1 / 35

Chapter 9 Review

Chapter 9 Review. Brady & Senese, 5th Ed. Periodic Trends in Lattice Energy. Coulomb’s Law. charge A X charge B. electrostatic force a. distance 2. energy = force X distance therefore. charge A X charge B. electrostatic energy a. distance.

vesna
Télécharger la présentation

Chapter 9 Review

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. Chapter 9 Review Brady & Senese, 5th Ed

  2. Periodic Trends in Lattice Energy Coulomb’s Law charge A X charge B electrostatic force a distance2 energy = force X distance therefore charge A X charge B electrostatic energy a distance cation charge X anion charge a DH0lattice electrostatic energy a cation radius + anion radius

  3. Figure 9.7 Trends in lattice energy.

  4. Covalent bond formation in H2. Figure 9.11

  5. Distribution of electron density of H2. Figure 9.12

  6. Strong covalent bonding forces within molecules Weak intermolecular forces between molecules Figure 9.14 Strong forces within molecules and weak forces between them.

  7. Figure 9.19 The Pauling electronegativity (EN) scale.

  8. Figure 9.20 Electronegativity and atomic size.

  9. PROBLEM: (a) Use a polar arrow to indicate the polarity of each bond: N-H, F-N, I-Cl. (b) Rank the following bonds in order of increasing polarity: H-N, H-O, H-C. SAMPLE PROBLEM 9.4 Determining Bond Polarity from EN Values PLAN: (a) Use Figure 9.19(button at right) to find EN values; the arrow should point toward the negative end. (b) Polarity increases across a period. SOLUTION: (a) The EN of N = 3.0, H = 2.1; F = 4.0; I = 2.5, Cl = 3.0 N - H F - N I - Cl (b) The order of increasing EN is C < N < O; all have an EN larger than that of H. H-C < H-N < H-O

  10. Figure 9.21 Electron density distributions in H2, F2, and HF.

  11. 3.0 DEN 2.0 0.0 Figure 9.22 Boundary ranges for classifying ionic character of chemical bonds.

  12. Figure 9.23 Percent ionic character of electronegativity difference (DEN).

  13. Figure 9.24 Properties of the Period 3 chlorides.

  14. Figure 9.25 Electron density distributions in bonds of the Period 3 chlorides.

  15. Steps for converting a molecular formula into a Lewis structure: • The atom requiring the most additional valence e- is placed centrally. If multiple atoms require the same number, place the larger one centrally. • Sum up the total valence e- • Connect the surrounding atoms to the central atom with single bonds. • Place the remaining e- as lone pairs (starting with the surrounding atoms) to satisfy the octet rule. • If you run out of e- before the octet rule has been satisfied (where applicable) convert one or more lone pairs to multiple bonds. Common octet rule exceptions: • H, Be and B form covalent compounds with less than 8 valence e-. They are stable with 2,4 and 6 valence e- respectively. • Elements in the 3rd and below may sometimes exceed the octet rule having as many as 10 or 12 valence e-. • Compounds with an odd number of valence e- are not able to satisfy the octet rule. These compounds are called radicals.

  16. Molecular formula For NF3 Atom placement : : N 5e- : F : : F : : Sum of valence e- F 7e- X 3 = 21e- N Total 26e- : F : : Remaining valence e- Lewis structure

  17. H H H H C C C C H H H H N : N : . . N : N : N : N : . . . . SAMPLE PROBLEM 10.3 Writing Lewis Structures for Molecules with Multiple Bonds. PROBLEM: Write Lewis structures for the following: (a) Ethylene (C2H4), the most important reactant in the manufacture of polymers (b) Nitrogen (N2), the most abundant atmospheric gas PLAN: For molecules with multiple bonds, there is a Step 5which follows the other steps in Lewis structure construction. If a central atom does not have 8e-, an octet, then two e- (either single or nonbonded pair)can be moved in to form a multiple bond. SOLUTION: (a) There are 2(4) + 4(1) = 12 valence e-. H can have only one bond per atom. : (b) N2 has 2(5) = 10 valence e-. Therefore a triple bond is required to make the octet around each N.

  18. A - central atom X -surrounding atom E -nonbonding valence electron-group integers VSEPR - Valence Shell Electron Pair Repulsion Theory Each group of valence electrons around a central atom is located as far away as possible from the others in order to maximize repulsions. These repulsions maximize the space that each object attached to the central atom occupies. The result is five electron-group arrangements of minimum energy seen in a large majority of molecules and polyatomic ions. The electron-groups are defining the object arrangement,but the molecular shape is defined by the relative positions of the atomic nuclei. Because valence electrons can be bonding or nonbonding, the same electron-group arrangement can give rise to different molecular shapes. AXmEn

  19. linear tetrahedral trigonal planar trigonal bipyramidal octahedral Figure 10.2 Electron-group repulsions and the five basic molecular shapes.

