Cobalt(III) trifluoroacetylacetonate, Co(tfac)₃, is a sixcoordinate, octahedral metal chelate in which three planar, bidentate tfac ligands are attached to a central Co atom:

(b) Diastereoisomers A and B have dipole moments of 6.5 D and 3.8 D, respectively. Which of your diastereoisomers is A and which is B?

Video transcript

Hi, everyone. Welcome back. We have a problem here with a really that starts off with a really long chemical name because we're dealing with a ligand with a long name. That's abbreviation is D TB P I. So with that in mind, Tris, open parentheses, two, open parentheses. Dieter Butyl phosphene, closed parentheses. One, phenyl indo closed parentheses. Rhodium two and its molecular formula is we have brackets with an overall charge of positive two Rh. And then in parentheses, D TB P I subscript three is an octahedral complex with three vent ligands surrounding the central metal atom. We're given sort of a shorthand structure here with a note that D TB P I is represented as an N and A P connected by an arc that arc representing the rest of the ligand. So it's a bate ligand, as our problem told us, the two binding atoms are different, one is nitrogen and one is phosphorus. And we have this shorthand structure here shows our rhodium with a nitrogen and phosphorus adjacent to each other, the 90 degree angle between them and brackets around that ligand with a subscript of three. Then our question says, which of the two dimers will have a larger dipole moment. So we have two, they're not an an tumor, they're not mirror images. So they're two different configurations around the rhodium. So in structure number one, when we look at this in a two dimensional way, we notice as we go around clockwise, we alternate NPNPNP and structure number 21 of those is in the opposite sort of direction orientation. So we have PNNPNP. The end result of that arrangement is in structure one, we have something analogous to a back isomer. It's not exactly the same because then you're talking about similar ligands on the same face of the octahedron. Whereas here the nitrogen and phosphorus are part of the same ligand. And we're talking about the fact that the nitrogen atoms are all on the same face of the octahedron. So not quite the same thing as a fat guy simer, but analogous in this case. And that's because nitrogen is more electron negative than phosphorus, the electrons will want to hang around the nitrogen more often. So in structure number one, when all these nitrogens are on the same face of the octahedron, imagine that face, our nitrogen is kind of going like like a tripod, see the two nitrogens in the square plane are projecting back to the back out of the screen in the back with one going down. So the electrons will all be sort of concentrated on that face of the octahedron or more properly. We should say the electrons are more attracted to one face of the octahedron that face where the nitrogens are. So that means the, the dipole moment will be in that direction of that base. In structure. Number two, we have a structure analogous to a mirr isomer because our nitrogens are all in one plane. While our phosphorus are all in another plane, the phosphorus are all in a square plane and then the nitrogens are in this plane that goes to the top and bottom of our octahedron and then projects out to the back, right? So now imagine our electrons sort of being more attracted to those nitrogens. But now the nitrogens are all around the median of this rhodium here. And in fact, two of the nitrogens are opposite from each other. So in this case, the dipole moment will be more spread out or more dispersed than in structure number one. So the answer to which dier will have a larger dipole moment that will be structure one since the dipole moment will be towards one face of the octahedron, whereas in structure two, that dipole moment will be more dispersed over a larger space. See you in the next video.