Give a valence bond description of the bonding in each of the following complexes. 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. 

(b) [NiBr₄]^2- (tetrahedral) 

Video transcript

Welcome back everyone. Our next question says, draw the orbital diagram where the un bonded metal ion and the metal ion in the tetra in the tetrahedral complex, which is, and the overall charge is negative two with a complex of cu and then the ligands in parentheses, CN subscript four. What is the hybridization used for bonding? How many unpaired electrons are present in the complex? Well, first off, we know already what the hybridization used for bonding is because we're told that it's a tetrahedral complex. So we know this will be sp three hybrid orbitals. So have a little jump ahead there and we can check that we're doing it the right way. So to draw an orbital diagram, of course, we need to start with how many electrons we have in our D orbital. So let's look at our complex here. We have a negative two overall charge. The ligand is four cyanide liens. So cyanide has a chart of negative one multiplied by four. So giving us a negative four charge from the ligands and therefore, our copper must have a charge of positive two. So we know we're looking at cu two plus as the ion that is in the complex. So we need to look at our orbitals, our electrons, excuse me. So the electron configuration for copper is our nearest noble gasses argon. So argon on brackets and then coppers, one of those non-standard arrangements, it has four s 1 3d 10. And again, it has that because just having the one s electron lets it have that very stable configuration with fully filled D orbitals, but we have copper two plus. So we've lost two electrons. So we have again, argon and brackets, we've lost our S electron and one of the D electrons and we have 3d 9. So let's look at our electron configuration for the un bonded metal ion. So cu two plus the argon in brackets. And then here's our 12345 3d orbitals. We have a tetrahedral complex. This means we have low or small delta, meaning we have high spin. So the differences in energy level between my different de orbitals aren't, don't matter in terms of how I fill them. Of course, with nine D electrons, it wouldn't matter anyway. But let's go ahead and fill them in 12345 and then the remaining four pair up. So we have four pairs and one unpaired electron in our un bonded metal ion. So that's the arrangement for the un bonded. I am to highlight that since that's one of the things we're looking for here and now we need the arrangement in the tetrahedral complex. So we will repeat our 5 3d electrons 3d orbitals, excuse me, with our nine D electrons in them, all the orbiters are full. So we're going to move on to the four S and the 34 p orbitals. When we look at our ligands, we know we have four ligands. So we get eight electrons in four pairs. So important to note that the li the electrons from the ligands are always in pairs. So we'll fill them in here, one pair in the four S orbital and then the three remaining pairs filling in the four p orbitals. So the orbitals used in bonding are the four S and four P. So we have sp three hybridization and we need to as we expect it and then how many unpaired electrons just one unpaired electron from the copper? So we'll highlight that because that was the other answer we were looking for. We've got our electron configuration for the metal ion in the tetrahedral complex. We know it uses sp three hybridization for bonding and has one unpaired electron. See you in the next video.