Write a hybridization and bonding scheme for each molecule or ion. Sketch the structure, including overlapping orbitals, and label all bonds using the notation shown in Examples 10.6 and 10.7. d. I₃^-

Video transcript

hey everyone in this example. We want to draw the molecule iodine, Penta chloride and show the orbital overlap and label of the hybridization of the bonds in this molecule. We also need to give the hybridization of the orbital's in the iodine chlorine single bonds. Our first step for this question is to determine the total number of valence electrons in this atom. And so we should recall that first. Looking at I dine on our periodic table, we would find it in group seven A. And so we would recall that because it's in group seven A. That corresponds to seven valence electrons. Now looking at chlorine, we want to recall that it's located in group seven a. As well. So it also corresponds to seven valence electrons. However, we do have the subscript five here. And so that means we're going to multiply seven times five and add it to the original seven electrons, seven valence electrons from our iodine. And this is going to give us a total of seven times five we know is 35-plus seven is going to give us a total of 42 valence electrons total. And this is for our lewis structure. So we want to go ahead and recognize that for quick lewis structure. We're gonna have ideas in the middle surrounded by five chlorine atoms. And we should have these connected by the bonds single bonds first to the iodine and recognize that iodine has 10 electrons in these five bonds that we've created which is okay because it's a it's past period three on our periodic table and any element past period three is able to have an expanded octet and now we're going to fill in the remaining electrons around our chlorine atoms. So we should have 45689, 10 11 12 14 18 2022 24 28 30. And then we're just left with two more electrons which we can actually go ahead and put on our idea as a lone pair. Okay now that we have our lewis structure completed to the correct structure, we want to go ahead and determine our electron domains or r regions of electrons and we should recall that to find the number of electron regions or electron domains. We're gonna take our number of sigma bonds added to our total number of lone pairs and I'll make that a capital L. So we should recall that sigma bonds are also known as single bonds because within every single bond we have sigma bonds. So we would go ahead and count our total number of sigma bonds. We would have 1234 and five. So we have five sigma's plus, we need to count the number of lone pairs and this is in corresponding to our central atom. So any time we want to find our number of electron regions or electron domains, we're focusing on our central atom here, which in this case is iodine. So we counted the number of sigma bonds attached to the central atom. And now we also would recognize that we have this lone pair that we added at the end to our iodine, our central atom. And so we would say that we have one lone pair added to the five sigma bonds. And this gives us a total number of six electron domains for our structure or six electron rich regions. And we can say therefore now that we know our total number of domains, we can say that therefore we will have the sp three D two hybridization for our central atom. And so this is the type of hybridized orbital are iodine will be in in the molecule. And because we know we have sp three D two hybridization, we would recognize that this corresponds to an octahedron geometry, which we should recall. Octahedron geometry is going to be shaped as follows. So something like that. And so now we can go into showing our orbital overlap diagram. So we're always going to start with the central atom with the hybridized orbital that it's contained in. And we determine that it's sp three D two which is an octahedron shape. So we will draw the orbital as follows. We're going to have sort of a flower shape that we're going to draw in and they should be pretty symmetrical in shape to one another. And sorry, I'm not the best artist but I'll do my best for you and we're gonna add in two more lobes over here. So this is going to be for our I dean central atom, the sp three D two hybridized orbital and we can go ahead and label this in just so it's very clear. And now we also want to be sure to include that lone pair that we added on our central atom to complete our valence. So we just draw it in in the bottom part of our hybridized orbital down here. Our next step is to focus on our five chlorine atoms. We want to recall that chlorine on the periodic table is located across period three in group seven A. And so we would therefore say that it has the S. P. We're sorry rather the P three p orbital presence because we would find it again across period three in group seven A. To land on our chlorine atom. And so we would be within the three P orbital. So we should recall that our p orbital's our dumbbell shaped roughly. So like this. And when we show them overlapping in our molecule here for iodine, Penta chloride, we need to make sure that we're overlapping it with our hybridized orbital parallel head to head to each of our hybridized orbital's here. So we would go ahead and let's make the first connection where we would have from, sorry. From the head here, we're going to make the beginning of our attachment and we're gonna complete our P orbital here and I'm actually going to make it a little bit better than that. Sorry. So we should have something like okay, that was worse. We should have something like this. As you can see, we've created an intersection here where the two orbital's are overlapping and we want to recall that within this overlapping we have the region of our electrons being shared in the covalin bond and opposite spins between these two different orbital's. And again, this is our three P orbital for our chlorine atom. And we're going to do that for each of the chlorine atom bonds that we have in this molecule. Or rather the chlorine iodine bonds in this molecule. So again, head to head connections. So we draw in our three p orbital and again we have our electrons here within the overlapping portion of our covalin bond between I dean and chlorine. And we should recognize that it is within this overlap between these two orbital's that our single bond or our sigma bond is created. So we can go ahead and also label these sigma bonds here because at these intersections or these overlapping between our sp three D two and R. S. P. We're sorry, three p orbital's here for chlorine. We are creating our sigma bonds. And so we have three more connections to make. So let's go ahead and do those. So we have our three p orbital overlapping. And sorry, this should be parallel again, we want to make sure everything is parallel. Three p orbital for our chlorine atom, we have our electrons here in opposite spins, forming our sigma bond connection between chlorine and iodine. And we're gonna do that again here. Head to head connection again for the three P orbital for chlorine. These are all for chlorine here because we only have chlorine and iding in our molecule, we have our electrons here and we have our sigma bond that has formed and then we just have one more to draw in. And sorry, that was a very poor sigma bond. And then lastly, we just have that last three p orbital to draw in. So head to head Overlap, three p. This is for chlorine and then we have our electrons here and this is a little bit bad to draw but you understand what I mean by now. And so we have successfully shown how we have the overlap between the hybridized sp three D two bonds and the three P orbital's. Or sorry, the hybridized sp three D two orbital's with the three p orbital's of our chlorine and iodine sigma bonds. However, we want to go ahead and clearly express the answer here. So we would say the hybridization of the iding chlorine bonds is going to be SP three D dash not three P, but just p. We would say for our chlorine atoms because ultimately they are in the p orbital. And so this would be this entire diagram here that we filled in as well as our distinction here for the ID in chlorine bonds would be our final answer to complete this example again, at every overlap of our sp three D two orbital and our three P orbital. We've created the sigma single bonds that make up our molecule, which is why we have a total of five single bonds in our molecule here. And because all of our sigma bonds are iodine chlorine single bonds, they have the hybridization of the combination of the orbital from iodine and the orbital from chlorine sp three D two for iodine and the P orbital for our chlorine atom. So this will complete this example as our final answer. I hope that everything I reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.