1 / 103

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. Review Session???. Some molecules have electron-dot structures that do not satisfy the octet rule. Some have an odd number of electrons, such as NO. Other molecules either have too few or too many electrons around the central atom.

glyn
Télécharger la présentation

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

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Final ExamF 12/119 am We have the room until 12 pm. Review Session???

  2. Some molecules have electron-dot structures that do not satisfy the octet rule. Some have an odd number of electrons, such as NO. Other molecules either have too few or too many electrons around the central atom.

  3. Elements that have too few electrons are in Groups IIA and IIIA. Be, B, and Al exhibit too few electrons around the central atom.

  4. There are many more examples of central atoms with more than an octet. Because elements of the third period and beyond have a d subshell, they can expand their valence electron configurations. S, P, Cl (as a central atom), and other elements in period 3 are examples of atoms in this situation. Elements in the second period, having only s and p subshells, are unable to do this.

  5. Give the Lewis formula of the IF5 molecule.

  6. Count the valence electrons in IF5: I 1(7) F 5(7) 42 valence electrons I is the central atom. Thirty-two electrons remain; they first complete F octets. The remaining electrons go on I.

  7. Bond length (or bond distance) is the distance between nuclei in a bond. Bond order is, defined in terms of the Lewis formula, the number of pairs of electrons in a bond. Bond length decreases as bond order increases.

  8. The Relation of Bond Order,Bond Length and Bond Energy Bond Bond Order Average Bond Average Bond Length (pm) Energy (kJ/mol) C O 1 143 358 C O 2 123 745 C O 3 113 1070 C C 1 154 347 C C 2 134 614 C C 3 121 839 N N 1 146 160 N N 2 122 418 N N 3 110 945 Table 9.4

  9. Consider the propylene molecule: 150 pm 134 pm One of the carbon–carbon bonds has a length of 150 pm; the other 134 pm. Identify each bond with a bond length. The shorter bond is the double bond; the longer bond is the single bond.

  10. Bond energy is the average enthalpy change for breaking the A—B bond in a molecule in the gas phase. Bond energy is a measure of bond strength: the larger the bond energy, the stronger the bond.

  11. Bond energies can be used to estimate the enthalpy change, DH, for a reaction. To do so, we imagine the reaction in two steps: breaking bonds and forming new bonds. DH = sum of the bond energies for bonds broken – sum of the bond energies for bonds formed When DH is negative, heat is released. When DH is positive, heat is absorbed.

  12. Estimate the enthalpy change for the following reaction, using bond energies: Bonds Broken: 1 C=C 602 kJ 1 Cl—Cl 240 kJ Absorbed 842 kJ Bonds Formed: 1 C—C 346 kJ 2 C—Cl 654 kJ Released 1000 kJ DH = 842 kJ – 1000 kJ DH = –158 kJ

  13. The formal charge on an atom in the Lewis formula is the hypothetical charge you obtain by assuming that bonding electrons are equally shared between bonded atoms and that the electrons of each lone pair belong completely to one atom.

  14. Formal charge = valence electrons on free atom – ½ (number of electrons in bonds) – (number of lone-pair electrons) The sum of the formal charges on the atoms equals the charge on the formula.

  15. Formal charges can help to determine the most likely electron-dot formula using three rules: 1. Whenever you can write several Lewis formulas for a molecule, choose the one having the lowest magnitudes of formal charges. 2. When two proposed Lewis formulas have the same magnitudes of formal charges, choose the one having the negative formal charge on the more electronegative atom. 3. When possible, choose Lewis formulas that do not have like charges on adjacent atoms.

  16. The left structure is better. • Compare the formal charges for the following electron-dot formulas of CO2. Formal charge = group number – (number of bond pairs) – (number of nonbonding electrons) For the left structure: For the right structure: C: 4 – 4 – 0 = 0 C: 4 – 4 – 0 = 0 O: 6 – 1 – 6 = –1 O: 6 – 2 – 4 = 0 O: 6 – 3 – 2 = +1

  17. Other Resources • Visit the student website at http://www.college.hmco.com/pic/ebbing9e

  18. Contents and Concepts Molecular Geometry and Directional Bonding We can predict the molecular geometry of a molecule—that is, its general shape as determined by the relative positions of atomic nuclei—with a simple model: the valence-shell electron-pair repulsion model. After exploring molecular geometry, we explain chemical bonding by means of valence bond theory, which gives us insights into why bonds form and why they have definite directions in space, giving particular molecular geometries.

  19. The Valence-Shell Electron-Pair Repulsion (VSEPR) Model • Dipole Moment and Molecular Geometry • Valence Bond Theory • Description of Multiple Bonding

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

  21. 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.

  22. 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).

  23. 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.

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

  25. 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.

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

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

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

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

  30. 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.

  31. 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).

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

  33. 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.

  34. 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.

  35. 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.

  36. 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.

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

  38. 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.

  39. 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.

  40. 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.

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

  42. 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.

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

  44. 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.

  45. 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.

  46. 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.

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

More Related