Problem

A molecule with the formula AB2, in which A and B…

01:45

Problem 106 Hard Difficulty

Draw the Lewis structures and predict the shape of each compound or ion:
(a) $\mathrm{CO}_{2}$
(b) $\mathrm{NO}_{2}^{-}$
(c) $\mathrm{SO}_{3}$
(d) $\mathrm{SO}_{3}^{2-}$

Answer

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Allea C.

University of Maryland - University College

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Video Transcript

this question asks you to draw the Louis structure and predict the shape of each compound or ion. Similar to previous questions will first draw the Louis structure. Carbon has four valence electrons. Oxygen has six there, two of them. So that's a total of 16 Valence electrons. But in carbon in the center and bonding to oxygen's use of force, we now have 12 valence electrons left over. Those 12 valence electrons could be used to satisfy the octet on all of the oxygen's. But then we wouldn't satisfy the octet on carbon, so one lone pair on each oxygen needs to be shared with carbon as a another bond. So there are two CEO double bonds because there are two electron groups surrounding carbon than this would have a electron group geometry that's linear. And because they're both bonding groups, both double bonds than the molecular structure would also be linear. The next one is zero to minus. Nitrogen has five valence electrons, Oxygen has six and there are two of them. Plus one more because it's negatively charged gives us a total of 18 valence electrons. If we bond to oxygen's to the central nitrogen, we've used up four. We have 14 valence electrons that air left. We could use 12 of them to satisfy the octet for oxygen, and then the remaining two could go on nitrogen. But nitrogen won't have an octet unless one of the oxygen's shares one of its lone pairs. It doesn't matter which one. Hopefully, you can see that there would be two resonant structures here. But this question doesn't ask us for resonant structures. Either of the two resonant structures will give you the same, um, molecular structure or shape of the molecule, so we'll just stick with this one. Put it in brackets and put the charge on the outside. We see that there are three electron group surrounding nitrogen, one that's a double bond, one this alone pair and one that's a single bond. Three electron groups results in an electron group geometry that is tribunal plainer. But when one of them is a lone pair in the molecular structure or the shape of the molecule inset being bent about 120 degrees, the next one is sulfur. Try fluoride. Sulfur has six valence electrons. Oxygen also has six, and there are three of them giving us a total of 24 valence electrons. If we take sulfur and put it in the middle with three oxygen's around it, we would have used up six valence electrons, so we have a total of 18 valence electrons left. We can use those 18 valence electrons to satisfy the octet for all of the oxygen's. But if we do that, sulfur would not have an octet. So one of the lone pairs on one of the oxygen's needs to be used to form a second bond that will give sulfur now in octet. Hopefully, you can see that this double bond could have been between any of the three oxygen's and sulfur. So there are three possible Lewis resonant structures, and it doesn't matter which resonant structure you choose. They will all give you the same shape of the molecule with three electron group surrounding sulfur one that's the double bond to that air. Single bonds. We end up with an electron group geometry that his tribunal plainer and because all the electron groups or bonding groups there are no lone pairs in the shape of the molecule or the molecular structure is also tribunal plainer. The last one, we have eso three to minus the soul fight. Ion sulfur has six valence electrons. Oxygen has six and there three of them. Plus, we have two additional electrons from what we had up here because of the to minus charge. So this then gives us a total of 26 valence electrons if we put sulfur in the middle and the three oxygen's around it, we've used up six valence electrons, so we have 20 left. We can use 18 of them to satisfy the octet for all of the oxygen's. And we don't have to double bond here because we have two remaining that could go is a lone pair on sulfur, however truthfully, so that we minimize the formal charge on sulfur, make it zero and make the formal charge on one of the oxygen's also zero. We would probably have one of these form of double bond. I'm not showing it here. It's not gonna matter when we determined the structure of the eye on. But if one of these lone pairs formed a double bond, then we would have three resonant structures where the double bond could be between any of the oxygen's and sulfur. Either way, we're gonna end up with four electron groups surrounding sulfur three. They're bonding. One that's alone. Pair for electron groups. Gives us an electron group geometry that is Tetra Hydro, all with one of them being alone. Pair, then the shape of the molecule or the molecular structure is tribunal parameter all.

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