Nickel(II) complexes with the formula NiX₂L₂, where X⁻ is Cl⁻ or N-bonded NCS⁻ and L is the monodentate triphenylphosphine ligand P(C₆H₅)₃, can be square planar or tetrahedral.

(a) Draw crystal field energy-level diagrams for a square planar and a tetrahedral nickel(II) complex, and show the population of the orbitals.

Video transcript

Hi, everyone. Welcome back. Here's our next question. Cobalt complexes characterized by the chemical formula CO X two L two where an X is CL minus or N bonded NCS minus and L is a mono dentate trenny phosphine ligand can exhibit either a tetrahedral or square planar geometry. What are the crystal field energy level diagrams that illustrate the distribution of electrons within the orbitals for tetrahedral and square pinar cx two L two complex. So in order to illustrate the distribution of electrons, first step is figure out how many D electrons we have on our cobalt. So we need to figure out its oxidation state. So let's take a look at our complex here. We have Cobalt, we have two of those X ligands and two of the elegans. Well, X we're told can be either chloride cl minus or N CS minus, either of these has a negative one charge. So for X two, we'll have a negative one charge multiplied by two since there are two of them or elegant is trenny phosphine. This is sometimes abbreviated PP three and it's a phosphorus atom bonded to three phenyl groups and there's a lone pair on that phosphorus that can participate in this bonding with the Cobalt, it's a neutral ligand. As you can imagine if you think of that structure. So it has a charge of zero, obviously won't matter. There's two of them, since the charge is zero, we can see that our complex overall has no charge. So we'd say that our overall charge of zero equals the oxidation state of Cobalt. So o period S period co plus our charge on X two, which is negative one multiplied by two plus zero for L two. So let's solve for the oxidation state of Cobalt, what equals zero plus two. So it must be positive too. So now we just need to look at the electron configuration of elemental Cobalt which when we look that up on our periodic table, we have argon as the closest noble gas. So A R and brackets 3d 7 four S two. So pretty straightforward, our Cobalt two plus must be argon 3d 7. We just get rid of the 24 S electrons. So we've got seven D electrons to distribute when we look at our energy level splitting. So let's think about our crystal field energy level splitting in tetrahedral and square planar geometry. So these are slightly different from our octahedral geometry. So for tetrahedral, we have the same sort of grouping of two orbitals and three orbitals, but they're reversed from octahedral. So we have three higher level energy orbitals which are the Xyxz and yz orbitals and then to lower level orbitals which are the Z squared and X squared minus Y squared. And it's important to note that tetrahedral complexes are always, always have a small delta. And that's due to the fact that none of these orbitals in the tetrahedral geometry point directly at the ligands. So the difference in energy between the higher and lower is always small. And so you're always going to have a high spin arrangement. So we have our seven D electrons and with our small delta, we fill in one electron in each of the five D orbitals first and then our remaining two go on those two lower energy levels. So here is our diagram for our tetrahedral geometry, we have four orbitals in two pairs in the two lower levels. And then each of the three upper levels has a single electron each. Now let's look at our square planar energy feel dry gram. So square planar, we have this sort of unusual arrangement. If we think about it, the square planar arrangement is very similar to our octahedral arrangement, but it's missing the two ligands on the z axis. So with that, that z squared energy level moves much lower down as compared to an octahedral arrangement. And we have the X squared minus y squared orbital at a very high level because that orbital will point directly at our four ligands. So that will have a lot of repulsion there So we have this very high X squared minus Y squared orbital and then we have a large gap, a large delta. And then we have our lower levels which are split with one, the XY. And then just below it, we have the Z squared level and then the lowest two RDXZ and YZ levels, all of those Z levels being down lower since we no longer have ligands on the Z axis. But these lower four energy levels have very small distances between them. So they will all fill in as if they had a small delta between them. And then there's this large gap before you jump up to the X squared minus Y squared level. So with our seven D electrons, we put one electron in each of our four lower energy levels. And before we'd even think about jumping up to that highest level, we go back and pair them in the lower levels. So we have three left. So three of our lower energy levels get a second electron in them and we end up with pairs in our three lowest energy levels, a single electron in that xy level and no electrons in that very highest energy level. So there's our square planar diagram which again, we have, that's a rather unusual splitting where you have those four lower orbitals with a small difference between them. And then a large gap before the highest energy level. See you in the next video