Download
final exam f 12 11 9 am we have the room until 12 pm n.
Skip this Video
Loading SlideShow in 5 Seconds..
Final Exam F 12/11 9 am We have the room until 12 pm. PowerPoint Presentation
Download Presentation
Final Exam F 12/11 9 am We have the room until 12 pm.

Final Exam F 12/11 9 am We have the room until 12 pm.

189 Views Download Presentation
Download Presentation

Final Exam F 12/11 9 am We have the room until 12 pm.

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. Final ExamF 12/119 am We have the room until 12 pm. Review Session R 12/10 2 pm SL 110

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

  3. 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.

  4. 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).

  5. 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.

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

  7. 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 pyramidal. The geometry is T-shaped.

  8. 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.

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

  11. 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.

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

  13. 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.

  14. 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.

  15. 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.

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

  17. 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.

  18. 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.

  19. 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.

  20. 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.

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

  22. 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.

  23. 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.

  24. 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.

  25. 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.

  26. 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

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

  28. 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.

  29. 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.

  30. 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.

  31. The sp Hybrid Orbitals in Gaseous BeCl2

  32. 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 sp2 orbitals • • one s orbital + three p orbitals gives four sp3 orbitals • • one s orbital + three p orbitals + one d orbital gives five sp3d orbitals • • one s orbital + three p orbitals + two d orbitals gives six sp3d2 orbitals

  33. Hybrid orbitals have definite directional characteristics, as described in Table 10.2.

  34. 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.

  35. Let’s look at the methane molecule, CH4. 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.

  36. First, the paired 2s electron is promoted to the unfilled orbital. Now each orbital has one electron. Second, these orbitals are hybridized, giving four sp3 hybrid orbitals.

  37. The sp3 Hybrid Orbitals in NH3 and H2O

  38. The sp3d Hybrid Orbitals in PCl5

  39. The sp3d2 Hybrid Orbitals in SF6 Sulfur Hexafluoride -- SF6

  40. Use valence bond theory to describe the bonding about an N atom in N2H4. 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 sp3 hybrid orbitals.

  41. 1s 1s 2s 2p sp3 • The orbital diagram of the ground-state N atom is • The sp3 hybridized N atom is • Consider one N in N2F4: the two N—F bonds are formed by the overlap of a half-filled sp3 orbital with a half-filled 2p orbital on F. The N—N bond forms from the overlap of a half-filled sp3 orbital on each. The lone pair occupies one sp3 orbital.