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Ch.23 - Transition Metals and Coordination Chemistry

Chapter 23, Problem 64d

Draw the crystal-field energy-level diagrams and show the placement of electrons for each of the following complexes:


d. [NiCl4]2+ (tetrahedral),

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All right. Hi everyone. So this question is asking us to draw the filled crystal field energy level diagram of the complex CC 42 negative which is tetrahedral. Now because we know that this particular complex is tetrahedral. I've already prefilled or I've already written out the template for that tetrahedral crystal field energy level diagram. In this particular case, we have three higher energy orbitals that's DXYDYZ and DXZ followed by two lower energy orbitals that's DX squared subtracted by Y squared and DZ squared. So before we can go ahead and fill it out with electrons, we have to know how many D electrons are in the transition metal that is inside of this complex in this case, Cobalt. So to understand that we first have to identify the oxidation number and then the electronic configuration. So as mentioned previously, our central metal which is a transition metal is cobalt and Cobalt has only four ligands, all of which are chloride ligands. Now recall that the oxidation state of Cobalt, as well as the sum of all Liggins charges must equal the overall charge of the complex. So what that means to say is that the sum of Cobalt oxidation state and the charges of all Liggins present must equal an overall charge of negative two. Now, the oxidation state of Cobalt is our unknown. So I'm going to replace this or substitute it as a variable. Let's say X and X is going to be added to the charges of all ligands. Now, in this case, we have four of the same type and the chloride ligand happens to have a charge of negative one. So X subtracted by or multiplied by negative one must equal negative two. I can simplify this expression which means that X subtracted by four equals negative two, solving for X which is the oxidation state of Cobalt gives us positive too. So now lets go ahead and recall the electronic configuration of Cobalt in the neutral state. When Cobalt is neutral, the electronic configuration of it is an Argon core, that's the noble gas core followed by 3d 7 four S two. However, to obtain an oxidation state of positive two, that means that two electrons must have been removed and they were removed from the highest energy orbital which in this case is four S. So after the removal of those highest energy electrons or rather electrons from the highest energy orbital, the electronic configuration of Cobalt to positive is argon 3d 7, meaning that there are seven electrons to be accommodated within the energy level diagram. So here recall that chloride is an example of a weak field ligand, so since we have a weak field ligand, there's a higher spin value or a higher spin, which means that all orbitals must be filled first and then electrons are added to pair those initial electrons. So in this case, I have a total of seven D electrons, the first five of them are going to go into each of the individual orbitals. This means that I have two more remaining, which means that the remaining two electrons are going to pair the lower energy orbitals or create pairs of electrons in the lower energy orbitals. This means that the three higher energy orbitals have an unpaired electron each and there you have it. So with that being said, thank you so very much for watching and I hope you found this helpful.