as are atomic orbitals. As an example, we’ll study hydrogen, H 2. Each hydrogen atom has one electron in a 1 s orbital, for a total of two electrons. When the two 1 s orbitals combine, they form two s molecular orbitals. 10 | 74

The bonding molecular orbital is lower in energy than the antibonding molecular orbital. The notation for a molecular orbital has the original atomic orbitals as subscript. The * indicates the antibonding molecular orbital. 10 | 75

The two electrons are placed in the lower-energy bonding molecular orbital with opposite spin. The ground-state molecular orbital electron configuration is (σ1 s)2. Excited states are also possible. 10 | 76

Bond order—the number of bonds that exist between two atoms—can be calculated using the equation nb = number of electrons in bonding orbitals na = number of electrons in antibonding orbitals 10 | 77

For H 2 (two electrons), the molecular electron configuration is (s 1 s)2. The bond order is ½(2 - 0) = 1. For H 2+ (one electron), the molecular electron configuration is (s 1 s)1. The bond order is ½(1 - 0) = ½. For He 2 (four electrons), the molecular electron configuration is (s 1 s)2 (s*1 s)2. The bond order is ½(2 - 2) = 0. This means no bond forms. 10 | 78

The strength of the interaction between two atomic orbitals to form molecular orbitals is determined by two factors: • The energy difference between the interacting orbitals • The magnitude of the overlap For the interaction to be strong, the energies of the two orbitals must be approximately equal and the overlap must be large. 10 | 79

Second-Period Homonuclear Diatomic Molecules In Li 2, the two 1 s atomic orbitals combine to form two s molecular orbitals: s 1 s and s*1 s. The two 2 s atomic orbitals combine to form two s molecular orbitals: s 2 s and s*2 s. Six electrons occupy the orbitals in order of increasing energy: 10 | 80

The molecular electron configuration is (s 1 s)2(s 1 s*)2(s 2 s)2 or KK(s 2 s)2 KK is the abbreviation for (s 1 s)2(s*1 s)2. The bond order is ½(4 - 2) = 1. 10 | 81

Be 2 has 8 electrons. The molecular orbital diagram is the same as for Li 2. The eight electrons occupy the orbitals in order of increasing energy: The molecular electron configuration is (s 1 s)2(s 2 s)2 or KK(s 2 s)2(s*2 s)2 The bond order is ½(4 - 4) = 0. 10 | 82

Molecular orbitals formed by the overlap of atomic p orbitals are of two types. When two p orbitals overlap end-to-end, two s molecular orbitals form: s 2 p and s*2 p. When two p orbitals overlap sideways, two p molecular orbitals form: p 2 p and p*2 p. Because two p orbitals on each atom overlap with two p orbitals on another atom, a total of four molecular orbitals form: two bonding and two antibonding. 10 | 83

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The relative energies of the molecular orbitals formed from second period atomic orbitals are shown here. The arrows are for N 2. 10 | 85

Because there are two bonding p orbitals, together they can hold four electrons. They are filled one electron per orbital before a second electron is added to the same orbital with opposite spin. 10 | 86

? Give the orbital diagram and electron configuration of the F 2 molecule. Is the molecular substance diamagnetic or paramagnetic? What is the order of the bond in F 2? 10 | 87

F 2 has 18 electrons. The KK shell holds 4 electrons so 14 remain. p 2 p The molecular electron configuration is KK(s 2 s)2(s*2 s)2(p 2 p)4(s 2 p)2 (p*2 p)4 The bond order is ½(8 - 6) = 1. The molecule is diamagnetic. 10 | 88

? A number of compounds of the nitrosonium ion, NO+, are known, including nitrosonium hydrogen sulfate, (NO+)(HSO 4 -). Use the molecular orbitals similar to those of homonuclear diatomic molecules and obtain the orbital diagram, electron configuration, bond order, and magnetic characteristics of the NO+ ion. Note: The stability of the ion results from the loss of an antibonding electron from NO. 10 | 89

NO+ has 14 electrons. The KK shell holds 4 electrons, leaving 10 electrons for bonding. p 2 p The molecular electron configuration is KK(s 2 s)2(s*2 s)2(p 2 p)4(s 2 p)2 The bond order is ½(8 - 2) = 3. The ion has a diamagnetic molecular orbital electron configuration. 10 | 90

Table 10. 3 (next slide) compares theoretical bond order to the experimental bond length, bond dissociation energy, and magnetic character for the second-period homonuclear diatomic molecules. 10 | 91

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Molecular orbital theory can describe delocalized bonding in terms of a single electron configuration. Besides having the conventional s bonds between two atoms, these molecules have molecular orbitals formed by more than two atomic orbitals. Ozone, O 3, is one example. 10 | 93

↔ Each oxygen has three localized electron pairs around it, suggesting that each atom uses sp 2 hybridized orbitals. 10 | 94

The overlap of one hybrid orbital on the central O atom with a hybrid orbital on each terminal O forms the two O—O s bonds. This leaves one hybrid orbital on the central O atom and two hybrid orbitals on the terminal O atoms. 10 | 95

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This leaves one unhybridized 2 p orbital on each of the three O atoms. Combining these three atomic orbitals forms three molecular orbitals: p antibonding, nonbonding, and p bonding. 10 | 97

All three of the molecular orbitals span the entire molecule. Both the bonding orbitals and the nonbonding orbitals (same energy as isolated atoms) are doubly occupied. This agrees with the resonance structure in which one delocalized pair is bonding and one is a lone pair (nonbonding) on the terminal atom. 10 | 98