Covalent Bonding And Molecular Geometr Essay Research

Covalent Bonding And Molecular Geometr Essay, Research Paper Covalent Bonding and Molecular Geometry Objective The objective of this exercise is to help in understanding the geometric relationships of atoms in simple molecules and the relationship of hybridization to the geometry present.

Covalent Bonding And Molecular Geometr Essay, Research Paper

Covalent Bonding and Molecular Geometry

Objective

The objective of this exercise is to help in understanding the geometric relationships of atoms in simple molecules and the relationship of hybridization to the geometry present.

Discussion

In the last 30 years, data obtained from spectrometric measurements, X﷓ray and electron diffraction studies, and other experiments have yielded precise information about bond distances, angles, and energies. In many cases, the data confirmed conclusions reached earlier. In other cases, valuable new insights were acquired. Structure theory has advanced far beyond the simple electron dot representations and now rests securely on the foundations of quantum and wave mechanics. Although problems involving only simple molecules can now be solved with mathematical rigor, approximations such as the valence bond theory and the molecular orbital theory are very successful in giving results that agree with experimental measurements.

This exercise will use valence bond theory or hybridization to look at the geometry formed from various hybridizations. You will use a framework model kit which gives the correct angles for the each of these hybridizations.

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 (sp3 hybridization)

CH4

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) ?

H3C﷓CH3

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?

CIH2C﷓CH2Cl

Make a model of 1,2﷓dichloroethane by replacing an H atom with a C1 atom on each C in your ethane model. Use a different color rod to show the C1 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 C1 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.

NH3,

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 angle

H2O

Use 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 (sp2 hybridization)

BF3

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 trifluoride. (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

NO31-

The BF3 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 SP2 hybridization what is the function of Pz ? Location of p bond compared to plane of the molecule__________Approximate O-N-O angle________

H2C=CH2

Use sp2 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 H1 and H3 with different color rods to represent two Cl atoms (cis-geometry.) Make a second model of ethene and replace H2 and H3 with two Cl’s (trans-geometry.)

Are the cis-and trans-models the same?

If you replaced H1 and H4 with Cl’s, would it be the cis-, the trans- or another geometry?

If you replaced H2 and H4 what is the geometry?

If you replaced H1 and H2 what is the geometry?

Draw planar structural diagrams for each of the following molecules. Your drawing should show correct bond angles.

H2C=CH2 H2C=CHCl cis-HCIC=CClH trans-HCIC-CClH C12C=CH2

Linear (sp hybridization)

BeH2.

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 dihydride.

Geometry Lewis dot diagram # of s bonds on Be_________# of non bonding pairs on Be__________ # atoms in one plane Approximate H-Be-H angle

H – C = C – H

Use 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 (sp3d2 hybridization)

SF6

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 hexafluoride.

Geometry Lewis dot diagram # of s bonds on S# of p bonds on S # of lone pairs on S Approximate F-S-F angle

BrF5

Remove one of the rods from the SF6 model to make a model for bromine pentafluoride.

Geometry Lewis dot diagram # of s bonds on Br# of p bonds on Br # of lone pairs on Br Approximate F-Br-F angle

XeF4.

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 tetrafluoride.

Geometry Lewis dot diagram # of s bonds on Xe# of p bonds on Xe # of lone pairs on XeApproximate F-Xe-F angle Lone pair to lone pair angle___________Why are they opposite, not adjacent?

Trigonal Bipyramid (sp3d hybridization)

PF5

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 pentafluoride. 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 ApproximateF1-P-F3 angleApproximate F3-P-F5 angle

SF4

Remove the F3 rod from the PF5 model to make a model of sulfur tetrafluoride.

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 VESPR 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 180o BeH2H﷓C=C﷓H

3 sp2 120o BF3, NO31-, H2C=CH2

4 sp3 109.5o CH4. NH3, H2O H3C﷓CH3, H2CIC﷓CClH2

5 sp3d 90, 120o PF5, SF4

6 sp3d2 900 SF6, BrF5, XeF4

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