1 / 21

Final Exam F 12/11 9 am We have the room until 12 pm.

Final Exam F 12/11 9 am We have the room until 12 pm. Review Session R 12/10 2 pm SL 110 Office hours next week: T 1:30 – 3 pm W 2-4 pm R 3-4 pm Or by appointment. Use valence bond theory to describe the bonding about an N atom in N 2 H 4 .

zytka
Télécharger la présentation

Final Exam F 12/11 9 am We have the room until 12 pm.

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. Final ExamF 12/119 am We have the room until 12 pm. Review Session R 12/10 2 pm SL 110 Office hours next week: T 1:30 – 3 pm W 2-4 pm R 3-4 pm Or by appointment

  2. Use valence bond theory to describe the bonding about an N atom in N2H4. The Lewis electron-dot structure shows three bonds and one lone pair around each N atom. They have a tetrahedral arrangement. A tetrahedral arrangement has sp3 hybrid orbitals.

  3. 1s 1s 2s 2p sp3 • The orbital diagram of the ground-state N atom is • The sp3 hybridized N atom is • Consider one N in N2F4: the two N—F bonds are formed by the overlap of a half-filled sp3 orbital with a half-filled 2p orbital on F. The N—N bond forms from the overlap of a half-filled sp3 orbital on each. The lone pair occupies one sp3 orbital.

  4. 3s 3s 3p 3p 3d 3d • Use valence bond theory to describe the bonding in the ClF2- ion. • The valence orbital diagram for the Cl- ion is • After the promotion to get two half-filled orbitals, the orbital diagram is

  5. sp3d 3d • The sp3d hybridized orbital diagram is • Two Cl—F bonds are formed from the overlap of two half-filled sp3d orbitals with half-filled 2p orbitals on the F atom. These use the axial positions of the trigonal bipyramid. • Three lone pairs occupy three sp3d orbitals. These are in the equatorial position of the trigonal bipyramid.

  6. One hybrid orbital is required for each bond (whether a single or a multiple bond) and for each lone pair. • Multiple bonding involves the overlap of one hybrid orbital and one (for a double bond) or two (for a triple bond) nonhybridized p orbitals.

  7. To describe a multiple bond, we need to distinguish between two kinds of bonds. • A s bond (sigma) has a cylindrical shape about the bond axis. It is formed either when two s orbitals overlap or with directional orbitals (p or hybrid), when they overlap along their axis. • A p bond (pi) has an electron distribution above and below the bond axis. It is formed by the sideways overlap of two parallel p orbitals. This overlap occurs when two parallel half-filled p orbitals are available after s bonds have formed.

  8. Figure 10.26: Sigma and pi bonds.

  9. Figure A illustrates the s bonds in C2H4. The top of Figure B shows the p orbital on each carbon at 90° to each other, with no overlap. The bottom of Figure B shows parallel p orbitals overlapping to form a p bond.

  10. In acetylene, C2H2, each C has two s bonds and two p bonds. The s bonds form using the sp hybrid orbital on C. This is shown in part A. The two p bonds form from the overlap of two sets of parallel p orbitals. This is illustrated in Figure B.

  11. Restricted Rotation of -Bonded Molecules A) Cis - 1,2 dichloroethylene B) trans - 1,2 dichloroethylene

  12. cis trans The description of a p bond helps to explain the cis-trans isomers of 1,2-dichloroethene. The overlap of the parallel p orbitals restricts the rotation around the C=C bond. This fixes the geometric positions of Cl: either on the same side (cis) or on different sides (trans) of the C=C bond.

  13. We can see this illustrated with two compounds: cis-1,2-dichloroethene trans-1,2-dichloroethene There is no net polarity; this is a nonpolar molecule. The net polarity is down; this is a polar molecule. Boiling point 60°C. Boiling point 48°C.

  14. One single bond and one triple bond requires two hybrid orbitals and two sets of two parallel p orbitals. That requires sp hybridization.

  15. 1s 2s 2p • Describe the bonding about the C atom in formaldehyde, CH2O, using valence bond theory. The electron arrangement is trigonal pyramidal using sp2 hybrid orbitals. The ground-state orbital diagram for C is

  16. 1s 1s 2s 2p sp2 2p • After promotion, the orbital diagram is • After hybridization, the orbital diagram is

  17. The C—H s bonds are formed from the overlap of two C sp2 hybrid orbitals with the 1s orbital on the H atoms. • The C—O s bond is formed from the overlap of one sp2 hybrid orbital and one O half-filled p orbital. • The C—O p bond is formed from the sideways overlap of the C 2p orbital and an O 2p orbital.

  18. That's all, folks!!

More Related