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Geometry Prediction

Geometry Prediction. Ray A. Gross, Jr. Prince George’s Community College April 18, 2007. Nature of Bonding. Central Atom (CA) Ligands (L) Lone Pairs (LPCA). G. N. Lewis Linus Pauling. 1930. 1931. Water (H 2 O). Sulfuric Acid (H 2 SO 4 ). Note: L = # s bonds to CA.

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Geometry Prediction

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  1. Geometry Prediction Ray A. Gross, Jr. Prince George’s Community College April 18, 2007

  2. Nature of Bonding • Central Atom (CA) • Ligands (L) • Lone Pairs (LPCA)

  3. G. N. LewisLinus Pauling 1930 1931

  4. Water (H2O)

  5. Sulfuric Acid (H2SO4) Note: L = # s bonds to CA.

  6. Nature of Bonding • Hybrid orbitals (HO) • p (unhybridized)orbitals HO = L+ LPCA

  7. Water HO = L + LPCA= 2 + 2 = 4

  8. Sulfuric Acid HO = L + LPCA = 4 + 0 = 4

  9. Hybridization tells us the number of hybrid orbitals (HO) • sp = two hybrid orbitals, or HO = 2 • sp3 = three hybrid orbitals, or HO = 3 • sp3d2= six hybrid orbitals, or HO = 6 If we know the hybridization, we know HO. If we know HO, we know the hybridization.

  10. Geometry determined by any two of HO, L and LPCA Water Example

  11. Geometry determined by any two of HO, L and LPCA Sulfuric Acid Example

  12. Gross Procedure: Determine Hybridization and Geometry by finding L and LPCA • Find L and LPCA • L + LPCA = HO • HO hybridization • HO and L  geometry

  13. Example: BH3 • Find L; L = 3 • Find LPCA; LPCA = ?

  14. Example: BH3 LPCA = half the CA’s electrons leftover after bonding.

  15. LPCA = half the CA’s electrons leftover after bonding. • Let ERL = CA’s electrons required by L • Let VE = CA’s valence electrons Then LPCA = ½(VE – ERL)

  16. LPCA = ½(VE – ERL) • For BH3: VE = 3, ERL = 3 (1 per H) LPCA = ½(3 – 3) = 0

  17. Example: BH3 • Find L; L = 3 • Find LPCA; LPCA = 0 • Thus, • HO = 3; hybridization = sp2 • L = 3; geometry = trigonal planar

  18. Example: NH3 Each H requires one of CA’s electrons.

  19. Example: NH3 • L = 3 • LPCA = ½(VE – ERL) = ½(5 – 3) = 1 • HO = L + LPCA = 3 + 1 = 4 = sp3 • HO = 4, L = 3 = trigonal pyramidal

  20. Example: H2SO4 Each O requires two electrons; each OH requires one electron.

  21. Example: H2SO4 = (HO)2SO2 • L = 4 • LPCA = ½(VE – ERL) • LPCA = ½[6 – (2 x 1) – (2 x 2) = 0 • HO = 4, L = 4 • sp3; tetrahedral

  22. Example: GeCl4-2 • VE = 4 + 2 = 6 • ERL = 4 x 1 = 4 • LPCA = ½(6 – 4) = 1 • L = 4 • HO = L + LPCA = 5 • HO = 5, L = 4 • sp3d, seesaw geometry

  23. Findings Available electrons, VEadj= VE adjusted for charge CH • VEadj = VE – CH Electrons Required by Ligands (ERL) • ERL = Normal Covalence for atoms • ERL = # for L atom next to CA to complete its octet. H and O as ligands • H and O peripheral atoms equal an OH unitary ligand.

  24. Five-step Procedure

  25. What is the geometry of HNO3? • N = CA; VEadj= 5 – 0 = 5 • L = OH and 2 O’s = 3 • ERL = 1 + 2 x 2 = 5 • LPCA = ½(5 – 5) = 0 • HO = 3; L = 3 • sp2; trigonal planar

  26. Demo GP

  27. Organic Procedure

  28. Demo GP

  29. Conclusions • GP combines inorganic and organic procedures. • GP gives fast, accurate results.

  30. Acknowledgements • Thanks to all who came to my last seminar. • I was inspired by it to do this work. RAG

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