Which of the following complexes are chiral?

(a) Pt(en)Cl₂

(b) cis-[Co(NH₃)₄Br₂]+

(c) cis-[Cr(en)₂(H₂O)₂]³⁺

(d) [Cr(C₂O₄)₃]^3-

Video transcript

Hi, everyone. Welcome back. Let's look at our next problem. Consider the complexes given below and we have four different complexes described by their formula and it says identify the complexes that are Cairo A one and two B two and three, C two and four or D three and four. Let's recall that chiral complexes are ones that have no plane of symmetry. So we're going to have perhaps mirror images that cannot be superimposed upon each other. I notice that we know we only have two that are ciral in every case. So let's look at our complexes and think about their structure starting with number one, we have in brackets with an overall charge of two minus N I and then in parentheses C 204 and then VR two. So C 204 is oxalate and it is a B dentate ligand. So we've got one of these bent ligands and then two bromides. So we only have four ligands. So this is a square planar complex. So we have nickel in the center, we have two bromides, but I am drawing on the right side, but of course, they could go anywhere and then we have the one oxalate, which is bate has two oxygens that bind the nickel and then they each have a carbon bound to them. And the two carbons are attached, that's what makes it sort of this ring structure. And then each carbon has a double bond to an o, the bros have to be next to each other. Since this is a bent ligand, there's no way to separate the two oxygens that are part of the oxalate. So as a square planar context complex, it's pretty easy to see that we have a plane of symmetry that runs through this complex. So it is a Cairo. So number one is a Cairo. So any answer choice with one in it, we will cross off. So choice A is now eliminated since that's one and two. So the square planar ones are pretty nice in terms of being able to tell by looking at it. Octahedron ones are a little trickier since you're in sort of three dimensions and you have things projecting out of the screen and behind the screen. So those can definitely be a little more complicated. So let's think about that. When we look at complex number two, in brackets with a chart of two plus, we have co and then in parentheses, lowercase E lowercase N subscript three. So our central atom here is Cobalt. So I'm going to move number two over to the side a little bit cobalt and then we have our ligand of en that's ethylene diamine, it's a bate ligand and we have three of them. So let's draw our, we have an octahedral structure. So we have this square plane that's projecting out of the screen in front and behind the screen in back and then two above and below. So we have our three ligands that are bent eight, they're connected to each other. So let's just, they're all the same. So let's fill them in here as ends connected by will draw a little arc between them. So they're just randomly drawing ends next to each other connected by their arcs. Well, here's where things get tricky with the octahedral complexes. And you always want to draw out the mirror image in these cases because since we're looking at this in the flat plane of the screen, even though it's a three dimensional object, this looks superficially superficially symmetrical. It looks like we can draw a plain horizontally through the middle of this molecule. But we actually cannot because of the way we have our square plane and they're projecting out of the page or out of the screen, I should say. So just we, as I said, we always want to draw the mirror image. So I'm going to draw dashed vertical line and I'm going to draw the mirror image of this molecule. So to nitrogens from one of our rings, one projecting behind one in front, central cobalt. Well, we'll go ahead and put nitrogens everywhere. Since we know all of these bonds are to nitrogens, and we just have to determine how the arcs are connected to make it a mirror image. So we have an arc from the top to the one, the ire protecting back on the left side. And then we have an arc from the, the two far right nitrogens on our flat screen. But it would be the two nitrogens in on the left side of the square plane. And then an arc from the bottom nitrogen up to the forward, projecting nitrogen on the left side. So that's our mirror image. Now, imagine trying to superimpose this mirror image on the right on top of the one on the left. So imagine kind of flipping it over as if you were, you were folding a piece of paper. Well, in that case, what's going to happen is that while your ring that's in the square plane will remain the same, the ones that are attached at the top and bottom will be going in opposite directions. So imagine again that right most image flipped over and then your arc from the top nitrogen instead of going to the projecting back nitrogen as the image on the left is it will be coming out to the projecting forward nitrogen. And the same thing will happen in reverse with the nitrogen on the bottom. So this is why it's always important to draw this mirror image as you're trying to wrap your mind around the 3d structure here. So this is a chiral molecule. So we know that number two is Cairo, so we eliminate anything that doesn't have number two in it, which would include choice D, which is just three and four. So let's move on to structure number three, this house sis and then within brackets of the charge of negative one cu then in parentheses, C 204 subscript two and then in parentheses, NH three subscript two. So we have a central copper and then when we glance at it, at first, we might think it has just four ligands, there's two oxalates and two ammonia. But let's recall that oxalate is bent. So we have six bonds. It's an octahedral arrangement. So we're definitely going to want to watch out for those mirror images, not being super imposable, but let's look at our structure here. So my copper get the six bonds sticking out. Well, the oxalates of course, will have to be on adjoining 90 degrees separated uh bonds. This is bate and then we have the two ammonia. This is a cy structure. So the two ammonia will need to be 90 degrees apart as opposed to 180 degrees apart. So in my structure ill, just draw the ammonia on the right hand side of our square planar uh bonds and then we'll draw on our two oxalates. They must be 90 degrees apart. We have two oxygens. So on the top and on the left rear projecting bond, so draw this that oxalate in there connecting the top and the rear left and then the other one would be connected to the bottom and the front board projecting right. And again, superficially looks like it does have a Plaintiff Symmetry with those NH threes next to each other. But again, we always have to think about that 3d structure. Draw yourself a mirror image to see if that's actually the case. So put my two ammonia next to my dash vertical line, we have my copper in the middle, add my other square planes and my up and down bonds, then add the oxygens for the top oxalate and then the oxygens for the bottom oxalate. And now I will imagine flipping this over. Is it super imposable? Well, once again, because I have this 3D structure and my bonds in my square plane are projecting behind and in front of the screen when I flip it over my bond that goes up from the copper center, actually, I think this will be better with let's try some highlighting on this one to make it a little more clear. So on the left hand structure, I have a bond from the oxygen up on the top and it shares the other part of the b dentate ligand is to the bond that's projecting back behind the screen on the left side of the square plain my structure on the right if I flipped it over again, like if I had a, a paper that I was folded along the vertical line and tried to superimpose that my oxygen bond at the top when I flip it over would be projecting forward out of the screen. So these are not super imposable mirror images. And therefore this is a chiral molecule. So number three is chiral. So I'm going to rule out choice C which is two and four. So by process of elimination, I've now come to my answer, which is choice B two and three to be thorough. I'll take a quick look at structure number four, which should be a chiral according to my process of elimination. And that is sis and then in brackets with a charge of plus one, I have fe in parentheses, H2O subscript four and then cl two. So I have a central iron, I have four water molecules as ligands and two chlorides. And I know this is a cyst arrangement. So my chlorides must be 90 degrees apart from each other. So I will put them on the right side of my square lane and then it says octahedral because there's six bonds and there's four waters filling in the other spots. So two waters on the other bonds of the square plane and then a water above and a water below. Well, here we don't have any of those um bent ligands or trident ligands that have complicated things as we've seen in octahedral complexes. We just said that these simple Monod ligands are chlorides on the same side. So here we do have a plane of symmetry that goes through as a horizontal line. You see that again, the mono dentate ligands make this a lot easier and the cyst arrangement makes it clear that this is an a chiral molecule as we expected. So once again, we have looked at our four complexes, the complexes that are chiral are in choice B two and three. See you in the next video.