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

(a) [CrF₆]^3-

(b) [V(H₂O)₆]³⁺

(c) [Fe(CN)₆]^3-

Video transcript

All right. Hi, everyone. So for this question, let's draw the crystal field energy level diagram and determine the number of unpaired electrons in each of the following complex ions. For part one, we have hex cyano mangan two. For part two, we have hex aqua chromium three. And for part three, we have hexaco Modin three. So notice how each of the given complex ions has six ligands around the central metal ion and all of the ligands here, cyanide, water and chloride happened to be mono dentate, which means that all of them attached to the central metal ion v one donor atom. So this means that this is going to be or all three parts, all three ions are going to have octahedral complexes. And because they are all octahedral complexes, if I scroll down here, I actually went ahead and drew the splitting diagram or the octahedral complex three times for each part, right, we have the three lower energy orbitals and the two higher energy orbitals. So in this case, I'm going to scroll up again because I want to go ahead and actually find the electronic configuration of each central metal first starting off with our hex cyano mangan three or sorry hex Ayano mangan 82 in the first part, right. So in this case, our central metal ion is going to be manganese too positive. And here we have six cyanide Liggins. Now, manganese, if you recall has an atomic number of 25 no, the atomic number is equal to the number of protons. However, in a neutral atom, the number of protons is equal to the number of electrons, which means that a neutral atom of manganese should have 25 electrons as well, which gives us an electronic configuration of Oregon four S 2 3d 5. However, we are dealing with an ion with a positive two charge. And because the charge is positive recall that electrons have to be removed from the highest energy level. In this case, two electrons are being removed, which means that the two electrons in the four s orbital are going to be removed, giving us an electronic configuration of argon 3d 5, which means that there are going to be five electrons in our D orbitals. So if I scroll down very briefly to go to the diagram that I do for part one here recall that cyanide is a strong field ligand that results in a low spin complex. And so what that means is that electrons in the lower energy D orbitals must be paired first before you can proceed to higher energy orbitals. So in this case, there are five D electrons, which I will first distribute among the three lower energy orbitals. And then the remaining two electrons have to be paired first in the lower energy orbitals before I can proceed to the higher energy one. So this means that two of my lower energy D orbitals end up paired leaving me with one unpaired electron or the first complex ion. So now we can go ahead and do the same thing with our second complex ion which is hex aqua chromium three. So in this case, for part two, here, our central metal ion is going to be chromium three positive. So recall that neutral chromium or chromium in general has an atomic number of 24 which means that it should have 24 electrons when neutral. This leads to a neutral electronic configuration of Aragon four S 1 3d 5. Now, it's worth noting here that chromium is an exception to the standard electronic configuration because a de orbital is more stable when it is half built. So now similar to our discussion with manganese because chromium has a positive charge. In this case, specifically a positive three charge, three electrons must be removed from the highest energy levels, which means that one electron gets removed from the four S one orbital or from the four S orbital and two are removed from the 3d orbital giving us an electronic configuration of Oregon 3d 3, which means that there are going to be three electrons to distribute among the D orbitals. Now, in this case, our only ligand for part two happens to be water, which is a weak field ligand, meaning that it will result in a high spin complex in which orbitals are half filled before they are paired. So among the three lower energy D orbitals, each one of them is going to receive one unpaired electron. So this means that for part two, we have three unpaired electrons. So last but not least is our third complex ion which is hex aloro moyen three with a central metal ion of molybdenum three positive. So here right, molybdenum has an atomic number of 42 which means that it has 42 electrons went in as neutral leading to an electronic configuration of krypton five S 14 D five. And molybdenum is similar to chromium in the sense that it is an exception to standard electronic configuration because the de orbital is more stable when it is half built. So here, right, to achieve the electronic configuration of our ion three electrons must be removed from the highest energy levels. So one electron gets removed from the five S orbital and two are removed from the four D orbital leading to an electronic configuration of krypton four D three and 3d electrons to distribute among the D orbitals. The chloride once again is a weak field ligand that results in a high spin complex, meaning that in this case orbitals have to be half filled first before they are paired up. So the 3d electrons are going to be distributed among the three lower energy d orbitals, meaning that the complex ion three also has three unpaired electrons and there you have it. So here is your answer for parts 12 and three. The nice thing was that all of them were octahedral complexes, which meant that all of them had the same diagram. And so with that being said, if you stuck around to the end of this video, thank you so very much for watching. And I hope we found this helpful.