Tell how many diastereoisomers are possible for each of the following complexes, and draw their structures. 

(c) [Cu(H₂O)₄Cl₂]⁺

(d) Ru(NH₃)₃I₃

Video transcript

Hi, everyone. Let's take a look at our next question. It says how many dia stereo isomers of? And then in brackets with a positive one charge we have fe and then BR two and then the ligand co in parentheses subscript four. And then of course, our positive one charge. How many di diao isomers of this complex exist? What are their structures? So first, let's think about the structure of this. What's the basic geometry of it? And therefore how many disaster isomers we can have? We call that this means you have all of the same Adams with the same bonding in terms of which kind of atom is bonded to which other kind of atom, but perhaps in different arrangements in space, they are non mirror images and non identical. We recall that the term for mirror image, non identical mirror images is an anti ierse. So we're not talking about mirror images, just different arrangements in space around a stereo center. So in this case, we have iron as our stereo center. And we see that we have six ligands, we have two bromides and four carneals. So we have iron center, six legends, the ligands are all mono dentate. And therefore, we know we have an octahedral geometry. So we know we have iron, we have fe in the center, we have the square planar arrangement of the four within a plane. So dash is showing bonds projecting behind the screen and wedges showing them projecting out and then bonds above and below the iron. So when we have this octahedral arrangement, and we have this situation where we have two of one kind of ligand and four of another, those two BROM means can be arranged so that they're next to each other or 90 degrees apart from each other, or they can be arranged so that they're opposite each other or 180 degrees apart from each other. And that would be the cis and trans arrangements. So we can see how it's the same uh atoms bonded to the same other type of atom so that these ligands are all attached to the iron with this iron in the center. But the way they're arranged in space around the iron is different. And that isn't a mirror image when you think of the 180 degrees or 90 degrees. So we'll look at the trans situation because that's a little bit simpler first. So in this case, I'll write our bromide ligands above and below the iron with the carbon lance arranged around the square plane. We know that our BROM means are 180 degrees apart. They're opposite from each other. This is our trans complex put brackets around it and give it its positive one charge. Now, do we need to write other arrangements of this with the bromine say opposite itself in the square plane in the square plane? No, we do not. As we can see, we have a plane of symmetry that I will draw in black here going in this case, horizontally through our complex. So we can rotate this around. If our bromine was in the square plane, rotating it around will make it identical to the version I've drawn. So we only need to draw one trans arrangement in this case, because any other version can be made identical to this just by rotation. So now let's draw our sis version. So a bracket iron in the center. In this case, we have two brom means 90 degrees apart oops and then the carbs in the other space. So in this case, I will draw the BROM means in the square plane with one projecting behind and one projecting in front and then our Carbone groups on the other side and then on top and bottom. So we see in this case, our brom means are indeed 90 degrees apart, finished drawing the bracket and our charge. And this would be the sis version, you can see that this is not a mirror image of our other version and it's non identical. And then if we look and then we need to think about. Well, can, would we put these BROM means in other locations? For instance, we can see pretty easily if we put them in other places in the square plane that would be identical by rotation. But what about the version if we put the bromine, one of the bromine's up on top and then the other bromine projecting behind the screen next to it again, 90 degrees, they're next to each other 90 degree angle between them at, on our carbons. But we can see that if we rotated this, so that those brom means were down in the square plane, we can just by rotation, make it identical to the version I just drew, which we should expect because again, the version we drew has a plane of symmetry going through it. It's a chiral, you're not going to get a different uh a different molecule just by altering which two bonds 90 degrees apart from each other you put it in. So I'm going to put a big X over the second cis structure I drew because it's identical. So we have two di stereo isomers of this molecule available. And we've drawn their structures trans and cysts, one with the two bromide ligands 180 degrees apart and one with the two bromide ligands 90 degrees apart. And again, any other versions of these can be made identical just by rotation. So they're not distinct dier isomers. So there's our two versions of this molecule. See you in the next video.