The molecule dimethylphosphinoethane [(CH₃)₂PCH₂CH₂P(CH₃)₂, which is abbreviated dmpe] is used as a ligand for some complexes that serve as catalysts. A complex that contains this ligand is Mo(CO)₄(dmpe) .

a. Draw the Lewis structure for dmpe, and compare it with ethylenediamine as a coordinating ligand.

b. What is the oxidation state of Mo in Na₂[Mo(CN)₂(CO)₂(dmpe)] ?

c. Sketch the structure of the [Mo(CN)₂(CO)₂(dmpe)]^2- ion, including all the possible isomers.

Video transcript

Hello everyone. Today, we have a foreign problem. Consider the ligand di di metox ethane or do you mean first provide the Lewis structure of DME? How is it similar with ethyl and diamine as a ligand? So based on its condensed structure, we can surmise that we have a methyl group which has the following struck configuration bound to an oxygen and that oxygen is bound to a methylene carbon, which is then also bound to a methylene carbon. And the second methylene carbon has a bond with oxygen which is bound to another methyl group. So to draw in the remaining electrons, we must make note or recall the, we must recall the octet rule that each of these atoms must add up to eight electrons with the exception of hydrogen and helium. So we see oxygen is bound to only two other groups. So it has a total of four electrons, but it needs eight. So we can actually add two moon pairs in each lark each oxygen and then each carbon already has one bond and each hydrogen has two electrons. Now we then need to calculate the total number of valence electrons. And so we can do that by taking the number of carbon atoms that we have in this molecule, which is four and multiplying by its individual valence electron number. And so, since carbon is in the is in group four, it has four valence electrons, we then add that to the two atoms of oxygen and oxygen is in group six a in the periodic table. So it has six valence electrons, then we finally add it to the 10 elec to the 10 atoms of hydrogen, which is in the first group. So it has a balanced electron value of one. When we add up these valence electrons, we get 38 valence electrons total. Now looking at our structure, the two oxygens of this DME have loan pairs that can be donated to a metal center. And because of these two donor atoms, DME is a by dentate Lagan just like just like our ethyl NN diamine, which can be rated as en. However, oxygen is more electron native than nitrogen, which makes it a weaker, which makes DME a weaker electron pair donor than the en. And so it's actually more likely. So it is more likely that DME is lower on the spectrum. And this is in comparison. True en so to then move on to our second part, we have to determine the occident state of niobium in that following compound. So to do this, we must look at each individual part. So we have potassium, potassium is in group one A on the per table and therefore has an oxidation state of positive one. Our chlorine is a group seven A element and therefore has an oxidation state of minus one. And then our carbon monoxide is neutral. So it has an oxidation number of zero and D MA is also neutral. So it has an oxidation number of oxidation state or a number of zero. So the total, see if we were to add this, we would have positive one minus one, giving us a total oxidation state of this complex ion of zero. Now, if we let X equal the oxidation state of niobium, we would have the total oxidation state of the ion which is zero equal to, we have the positive one from the potassium and we have two of those. So we multiply by two, we add X for the niobium and then we add the minus one from the oxidation state of chlorine and there are two of those. So we multiply that by two and then we just have a plus zero plus zero because the DME and the carbon monoxide are neutral. So if we were to solve for our X, we would have zero is equal to two plus X minus two giving us an oxidation state of niobium as equaling zero. And for our last and final portion, this is we were to draw the structure of that complex ion and to include isomers if necessary So if we wanted to, we could represent DME as the following and that we need to draw where the two coins are on opposite sides. So we would have then our niobium, which is the central metal ion and it will be bound to chlorine on opposite sides are 180 degrees apart from each other. And then if we were to fill out the rest of the structures, we would have our dear me bound to the niobium. And then we would also have our carbon monoxide, both of them bound at tuna alum. Now, there is actually a plane of symmetry in this molecule which bisects it in half and each half is equal. So this has no stereo isomerism. Now, if we were to redraw this, such that we have our niobium in the middle or in the center, but we had our chlorine at 90 degrees. So we had our chlorine at the top, but then our second chlorine was closer at 90 degrees. And then we kept our DME in the same position. But then now we had our carbon monoxide adjacent to one another. There is no plano symmetry. So this compound, the second one that we drew does exhibit stereo. So is, and so because they are non impo non superimposable mirror images, we think that this is an optical isomer. So its isomer, if we were to draw it in a different location would be the following. And so with that, we have finally solved the problem overall, I hope is helped. And until next time.