Tris(2-aminoethyl)amine, abbreviated tren, is the tetradentate ligand N(CH2CH2NH2)3. Using to represent each of the three NCH2CH2NH2 segments of the ligand, sketch all possible isomers of the octahedral complex [Co(tren)BrCl]+.

Video transcript

Welcome everyone. Our next problem says tri ethylene tetra amine abbreviated try and is the tetra dentate ligand. And we have in brackets CH two NHCH two CH two NH two. And then all of that in brackets subscript two, draw all possible isomers of the octahedral complex. We have chromium with trying cl two, all that in brackets with a charge of two plus use. And there's two NS connected by an arch to represent each of the four NH two CH two ch two NH two segments of the ligand. So we've got a lot of words here, but let's focus in on our molecule and we need to draw the possible isomers of what it would look like. We have an octahedral complex and we know that we have one of the trian, which is a tetra dentate ligand. So it will take up four of the bonds and that will be done with its four nitrogen molecules. And then it has two other spots open for the two chloride ions. So let's look at our basic oc- octahedral structure with our chromium atom in the middle. We have the chromium atom in the middle with two bonds projecting above and below. And then we have our square plane and we draw sort of dash lines, meaning imagine these bonds sort of going behind the paper and the wedges coming out in front of the paper to kind of imagine that 3d structure. So again, we have our try and it has four connections using up four of these bonds. So let's imagine the different ways it can be arranged here because we're going to have isomer is based on those different arrangements. So it can be attached with all in the same plane, with two nitrogens in the same plane and two going above and below or with three nitrogens in the same plane and one going above or below. So let's just draw these to see how they would work. And we also would want to look at their mirror images to see if we have chiral or non chiral molecules. So let's start with putting our four nitrogens around this plane in the middle our square plane. So put nitrogens around our square plane and connect them with their rings. So we have three rings going around this plane. And then of course, above and below, we have our other two ligands, the chloride ions. So chlorine on top and bottom well with the same atoms on top and bottom. And then the four nitrogens all the way around. If we drop a mirror image of this isomer, it's going to be exactly the same, it's super imposable. So this one doesn't have a mirror image isomer. So we've got this one by itself. Now let's draw an arrangement where we have three nitrogens in this plane. So 123 and then the fourth one up on top. So let's draw our little connecting rings there. And then we have a chlorine or chloride on the bottom and a chloride on one of the four bonds in the plane. So let's look at the mirror image of this because this is not a symmetrical molecule, there's no plane of symmetry since we've got our chloride and our nitrogens. But if we try and superimpose this one on its mirror image, it won't match up exactly. Because if we imagine sort of flipping this over the chloride will then be coming out as a wedge and the nitrogen projecting behind our paper if we imagine our screen as a paper here. So these are two isomers since are non superimposable mirror images. And finally, we've got one more arrangement where we have just two nitrogens in our square plane. So I'm just going to scroll up a little bit here to get that up or we can see it. So I'm going to draw two nitrogens in the square plane and then a nitrogen up top and a nitrogen below, we connect them with their rings. And then we've got chloride ions on the other side of the plane. So let's draw the mirror image to see if it would be the same or different. And this one's definitely trickier because it looks superficially symmetrical uh with the two chlorides next to each other and then the four nitrogens. But just imagine sort of flipping over as if you were turning the pages of a book, the molecules and trying to superimpose them on each other. While the atoms would each be in the same location, the nitrogen that would then be projecting out into the front would have a bond to the nitrogen on top. While the nitrogen projecting to the back would have a bond to the nitrogen on the bottom, which is reversed from the original form of the molecule. So these would also be additional stereo isomers, they're not chiral, you can't soon impose them on each other. So in the end, when we look at our problem, it says to draw all possible isomers. So we have here are five possible isomers, two mirror images, each one with three nitrogens in the square plane, one with two and then one that's is symmetrical. It's a superimposable mirror image and that's the one with the four nitrogens in the square plane. So we have a total of five possible isomers for our molecule here. See you in the next video.