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Chapter 10 Molecular Geometry and Chemical Bonding Theory PowerPoint Presentation
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Chapter 10 Molecular Geometry and Chemical Bonding Theory
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  1. Chapter 10Molecular Geometry and Chemical Bonding Theory

  2. 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.

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

  4. 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.

  5. 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).

  6. 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.

  7. These arrangements are illustrated below with balloons and models of molecules for each.

  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.

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

  10. Molecular geometries with a tetrahedral electron-pair arrangement are illustrated below.

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

  12. Molecular geometries for the octahedral electron-pair arrangement are shown below.

  13. 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.

  14. Predicting Molecular Geometry Using VSEPR • Write the electron-dot formula from the formula. • Based on the electron-dot formula, determine the number of electron pairs around the central atom (including bonding and nonbonding pairs). • Determine the arrangement of the electron pairs about the central atom (Figure 10.3). • Obtain the molecular geometry from the directions of the bonding pairs for this arrangement (Figure 10.4).

  15. Use the VSEPR model to predict the geometries of the following molecules: • a. AsF3 • b. PH4+ • c. BCl3

  16. AsF3 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.

  17. PH4+ 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.

  18. BCl3 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.

  19. 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.

  20. Using the VSEPR model, predict the geometry of the following species: • a. ICl3 • b. ICl4-

  21. ICl3 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.

  22. ICl4- 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.

  23. 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.

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

  25. 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.

  26. Dipole Moment A quantitative measure of the degree of charge separation in a molecule.

  27. 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.

  28. 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.

  29. 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 d- charge, while H has the d+ charge.

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

  31. 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.

  32. To illustrate this process, we use arrows with a + on one end of the arrow. We’ll look at CO2 and H2O. CO2 is linear, and H2O is bent. The vectors add to zero (cancel) for CO2. Its dipole moment is zero. For H2O, a net vector points up. Water has a dipole moment.

  33. The relationship between molecular geometry and dipole moment is summarized in Table 10.1.

  34. 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.

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

  36. The formula AX3 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.

  37. Which of the following molecules would be expected to have a zero dipole moment? • a. GeF4 • b. SF2 • c. XeF2 • d. AsF3

  38. GeF4: 1(4) + 4(7) = 32 valence electrons. Ge is the central atom. 8 electrons are bonding; 24 are nonbonding. Tetrahedral molecular geometry. GeF4 is nonpolar and has a zero dipole moment.

  39. SF2: 1(6) + 2(7) = 20 valence electrons. S is the central atom. 4 electrons are bonding; 16 are nonbonding.Bent molecular geometry. SF2 is polar and has a nonzero dipole moment.

  40. XeF2: 1(8) + 2(7) = 22 valence electrons. Xe is the central atom. 4 electrons are bonding; 18 are nonbonding.Linear molecular geometry. XeF2 is nonpolar and has a zero dipole moment.

  41. AsF3: 1(5) + 3(7) = 26 valence electrons. As is the central atom. 6 electrons are bonding; 20 are nonbonding.Trigonal pyramidal molecular geometry. AsF3 is polar and has a nonzero dipole moment.

  42. Which of the following molecules would be expected to have a zero dipole moment? • a. GeF4 tetrahedral molecular geometryzero dipole moment • b. SF2 bent molecular geometry nonzero dipole moment • c. XeF2 linear molecular geometryzero dipole moment • d. AsF3 trigonal pyramidal molecular geometry nonzero dipole moment

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

  44. 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.

  45. 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.

  46. 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.