Chemical Bonding - PowerPoint PPT Presentation

chemical bonding n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Chemical Bonding PowerPoint Presentation
Download Presentation
Chemical Bonding

play fullscreen
1 / 17
Chemical Bonding
243 Views
Download Presentation
rudolf
Download Presentation

Chemical Bonding

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. Chemical Bonding Sigma and Pi Bonds

  2. What is a Sigma Bond? • This is a covalent bond in which the electron cloud overlap is directly between the nuclei of the atoms involved in the bonding. • A sigma bond will be formed using electrons in s orbitals, p orbitals, or hybrid orbitals. • A sigma bond will always be a single bond (as opposed to a double or triple bond).

  3. Sigma Bond – Example #1 H2 This molecule has the two hydrogen atoms having 1 electron in an “s” orbital. Remember that the s orbital is spherical. The atom on the right has been made “clear” to show the cloud overlap. Note how the overlap of the two atom’s clouds is directly between the nuclei of the atoms. The bond in this example is characterized as a sigma bond.

  4. Sigma Bond – Example #2 Remember that the valence of chlorine = 7 so that the bond between the atoms is formed by the overlap of p orbitals. Remember also that the shape of a p orbital is “figure-8”. Cl2 Cl Cl The p-orbital of the Cl atom on the right. The p-orbital of the Cl atom on the left. Note again how the overlap is directly between the nuclei of the two atoms.

  5. Sigma Bond – Example #3 The hydrogen atom has a bonding electron in an s-orbital and the chlorine atom has its unpaired electron in a p-orbital. HCl H Cl Again, the overlap is directly between the two nuclei. This bond is characterized as a sigma bond.

  6. Sigma Bond – Example #4 Observe that the Carbon atom is in sp3 hybridization and is showing a tetrahedral geometry. There are 4 bonds, but each is a sigma bond. CH4

  7. The BCl3 molecule has the Boron atom in sp2 hybridization, creating a trigonal planar geometry. Each of the Chlorine atoms is using a half-filled p-orbital for bonding. Believe it or not, the molecule is essentially “flat”. Cl B Cl Cl

  8. Pi Bonds • These are the 2nd and 3rd bonds that form in the event of double and/or triple bonds. • You cannot have a pi bond until there is a sigma bond. • A pi bond can only be formed by a p-orbital that is not involved in hydridization. • The overlap of the orbitals in a pi bond is NOT directly between the nuclei of the atoms involved in the bonding. • Because there are two “lobes” to a p-orbital, a pi bond actually has two “half-sections”

  9. H H H C C H H H Consider the molecule C2H6 Both of the Carbon atoms will be in sp3 hybridization. Therefore the molecule will actually appear to be two tetrahedrons connected together with a spherical s-orbital attached at each of the remaining corners.

  10. But what about C2H4 ??? H H H C C H • Because each Carbon atom MUST form 4 bonds, a double bond is formed between the two carbon atoms. • Now it gets strange…. • The pi-bond is the second bond in the double bond system. The first bond is a sigma bond. • Since the pi-bond cannot be a hybrid orbital, each Carbon atom is actuallyusing three hybrid orbitals – that means that carbon is in sp2 hybridization and that the geometry of the hybrids around each carbon atom is trigonal planar. (which would result in a 120o bond angle)

  11. C C Here is what the hybridized carbons will look like… Each of the bonds in this graphic is a sigma bond. In the next slide, the p-orbital of each carbon atom will be added to show how the pi-bond is formed.

  12. C C The red colored shapes represent the p-orbitals that were not involved in the hybridization. Notice that they extend above and below the area that directly joins the nuclei of the carbon atoms. The pi-bond has two halves, one above the plane between the carbon nuclei and the other below that plane. A double bond starts with a sigma bond that usually involves hybrid orbitals, then adds a two-part overlap pi-bond that is formed between the p-orbitals of the adjacent atoms.

  13. Consider the molecule C2H2 (acetylene – used for welding) Its bonding will look like this: Now for a triple bond: H C C H Note the triple bond between the carbon atoms. There are three sigma bonds in the molecule, one connecting each of the hydrogens to a carbon and another connecting the carbons to each other. But there are also two pi bonds, the 2nd and 3rd bonds between the carbon atoms.

  14. C C The sigma bonding will be oriented like this: Each carbon atom is using sp hybrids for the sigma bonds. Therefore the molecule is going to be linear in shape. The next slide shows the un-hybridized p orbitals on each carbon atom. They will form the pi bonds.

  15. C C One of the pi bonds involves the top and bottom overlap of the red colored p-orbitals (like in the C2H4 molecule) and the other pi bond involves the front and back overlap of the blue colored p-orbitals. The lines illustrating the overlapping are deliberately not drawn – they would clutter up the image.

  16. Really cool molecule – CO2 The molecule is linear in shape with double bonds between the carbon atom in the center and the oxygen atoms on either side. The bonding would be represented like this: O C O But the double bond on each side requires an overlap of a p-orbital from each atom. Remember that the p-orbitals are aligned with the axes of a 3-dimensional graphing system and you get the pi bonds forming as illustrated on the following slide.

  17. O C O Note the pi bond overlap is on the “Y” axis on one side of the molecule and on the “Z” axis on the other side. But, since the geometry (shape) of the molecule is determined by the orientation of the sigma bonds, the molecule is linear.