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Ch.8 - Basic Concepts of Chemical Bonding

Chapter 8, Problem 66

(a) Describe the molecule xenon trioxide, XeO3, using four possible Lewis structures, one each with zero, one, two, or three Xe¬O double bonds. (b) Do any of these resonance structures satisfy the octet rule for every atom in the molecule? (c) Do any of the four Lewis structures have multiple resonance structures? If so, how many resonance structures do you find? (d) Which of the Lewis structures in part (a) yields the most favorable formal charges for the molecule?

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Hello everyone today. We are being given the following problem. The molecule sulfur trioxide can be described using four possible leWIS structures. Each with a different number of sulfur oxide, oxygen double bonds, 012 or three draw the leWIS structures. So the first thing I wanna do is they want to gather how many valence electrons we will be working with. So first we have a sulfur atom and we know that sulfur is in group six A. And therefore has six valence electrons. And since there's only one sulfur atom here, We just leave the six surveillance electrons as is for oxygen. However, that is also in group six a. However, there are three of them Making this three times six giving us a total of 18 valence electrons. For oxygen We add up these valence electrons and we have valence electrons to use. And so until we see for our first structure here that one we have one sulfur surrounded by three oxygen's here. So we could draw one sulfur here, one oxygen there, one there and one there and each bond line accounts for two electrons. And so since we have three bond lines, that's already six electrons. And we can go ahead and start filling out the outside atoms to fill their octet. And so for the oxygen on the left we have 123456. For this oxygen over here we have 1234 and for the oxygen down here we're gonna have 123456. Now earlier we said that oxygen has a valence Electron number of six due to being in group six, a notice how this oxygen here has seven valence electrons. This one and this one have seven valence electrons. The one on the left and the one on the bottom. Therefore we must give the one on the left here a negative charge and the one here a negative charge at the bottom. And so when two negative charges are added together, that gives an overall negative charge of -2. And so that's gonna be our first Lewis structure. We have three more to draw For our 2nd Lewis structure. It's gonna be the same arrangement of atoms on sulfur. One oxygen, one oxygen and one oxygen. However, we noted that there were going to be two or one double bond present. And so when we do that, we're gonna go ahead and just put it on this sulfur oxygen bond here, we can do it in anyone, it doesn't matter. We can start filling in our outside of electrons. And so we have 1234 to give us six valence electrons. There 1234 that gives us five valence electrons. And then we have 123456 valence electrons left to give. And we notice that we have two more additional ones that we need to give. And so we're gonna go ahead and put that on these sulfur in the middle. And the special thing about sulfur is that it can expand its octet due to being in a period of three. And so since we only have one negative charge here, we can go ahead and sign this a negative charge there. And we're going to go ahead and do the rest for our 3rd and 4th structure. So for our third structure, similar orientation of atoms or some arrangements. Now, we're gonna Have two double bonds here in one single bond in the left, filling in our valence electrons. We have 123456. Notice we have an extra electron here. So that gets a negative charge and the other oxygen's are neutral. And so is the sulfur here. So this gets a negative one charge overall. And so for our last and final structure, we're going to draw the same arrangements. But now we're going to be working with three double bonds 1234. And now we fill in our outside valence electrons. And notice how this isn't. This is actually a neutral molecule because the valence electron is fulfilled and there are no charges. And with that we have our final answer. I hope this helped. And until next time
Related Practice
Textbook Question

Draw the Lewis structures for each of the following molecules or ions. Identify instances where the octet rule is not obeyed; state which atom in each compound does not follow the octet rule; and state how many electrons surround these atoms: (a) PF6-, (b) BeCl2, (c) NH3, (d) XeF2O (the Xe is the central atom), (e) SO42- .

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Textbook Question

In the vapor phase, BeCl2 exists as a discrete molecule. (a) Draw the Lewis structure of this molecule, using only single bonds. Does this Lewis structure satisfy the octet rule?

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Textbook Question

In the vapor phase, BeCl2 exists as a discrete molecule. (c) On the basis of the formal charges, which Lewis structure is expected to be dominant for BeCl2?

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Textbook Question

There are many Lewis structures you could draw for sulfuric acid, H2SO4 (each H is bonded to an O). (a) What Lewis structure(s) would you draw to satisfy the octet rule?

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Textbook Question

There are many Lewis structures you could draw for sulfuric acid, H2SO4 (each H is bonded to an O). (b) What Lewis structure(s) would you draw to minimize formal charge?

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Textbook Question

Some chemists believe that satisfaction of the octet rule should be the top criterion for choosing the dominant Lewis structure of a molecule or ion. Other chemists believe that achieving the best formal charges should be the top criterion. Consider the dihydrogen phosphate ion, H2PO4-, in which the H atoms are bonded to O atoms. (b) What is the predicted dominant Lewis structure if achieving the best formal charges is the top criterion?

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