For each of the following complexes, describe the bonding using valence bond theory. Include orbital diagrams for the free metal ion and the metal ion in the complex. Indicate which hybrid orbitals the metal ion uses for bonding, and specify the number of unpaired electrons. 

(a) [AuCl4]2 (square planar)

Video transcript

Hi, everybody. Let's take a look at our next problem describe the bonding in and we have in brackets ZN and then in parentheses, co subscript two and then outside the brackets two plus as the charge, using the valence bond theory, draw the orbital diagrams for both the free and complex metal ions find out how many unpaired electrons there are and which hybrid orbital is utilized to form the bonds in the complex. Well, first, let's take a look at the complex. We have, we have our metal ion zinc and then we have two co ligands. We've only got two ligands. So we know that must be an SS P hybrid orbital. So we kind of already know that ahead of time, but we'll need to draw that orbital diagram anyway, as that's what our problem asks us. So first, we need to know how many electrons we're dealing with how many D electrons. So we have zinc and we need to find out what charge it has. So let's look at our complex here. We have zinc and then two of these co ligands and this overall positive two charge our ligands have are neutral. They have a zero charge. So zinc must have a positive two charge to make that overall charge. So we have ZN two plus as our metal ion. So first, we need to look up zinc in its neutral state in our periodic table. And zinc is the closest noble gasses argon. So we have a R and brackets and then four S two, 3d 10, it's got that full D she, we're going to remove two electrons. So zinc, two plus is argon and it's going to lose the 24 S electrons and just be 3d 10. So first, we need to draw the orbital diagram for the free ion. So zinc two plus, we have argon and brackets and then we'll draw our five D orbitals. So five lines there labeled 3D and we have 10 D electrons. So that's very simple. Normally, I would draw these electrons one at a time to emphasize, filling in first one electron and then pairing it. But I know I have all 10. So that's why I wrote them all putting two in each uh spot. So we've got a full D 3d shell here. We have two electrons in every one of the D orbitals. So that's our free I am here. And let's think about our complex. So for our complex, we start with our 3d electrons from our zinc. So we'll put in our 3d orbitals which as we know, have all 10 electrons in there, there's 3d So for complex, we need to use the four S and four P orbitals, which we know are empty and we have two of ligands. And so we get so two ligands, we get two pairs of electrons that have to be given as pairs, you can't separate them. So one pair, I'm going to put brackets around my or open bracket before my four S, one pair of electrons goes in there. And then the second pair of electrons goes in our first four P orbital. So you can see we just have electrons in the four S and one of the four P electrons, four P orbitals, excuse me. So are hybridized. Our hybrid orbital is sp So our last question here is how many unpaired electrons are there? So there are zero unpaired electrons since we have 10 D electrons. And of course, our ligand electrons come in pairs. So again, we needed to draw the orbital diagrams for a free ion. So we had that for 10 3d electrons. And with our complex, we added two electrons in four S two electrons in four P to make an sp hybridized orbitals. And we have zero unpaired electrons. See you in the next video.