Write a Lewis structure that obeys the octet rule for each molecu...

Question

Write a Lewis structure that obeys the octet rule for each molecule
or ion. Include resonance structures if necessary and assign
formal charges to each atom.
a. SeO2

Accepted Answer

welcome back everyone In this example, we have the S. N. O minus ion and the nso minus an ion as too simple ions from a large family of compounds called the NSO compounds, These compounds are mainly found in geological formations within asphalt teens and other mixtures. The S. N. O minus ion is more stable than the nso minus ion. We need to draw lewiS structures for both ions, including anyway residents forms and using formal charges to determine why the snl minus ion is more stable than the Nso minus ion. So what we're going to begin with is drawing out our Lewis structure of the S. N. O minus an eye on. We want to recall that before drawing lewis structures. We need to calculate total valence electrons. So beginning with sulfur and oxygen, we would recall that their location on our periodic table is in group six A. And so we would recall that that corresponds to a number of valence electrons being six for each of these atoms. Moving on to our nitrogen atom. We recall that that group five A. On the periodic table corresponding to five valence electrons. We also recognize that because we have this minus one and in charge, this means that we gain one electron. So we would add one electron to our total and we would get a sum equal to a total of 18 valence electrons. So drawing out our structure according to our formula, we have nitrogen in the middle surrounded by sulfur and oxygen and beginning with filling in our valence electrons. We recall that beginning with oxygen, it prefers to have a total of six. So we have 1234 pairing them up now five and six. However, according to our formula, we need to have one more extra electron which is assigned to our oxygen atom. And so we would use that here and I use the color red to represent that gain of that extra valence electron. And so oxygen now has seven valence electrons on itself. So we would say 18 minus and sorry about that. We have 18 minus seven of our valence electrons. Use so far. This leaves us with 11 valence electrons left where we would fill in on nitrogen, which we recall, prefers to have a total of five. So we have 12345 valence electrons where we can recognize that we can use each of these valence electrons between nitrogen oxygen to form our first covalin bond because they're both non metal items. So taking 11 minus R five valence electrons. We've used we now have a difference of value of six valence electrons left and that would be perfect for our valence electron requirement for sulfur. So we have £1234 now, five and six. Now we don't want any radicals. So we're going to move this valence electron here next to this one here, pairing it up so that we can use it to form with this extra electron on nitrogen, a double bond here. So we have to double or two bonds here which forms our double bond between sulfur and nitrogen. And this keeps both of our atoms happy and stable because nitrogen has its bonding preference of three bonds and it's one lone pair, meaning it has a formal charge of zero and sulfur has its six fans electrons. Where two of its fans electrons are shared in a co valent bond here. And so sulfur also has a formal charge of zero. Recall that to calculate formal charge, we take our valence electrons Subtracted from our non bonding electrons and then subtracting that from our bonding electrons. And so what this means is that because we recognize oxygen has that extra valence electron being its seventh fans electron. It has a formal charge of -1. So now we want to go ahead and show our residence forms which we will have because we do have that negative charge which can be conjugated as a double bond on our structure here. And so we would have our minus charge. Move to this area to form a double bond with nitrogen. Where now we would have to have this double bond here fall back on our silver atom as a lone pair. And so let's actually draw that properly. So it would just go in this area since we have no lone pairs here. So to draw our residents form, We would have our sulfur nitrogen and oxygen still. However, now we have a double bond going towards our oxygen atom where we have just a single bond between our nitrogen and sulfur atom and our sulfur atom now has a total of three lone pairs around itself which gives it that -1 and ion charge as we had before. Now filling in the rest of our valence electrons. We still have that lone pair on nitrogen and oxygen will now have what sulfur had originally which was two lone pairs. And so we want to place this in brackets because these are our two different resonance forms of our lewis structure of S. N. O minus. Moving on to our next ion according to the prompt, that is the N. S. O. Minus ion. So calculating total valence electrons, we would still get the same amount where we have a total of 18 valence electrons because these are the same atoms still nitrogen, still sulfur and still oxygen and we still have that minus one charge. So we still have 18 valence electrons to use. However, this time our central atom is sulfur surrounded by nitrogen and oxygen so filling in our valence electrons. Beginning with our oxygen atom, we recall it 16. So we have 1234 pairing them up five and six. And we want to recall that our according to our given formula r minus charges on our oxygen atom. And so we would fill in 1/7 electron on oxygen so that it will have that minus one formal charge. Moving on to our adam sulfur, we've used a total of seven of Iran's electrons leaving us with again 11. We have sulfur which we recall prefers also to have six valence electrons. So filling them in, we have 1234 pairing them up five and six. So we would say minus £6 electrons. This leaves us with a total of £5 electrons to fill in for our nitrogen. So we would have 1234 pairing them up. Now we would have five. Now, what we should recognize before we make our bond connections is that oxygen is not a period three elements. It is in period one or sorry, period two and so on our product tables we recall that period three elements can have expanded octet. Right now, oxygen has a total of 246 and then 78. When we form a co valent bond between sulfur and oxygen here, electrons, meaning