CHE 375 F11 HW 8

CHE 375 F11 HW 8 - O O H H O H H In class we discussed the...

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CHE 375 Fall 11 Homework 8 Name: 1. Our textbook (Miessler/Tarr, 4 th ed.) states on p. 95 that the molecule IOF 3 has no symmetry (other than E ). Its point group is C 1 . Determine the VSEPR structure of IOF 3 and check whether you come to the same conclusion. 2. Determine the point groups of the following structures: Show the characteristic symmetry elements. (Those needed to determine the point groups unambiguously.): SF 5 Cl (determine and drawVSEPR structure) This molecule is a biphenyl derivative. Note, the central C-C bond, the carbon atoms in the para positios of the phenyl rings (top and bottom carbon atoms) as well as the two sulfur atoms all lie in the plane of the paper. The planar rings are twisted relative to each other as indicated by the bold lines. S S Hydrogen peroxide (H 2 O 2 ) in its most stable conformation (shown in two perspectives):
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Unformatted text preview: O O H H O H H In class we discussed the symmetry of the allene molecule. For your reference, the molecule is reproduced in three perspectives. In the structure below the hydrogen atoms have been replaced by hydroxyl groups. The hydroxyl groups are all arranged so that they form the same dihedral angle with the central axis, as indicated in the drawing. C C C H H H H C C C O O O O C C C H H H H H H H H C O O O O H H H H C H H H H Determine the point group of the S 8 molecule. Does this molecule have an S 8 symmetry element? S S S S S S S S The triphenylboron molecule is shown in three conformations. In each case, the three phenyl rings are oriented relative to the central axis in the same way (perpendicular, twisted, and parallel). [Think about the same conformations in triphenylamine.] B B B...
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CHE 375 F11 HW 8 - O O H H O H H In class we discussed the...

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