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Chapter 10

Chapter 10. The Shapes of Molecules. Lewis Structures. Mostly using the Octet Rule, but there are exceptions The best rules are in your handouts as well as in your lecture notes and following. Use them!. THE BEST RULES for writing Lewis Electron Dot structures for covalent compounds:.

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Chapter 10

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  1. Chapter 10 The Shapes of Molecules

  2. Lewis Structures • Mostly using the Octet Rule, but there are exceptions • The best rules are in your handouts as well as in your lecture notes and following. Use them!

  3. THE BEST RULES for writing Lewis Electron Dot structures for covalent compounds: 1. Write the Lewis structure for each atom, then total the number of valence electrons. DO NOT worry about which electrons go with which atom! For example, CO2 would have a total of 16 valence electrons. 2. Put the atom that is "leftist &/or lowest" as the central atom. Arrange the other atoms around it. Then draw one single covalent bond of 2 electrons from the central atom to each of the other atoms around it. Subtract the number of electrons in bonds from the total found in step 1. In the CO2 example, C is central and O’s are terminal. The electron counting is: 16 - 4 = 12. 3. Put the remaining electrons around the terminal atoms as lone pairs of electrons until they are “happy,” except for H. (Why?) If there are still some electrons left they go on the central atom.

  4. THE BEST RULES for writing Lewis Electron Dot structures for covalent compounds: 4. See if each atom has 8 electrons around it. If not, you take one lone pair on a terminal atom and make it a bond pair, then another, etc., until all the atoms are happy. (This is what makes multiple bonds, double or triple is possible, quadruple is not!) In CO2 our O's are happy but C is not. Take a LP from each terminal O and make it a bond pair. 5. Check again that you have a total of 16 electrons and that all atoms “think” that they have 8 electrons. If it is a polyatomic ion, you must account for the charge by adding in extra electrons in step 1 if it is an anion, or subtracting if it is a cation, like NH4+. The bracket must be placed around the structure and the charge indicated.

  5. PROBLEM: Write a Lewis structure for CCl2F2. : : : Cl : : : : Cl C F : : : : F : SAMPLE PROBLEM 10.1 Writing Lewis Structures for Molecules with One Central Atom SOLUTION: Cl Steps 1, 2 & 3: There are 32 total valence e-s. Put C in center & arrange other terminal atoms around it. Each bond is 2 e-s, or 8 e-s. 32-8=24 e-s to go as LP’s. There is room enough for all 24 e’s, so none go on the central atom. Cl C F F Steps 4 & 5: Each atom has 8 e-s, and the total is still 32. Note that bonds can be shown as lines, but lone pairs of electrons are always dots!

  6. More Complex Lewis Structures Many compounds have more than one central atom, especially organic compounds where all the carbon atoms are “central” and all H’s are terminal. With multiple bonds, consider how many single bonds each atom would usually “like” to make: C = 4, O = 2, N = 3, H and F always = 1, halogens “like” 1 H2O2 has two O atoms as central. Draw it that way and distribute the H’s equally around the two central atoms. Keep using the rules! .. .. H-O-O-H . . . .

  7. PROBLEM: Write the Lewis structure for methanol (molecular formula CH3OH). SAMPLE PROBLEM 10.2 Writing Lewis Structure for Molecules with More than One Central Atom SOLUTION: Hydrogen can have only one bond so C and O must be next to each other with H’s around them. There are 4(1) + 4 + 6 = 14 valence e-s. C likes to have 4 bonds and O likes to have 2, therefore O gets the 2 Lone Pairs of e-s. H : H C O H : H

  8. SAMPLE PROBLEM 10.3 Writing Lewis Structures for Molecules with Multiple Bonds. PROBLEM: Write Lewis structure for Ethene (C2H4), the most important reactant in the manufacture of polymers and also for N2. (a) There are 2(4) + 4(1) = 12 valence e-. H can have only one bond per atom. C wants to make 4 bonds. SOLUTION: H H H H : C C C C H H H H

  9. Diatomic nitrogen N2 has 10 valence e-s N-N 10-2=8 e-s remain .. .. :N-N: 8 – 8 = 0 e-s, but neither N happy Move one LP from each N in to become a shared pair or covalent bond: :N=N: Triple bond makes N happy

