In acidic aqueous solution, the complex trans-[Co(en)₂Cl₁]²⁺(aq) undergoes the following substitution reaction:

trans-[Co(en)₂Cl₁]⁺(aq) + H₂O(l) → trans-[Co(en)₂(H₂O)Cl]²⁺(aq) + Cl^–(aq)

The reaction is first order in trans-[Co(en)₂Cl₂]⁺(aq), and the rate constant at 25°C is 3.2×10^–5 s^–1.

e. Draw a crystal field energy-level diagram for ​trans-[Co(en)₂Cl₂]⁺ that takes account of the fact that Cl^– is a weaker-field ligand than ethylenediamine.

Video transcript

All right. Hi, everyone. So for this question, let's consider the following substitution reaction of the complex trans ruc two by purine two predict the crystal field splitting diagram for the reactant trans ruc two bid two. Given that chloride is a much more weaker field ligand than bid or B IP Y. So based on the given information, right, our two lids are chloride and bi purity and it just so happens that in the reactant we are looking at we have two of each right. Now, recall the bine is a B dentate ligand, meaning that there are two areas of attachment to the central meadow ion. Now, chloride on the other hand happens to be mono dentate. So if we consider the co ordination number of this complex, then each B purine ligand contributes a coordinating number of two because it is bic whereas chloride contributes a coordinating number of one. So when we add together those co-ordination numbers, that is two for every biop purity ligand, so that's two multiplied by two, add it to one multiplied by the number of chlorides, which happens to be two, right? So this gives us a total co ordination number of six, which normally would correspond to an octahedral complex. However, let's consider the structure of the reagent given as well or the reactive. So if I scroll down here, right, just for the record, the central metal ion happens to be rheum two positive. So if I scroll down here and placed ruthenium in the center, right? Because we're looking at are reacting which is in the transfiguration. Then this means that chloride, which I am representing as black circles here, my two chloride ligands are going to be on opposite sides of each other. As will the two by purine ligands, which I will represent as white circles connected to each other. Now, if you recall the shapes of all of the D orbitals in a standard octahedral complex, notice how the two chloride ligands point directly at the DZ squared orbital. Whereas the two B purine ligands point directly at the DX squared subtracted by y squared orbital. Now, what this means is that each or these respective orbitals, namely the DZ squared and DX squared subtracted by Y squared experience a greater repulsion because of the presence of these ligands pointing at them. So therefore, the DZ squared and the DX squared subtracted by y squared orbitals are going to have a higher energy level than the remaining 3d orbitals which in this case point in between the legates. Now recall that strong field ligands cause a greater splitting between the de orbitals. So because B pyridine, which in this case is the stronger field ligand points directly at the DX squared subtracted by y squared orbitals. Those orbitals will have a higher energy than even the DZ squared orbital. This means that in the crystal field splitting diagram, right, the TX squared subtracted by Y squared orbital is going to appear highest followed by oops the DZ squared orbital. Now both of these orbitals are going to have a higher energy than the remaining three, which is DXY DXZ and also Dyz. Now the reason why the DXY orbital has a higher energy than DXZ and DYZ is because the DXY orbital directly points in between the two ligands themselves, which means that its going to have a slightly higher energy. So here is the crystal field splitting diagram. Now the question is to distribute the electrons among each respective orbital. So to do that, we have to consider the electronic configuration of ruthenium to positive, which is the central metal ion. Now recall that it is often easiest to consider the electronic configuration of the neutral atom before you consider ions they recalled ruin, recall that ruthenium has an atomic number of 44 which is also equal to the number of protons. Now, when an atom is neutral, the number of protons is equal to the number of electrons, which means that in a neutral atom of ruthenium, there are 44 electrons. This means that the electronic configuration of rhenium when it is neutral is krypton five S one four D seven. Because recall that rhenium is an exception to standard electronic configuration. So now when considering the electronic configuration of an ion, in this case, a positively charged ion electrons must be removed from the highest energy levels. In this case, to achieve a positive two charge, two electrons must be removed from the highest energy levels. In this case, one from the five S orbital and one more from the four D orbital giving us an electronic configuration or ruthenium two positive of krypton four D six. This means that there are six D electrons to distribute. Now because the complex does contain some strong field ligands, then this is going to result in a low spin complex in which electrons in lower energy D orbitals have to be paired first before occupying higher energy orbitals. So in this case, I'm going to start off at the lowest energy orbitals which are DXZ and Dyz. Now each orbital has to have an unpaired electron before electrons can be paired. Which means that here I distribute four electrons in the DXZ and the Dyz orbitals. And then the next energy level is DXY, which receives the remaining two electrons. And there you have it here is your crystal field splitting diagram with all the electrons or this complex. And so with that being said, thank you so very much for watching. And I hope you found this helpful