Drawing the Lewis Structure for XeO3
Transcript: This is Dr. B. Let's do the XeO3 Lewis structure: xenon trioxide. Xenon is in group 8, 8 valence electrons. Oxygen, 6, we've got three of them. Eight plus eighteen: 26 total valence electrons. We'll put Xe at the center and then let's put the Oxygens around it. So we've got 26. Put it between atoms: 2, 4, 6; and then around the Oxygens, 8, 10, 12, 14, 16, 18, 20, 22, 24, and then we've got room here to complete that octet with 26. We've used all the valence electrons, but there's a problem, and the problem is this: Xenon often will have more than eight valence electrons around it. So we really need to check our formal charges to see that, you know, is this the best structure for XeO3. So let's do that, let's check the formal charges. So let's look at the formal charges. For Oxygen right here, and all of these Oxygens are the same, they're symmetrical, so we only need to do one. On the periodic table, 6 valence electrons, minus nonbonding, that's these ones right here, we have 6 of those. And then we have two bonding, we divide that by 2. Six minus 6 minus 1 is negative one. So we can put a negative one as the formal charge on the Oxygens. For the Xenon, it has 8 valence electrons minus the two nonbonding. And then the six bonding valence electrons divided by 2. Eight minus 2 minus 3 is plus 3. Since these aren't zero, I'm thinking this is probably not the best structure for XeO3. So let's try something else. When I see this +3 right here, that makes me think I'm going to need three double bonds in here to equalize that out. So let's try that. So for each Oxygen, I moved the electrons out here in to form a double bond. I've used the same number of valence electrons, it's just now they're sharing. So let's calculate the formal charges for this new XeO3 Lewis structure. Oxygen, we have 6 valence electrons, looking at the periodic table. Nonbonding, those right there aren't involved in bonds, we have 4. And bonding, we have those 4, which we divide by 2. Six minus 4 minus 2 is zero. For the Xenon, we have 8 valence electrons. We have 2 that are nonbonding, and then we have 4 plus 4 plus 4: 12 bonding electrons, which we divide by 2. Eight minus 2 minus 6 is zero. Because these formal charges are zero, that tells me that this right here is going to be the best Lewis structure for XeO3. Even though the other one seemed to satisfy the octets, the formal charges really didn't work out. They should be close to zero or zero. That's going to be the best Lewis structure for XeO3. This is Dr. B., and thanks for watching. |
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