C H Bond On One C Atom example essay topic
This exercise will use valence bond theory or hybridization to look at the geometry formed from various hybridization's. You will use a framework model kit which gives the correct angles for the each of these hybridization's. The first bond formed between any two atoms is always a sigma (s) bond (one that is symmetric about the bond axis). Additional bonds between the same two atoms will be pi (p) bonds (perpendicular to the bond axis). It is the sigma bonds and any lone pairs of electrons occupying the sigma hybrid orbitals that determine the geometry of a molecule. Pi bonds are always perpendicular to the sigma bonds and follow the geometry formed by the sigma bonding.
Procedure Check out a molecular model kit from the stockroom. Read the kit directions to see which framework center is used for each hybridization. Tetrahedral (sp 3 hybridization) CH 4 Construct a model of methane using a tetrahedral center (4 prongs) and four rods of the same color to show how the 4 H's are attached. Geometry Lewis dot diagram # of's bonds on C Approximate H-C-H angel Max # atoms (incl. C) in one plane Is there a mirror plane (divides the molecule in equal halves)? H 3 C CH 3 Construct a model of ethane using a tetrahedral center for each C and the same color rods for all 6 H's with a C C bond present.
Geometry Lewis dot diagram # of's bonds on each C Approximate H-C-H angle Approximate H-C-H angle The C-C bond is a single bond and has free rotation about it. Arrange the ethane molecule so that each C H bond on one C atom is exactly parallel to a C H bond on the second C atom. (This is the eclipsed position.) View this arrangement by looking along the C C bond such that the atoms on the front C blank out those on the back C. From this view the model is represented as in I (a Newman projection.) Now rotate the front carbon 60 degrees clockwise producing II. Which of the above arrangements allows more space per atom? Which arrangement would you expect to be more stable?
CIH 2 C CH 2 Cl Make a model of 1, 2 by replacing an H atom with a C 1 atom on each C in your ethane model. Use a different color rod to show the C 1 atoms. The eclipsed Newman projection is given in 111. Rotate the front C by 60 degrees and show the result on IV, rotate another 60 degrees and show the result on V, etc., until you have completed a 360 degree rotation about the front C. Remember that a C 1 atom is much larger than an H atom Which of the structures would you expect to be the most stable? Explain why. Which of the structures would you expect to be the least stable?
Explain why. NH 3, Use a tetrahedral center for the N atom and three rods of the same color for the 3 H's to make a model of ammonia. Geometry Lewis dot diagram # of's bonds on N # on non-bonding pairs on N Approximate H-N-H angleH 2 OUse a tetrahedral center for the O atom and two rods of the same color for the 2 H's to make a model of water. Sketch the geometry Draw a dot diagram # of's bonds on O # on non-bonding pairs on O Approximate H-O-H angle Trigonal Planar (sp 2 hybridization) BF 3 Use a trigonal planar center (has 3 or 5 prongs depending on the model kit) and three rods of the same color for the three F's to make a model of boron tri fluoride. (If you have a model kit without a 3 prong center, use the 5 prong center and put the three Fs on the 3 equivalent positions.) Geometry Lewis dot diagram # of's bonds on B # atoms in one plane Approximate F-B-F angle NO 31-The BF 3 model is the same model needed for the nitrate ion. Geometry Lewis dot diagram Total # of bonds on N # of's bonds on N # of p bonds on N If Px & Py are used for SP 2 hybridization what is the function of Pz?
Location of p bond compared to plane of the molecule Approximate O-N-O angle H 2 C = CH 2 Use sp 2 centers for both C atoms and four rods of the same color for the four H's to make a model of ethene, which has a C = C bond. This is a planar molecule. If your model kit has a 5 prong center, align the 4th and 5th positions on the two C atoms parallel with each other. With a 3 prong center, put a rod through the hole in the center of each and align these rods parallel. These represent the Pz orbital on each C atom which overlap and form the pi bond between the carbons. In contrast to single bonds (sigma only) which have free rotation, there is no rotation around bonds that contain pi bonds.
