Draw the structure of all isomers of the octahedral complex [NbX₂Cl₄]^- (X- = NCS^-), and identify those that are linkage isomers.

Video transcript

Hi, everyone. Let's take a look at our next question. It says, consider the octahedral complex and we have a complex with a charge of positive one Cobalt as the metal. And then X subscript two NH three in parenthesis, subscript four. So Cobalt with two X ligands and four ammonia ligands where X is N minus give the structures of all possible isomers for the given complex and identify the linkage isomers. So let's think about the complex we have here and recall what a linkage isomer is. So a linkage isomer occurs when a ligand has two possible electron donors. So either of them can bond with a central metal atom. So in this case, our the R ligand with two possible electron donors is X which is OCN minus because both oxygen and nitrogen are good electron donors. So you could have a situation where the oxygen atom links to the central metal metal atom, excuse me, which will give you the ligand cyan or in the other order with the metal bonded to nitrogen. So NCO minus which will be the ligand isocyanate. So our ligands can exi our ligand X can exist in these two forms and we know we have two of them attached to our metal atoms. So we could have a situation where we have two with the o linkage, two with the N linkage or one of each. So we know we're going to have those three different possibilities for linkage isomers. But then we also, when we look at the structure of our complex, we have six ligands. So we have an octahedral geometry and our formula is in the form M A subscript four B subscript two. And whenever you have that, you know, you have the possibility of dimers because those two B ligands can be an assist or trans arrangement. So in this case, we've got our two X ligands that can be attached one of two ways and they can be an assessor trans arrangement. So let's draw these. So I have my central Cobalt atom and then I now have my octahedral bonds. So I'll start with putting, I'll start with OCN minus the cyan eight. So I will put an N ligand on the top and on the bottom. So this would be a trans arrangement and then my ammonia Adams around the square plain. And then of course, another possible isomer would be this same group of ligands. So same linkage on our OCN, but in the CIS arrangement. So I'll put that in. So here I have my OCN ligands next to each other with a 90 degree distance between the two bonds as opposed to the 180 degree distance. So this is going to be a cis arrangement. And then I've added in the fact that these are ions with a charge of positive one. So I've added brackets and that positive one charge. So these two are not linkage isomers because the two ligands are both linked, the metals linked to the oxygen atom in both, but these are ditherers. So now let's change up one of our OCN ligands to the other linkage. So this next one will be a different linkage isomer. So here we still have our two ligands, our two X ligands 180 degrees apart. But I've changed the bottom one to the metal bonded to the nitrogen. So our two in the top row are linkage isomers. Now, this molecule I've just drawn is in the trans arrangement and I can also draw its cis dite. So I'll do that. So these two that I've just drawn top and bottom are not linkage isomers because they both have the same uh combination, one with the cobalt bonded to the oxygen, one with the cobalt bonded to the nitrogen, but they are ditherers because one has a trans arrangement and one has a cyst arrangement. So you can see this pattern here. The last one left or the last two left to draw, I should say would be where both of our X ligands have the cobalt bonded to the nitrogen. So you can see again, I have both of the X ligands. Now in the form NCO with the bond of the nitrogen, and I've drawn both the trans and cis arrangement of those. So we now have two rows of three where all the ones in the top row are linkage isomers since they differ in how the ligands are attached to the Cobalt by which donor atom. And then the three on the bottom are linkage isomers. But the bottom and top rows in relation to each other are ditherers. So this would be all possible isomers. Any other way we can draw any other configuration would be a configuration that due to the plane of symmetry in our molecule, you could rotate it to be the same. So once again, here are all the possible isomers, two batches of three linkage isomers, each of which exists in assists and trans arrangement. See you in the next video.