The reaction of three molecules of fluorine gas with a Xe atom produces the substance xenon hexafluoride, XeF6:
Xe(g) + 3 F2(g) 
XeF6(s)
(a) Draw a Lewis structure for XeF6. (b) If you try to use the VSEPR model to predict the molecular geometry of XeF6, you run into a problem. What is it? (c) What could you do to resolve the difficulty in part (b)? (d) The molecule IF7 has a pentagonal-bipyramidal structure (five equatorial fluorine atoms at the vertices of a regular pentagon and two axial fluorine atoms). Based on the structure of IF7, suggest a structure for XeF6.
The reaction of three molecules of fluorine gas with a Xe atom produces the substance xenon hexafluoride, XeF6:
Xe(g) + 3 F2(g) XeF6(s)
(a) Draw a Lewis structure for XeF6. (b) If you try to use the VSEPR model to predict the molecular geometry of XeF6, you run into a problem. What is it? (c) What could you do to resolve the difficulty in part (b)? (d) The molecule IF7 has a pentagonal-bipyramidal structure (five equatorial fluorine atoms at the vertices of a regular pentagon and two axial fluorine atoms). Based on the structure of IF7, suggest a structure for XeF6.
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Related questions
Experiment 1:
Molecular Models of Neutral Molecules
In this experiment, you will predict the three-dimensional geometry of a series of neutral molecules using the VSEPR theory.
Procedures
Part 1: Use your Periodic Table of Elements to determine the elemental symbol, group number and valence electrons for the elements listed in Table 1.Ā
Table 1: Lewis Dot Structures
Ā
|
Element |
Symbol |
Group Number |
Valence Electrons |
Lewis Dot Structure |
|
Hydrogen |
H |
Ā | Ā | Ā |
|
Carbon |
C |
Ā | Ā | Ā |
|
Chlorine |
Cl |
Ā | Ā | Ā |
|
Aluminum |
Ā | Ā | Ā | Ā |
|
Oxygen |
O |
Ā | Ā | Ā |
|
Fluorine |
F |
Ā | Ā | Ā |
|
Neon |
Ne |
Ā | Ā | Ā |
|
Nitrogen |
N |
Ā | Ā |
Part 2: Construct the three-dimensional geometry for the molecules listed in Table 2.Ā
Table 2: Molecular Models
Ā
|
Molecule (central atom is underlined) |
Ā |
Shape Name Bond Angle |
Ā |
Bond Angles (more than one may be possible) |
|
CO2 |
Ā |
Linear |
Ā | Ā |
|
AlCl3 |
Ā |
Trigonal Planar |
Ā | Ā |
|
H2O |
Ā |
Bent |
Ā | Ā |
|
NH3 |
Ā |
Trigonal pyramidal |
Ā | Ā |
|
CCl4 |
Ā |
Tetrahedral |
Ā | Ā |
|
PCl5 |
Ā |
Trigonal bipyramidal |
Ā | Ā |
|
SF6 |
Ā |
Octahedral |
Ā |
Post-Lab Questions:
1. What is the molecular geometry (shape) of the methane molecule (CH4)? Explain completely why all the bond angles in methane are equal.
2. What is the molecular geometry (shape) of the chloromethane molecule (CH3Cl)? Explain completely why NOT all of the bond angles in chloromethane are equal.
3. Chloromethane can be thought of as resulting from methane via the replacement of one Hydrogen atom with a Chlorine atom. Subsequent replacements of H with Cl yield dichloromethane (CH2Cl2), trichlormethane (CHCl3) and tetrachlormethane (CCl4). Which of these three previous molecules will have all bond angles equal to each other? Explain completely.
4. Sketch the molecular shape of the compounds in the table below. Be sure to identify the bond angles. H2 Cl2 SO2 NH3 MoF6