it fulfills the octet rule. So we can't have this extra electron here. So we're going to have to move it to this side of our sulfur where we can form with this film solectron to this film's electron. A Covalin bond between nitrogen and sulfur. And we can then move this valence electron to the nitrogen atom where it can form a covalin bond with sulfur here. We can now move this to be paired up with this lone pair. And so this is going to create a formal charge of nitrogen where we recognize it originally wants to have five valence electrons, but right now it has a total of 123456 valence electrons, meaning it will have a formal charge of now minus one as well as the oxygen atom here and then going to ourselves for adam we would say that moving those valence electrons around, we have a total of 1234 valence electrons where it prefers to have six. And so I'm sorry about that. That would be 12345 valence electrons and it prefers to have six, which would leave us a difference of one. And so this would be a plus one formal charge here since it has one less than it would prefer. So it's missing one valence electron. And so now we want to go ahead and show our formal charge how they can congregate to give us our resonance forms of our nso minus an eye on. So this is our first residence form. So, drawing our second residence form, we again have nitrogen surrounding sulfur as well as oxygen where are negative charge can now form a double bond here between sulfur and oxygen. And then our double bond on, nitrogen can move to nitrogen as a lone pair instead so that it would leave a single bond between nitrogen and sulfur. And so that would form the following resident structure. Whereas we stated, we have a single bond between nitrogen and sulfur. It now will have three lone pairs. Since that double bond is now moved as a lone pair on nitrogen, Which will now give nitrogen a new formal charge where it has a total of 1234567 valence electrons, meaning it has a formal charge of having two extra than it would like because nitrogen prefers to have five valence electrons. So it has a formal charge of -2. Moving on to our oxygen atom, we now have a double bond between our oxygen atom where oxygen will now only be left with two lone pairs And our sulfur atom still only has that one lone pair. So our oxygen atom is back to being stable and happy with its six total valence electrons. Where as our sulfur atom still has that plus one can be in charge because it only has a total of +12345 valence electrons. Where it prefers to be like oxygen having six. So it's missing one valence electron. So this is our second resonance form and now we can actually draw a third resonance form. So we'll draw that below here. So this is residents form number two and now moving on to residents form number three, we still have nitrogen surrounding sulfur and oxygen surrounding sulfur where we can show one of these valence electrons forming a double bond between our nitrogen and sulfur atom. And so that would give us the following structure where we have a double bond between nitrogen and sulfur, we have just two lone pairs on nitrogen, which would yield again a charge, a formal charge of minus one because it only has a total of six valence electrons, which is one more than I would prefer to have. And then our oxygen and sulfur atom still share that double bond between each other. We still have one lone pair on sulfur and we still have oxygen with two lone pairs, meaning our sulfur atom in this structure will have a total of 123456 of its preferred valence electrons, meaning it has a net zero formal charge on sulfur. So it's happy and neutral, just like oxygen. And to get to our previous residents form, we would just have this double bond congregate back to our nitrogen atom as a lone pair. And so this would complete the three resonance forms of the nso minus an ion. Now, out of all these residents forms, we would recognize that the most stable is going to be the one where our negative charge is on our oxygen atom. So this is the most stable. And we recognize this because based on our electro negativity trend on the periodic table, which we recall increases from left to right, going upward on a product table, oxygen is the most or more electro negative atom compared to nitrogen and sulfur. And so it can handle having this negative charge on itself, having more electrons than it prefers to have better than nitrogen oxygen have. And so this is our most stable resonance form now going up to R S N. O minus an ion. We also would confirm that this first form is the most stable resonance form because this is the form where are minus one charges on our oxygen atom. So it can handle it the best better than sulfur nitrogen because oxygen is more electro negative. Now we want to recognize that based on net formal charge, we would have a total of minus one plus minus one which would give us minus two and then plus one is just minus one. So we have a net charge of minus one for our first residence form. In this case we also have a net charge of minus one for the second residence form because we would have minus two plus one and then for a third residence where we still have a minus one net charge. So that would give us a total of minus three for all of our residents forms here for nso minus, whereas for S. N. O minus we have a net formal charge of minus one for the first residents form and minus one for the second residence form being a formal or sorry a total of net formal charge being minus two for the two residents forms of S. N. O minus. And so this means that we have the least. Or rather, let's just say we have minimal formal charges in its resonance forms. So let's say in S N O minus resonance forms. And so therefore we would say that S N O minus is more stable than the N S. O minus ion. And so because these resonance forms of the S N O minus ion have minimal formal charge compared to the nso minus ion, we would say that the snl minus ion is more stable than Nso minus. And so this would be our final answer highlighted in yellow to complete this example. So I hope that everything I read was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

Write a Lewis structure that obeys the octet rule for each molecule or ion. Include resonance structures if necessary and assign formal charges to each atom. a. SeO2

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