Draw a crystal field energy-level diagram, assign the electrons to orbitals, and predict the number of unpaired electrons for each of the following.

(a) [Cu(en)₃]²⁺

(b) [FeF₆]^2-

(c) [Co(en)₃]³⁺ (low spin) 

Video transcript

All right. Hi, everyone. So this question says that for each of the complex ions given below, draw the splitting energy diagram and determine the number of unpaired electrons. For part one, we have Hexes Ciano vanity two. For part two, we have Tris ethylene diamine ruthenium two and part three, we have Hexaco rotate three. So before we go ahead and we talk about each of these complexes individually, I want to point something out with respect to their geometry, specifically their co ordination number. So ions one and two, right, both contain mono dentate ligands, specifically cyanide and chloride respectively. So the co ordination number for both of them is going to be six because there are six mono dentate ligands. The second complex ion though contains three ethylene diomed ligands, but each ethylene diomed present is bent. So because of that, its co ordination number is still six. So what that means is that all three of these compounds exhibit octahedron geometry and therefore, their energy diagram is going to be the same. So if I scroll down here very quickly, you can see that I've gone ahead and drawn blank splitting energy diagrams or all three octahedral complexes. So the question is how many electrons are going to go in each of these diagrams? So now let's go ahead and start with that in part one, part one being hexaco vanity two. So the central metal ion here is vanadium two positive recall that vanadium has an atomic number of 23 which corresponds to the number of protons. However, recall that when an atom is neutral, the number of protons and the number of electrons are equal. So in the neutral state vanadium has 23 electrons as well. This results in an electronic configuration of an argon core followed by four S 2 3d 3. But here we have vanadium with a positive two charge, which therefore means that two electrons must be removed from the highest energy level. In this case, the four s orbital. So when removing those two electrons, the final electronic configuration becomes argon 3d 3. So now it's these three outermost electrons that are going to be used to fill in the splitting energy diagram scrolling down for a second to see the diagram itself recall that cyanide is a strong field ligand, which is going to result in a low spin complex. And in the case of a low spin complex, the electrons in the three lower energy D orbitals must be paired first before the higher energy orbitals can be occupied. So in this case, I would place one electron in each of the three lower energy D orbitals, which means that all three are unpaired in this case. So that settles part one. So now lets go ahead and start with the second complex ion right after I go ahead and make some space though. There we go. I figured it would be easier to write R part one here towards the top of the screen. So now for part two, we have tris ethylene diamine rhenium too, which means that our central metal ion is going to be rum two positive. Now here, because rhenium has an atomic number of 44 it's going to have 44 protons as well as 44 electrons when its neutral. This results in an electronic configuration in the neutral state of krypton four D five. Wait not in the neutral state, excuse me, in the neutral state. The electronic configuration of ruthenium is krypton five S 14 D seven recall that it's an exception from the standard electronic configuration. So now with this in mind to achieve a charge of positive 22 electrons must have been removed from the highest energy orbital. So one electron gets removed from the five S orbital and one more gets removed from the four D orbital. This results in a configuration of krypton four D six. So the six outermost electrons will be used in the splitting energy diagram. Once again, Ethylene diamine is an example of a strong field ligand, which is going to result in a low spin complex. So the lower energy orbitals must be paired before the higher energy orbitals can be occupied. So scrolling down here to my second diagram, I would start with adding one electron to each of the three lower energy orbitals and then proceeding to pair them. And in this case, it just so happens that there are zero unpaired electrons. So last but not least for part three here we have Hexaco rotate three. So for part three, our central metal ion is rhodium three positive. When radium is neutral, it's going to have 45 electrons since the atomic number is 45. This results in an electronic configuration of krypton five S 14 D eight rhodium is also an exception from standard electronic configuration. So to achieve a charge of positive three rhodium must have lost three electrons, one from the five S orbital and two more from the four D orbital. This results in an electronic configuration of Krypton four D six. However, chloride is an example of a weak field ligand which is going to result in a high spin complex because of this, all of the orbitals must be half filled before they can be paired up. So in this case, there are six electrons to distribute in our energy diagram here. But in the case of a high spin complex me actually open up some space here. But in the case of a high spin complex, five electrons are going to go into each of the d orbitals, which means that the final electron creates only one pair. And since there's only one pair, there are a total of four unpaired electrons and there you have it. So here Weve arrived at our final answer. So if you watch this video all the way through the end, thank you so very much. I appreciate it. And I hope you found this helpful.