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?

Video transcript

Hello everyone today we are being asked to determine the lewis structure that would be dominant for boron chloride. And so the first thing that we want to make note of is the formal charges of each atom present. And so the dominant form Will essentially have zero charges present on the molecule. And so before we label our formal charges, we can denote this first structure is A. The second one is B, third one is C. And the third one and the fourth one as D. We can then go ahead and label the different atoms present. So this can be chlorine one. This can be chlorine too. For B. This will be chlorine one as well because it has a similar form as the one in answer choice A. And then this chlorine at the bottom for B will be to moving on to see This first chlorine will be chlorine too. And the chlorine on the right is one and four D. All of these chlorine are the exact same. So they can just be known as chlorine one. And of course our boron is going to be a little different here as it only has three individual lines to it. So we're gonna name it one notice how we didn't label the other chlorine in the other answer choices. And that's because they're similar to the ones that we are labeled. So for example for a we already labeled the first chlorine chlorine one. That's similar to the chlorine at the bottom here as it has six electrons around it with one bond line present. And so if we were to calculate formal charges, let's go with answer choice a first for our first chlorine, we're going to say that it has seven valence electrons because it's in group seven, there's one bond line that's attached to it and six individual electrons around it, giving us a formal charge of zero. The formal charge of that original boron is going to be three. Well the valence electrons will be three because bronze in the third group, there's four bond lines around it and zero electrons giving us a formal charge of negative one. And the formal charge of that chlorine to will be seven valence electrons minus the two bond lines attached to it minus the four individual electrons that we see giving us a formal charge of positive one. We can't rule it out just yet because we have to explore other options. So let's go to be, we have our formal charge of our first chloride. Which will be similar to the one of that in a which will be zero. Our formal charge of our second chlorine. So this formal charge of our chlorine to will be similar to that of answer choice A as well. And that'll be positive one. Now note how this chlorine here only has four individual lines here. It's very different. So we're gonna label this chlorine three And that'll be the only instance of that. And so the formal charge of our chlorine, three is going to be r seven valence electrons because chlorine is in the seventh group of the periodic table, minus our how many lines around it. One minus how many electrons to see around it, giving us a formal charge of positive two. Moving on to see the process is going to be very similar with our first chlorine following the same principles as the other ones. Formal charge of zero. The formal charge of our chlorine to Is going to follow once again, it'll be the formal charge of positive one. The formal charge of that boron is going to be minus one. And for the previous answer choice B the formal charge of that boron is going to be similar as well, which would be negative one. And so moving on to D we have to see which one will have a formal charge of zero. So that formal charge of that first chlorine will be zero as before. And the formal charge of that boron call it boron one will be three valence electrons minus how many lines around it? Minus how many electrons around it giving us zero. And so we see that we have an overall formal charge of zero for answer choice D which as we said before, the dominant form will have zero charges. And so with that our answer choice is D or the bottom right answer. I hope this helped. And until next time