Construct the octahedral crystal-field splitting diagram for the metal in each species.

Hello, everyone. Today, we have the following problem which among the following diagrams represents the crystal field splitting for the following complex. So we are given this cobalt and metal containing compound and we are being asked to fill out a crystal field splitting energy diagram. And so that is going to include the following. So we will first have our, the orbitals present, we will have our D orbital for our xy axis, our D orbital for our xx axis nrd orbital for ryz axis at the lower energy. And then on a higher energy level, we will have our D orbital X squared and then we will have our DX squared Y squared. So if we look at this compound, we know that our central metal is our cobalt. And then we see that our, our ligand will be these six cyano groups. So we have six of these cyano groups. And so this means that our co-ordination number, our co-ordination number for this complex is equal to six. And this gives us the configuration of an octahedral geometry. So to determine the number of electrons in the central metal ion, we first make note of the neutral metal, which is cobalt. And if we were to ter to determine the electron configuration, the noble gas electron configuration, where we take the most recent preceding or the closest preceding noble gas, we put it in brackets. So for this example, this will be argo and then we continue into the next row in the S row. So we have four S two and then we have 3d 7. And that is a configuration for Cobalt. However, we have, we have to make note that we will have our charge of positive three. So for our Cobalt of positive three and it's positive three because our overall charge is negative three. So we have to stabilize that and the charge for cyanide is minus one. So our cobalt must be plus three. So because we have plus three, we have to remove three electrons from our highest energy level. And so we will first remove them from our four S two orbital and then we will remove one from our d orbital such that our electron configuration is argon 3d 6. And as such, we will just fill up the, the orbitals. So we have to follow half off balls and Hans rule such that we pair up our electrons and the lowest energy orbitals first, we write one of them and then we pair them up afterwards. And so we filled out our six electrons giving us an answer of answer choice. A and with that, we have solved the problem overall. I hope this helped. And until next time.