The reaction of the octahedral complex Co(NH3)3(NO2)3 with HCl yields a complex [Co(NH3)3(H2O)Cl2]+ in which the two chloride ligands are trans to one another.

(a) Draw the two possible stereoisomers of the starting material [Co(NH3)3(NO2)3]. (All three NO2- ligands are bonded to Co through the N atom.)

(b) Assuming that the NH3 groups remain in place, which of the two starting isomers could give rise to the observed product?

Video transcript

Hi, everybody. Let's move on to our next problem. It says the octahedral complex and our complex is sc or scandium with three water ligands and three cyanide ligands reacts with HF hydrogen fluoride and produces a complex overall complex having a charge of positive one, which is scandium with four water ligands and two fluoride ligands. The two F ligands are trans to one another. Then it says a sketch, the two possible stereo isomers of the starting material and then b identify which of the two starting isomers could produce the observed product assuming that the cyanide groups remain in place. So let's look at our initial molecule. It has this scandium in the center and we know it's octahedral. So I've sketched two octahedral structures and we know we have three of one ligand and three of the other kind. So there's two different arrangements as our problem tells us, one is where all three of the one ligand are in the same plane. So it's often easier to see in the square plane part of portion of the molecule. So I'm going to write my three waters all within the square plane. So I have two on the left side and one on the right. But I could obviously write it either way and then fill the cyanide in the other open spaces. So, although it's a little trickier to see, you can see that this arrangement also puts the cyanide in the same plane. You've got two top and bottom and you could make a plane going through the one on the front, right. So we have this arrangement where three ligands are all on the same plane. Our other arrangement would be where they are all three adjacent to each other. And therefore on the same face of the octahedron, but not in the same plane. So let's draw our water molecules, our water ligands, excuse me, all in a row going clockwise from top and then in the square plane back right front, right, and then we'll fill in our cyanide and the other spaces. So they also are all three in a row and you can see where they're not in the same plane, but they're all sort of on the same face of our octahedron. Actually, I'll draw that in red. So it's not so confusing, it doesn't look like bonds, but we've got this sort of triangular face of the octahedron here. Sorry, that was a little tricky to drop. So here's our two isomers. Now, we need to think about which of these would lead to the final product. Well, let's think about our molecules here in our original molecule, we have three water ligands and three cyanide ligands. And that goes to a product where we now have four water ligands and two fluoride ligands. So in that case, two of the cyanide ligands were replaced, buy fluoride and one cyanide ligand was replaced by water. So since we know that the fluoride has to be trans to each to each other, we need to look for our original molecule that has two cyanide ligands 180 degrees apart. So when we do that, we see that our original arrangement. So the mole first molecule I drew in which we have all three of the ligands in the same plane of the molecule has two of the ligands being in the same place. So number one, if I number them one and two, so isomer, number one, I'll say same lane could lead to the product because again, we need those cyanide to be replaced two, just two of the cyanide will be replaced by fluorides. And so we need the cyanide in the original molecule to be 180 degrees apart. So there's our two isomers and we know that I or number one where they're all on the same plane is the one that could lead to the given product arrangement. See you in the next video.