Although Cl^- is a weak-field ligand and CN^- is a strong field ligand, [CrCl₆]^3- and [Cr(CN)₆]^3- exhibit approximately the same amount of paramagnetism. Explain.

Video transcript

Hi, everybody. Let's look at our next question. It says bromide br minus ion is a weak field ligand and nitrite. No two minus ion is a strong field. Ligand. Explain why the complex ions we have a complex with a negative four overall charge of copper with six bromide ligands. And another complex, an overall charge of negative four copper with six nitrate ligands exhibit roughly equivalent paramagnetism. A both complexes will exhibit roughly equivalent paramagnetism because the electrons occupy the D orbitals the same way b both complexes will exhibit roughly equivalent equivalent paramagnetism because the orbitals are fully filled. See, they both will exhibit this. I'm going to paraphrase that because all the electrons in the D orbitals are paired or D both will exhibit this because all the electrons in the D orbitals are unpaired. Well, to evaluate this, we need to look at what the electron configurations of, of are of these complexes. So we've started with copper. So let's start with just neutral copper and its electron configuration are gone as the nearest noble gas and the non-standard equi the non standard excuse me configuration of four S 1 3d 10. Again, copper is a good one to memorize because that 3d 10 arrangement is that much more stable that it's worth having the single s electron. So good to know this one. Now, let's look at the oxidation state of copper in our complexes. So we have RCUBR six with a negative four charge, the bromide ions of course, have a negative one charged multiplied by six total negative six. So the copper must be positive too. We can see that our other complex will be the same thing because nitrite also is a negative one charge. So both we'll have this cu two plus. So we missing two electrons. So we have argon and brackets, it will lose its s electron and one of its D electrons and have 3d 9. So we have the same number of D electrons. We can go ahead and eliminate choice B because our D orbitals are not fully filled, we have nine D electrons. We also know that they will not have all the electrons paired, which is choice C because we have an odd number of electrons, they can't all be paired. So we have our choice of the electrons occupy the de orbitals the same way or that they're all unpaired. Well, nine D electrons are not all going to be unpaired because we're only going to have one unpaired since we can only have five D orbitals, each with a pair of electrons. So choice D, we can eliminate only one unpaired and I'll go ahead and draw out the orbital diagram. So we can see this clearly. But our answer will indeed be choice a they both occupy the D orbitals the same way because even though one has a large field splitting and one has a small, because we have those nine electrons, we only have one left unpaired. So it's going to occupy that higher energy level no matter what. So no matter what order it was filled in the final arrangement is the same for both. So right, final arrangement is the same and we can draw this out and see this, we have an octahedral arrangement with six ligands. So our splitting will have two higher energy levels and three lower. So with a small delta and then we'll draw one with a larger delta. I don't know how much different that actually is, but I've written large and small. So we know which is which. So with a small delta, we would fill in all 51 electron in all five first and they go through the other four. But of course, we end up with just the one unpaired with a large delta, we would fill in 123 and the lower three, then pair them up and have six down below. But again, we have three left, one in each of the higher levels. And then we've got a pair of one. So even though filled in a different order, the final arrangement is the same for both. So the reason why both of these complexes exhibit equivalent paramagnetism, despite one being a weak field ligand and one being a strong field ligand is choice a both complexes will exhibit roughly equivalent to paramagnetism because the electrons occupy the D orbitals the same way. See you in the next video.