Draw a Lewis structure for each of the following:
b. CO

Question

Draw a Lewis structure for each of the following:

b. CO₃²⁻

Accepted Answer

Hi everybody. And welcome back. Our next problem says, draw the Lewis structure or clo two minus. So let's think through the steps of drawing a Lewis structure. And the first one of course is to figure out how many valence electrons do I need to place when drawing this. So I need to think about the fact that I have chlorine, which is in group seven A oxygen, which is in group six A and I have a negative one charge on my molecule. So there'll be one extra electron giving that negative charge. So when I calculate how many valence electrons I have. So valence electrons I have seven for my one chlorine atom, there's two oxygen atoms. So two multiplied by six for oxygen plus one for the negative charge. So seven plus 12 plus one, I have 20 valence electrons total. So let's draw a structure here. So we'll start by drawing a sigma bond structure, sort of a tentative starting structure or framework. Well, I've got two oxygens, one chlorine. How would I think about putting them together? Well, what I want to think about is the fact that chlorine is the most electron atom we're used to thinking of oxygen as being the more electron. But in this case, it's with a chlorine, which is even more electron. So it makes sense to put chlorine in the middle. So we'll put oxygen with a bond to the chlorine with a bond to the oxygen. This especially is a bit unusual because we're usually used to seeing halogen atoms just having one bond to fill in that one missing electron. So definitely a weird one that we're dealing with here. Then we have 20 valence electrons total. Well, how do we calculate the number of valence electrons from a structure? So the case of a structure, we use the formula that for valence electrons, it's the number of bonds plus the number of lone pairs and in that sum multiplied by two. So in this case, the way we started, we have two bonds here. So that would be four valance electrons accounted for just with bonds, we still have 16 more to go. Well, first of all, all, let's add lone pairs to give everybody a complete octet. So our oxygens at the end, which just have the one bond will get three lone pairs. The chlorine in the middle has two bonds. So it needs two lone pairs to give it a complete octet. And then we'll put brackets here because around the whole molecule because it has a negative one charge. So let's look at does this fulfill our requirements? Do we have all of our electrons accounted for. So for the valence electrons in this structure, we have two bonds. And then how many lone pairs do we have? Well, we've got three on each of the oxygens and two on the chlorine. So we have eight lone pairs. Notice this is number of lone pairs, not numbers of electrons. So we have two plus eight multiplied by two equals 20. So yes, this structure gives us all of the valence electrons we're supposed to have. So our last step is to calculate the formal charge on each electron. So the formula we use for calculating formal charge is the number of valence electrons. Each atom has minus the number of non bonding electrons. Now notice this is number of individual electrons, not number of lone pairs. So that's important don't get those mixed up minus the number of bonds. So the oxygens will obviously be the same. So for each oxygen, they have six valence electrons, each of them has three lone pairs. So minus six non bonding electrons, and then the number of bonds is one, so six minus six minus one. So we're left with a negative one formal charge on each oxygen. So we'll write that on our structure. And then for chlorine, it has seven valence electrodes, there are two lone pairs. So minus four non bonding electrons minus two bonds. So seven minus four minus two. So we end up with a positive one charge formal charge, we'll write that up there. Well, there's kind of two issues with this. The main issue is when we look at Lewis structures, we often can draw more than one to see which one is the most likely the more correct one, we want to minimize the number of formal charges or the total formal charges on the structure. The way we've drawn it, we have a formal charge on all three of the atoms. Now, we will need a formal charge on at least one because we have this negative one overall charge. Um But do we need a formal charge on every single one? So we need to redraw the structure to see if we can make a better structure with fewer formal charges. So I put a little arrow showing want to minimize the formal charges. So redraw the structure. Another thing that might ps in this might not be the best structure. This leaves a positive formal charge on the most electron atom negative charges on the less electron atoms. So that kind of doesn't pass the gut check basically for me. So let's scroll down and see if we can come up with a better solution. So what could we do? Well, the first approach should be let's throw a double bond in somewhere and see if that improves the situation. So let's redraw it. We'll put oxygen single bond into chlorine. And now let's put a double bond in there between chlorine and a second oxygen again, putting brackets and that negative one charge. And now let's fill in the octets here. So the first oxygen has three lone pairs, just as in the original structure. The chlorine now has three bonds. So it only gets one lone pair to give it a complete octet. And then our second oxygen has a double bond. So it has just two lone pairs. So now everyone here has a complete octet. Let's check our valence electrons. So for valence electrons are the structure that will equal the number of bonds, which is three total plus the number of lone pairs, we've got three plus one plus two, we have six lone pairs, all that multiplied by two. So nine multiplied by two gives us only 18 valence electrons. So this can't be correct. We still need two more accounted for. Well, where are we gonna put them? Everybody has a complete octet. Well, let's recall that chlorine is in the third row of the periodic table. And what does that mean? Well, in the case of chlorine, anything in row three or above has D orbitals, chlorine's got empty D orbitals. So that means it is OK to put an expanded octet on chlorine. So I can have an expanded octet. So that means we can give it more than eight valence electrons. So that's what we will do with our extra two valence electrons. So we'll go ahead and put them as another long pair on chlorine. So that will give us our 20 valence electrons. Wait. And now we just need to look at the formal charges to see. Does this make the situation any better? So now our oxygens have two different formal charges. The one on the left has a negative one formal charge just as it did before. It's in the same situation. Fix that a little because I actually accidentally covered up my own hair. So let's look at the chlorine. So it has seven valence electrons. It has and you know what, I'm going to highlight that formula. So I can find it in among all everything else formula for the formal charge. So seven valence electrons minus we have two lone pairs. So that gives us four non bonding electrons and now it has three bonds. So seven minus four minus three now has a formal charge of zero. The knee is second oxygen, six valence electrons minus we have two lone pairs. So four non bonding electrons minus two bonds, six minus four minus 20. So this gives us zero formal charge on the chlorine and one of the oxygens, the unavoidable negative one formal charge on the oxygen since we have that negative one charge. So this is a better Lewis structure. So that will be our final answer. Just highlight that in yellow to pick it out. And we have in brackets with a negative one charge oxygen with three lone pairs a single bond to chlorine with two lone pairs and a double bond to another oxygen with two lone pairs giving a formal charge of negative one to the oxygen that's in a single bond with the chlorine. See you in the next video.

Draw a Lewis structure for each of the following:
b. CO₃²⁻

Key Concepts

Lewis Structures

Formal Charge

Resonance Structures

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