1 / 19

COVALENT BONDING: ORBITALS

COVALENT BONDING: ORBITALS. HYBRIDIZATION (9.1). HYBRIDIZATION. Consider methane,CH 4 C has 4 valence electrons 1s 2 2s 2 2p 2

teegan-chandler
Télécharger la présentation

COVALENT BONDING: ORBITALS

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. COVALENT BONDING: ORBITALS HYBRIDIZATION (9.1)

  2. HYBRIDIZATION • Consider methane,CH4 • C has 4 valence electrons 1s2 2s2 2p2 • This suggests that there might be two kinds of C-H bonds: one involving a 2s e- on carbon pairing with the 1s on H and the other involving a 2p e- on carbon pairing with the 1s on H. • Expt evidence confirms that the four C-H bonds in CH4 are identical and that CH4is tetrahedral.

  3. Figure 9.1 a & b a) The Lewis Structure of the Methane Molecule b) The Tetrahedral Molecular Geometry of the Methane Molecule

  4. HYBRIDIZATION (2) • To resolve this conflict, promote a 2s electron to the empty 2p orbital, then mix or hybridize the 2s (1) and 2p (3) orbitals to form four identical hybrid AOs named sp3 • These hybrid atomic orbitals overlap with the 1s orbital on hydrogen to form the covalent C-H bond (sp3 – 1s). • Why do hybrids form? To minimize total energy.

  5. HYBRIDIZATION (3) • Using the VSEPR rules, C has four covalent bonds and has tetrahedral molecular geometry. The H-C-H bond angle = 109.5o. This agrees with exptal measurements (4 identical C-H bonds). • Hybridization integrates electron configurations with expt measurements. • Other hybrids: sp2 (3 e pairs), sp (2), dsp3 (5), d2sp3 (6)

  6. Figure 9.24 Relationship of the Number of Effective Pairs, Their Spatial Arrangement, and the Hybrid Orbital Set Required

  7. Figure 9.8 The Hybridization of the s, px, and py Atomic Orbitals Results in the Formation of Three sp2 Orbtitals Centered in the xy Plane

  8. Figure 9.9 An Orbital Energy-Level Diagram for sp2 Hybridization

  9. Figure 9.11 The Sigma Bonds in Ethylene

  10. Figure 9.13 (a)The Orbitals Used to Form the Bonds in Ethylene (b) The Lewis Structure for Ethylene

  11. Figure 9.16 The Orbital Energy-Level Diagram for the Formation of sp Hybrid Orbitals on Carbon

  12. Figure 9.20 a-d (a) The sp hybridized N atom (b) The  bonds in the N2 molecule (c) The two pi bonds in N2 are formed when electron pairs are shared between two sets of parallel p orbitals (d) The total bonding picture of N2

  13. Figure 9.21 A Set of dsp3 Hybrid Orbitals on Phosphorus Atom

  14. Figure 9.23 An Octahedral Set of d2sp3 Orbitals on Sulfur Atom

  15. Figure 9.24 Relationship of the Number of Effective Pairs, Their Spatial Arrangement, and the Hybrid Orbital Set Required

  16. HYBRIDS AND MOLECULAR STRUCTURE • Write Lewis structure and use VSEPR method to predict e pair geometry • Select hybridization scheme this is consistent with VSEPR prediction (Fig 9.24) • Identify orbital overlap • Form multiple bonds if needed • Determine molecular geometry

  17. HYBRIDS AND MULTIPLE BONDS (1) • Use Valence Bond method to determine 3-dimensional structure of hydrocarbons with double and triple bonds (planar) • Sigma () or end-to-end orbital overlap bond • Pi () or side-by-side orbital overlap bond • Geometric isomers (2-butene) • Benzene and other aromatic compounds

  18. HYBRIDS AND MULTIPLE BONDS (2) • A single bond has one sigma bond. • A double bond has one sigma bond and one pi bond. • A triple bond has one sigma bond and two pi bonds.

  19. Problems • 16, 18, 22, 24, 28, 32, 33

More Related