1 / 15

Dr. C. Yau Spring 2014

Chemical Bonding: Valence Bond Theory “in a nutshell” Chapter 10 Section 4 through 6 of Jespersen 6 th Ed). Dr. C. Yau Spring 2014. 1. 1. 1. VSEPR Theory. You had previously learned how to predict the molecular geometry of a species from examining its Lewis structure.

arva
Download Presentation

Dr. C. Yau Spring 2014

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. Chemical Bonding:Valence Bond Theory“in a nutshell”Chapter 10 Section 4 through 6of Jespersen 6th Ed) Dr. C. Yau Spring 2014 1 1 1

  2. VSEPR Theory • You had previously learned how to predict the molecular geometry of a species from examining its Lewis structure. • It utilizes the concept of repulsion amongst the charge clouds of the central atoms. • This was the Valence Shell Electron Pair Repulsion Theory (VSEPR Theory). • It does not explain how a bond is formed and how bonding relates to the s, p, d orbitals. 2 2 2

  3. 1s 1s Valence Bond Theory (VB Theory) The VB Theory explains bonding as an overlap of valence orbitals. H2 bonds are due to the overlap of their 1s valence orbitals. A B 2 separate H atoms H2 molecule with covalent bond due to overlap of the 1s orbitals. 3 3 3

  4. Hybridization in VB Theory • Atomic orbitals are mixed to allow formation of bonds that have realistic bond angles • The newly mixed orbitals that result are called “hybrid orbitals” with specified shapes: Review: # charge clouds Hybridization Bond Angles 2 sp 180o 3 sp2 120o 4 sp3 109.5o 5 sp3d 120o & 90o 6 sp3d2 90o & 180o 4 4

  5. How are sp3 hybrid orbitals formed? 4 orbitals 4 orbitals __ __ __ __ p s __ __ __ __ four sp3 hybrid orbitals 109.5o If we take s and all threep we form foursp3hybrid orbitals. These hybrid orbitals are “degenerate.” That is, they are of the same E, higher than s but lower than p. Note that # orbitals is conserved. 5

  6. How are sp2 hybrid orbitals formed? __ p __ ____ __ p s __ __ __ three sp2 hybrid orbitals p sp2 sp2 If we take s and just twop orbitals, we form threesp2hybrid orbitals leaving one pure p untouched. sp2 These three sp2 hybrid orbitals are planar with angles of 120o. The un-hybridized p (pure p) are used in double and triple bonds. 6

  7. How are sp hybrid orbitals formed? __ __ two p ______ __ p s __ __ two sp hybrid orbitals p If we take s and just onep orbital, we form twosp hybrid orbitals leaving two pure p untouched. p sp sp These two sp hybrid orbitals are linear with angles of 180o. 7

  8. Hybrid Orbitals Involving d-Orbitals Hybrids from one s and three p can only make a maximum of 4 hybrid orbitals. When we run out of p orbitals, we start using d orbitals. Thus we have sp3d hybrid orbitals and sp3d2 hybrid orbitals (but only for elements larger than Ne, beyond Period 2). Remember that elements smaller than Ne do not have d-orbitals and therefore cannot have sp3d and sp3d2 hybrids. 8 8

  9. Two types of bonds result from orbital overlap: sigma () bonds from head-on overlap lie along the bond axis account for the first bond pi ()bonds pi bonds are perpendicular to bond axis account for the second and third bonds in a multiple bond Bonding Types 9 9

  10. Sigma and Pi Bonding Given the structural formula of a compound, be able to specify the hybridization of each atom, state the bond angles and determine the # of sigma and pi bonds. 10 10

  11. KNOW THIS WELL! X X single bond =  bond X X double bond =  bond +  bond X X triple bond =  bond + two  bonds 11 11

  12. 4 Always start with hybridization. Give the hybridization of each C and each O. Give the bond angles. How many  bonds are there? How many  bonds are there? Ans. C1 = sp C2 = sp C3 =sp2 C4 = sp3 O = sp3 C1-C2-C3 = 180o C2-C3-O = 120o C2-C3=O = 120o O-C4-H = 109.5o 1 2 3 9  bonds 3  bonds 12 12

  13. Each C is sp3. Chemical Bonding in C2H6 The C−C bond is a  bond. It is a “head-to-head” overlap.

  14. Each C is sp2, with one pure p. Chemical Bonding in C2H4 The C=C bond is made of  bond and a  bond. The  bond consists of overlap of parallel p orbitals.

  15. Each C is sp, with two pure p. Chemical Bonding in C2H2 The CC bond is made of one  bond and two  bond. The  bond consists of overlap of parallel p orbitals. Practice Exercise 10.20 & 10.21 on p. 444 & ques on p.467 #10.108 (ans at back of book)

More Related