  10. Lewis Structures PRACTICE: OF2, H2CO, NO2, O3, I3-, HCOOH, CH3CH2OH, SF4, NO, C6H6 in a ring EXCEPTIONS: H2 has only two e-s each - not an octet. Why is that OK? I3- and SF4 have EXPANDED valence shell - central atom has 10 e-s NO has a lone e- called a FREE RADICAL, makes it a very reactive compound

  11. Lewis Structures O3 and C6H6 can be written in two different ways: which is the correct way? BOTH ARE! RESONANCE STRUCTURES: resonance is an OLD concept: e-s don't bounce around bond to bond, but are continually being shared.

  12. is used to indicate that resonance occurs. Resonance: Delocalized Electron-Pair Bonding O3 can be drawn in 2 ways - Neither structure is actually correct but can be drawn to represent a structure which is a hybrid of the two - a resonance structure. Resonance structures have the same relative atom placement but a difference in the locations of bonding and nonbonding electron pairs.

  13. SAMPLE PROBLEM 10.4 Writing Resonance Structures PROBLEM: Write resonance structures for the nitrate ion, NO3-. SOLUTION: Nitrate has 1(5) + 3(6) + 1 = 24 valence e- 24 – 6 = 18 18 – 18 = 0 N does not have an octet; a pair of e- will move in to form a double bond.

  14. For OC For OA # valence e- = 6 # valence e- = 6 # nonbonding e- = 6 # nonbonding e- = 4 # bonding e- = 2 X 1/2 = 1 # bonding e- = 4 X 1/2 = 2 For OB Formal charge = -1 Formal charge = 0 # valence e- = 6 # nonbonding e- = 2 # bonding e- = 6 X 1/2 = 3 Formal charge = +1 Formal Charge: Selecting the Best Resonance Structure An atom “owns” all of its nonbonding electrons and half of its bonding electrons. Formal charge is the charge an atom would have if the bonding electrons were shared equally. Formal charge of atom = # valence e- - (# unshared electrons + 1/2 # shared electrons)

  15. Three criteria for choosing the more important resonance structure: Resonance (continued) Smaller formal charges (either positive or negative) are preferable to larger charges. Avoid like charges (+ + or - - ) on adjacent atoms. A more negative formal charge should exist on an atom with a larger EN value.

  16. A B C Resonance (continued) EXAMPLE: NCO- has 3 possible resonance forms - formal charges -2 0 +1 -1 0 0 0 0 -1 Forms B and C have negative formal charges on N and O; this makes them more preferred than form A. Form C has a negative charge on O which is the more electronegative element, therefore form C contributes the most to the resonance hybrid.

  17. PROBLEM: Write Lewis structures for (a) H3PO4 (pick the most likely structure); (b) BFCl2. SAMPLE PROBLEM 10.5 Writing Lewis Structures for Octet Rule Exceptions SOLUTION: (a) H3PO4 has two resonance forms and formal charges indicate the more important form. -1 (b) BFCl2 will have only 1 Lewis structure. B is e- deficient! 0 +1 0 0 0 0 0 0 0 0 0 0 0 more stable 0 0 Lower formal charges (plus O likes to make 2 bonds). P has expanded valence shell.

  18. Localized electron bonding model: Assumes that a molecule is composed of atoms that are bound together by sharing pairs of e-s using the atomic orbitals of the bound atoms - shared pairs are called bonding pairs - those found on one atom are called lone pairs This basic model was seen with our drawing of Lewis structures It is also the basis of VSEPR Theory

  19. VSEPR THEORY: VALENCE SHELL ELECTRON PAIR REPULSION THEORY Methane’s Lewis dot structure is 2-D and looks flat with 90o bond angles. Methane in reality, in 3-D, is a regular tetrahedron with bond angles = 109° Are CH3Cl and CH2Cl2 equivalent geometric structures? Molecular Shape = function of repulsion of electron pairs in valence shell of central atom(s) Electron pairs try to arrange themselves as far apart as possible to minimize repulsions • this includes LP's as well as BP's - to minimize electrostatic repulsion.

