Which of the following complexes can exist as diastereoisomers?

 

(a) [Cr(NH₃)₂Cl₄]^-

(b) [Co(NH₃)₅Br]²⁺

(c) [MnCl₂Br₂]^2- (tetrahedral)

(d) [Pt(NH₃)₂Br₂]^2-

Video transcript

Hi, everyone. Here's our next problem. It says, consider the complexes given below and we have four different complexes with metal centers listed. And then it says identify the complexes that have di stereo isomers to a one and two B one and three C two and three or D three and four. So let's return to our complexes and just go through each one and think about which have di stereo isomers. Remembering that these are non mirror image, non identical isomers. So choice one, we're told that we have a tetrahedral geometry and our complex is a palladium PD with two carbel co ligands and two cyano CN ligands. So for number one, we have our palladium in the center, we have tetrahedral geometry. So we usually draw one atom, one bond projecting up and then we have one in the plane of the, the screen. So the first two on the plane of the screen and then one with a wedge coming out of the screen and one with a dash line, meaning it projects behind the screen. Remember that tetrahedral arrangement, all of these bond angles angles are the same. So all bond angles the same all about 100 and nine degrees apart from each other. So you can imagine it is sort of a pyramid. You have one bond projecting upward and then imagine kind of a tripod with three legs. So we have four ligands, 2 CO2. So let's just put them on here. I've put one, not CO2, excuse me, two co ligands, one co ligand on the top and I'm just putting one on the right. So the one in the plane of the paper and then my cyano liens on the ones projecting in front and behind. Oh I wrote all bond angles. Plenty wrote all bond angles equal. Now, we might be tempted to think. Well, could I have a cis and trans arrangement here? I have my two ligands are each sort of next to each other. But this is where it can be confusing when you're looking at a structure drawn in a two D arrangement, try to represent a 3D arrangement because if we drew our structure, for instance, let me just draw over here, same thing, draw. So that kind of looks on paper that it's trans. So I'm drawing the uh carbon monoxide on the top as in my original structure. And then I'll draw another one projecting out in front of the plane of the paper and then I'll draw a cyanide in the plane of the paper and another one projecting behind. So in my two D drawing, it looks as if my co ligands are opposite to each other as in a trans arrangement, but that we only see in a square planar arrangement. And we can see how that happens if we imagine rotating our molecule. So I'm going to use different highlighting to emphasize to keep track of our different ligands. So in my first structure, the co on top I will highlight in this pinkish red color, then the co that looks like it's next to it in the plane of the paper in blue, the CN that's projecting out from the plane of the screen. Excuse me, not paper is in green and the CN projecting behind the plane of the screen is in yellow. So I'm going to imagine as if this were a top imagine holding by the top bond and spinning it like a top and everybody is spinning around one turn. Well, we can see if I do that but then I end up my co on top. The pink one is still where it was because that was sort of my axis of rotation. My yellow CN has gone from being behind the plane of the screen rotating round to being in the plane of the screen which corresponds with the second drawing I made. Then my co ligand that was in blue that was in the plane of the screen has now been rotated to be projecting out from the screen on that wedge uh bond. And then the CN that was in green has rotated around and now it is projecting behind the screen. So you can see how just by rotation I made these two the same. And that's because again, all bond angles here are equal. It's just the fault of the fact that we're drawing this in two dimensions that makes it look like we can have a sits and trans arrangement. We can't really with a tetrahedral complex where we have this plane of symmetry, which we do. So I will draw in black on my original that we do have a plane of symmetry in this molecule since we have two sets of two identical ligands. So this plane of symmetry can run between the cyanide ligands and between the carbon ligands. So again, that plane of symmetry means just by rotation, I can make those versions identical. So number one does not have disaster isomers. So I will write no next to compound number one, which again, we might expect with this tetrahedral arrangement with two pairs of identical ligands. So let's eliminate choice. A which is one and two and B which is one and three. So we're now down to two answer choices. So now we'll move on to complex number two. This one has a nickel as our central atom, a plus one charge overall. In the entire complex, we have five water li ligands and one thio cyano ligand SCN. So we have six bonds total six ligands giving us an octahedral geometry jot down octahedral and with octahedral and square planar