Draw all possible diastereoisomers of [Cr(C₂O₄)₂(H₂O)₂]^-. Which can exist as a pair of enantiomers?

Video transcript

Hi, everybody. Welcome back. Our next problem says, true or false. The given image shows all of the dia stereo isomers of and we have a complex ion here. So all within brackets with a charge two plus, we have co and in parentheses, lowercase E lowercase N subscript two. And then in parentheses, NH three subscript two with accompanying an anti American pair is clearly identified. So we have two different and anti immerse here, both showing of course, Cobalt as our central atom, we have our ligand of N, the en which is ethylene diamine, it's by dentate. So there's two of these en ligands and they're both by dentate. So they have two nitrogens that can form bonds. And then our other two ligands here are both amino groups NH three. So first, we'd put on our ligands that are by the bent take ligands. And we see we have two possible arrangements indicated here. One where we have the en ligand going from one bond within the square plane. Don't forget the octahedral arrangements have to the square plane in the middle, projecting out and projecting back from our screen and then two groups projecting up on top and bottom. So our first arrangement shows our N molecule going from one nitrogen in the square plane up to the top and then another one going from another adjoining nitrogen in the square plane down to the bottom. Our other drawing shows all of the en molecules in the same square plane. So they're both by dentate, both of them have their two nitrogens in that middle square plane. And then the NHC groups on top and bottom. So it's pretty easy to see that these are not superimposable on each other the way these are arranged, so they do not equal each other. So we definitely have at least two here. And if we try and rotate our second molecule where they're all within the square plane where this is clearly a symmetrical molecule and any rotation there will just be super imposable in the original or you can duplicate it just by rotating. However, our first molecule is a little bit different this connections on top and bottom. And what we have is a situation where we could continuously rotate this around. Now it is symmetrical. So our different rotations won't result in uns superimposable structures. So let's just put up those different rotations. And we can see that this second arrangement does have a plane of sympathy. And thus all of these different four structures I have drawn are essentially rotations of the first one. They can all be superimposed on the original structure so a little complex to think through. But again, if we focus on the fact that there's a plane of symmetry, we can each one represents a rotation around. So our given image does show all of the diary isomers, just these two versions, one with all the nitrogens in that central plane, one with them in different planes. And the pears are not labeled because since an an tumors are mirror images, we don't have any mirror images here because the first one has these sort of five rotational images. And the second one is also symmetrical. So we have no and anti American Paris. So the answer here since it's a true or false question is true because our image does show all of the dier isomers and we can't label any in anti American Paris because there are none. See you in the next video.