University:
Winston-Salem Forsyth County Schools
Course:
AP Chemistry
Academic year:

2014
Views:

334

Pages:

1
Author:

horselingd3sd

Unit 5 Homework Answers Lewis Structure Practice #1 and #2 VSEPR Theory Practice #1 1. BeCl2 – linear 2. Phosphorus tribromide – trigonal pyramidal 3. 3. Boron triiodide – trigonal planar 4. ClO2+ - bent 5. Oxygen difluoride – bent 6. Sulfate ion – tetrahedral VSEPR Theory Practice #2 1. Ammonium ion – tetrahedral 2. Carbonate ion – trigonal planar 3. Selenium hexachloride – octahedral 4. Silane, SiH4 – tetrahedral 5. Arsenic pentabromide – trigonal bipyramidal 6. Carbon dioxide – linear Intermolecular Forces Practice – Remember that in order to receive full credit for an answer on homework or the test, you must show the Lewis structure, molecular geometry (give the name of the shape), and give a few sentences justifying why your answer is correct. 1. PCl3 has LDF’s and dipole-dipole, while PCl5 has only LDF’s. Dipole-dipole IMF’s are stronger than LDF’s, so PCl3 has a higher melting point than PCl5. 2. Both molecules have only LDF’s, but silicon tetraiodide has more electrons (not valence electrons – TOTAL electrons) than silicon tetrabromide. This means that the LDF’s between silicon tetraiodide molecules are stronger than the LDF’s between silicon tetrabromide molecules. This means that silicon tetraiodide has a higher melting point than silicon tetrabromide, so it would be a solid at room temperature and silicon tetrabromide would be a liquid at room temperature. 3. The melting point of ammonia is higher than the melting point of methane because methane has only LDF’s. Ammonia has LDF’s, dipole-dipole and hydrogen bonding. Hydrogen bonding and dipole-dipole IMF’s are stronger than LDF’s, so ammonia has a higher melting point than methane. 4. Methylene chloride has LDF’s and dipole-dipole forces (different atoms on the ends), while methane has only LDF’s. Water also has dipole-dipole forces, so methylene chloride would be more soluble in water than methane because polar compounds tend to dissolve in polar compounds (or you could say that “like dissolves like”).