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Problem 167 says to predict and draw the possible molecular structures for the following molecules.
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The problem also says that there are several different configurations that can be drawn with the same central atom and to account for those structures.
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So for the top one here, we have a central bromine bound by a fluorine and two iodine groups.
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So keep in mind, of course, that the first thing we should always think about when drawing a lewis structure, is the number of valence electrons we need to account for.
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So since all four of these atoms are halogens, all of them have seven valence electrons, giving us a total of 28 that we need to think about.
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So the first possible configuration here is the two fluorians at the top, sorry, the fluorine at the top of the bromine here and the two iodines at the bottom.
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Now because again these are halogens, we can give each the fluorine and the iodines three lone pairs of electrons.
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And by accounting for these eight electrons for each, the fluorine and the two iodines, so far that's 24 total electrons.
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So we need to add four more, and that can be done by giving the bromine two lone pairs.
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Now, as you can see, the iodines here are 90 degrees apart, but you can imagine a scenario where instead of being separated by 90 degrees, the iodines are separated by 180 degrees.
02:08
And so as you can see here, the other possible structure for this particular molecule is simply flipping this drawing upside down.
02:25
And then the other possibility is separating the two iodines with the fluorine here.
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So for part b, we can go about doing the same thing.
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I'll draw this over here to give us some space for part c.
02:42
We have a xenon in the center here, and we can bind to that the two oxygens and the two fluorines.
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So just like before, we should count up the electrons that we need to think about...