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Topic # 8 Chemical Bonding

Topic # 8 Chemical Bonding. Bonding. Intra. Inter. Within compounds STRONG. between molecules of a compound WEAK. Intra. Ionic. Covalent. Strong electrostatic forces between ions Betw. Metal & Nonmetal Crystal lattice is formed.

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Topic # 8 Chemical Bonding

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  1. Topic # 8 Chemical Bonding

  2. Bonding Intra Inter Within compounds STRONG between molecules of a compound WEAK

  3. Intra Ionic Covalent Strong electrostatic forces between ions Betw. Metal & Nonmetal Crystal lattice is formed Want to share electrons to achieve full outershell Molecules are formed. Betw. Two non-metals

  4. Lewis Structures Lewis structures use dots to represent outer shell electrons in atoms when they form molecules.

  5. Rules for Drawing Lewis Structures • Calculate the total # of outer shell electrons • Decide which atom is central (least electronegative) adding electrons to fill central atoms’ octet. • Add electrons to give outside atoms full shells, any extra electrons, add to central atom. • Form multiple bonds so central has octet • Draw lines to represent one shared pair of electrons

  6. Exceptions to Octet Rule BF3 F B F F Electron Deficit Atoms

  7. F B F F N H H H Electron Deficient Atom: Boron does not have an octet These structures tend to attract other species with a non-bonded pair of electrons!

  8. F F F Some nonmetals have more than Octet S F F F I3- I I I

  9. Formal Charge Using oxidation numbers for assigning charges in molecules may be useful for redox reactions but can lead to an over estimation of actual charges. A more accurate charge assessment can be used in considering Lewis structures. FC is the difference between the number of valence electrons on the free atom and the number of valence electrons assigned to the atom in the molecule.

  10. Calculating Formal Charge = (# valence electrons) -( # of Unshared electrons ) - ½(shared electrons)

  11. N = 5 – 1 – ½(6) = +1 O = 6 -  4 – 1/2(4) = 0 double bonded oxygen O = 6 – 6 – ½(2) = -1           

  12. Which structure is correct? The one with no formal charges, or the smallest formal charges! (assign formal charges, now, to figure it out!)

  13. Two different structures can be drawn. Resonance structures O - S = O O = S - O O S O This results in an average of 1.5 bonds between each S and O.

  14. Resonance structures • Benzene, C6H6, is another example of a compound for which resonance structure must be written. • All of the bonds are the same length. or

  15. Structure for Ozone

  16. Resonance Structure for NO3- Arrows do NOT mean that the ion ‘flips’ from one to another, the actual structure is an average of the three!

  17. Bond Length The distance between two nuclei that includes the shared pair of electrons. The actual measured distance between the two nuclei is called the bond length. Includes the centers of + charge with a shared electrons in between. Double and Triple bonds make this length shorter than a single bond. However, in Nitrate Polyatomic Ion, the bond lengths are the SAME between all three nitrogen-oxygen!

  18. Bond Types: Covalent Polar Covalent Polar Dative or Co-ordinate Bonding………

  19. α + + α- covalent - Polar Covalent - + - + Ionic

  20. How can the degree of polarity be predicted? Use Electronegativites to make predictions about the type of bonding in a compound. The greater the difference in electronegativity of the two bonding elements, the greater is the chance for ionic bonding. The electronegativity difference between two covalently bonded atoms determines the polarity of the bond: The larger the difference, the more polar is the covalent bond. If the difference is zero or very small, the bond is essentially non-polar.

  21. We can use the difference in electronegativity between two atoms to gauge the polarity of the bonding between them

  22. Look at charts available on worksheets Table of Percent Ionic Character Topic 8 Bonding (Includes acceptable range)

  23. Dative or Co-ordinate Bonding: Where one atom does not offer any electrons within the bond. Electrons are only offered by one atom. Oxygen commonly uses Dative Bonding.

  24. Dipole Moments A method of describing charge concentrations. If an molecule has a clear center of positive charge away from a center of negative charge, it has a measured dipole moment.

  25. (Molecule in electrical field) Arrow starts at LOWER electronegative atom!

