Chapter 10 Molecular Geometry and Chemical Bonding Theory. Copyright Cengage Learning. All rights reserved. 10 1
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1 Chapter 10 Molecular Geometry and Chemical Bonding Theory Copyright Cengage Learning. All rights reserved. 10 1
2 Molecular geometry is the general shape of a molecule, as determined by the relative positions of the atomic nuclei. Copyright Cengage Learning. All rights reserved. 10 2
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. Copyright Cengage Learning. All rights reserved. 10 3
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 apart in three dimensions (a tetrahedral arrangment). Copyright Cengage Learning. All rights reserved. 10 4
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. Copyright Cengage Learning. All rights reserved. 10 5
6 Copyright Cengage Learning. All rights reserved. 10 6
7 These arrangements are illustrated below with balloons and models of molecules for each. Copyright Cengage Learning. All rights reserved. 10 7
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. Copyright Cengage Learning. All rights reserved. 10 8
9 The diagrams below illustrate molecular geometry and the impact of lone pairs on it for linear and trigonal planar electron-pair arrangements. Copyright Cengage Learning. All rights reserved. 10 9
10 Molecular geometries with a tetrahedral electronpair arrangement are illustrated below. Copyright Cengage Learning. All rights reserved
11 Molecular geometries for the trigonal bipyramidal electron-pair arrangement are shown on the next slide. Copyright Cengage Learning. All rights reserved
12 Copyright Cengage Learning. All rights reserved
13 Molecular geometries for the octahedral electronpair arrangement are shown below. Copyright Cengage Learning. All rights reserved
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. Copyright Cengage Learning. All rights reserved
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). Copyright Cengage Learning. All rights reserved
16 ? Use the VSEPR model to predict the geometries of the following molecules: a. AsF 3 b. PH 4 + c. BCl 3 Copyright Cengage Learning. All rights reserved
17 AsF 3 has 1(5) + 3(7) = 26 valence electrons; As is the central atom. The electron-dot formula is F As F F 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. Copyright Cengage Learning. All rights reserved
18 PH 4+ has 1(5) + 4(1) 1 = 9 valence electrons; P is the central atom. The electron-dot formula is H + H P H H 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. Copyright Cengage Learning. All rights reserved
19 BCl 3 has 1(3) + 3(7) = 24 valence electrons; B is the central atom. The electron-dot formula is Cl B Cl Cl 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. Copyright Cengage Learning. All rights reserved
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. Copyright Cengage Learning. All rights reserved
21 ? Using the VSEPR model, predict the geometry of the following species: a. ICl 3 b. ICl 4 - Copyright Cengage Learning. All rights reserved
22 ICl 3 has 1(7) + 3(7) = 28 valence electrons. I is the central atom. The electron-dot formula is Cl Cl I There are five regions: three bonding and two lone pairs. The electron-pair arrangement is trigonal bipyramidal. The geometry is T-shaped. Copyright Cengage Learning. All rights reserved Cl
23 ICl 4- has 1(7) + 4(7) + 1 = 36 valence electrons. I is the central element. The electron-dot formula is Cl - Cl I Cl Cl There are six regions around I: four bonding and two lone pairs. The electron-pair arrangement is octahedral. The geometry is square planar. Copyright Cengage Learning. All rights reserved
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. Copyright Cengage Learning. All rights reserved
25 The impact of lone pair(s) on bond angle for tetrahedral electron-pair arrangements has been experimentally determined. Copyright Cengage Learning. All rights reserved
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. Copyright Cengage Learning. All rights reserved
27 Dipole Moment A quantitative measure of the degree of charge separation in a molecule. Copyright Cengage Learning. All rights reserved
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. Copyright Cengage Learning. All rights reserved
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. Copyright Cengage Learning. All rights reserved
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 δ+ and δ- to indicate partial charges. Because Cl is more electronegative than H, it has the δ- charge, while H has the δ+ charge. δ+ δ- H Cl Copyright Cengage Learning. All rights reserved
31 The figure below shows the orbitals involved in HCl bond: the H 1s and the Cl 3p. Copyright Cengage Learning. All rights reserved
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 δ+ atoms and go to δ- 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. Copyright Cengage Learning. All rights reserved
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. O + + C O + H + O + H 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. Copyright Cengage Learning. All rights reserved
34 The relationship between molecular geometry and dipole moment is summarized in Table Copyright Cengage Learning. All rights reserved
35 Copyright Cengage Learning. All rights reserved
36 Polar molecules experience attractive forces between molecules; in response, they orient themselves in a δ+ to δ- 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. Copyright Cengage Learning. All rights reserved
37 We can see this illustrated with two compounds: cis-1,2-dichloroethene H Cl + C C H Cl trans-1,2-dichloroethene H Cl + + C + C + Cl H 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. Copyright Cengage Learning. All rights reserved
38 Copyright Cengage Learning. All rights reserved
39 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 T- shaped. Copyright Cengage Learning. All rights reserved
40 ? Which of the following molecules would be expected to have a zero dipole moment? a. GeF 4 b. SF 2 c. XeF 2 d. AsF 3 Copyright Cengage Learning. All rights reserved
41 GeF 4 : 1(4) + 4(7) = 32 valence electrons. Ge is the central atom. 8 electrons are bonding; 24 are nonbonding. Tetrahedral molecular geometry. F GeF 4 is nonpolar and has a zero dipole moment. F Ge F F Copyright Cengage Learning. All rights reserved
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. F S F SF 2 is polar and has a nonzero dipole moment. Copyright Cengage Learning. All rights reserved
43 XeF 2 : 1(8) + 2(7) = 22 valence electrons. Xe is the central atom. 4 electrons are bonding; 18 are nonbonding. Linear molecular geometry. F Xe F XeF 2 is nonpolar and has a zero dipole moment. Copyright Cengage Learning. All rights reserved
44 AsF 3 : 1(5) + 3(7) = 26 valence electrons. As is the central atom. 6 electrons are bonding; 20 are nonbonding. Trigonal pyramidal molecular geometry. F As F AsF 3 is polar and has a nonzero dipole moment. F Copyright Cengage Learning. All rights reserved
45 Which of the following molecules would be expected to have a zero dipole moment? a. GeF 4 tetrahedral molecular geometry zero dipole moment b. SF 2 bent molecular geometry nonzero dipole moment c. XeF 2 linear molecular geometry zero dipole moment d. AsF 3 trigonal pyramidal molecular geometry nonzero dipole moment Copyright Cengage Learning. All rights reserved
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