geometry, we definitely needed to be on the alert for diasa isomers cysts and trans possibilities. However, when we draw number two, so here's our nickel and we'll draw our octahedral geometry with a bond above and below. And then a square planar arrangement in the middle with wedges and dashes representing projecting in and out of the plane of the screen. So now I'll just put a water on, I'm going to put it on top and then I will fill in my tho cyano as one of the back, right bonds SCN and then just go in filling in waters around my molecule. However, when I have only one of one type of ligand and then five others, I will not have dias isomers. Again, this is we have this plane of symmetry. So every different arrangement, we could draw that plane of symmetry running through our different ligand. Every different arrangement we draw could be changed to our original version by rotation, just put brackets around with my plus one charge to be proper here. And we'll just have a quick demonstration of that. Let's imagine rotating my whole structure using the I'll highlight and bright pink hair. I'm going to use my two waters that are top left and bottom right of my square plane as sort of an axis of rotation. And I'm going to rotate the whole thing so that my SCN group moves to the top position. And then again, this imagine this rotating around sort of like a wheel with these ligands switching position. And when I draw the result and structure, you can see I've stuck the water. Now the water is on a different bond. But you saw, I achieved that just by rotating the whole molecule and I could go on and do this and put the, not the water, excuse me, the SCN, I could go on and do this and put the SCN on each of these different bonds. Oops I left out of water there and just by rotation change it around. So this does not have disaster isomers which again, we'd expect since we have that five of one ligand, one of another. So choice number two or complex number two does not have ster isomers. So that's going to go ahead and eliminate. So C which is two and three. So just by elimination, I'm down to only one answer. Choice D three and four. So I have my answer. If I were on a test, I wouldn't even look at my other molecules just to be thorough. We'll take a quick look. We have complex number three, we're told is a square planar geometry. So we should be alert to the possibility of cysts and trans isomers. And indeed, that will be possible because we have a copper with two nitrate ligands and two ammonia ligands and the overall charge of two plus. So number three, I'll just draw very quickly because we have two of each, we have a square planar geometry. So I could put my nitrates 90 degrees apart from each other. Actually, you know, it's two, it's two of each kind of ligand. So they're each 90 degrees apart. So that would be my sis arrangement and then I have my trans arrangement where the similar ligands or the same ligands will be 180 degrees apart. And that will be my trans version with the identical Liggins being 180 degrees apart. I had to quickly edit my drawing because I had just put lines for the bonds instead of my proper wedges and dashes to represent that three dimensional space. So we do have ster isomers there, which again, we should expect with square planar geometry and two groups of two identical ligands. Finally, number four, we have an iron complex with a negative one charge. We have four cyano ligands and two Carbone ligands. So we have octahedral geometry. So we definitely should be on the alert for cis and trans arrangements. So number four, we have iron in the center and let's put our two carbons on top and bottom to make a trans arrangement 180 degrees between them and then put our cyano groups in our square plane. Oops. OK. So we've made a trans arrangement here and then, so let's add brackets and give it its negative charge. And then we have a cis arrangement where we'll put our co twos. Yep. Sorry. Just realized I had drawn my carbonel lids as CO2. It's two co ligands, not CO2. So, getting a little confuse there and remembering to draw my dashes and wedges here, add my cyanide ligands. And again, it's just for convenience that I'm writing the C OS in the square plane because it's easy to think of the cysts that way. But as we know, we can rotate that molecule around and put the C OS anywhere here as long as they have a 90 degree angle between them. So once again, this molecule with its octahedral arrangement and two similar ligands, four of the same ligand two and four can have cysts and transversion and therefore will have dier isomers. So sort of the summary of this all is square planar geometry, octahedral geometry definitely be on the alert for those, those cysts and trans possibilities, especially when you have something like two of one type of ligand, two of the other two of one type of ligand four of the other. Uh keeping in mind that when you have just one and five or one and three, you will not have to isomers. So our answer here again, as to the complexes that have dias isomers choice D three and four, see you in the next video