  20. Figure 10.3 The single molecular shape of the linear electron-group arrangement. Examples: CS2, HCN, BeF2

  21. Class Shape Figure 10.4 The two molecular shapes of the trigonal planar electron-group arrangement. Examples: SO2, O3, PbCl2, SnBr2 Examples: SO3, BF3, NO3-, CO32-

  22. 1220 Effect of Double Bonds 1160 real Effect of Nonbonding(Lone) Pairs Factors Affecting Actual Bond Angles Bond angles are consistent with theoretical angles when the atoms attached to the central atom are the same and when all electrons are bonding electrons of the same order. 1200 larger EN 1200 ideal greater electron density Lone pairs repel bonding pairs more strongly than bonding pairs repel each other. 950

  23. The three molecular shapes of the tetrahedral electron-group arrangement. Figure 10.5 Examples: CH4, SiCl4, SO42-, ClO4- NH3 PF3 ClO3 H3O+ H2O OF2 SCl2

  24. Lewis structures and molecular shapes. Figure 10.6

  25. Figure 10.7 The four molecular shapes of the trigonal bipyramidal electron-group arrangement. PF5 AsF5 SOF4 SF4 XeO2F2 IF4+ IO2F2- XeF2 I3- IF2- ClF3 BrF3

  26. Figure 10.8 The three molecular shapes of the octahedral electron-group arrangement. SF6 IOF5 BrF5 TeF5- XeOF4 XeF4 ICl4-

  27. A summary of common molecular shapes with two to six electron groups. Figure 10.9

  28. PROBLEM: Draw the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) PF3 and (b)COCl2. SOLUTION: (a) For PF3 - there are 26 valence electrons, 1 nonbonding pair SAMPLE PROBLEM 10.6 Predicting Molecular Shapes with Two, Three, or Four Electron Groups The shape is based upon the tetrahedral arrangement. The F-P-F bond angles should be <109.50 due to the repulsion of the nonbonding electron pair. The final shape is trigonal pyramidal. <109.50 The type of shape is AX3E

  29. 124.50 1110 SAMPLE PROBLEM 10.6 Predicting Molecular Shapes with Two, Three, or Four Electron Groups continued (b) For COCl2, C has the lowest EN and will be the center atom. There are 24 valence e-, 3 atoms attached to the center atom. C does not have an octet; a pair of nonbonding electrons will move in from the O to make a double bond. Type AX3 The shape for an atom with three atom attachments and no nonbonding pairs on the central atom is trigonal planar. The Cl-C-Cl bond angle will be less than 1200 due to the electron density of the C=O.

  30. PROBLEM: Determine the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) SbF5 and (b) BrF5. SOLUTION: (a) SbF5 - 40 valence e-; all electrons around central atom will be in bonding pairs; shape is AX5 - trigonal bipyramidal. SAMPLE PROBLEM 10.7 Predicting Molecular Shapes with Five or Six Electron Groups (b) BrF5 - 42 valence e-; 5 bonding pairs and 1 nonbonding pair on central atom. Shape is AX5E, square pyramidal.

  31. PROBLEM: Determine the shape around each of the central atoms in acetone, (CH3)2C=O. PLAN: Find the shape of one atom at a time after writing the Lewis structure. tetrahedral tetrahedral trigonal planar >1200 <1200 SAMPLE PROBLEM 10.8 Predicting Molecular Shapes with More Than One Central Atom SOLUTION:

  32. ethanol ethane CH3CH2OH CH3CH3 The tetrahedral centers of ethane and ethanol. Figure 10.11

  33. Electric field OFF Electric field ON The orientation of polar molecules in an electric field.

  34. PROBLEM: From electronegativity (EN) values (button) and their periodic trends, predict whether each of the following molecules is polar and show the direction of bond dipoles and the overall molecular dipole when applicable: PLAN: Draw the shape, find the EN values and combine the concepts to determine the polarity. SOLUTION: (a) NH3 SAMPLE PROBLEM 10.9 Predicting the Polarity of Molecules (a) Ammonia, NH3 (b) Boron trifluoride, BF3 (c) Carbonyl sulfide, COS (atom sequence SCO) The dipoles reinforce each other, so the overall molecule is definitely polar. ENN = 3.0 ENH = 2.1 molecular dipole bond dipoles

  35. SAMPLE PROBLEM 10.10 Predicting the Polarity of Molecules continued (b) BF3 has 24 valence e- and all electrons around the B will be involved in bonds. The shape is AX3, trigonal planar. F (EN 4.0) is more electronegative than B (EN 2.0) and all of the dipoles will be directed from B to F. Because all are at the same angle and of the same magnitude, the molecule is nonpolar. 1200 (c) COS is linear. C and S have the same EN (2.0) but the C=O bond is quite polar(DEN) so the molecule is polar overall.

More Related