  26. These dipoles cancel out, showing no effective charge on molecule.

  27. VSEPR Shapes of Molecules! Covalent: determined by the number of electron pairs around a central atom Remember: electron pairs REPEL each other! They will try to get away from each other! Valence Shell Electron Pair Repulsion Theory lists a set of shapes per # of electron pairs.

  28. What Is VSEPR? The Valence Shell Electron Pair Repulsion (VSEPR) model: *is based on the number of regions of high electron density around a central atom. *can be used to predict structures of molecules or ions that contain only non-metals by minimizing the electrostatic repulsion between the regions of high electron density. *can also be used to predict structures of molecules or ions that contain multiple bonds or unpaired electrons. *does fail in some cases.

  29. VSEPR Rules 1.Draw the Lewis structure for the molecule or ion. 2. Count the total number of regions of high electron density (bonding and unshared electron pairs) around the central atom. a. Double and triple bonds count as ONE REGION OF HIGH ELECTRON DENSITY. b. An unpaired electron counts as ONE REGION OF HIGH ELECTRON DENSITY. c. For molecules or ions that have resonance structures, you may use any one of the resonance structures.

  30. 3. Identify the most stable arrangement of the regions of high electron density as ONE of the following: linear trigonal planar tetrahedral trigonal bipyramidal octahedral

  31. Observe Chart on page 10 in Dingle Notes

  32. Lesson 14 used VSEPR to describe molecular geometry. Water was described as linear, 109.5o

  33. Why isn’t waters’ structure Tetrahedral with bond angles of 109o ? Because of the two lone pairs of electrons Reduces the bond angles

  34. A step beyond VSEPR The localized Electron Model Electron Cloud Hybridization * describes how the role of orbitals can mold the shape of molecules sp3, sp2, sp, sp3d, sp3d2 and sigma and pi bonds

  35. How does hybridization happen?

  36. Carbon in methane: CH4 Consider Carbon: 1s2 2s2 2p2 (1s2 electrons don’t enter into the bonding) Bonding electrons: 2s2 2px1 2py1 2pz0 Since the first two electrons are in s orbitals, the other two electrons are in p orbitals, the shape of the clouds and the bonding angles should be different. But all four bond angles are actually the same!

  37. 2s2 2px1 2py1 2pz0 Carbon combines the s and three p orbitals into four new orbitals, so electron can be alone: sp3 sp3 sp3 sp3 If this arrangement is true, then the bond angles would all the same: 109o, perfect tetrahedral. Carbon’s new electron configuration is: 1s2 (sp3)1 (sp3)1 (sp3)1 (sp3)1

  38. NH3 Ammonia HAS bond angles of 107o Nitrogen’s electron configuration: 1s2 2s2 2px1 2py1 2pz1 If hydrogen bonded with each of the p’s, normally, then an angle of 90owould be expected. So, there’s hybridization goin’ on… N’s NEW electron configuration: 1s2(sp3)2 (sp3)1 (sp3)1 (sp3)1 (takes up more room!) unshared pair !

  39. H2S bond angles are close to 90o, what orbitals are involved? Sulfur: 1s2 2s2 2p6 3s2 3p4 3p4 = 3px2 3py1 3pz1 Hydrogen’s 1s1’s must bond in the two p’s of sulfur and since the angles are close to 90o, safe to say NO hybridization is goin’ on.

  40. Sigma and Pi Bonds Double and Triple Bonds provide some problems The REAL structure of a double bond is more complex and changes hybridization!

  41. Double bonds always consists of one pi and sigma bond pi Bond never hybridizes Sigma Bond

  42. sp2 hybridzation Ethylene C2H4 H2C = CH2 2s2 2px1 2py1 2pz0 Carbon: 1s2 Double bond counts as only one bond. The pi bond is NOT counted in hybridization! so there are only 3 bonds to consider! only the 2s and two p orbitals are used! 2s and 2px and 2py = sp2

  43. no pi bond shown here.

  44. Trigonal planar arrangement = 120o

  45. sp hybridization Carbon Dioxide: CO2 O = C = O Only two orbitals are affected and will hybridize 2s2 2px1 2py1 2pz0 Carbon: 1s2 Turns into: 1s2 (sp)1 (sp)1 2py1 2pz1 pi bonds sigma bonds

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