The percent iron in iron ore can be determined by dissolving the ore in acid, then reducing the iron to Fe2+, and finally titrating the Fe2+ with aqueous KMnO4. The reaction products are Fe2+ and Mn2+.
(c) Draw a crystal field energy-level diagram for the reactants and products, MnO4-, 3Fe1H2O2642+, 3Fe1H2O2643+, and 3Mn1H2O2642+, and predict the number of unpaired electrons for each.

Question

The percent iron in iron ore can be determined by dissolving the ore in acid, then reducing the iron to Fe2+, and finally titrating the Fe2+ with aqueous KMnO4. The reaction products are Fe2+ and Mn2+.

(c) Draw a crystal field energy-level diagram for the reactants and products, MnO4-, 3Fe1H2O2642+, 3Fe1H2O2643+, and 3Mn1H2O2642+, and predict the number of unpaired electrons for each.

Accepted Answer

Welcome back everyone. The next problem says ruthenium or is dissolved in acid to reduce ruthenium to ru two plus, then titrated with KMNO four to produce RU three plus and MN two plus. What is the corresponding crystal field energy level diagram for MNO four minus the complex ru h2o 6 with a charge of positive two, the complex ru h2o 6 with a charge of positive three and the complex MN H2O 6 with a charge of positive two. What is the number of unpaired electrons for each? So we have four complexes here that we need to find a crystal field energy level diagram. For the first step we want to do is since we have multiple complexes of rhenium and manganese is look up the electron configuration for both of those elements. So we have the base we're starting from. So we have ruthenium, the nearest noble gas is krypton. So Kr in brackets, it has a non-standard or exceptional arrangement, we should say of five S one, four D seven and then manganese MN, its closest noble gases, argon A R and its configuration is four S two 3d 5. So we've got those to refer to. So let's work our way through our different compounds. So MNO four minus so negative one charge, let's calculate the oxidation state of manganese here. So uh the oxygen here will have its usual favorite oxidation state of negative two multiplied by four. So that will be negative eight for the oxygen here, our overall charge is negative one. So our manganese myths have an oxidation state of positive seven. So we have MN seven plus. So it's lost seven of its valence electrons. And that is all it has, it has two S electrons and five D so mn seven plus as the same electron configuration as argon and it has zero D electrons. So it would have zero unpaired electrons since it has none doesn't really have a crystal field energy light diagram because there are zero B electrons. So we'll just circle that we would note if it did have a diagram, it only has four ligands, four oxygens. So it, in this case, has a tetrahedral arrangement. So it would have that energy level splitting with three higher energy levels and two lower. So opposite to that of octahedral complexes. So we're dealt, we've dealt with the MNO four, let's move on to our complex that has ruthenium with six water ligands and a positive two charge, right? Are you h2o 6 with a positive two, water is a neutral ligand. So the ruthenium must be in the form of ru two plus which we know, actually, since our problem tells us that we've used acid to reduce ruthenium to ru two plus initially. So it's lost two of its valence electrons. So its configuration is krypton and it's lost its one s electron and one of its D electrons. So it's four D six, we know that water is a weak field ligand. So when we look at that splitting, we'll draw the orbitals two higher three lower, this is a small delta between them. So when we go to fill in our six electrons, we put three on the bottom and then we continue filling them in on the top with that small delta. Now we have five and the sixth one gets paired up down in a lower level. So we end up with four unpaired electrons. So there's our diagram with four unpaired. Now we move on to our other ruthenium complex which as you can see, since again, water being a neutral ligand, the complex has an overall charge of positive three. So we must have ru three plus here. So we're missing one more electron. So we're at four D five again seem ligand. So a small delta. So our five electrons fill in one in each the orbital. So we have five unpaired electrons as none have to double up there. So again, we have our diagram with one electron in each orbital and five unpaired. Finally, we move on to our last complex, the manganese complex where we have our water ligand, the overall charge is positive two. So we know that manganese must have a charge of positive two. This configuration would then be argo and it will be missing its two S electrons leaving 3d phi again, we have water as our ligand. So is small delta. So again, five D electrons just like our previous complex one in each. So again, five unpaired electrons. So we highlight our energy level diagram and five unpaired. So we've worked our way through each of our complexes and found our different diagrams. We're calling that our first is tetrahedral though. Again, there's zero D electrons there and our other three with six ligands are octahedral. See you in the next video.

Verified Solution

Key Concepts

Crystal Field Theory

Titration and Redox Reactions

Unpaired Electrons and Magnetic Properties