You would have to break the pi bond in order to rotate about such a bond and that requires too much energy for rotation to occur. (Rods through holes in center of 3 prong framework) (With the 3 prong centers, now that you realize there is no rotation about the C = C bond and why, remove the rods through the center holes to make it easier to see the planar nature of the molecule.) Geometry Lewis dot diagram # of's bonds on C# of p bonds on C # atoms of ethene in one plane Approximate H-C-H angle Approximate H-C-C angle Consider the H numbers in the drawing above. Replace H 1 and H 3 with different color rods to represent two Cl atoms (cis-geometry.) Make a second model of ethene and replace H 2 and H 3 with two Cl's (trans-geometry.) Are the cis-and trans-models the same? If you replaced H 1 and H 4 with Cl's, would it be the cis-, the trans- or another geometry?
If you replaced H 2 and H 4 what is the geometry? If you replaced H 1 and H 2 what is the geometry? Draw planar structural diagrams for each of the following molecules. Your drawing should show correct bond angles. H 2 C = CH 2 H 2 C = C HCl cis-HCIC = CClH trans-HCIC-CClH C 12 C = CH 2 Linear (sp hybridization) BeH 2. Use the center with 6 prongs for the Be atom and put two rods of the same color opposite each other then make a model of beryllium di hydride.
Geometry Lewis dot diagram # of's bonds on Be # of non bonding pairs on Be # atoms in one plane Approximate H-Be-H angleH - C = C - HUse two of the 6 prong centers for the carbon atoms and two same color rods for the H's to make a model of ethyne. This is a linear molecule. Geometry Lewis dot diagram # of's bonds on C # of p bonds on C # atoms in one plane Approximate H-C-C angle If bond axis is x, what are along y and z? Is there a free rotation along C-C bond?
Why? Octahedral (sp 3d 2 hybridization) SF 6 Use the 6 prong center for the S atom and rods of the same color for the 6 F's to make a model of sulfur. Geometry Lewis dot diagram # of's bonds on S# of p bonds on S # of lone pairs on S Approximate F-S-F angle BrF 5 Remove one of the rods from the SF 6 model to make a model for bromine. Geometry Lewis dot diagram # of's bonds on Br# of p bonds on Br # of lone pairs on Br Approximate F-Br-F angleXeF 4. Use a 6 prong center for the Xe and rods of the same color for the 4 F's. Place all the F's in the same plane around the Xe to make a model of xenon tetra fluoride.
Geometry Lewis dot diagram # of's bonds on Xe# of p bonds on Xe # of lone pairs on Xe Approximate F-Xe-F angle Lone pair to lone pair angle Why are they opposite, not adjacent? Trigonal Bi pyramid (sp 3d hybridization) PF 5 Use the 5 prong center if your set has one and put five same color rods on for the F's to make a model of phosphorous. If your set does not have a 5 prong center, use the 3 prong center and put a rod through the hole in the middle to make the 4th and 5th positions. Geometry Lewis dot diagram # of's bonds on P# of p bonds on S # of lone pairs on P Approximate 1-P-F 3 angle Approximate F 3-P-F 5 angles 4 Remove the F 3 rod from the PF 5 model to make a model of sulfur tetra fluoride.
Geometry Lewis dot diagram # of's bonds on S# of p bonds on S # of lone pairs on S Approximate F-S-F angle Why are two lone pairs in equatorial position? Summary When we consider the models we have made, we see that they all obey the VESPER rules (see your textbook). These are summarized in the table below. Total #'s bonds plus lone pairs Hybridization on central atom Bond angle Our examples 2 sp 180 o BeH 2 H C = C H 3 sp 2 120 o BF 3, NO 31-, H 2 C = CH 24 sp 3 109.5 o CH 4.