  20. VSEPR THEORY: VALENCE SHELL ELECTRON PAIR REPULSION THEORY This leads to five major families of e- pair arrangements: linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral. When drawing these we use AXbEc where:  A= central atom X (or B) = surrounding atom(s) E = nonbonding valence e- group (lone pair)

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

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

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

  24. 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

  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. Atomic bond angles: Bond angles are important in determining a compound's chemical behavior If all "ligands" are equal, each structural shape has specific bond angles Bond angles are affected if there's a multiple bond, a variety of ligands, and by lone pairs LP's affect bond angles and the shape of the molecule Stronger electrostatic repulsions: LP-LP > LP-BP > BP-BP

  28. “Electron Pair Families - Minor” TRIGONAL PLANARGROUP: - each has 3 e- pairs around central atom, at corners of a trigonal planar arrangement - includes AX3 (trigonal planar), AX2E (bent) LINEAR GROUP: - each has 2 e- pairs around central atom in linear arrangement - includes AX2 (linear) and ABX (linear)

  29. “Electron Pair Families” TETRAHEDRAL GROUP: - each has 4 e- pairs around central atom, at corners of a tetrahedron - includes AX4 (tetrahedral), AX3E (trig pyr), AX2E2 (bent) TRIGONAL BIPYRAMIDAL GROUP: - each has 5 e- pairs around central atom, 3 on equatorial and 2 on axial - includes AX5 (trig bipyr), AX4E (see-saw), AX3E2 (T-shaped), AX2E3 (linear)

  30. “Electron Pair Families” OCTAHEDRAL GROUP: - each has 6 e- pairs around central atom, at equal angles and distances - includes AX6 (octahedral), AX5E (square pyramidal), AX4E2 (square planar), and AX3E3 (T-shaped).

  31. Bonding and Shape Multiple Bonds: a double or triple bonded molecule has a predictable shape because they are counted as one group of e-s, connecting B to A. Because there are more e-s, they will repel more than single BP's. Roughly LP > TBP > DBP > BP If a molecule has a resonance structure, bonds sharing the resonance are equal in repelling LP's or BP's

  32. Figure 10.9 from 4th ed. A summary of common molecular shapes with two to six electron groups.

  33. See Figure 10.1 The steps in determining a molecular shape. Figure 10.10 Molecular formula Step 1 Lewis structure Count all e- groups around central atom (A) Step 2 Electron-group arrangement Note lone pairs and double bonds Step 3 Count bonding and nonbonding e- groups separately. Bond angles Step 4 Molecular shape (AXmEn)

  34. 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 When I ask for molecular geometry, write the name not the type!

  35. 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.

  36. 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.

  37. Multiple Central Atoms Molecules with more than one central atoms, like hydrocarbons, alcohols, thiosulfate ion, etc: apply VSEPR to each central atom and determine the shape around it Try methanol CH3OH (Note: methanol is organic and the formula is written to help perceive the structure, so the formula tells you that C has 3 H’s with it and O has 1 H.)

  38. 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:

  39. ethanol ethane CH3CH2OH CH3CH3 The tetrahedral centers of ethane and ethanol. Figure 10.10

  40. You practice finding Lewis structure, e- pair arrangement and molecular geometry • Try CO2, SO2, SO3, SF4, SF6, PCl3, PCl5, XeF4, I3-

  41. Dipole Moments & Electronegativity Refer to handout and recall electronegativity and polar covalent bonding from last chapter If a bond is polar covalent, one end is slightly negative and the other is slightly positive Leads to a dipole moment, which can respond to an external electrical field

  42. Electric field OFF Electric field ON The orientation of polar molecules in an electric field. Figure 10.11

  43. Molecular Dipole Moment DIPOLE MOMENT: polar covalent bonds cause a molecule to have a dipole moment, unless the polarity vectors cancel each other in a symmetrical arrangement (see handout) Nonpolar (CCl4, PCl5, SF6) If a molecule is not symmetric, it will probably have a dipole moment (BrF3, OF2, NH3, SF4)

  44. 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

  45. 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.

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