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Covalent Bonding: Orbitals

Covalent Bonding: Orbitals. Chapter 09. The four bonds around C are of equal length and Energy. Can you explain this based on your knowledge of electron energy levels?. 6 C 1s 2 2s 2 2p 2. Bonding from s is Different from bonding From p. In addition the Angles should be within

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Covalent Bonding: Orbitals

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  1. Covalent Bonding: Orbitals Chapter 09

  2. The four bonds around C are of equal length and Energy

  3. Can you explain this based on your knowledge of electron energy levels? 6C 1s2 2s2 2p2 Bonding from s is Different from bonding From p. In addition the Angles should be within 900!

  4. How to generate four equal orbitals?

  5. Hint: A key to wave mechanics is superposition Which is creating new waves from interference of old ones

  6. Let’s do some mixing

  7. Cross section of an sp3orbital.

  8. An energy-level diagram showing the formation of four sp3 orbitals.

  9. px py pz s Hybridization Ground state of C 1s2 2s2 2p2 Promote electron at n=2 2s12p3 Hybridize at n=2 sp3 sp3 sp3 sp3 s px py pz sp3 sp3 sp3 sp3 4sp3 orbitals of equal length, energy and in tetrahedral shape

  10. Hybridization • The mixing of atomic orbitals to form special orbitals for bonding. • The atoms are responding as needed to give the minimum energyfor the molecule.

  11. Valence Bond Theory and NH3 N – 1s22s22p3 3 H – 1s1 2s 2px 2py 2pz If use the 3 2p orbitals predict 900 Actual H-N-H bond angle is 107.30 How do you explain this?

  12. Consider the n=2 for N Original 2s 2px 2py 2pz Mix 1s and 3p And generate four Equivalent sp3 Hybridized orbitals sp3 sp3 sp3 sp3 1 sp3 lone pair 3 bonding orbitals Which can accommodate The 1s1 electron from hydrogen

  13. The nitrogen atom in ammonia is sp3 hybridized.

  14. An orbital energy-level diagram for sp2 hybridization. Note that one p orbital remains unchanged.

  15. When an s and two p orbitals are mixed to form a set of three sp2 orbitals, one p orbital remains unchanged and is perpendicular to the plane of the hybrid orbitals.

  16. Figure 9.13: (a) The orbitals used to form the bonds in ethylene. (b) The Lewis structure for ethylene.

  17. Pi bond (p) – electron density above and below plane of nuclei of the bonding atoms Sigma bond (s) – electron density between the 2 atoms

  18. The s bonds in ethylene. Note that for each bond the shared electron pair occupies the region directly between the atoms.

  19. A sigma () bondcenters along the internuclear axis. • A pi () bondoccupies the space above and below the internuclear axis.

  20. The orbitals used • to form the bonds • in ethylene. (b) The Lewis structure for ethylene.

  21. The orbital energy-level diagram for the formation of sp hybrid orbitals on carbon.

  22. When one s orbital and one p orbital are hybridized, a set of two sp orbitals oriented at 180 degrees results.

  23. The orbitals of an sp hybridized carbon atom.

  24. The orbital arrangement for an sp2 hybridized oxygen atom to form CO2. 8O 1s22s22p4

  25. The hybrid orbitals in the CO2 molecule.

  26. (a) The orbitals used to form the bonds in carbon dioxide. Note that the carbon-oxygen double bonds each consist of one s bond and one p bond. (b) The Lewis structure for carbon dioxide. 2sp orbitals from C to form two double bonds

  27. (a) An sp hybridized nitrogen atom. (b) The s bond in the N2 molecule. (c) The two p bonds in N2 are formed when electron pairs are shared between two sets of parallel p orbitals. (d) The total bonding picture for N2.

  28. H H C H C O O H Sigma (s) and Pi Bonds (p) 1 sigma bond Single bond 1 sigma bond and 1 pi bond Double bond Triple bond 1 sigma bond and 2 pi bonds How many s and p bonds are in the acetic acid (vinegar) molecule CH3COOH? s bonds = 6 + 1 = 7 p bonds = 1 10.5

  29. How to generate more than 4 bonds? • Remember the PCl5, SF6, etc… • The s and 3p can generate 4 orbitals • Include d orbitals to generate more!

  30. s p p p d d d d d Hybridize 1s and 3p and 1d Result in 5 dsp3 orbitals 1 2 3 4 5 dsp3 dsp3 dsp3 dsp3 dsp3

  31. A set of dsp3 hybrid orbitals on a phosphorus atom. Note that the set of five dsp3 orbitals has a trigonal bipyramidal arrangement. (Each dsp3 orbital also has a small lobe that is not shown in this diagram.)

  32. (a) The PCl5 molecule. (b) The orbitals used to form the bonds in PCl5. The phosphorus uses a set of five dsp3 orbitals to share electron pairs with sp3 orbitals on the five chlorine atoms. The other sp3 orbitals on each chlorine atom hold lone pairs.

  33. How to form 6 orbitals? s p p p d d d d d Hybridize 1s and 3p and 2d Result in 6 d2sp3 orbitals 1 2 3 4 5 6 d2sp3 ds2p3 d2sp3 d2sp3 d2sp3 d2sp3

  34. The relationship of the number of effective pairs, their spatial arrangement, and the hybrid orbital set required.

  35. The Localized Electron Model • Draw the Lewis structure(s) • Determine the arrangement of electron pairs (VSEPR model). • Specify the necessary hybrid orbitals.

  36. Molecular Orbitals (MO) • Analagous to atomic orbitals for atoms, MOs are the quantum mechanical solutions to the organization of valence electrons in molecules. Remember bonds are waves and wave may be arranged in constructive or destructive interference.

  37. Types of MOs • bonding: lower in energy than the atomic orbitals from which it is composed. • antibonding: higher in energy (unstable) than the atomic orbitals from which it is composed.

  38. The combination of hydrogen 1s atomic orbitals to form molecular orbitals.

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