Four-coordinate metals can have either a tetrahedral or a square-planar geometry; both possibilities are shown here for [PtCl₂(NH₃)₂].

a. What is the name of this molecule?

b. Would the tetrahedral molecule have a geometric isomer?

c. Would the tetrahedral molecule be diamagnetic or paramagnetic?

d. Would the square-planar molecule have a geometric isomer?

Video transcript

All right. Hello everyone. So this question says that Nibr two NH 32 is a four coordinate metal complex. The two possible geometries ATS tetrahedral or square planar are represented by the models below. Part one says to provide the name of the molecule. Part two says to identify if the square planar molecule has a geometric isomer. Part three says to identify if the tetrahedral molecule has a geometric isomer. And part four says to classify the tetrahedral molecule as diamagnetic or paramagnetic. All right. So first, let's go ahead and provide the name of the molecule based on the central metal and all ligands present. First, our central metal is nickel, which is in fact a transition metal. This means that we're going to have to specify the oxidation state of nickel in this complex. But for now, let's go ahead and talk about identifying our ligands. In this particular case, we have two ligands of which we have two of each. So for example, we have two NH three ligands that's ammonia and two bro meats as like now, when bromine is discussed as a ligand, it's referred to as bromo. And because there are two in the same complex. That name is di bromo. Similarly, because we have two NH three ligands in this complex in this molecule, our name is going to be die Amy. And so now that we've identified the names of our ligands, they're to be placed before the name of the central metal in alphabetical order depending on their ligand name. So for example, because the A in amine goes before the B in promo die, amine gets written first. But now let's go ahead and talk about the oxidation state of nickel. Recall that the sum of the oxidation state of the central metal, as well as the charges of all ligands is going to equal the overall charge of the complex. This complex is neutral overall. This means that the oxidation state of nickel, as well as the charges of any ligands present, the sum of these things should ultimately equal the charge of the complex which is zero. So here, because we're solving for the oxidation state of nickel, I'm going to replace that with a variable X. And as for our liens, we have four in total two of which are bromo ligands which each have a charge of negative one. So its two multiplied by negative one. And we also have two amine ligands which are neutral, let's two ligands multiply by their charge of zero. For that term, all of which ultimately equals the charge of the complex which is zero. And so simplifying this expression somewhat X subtracted by two is equal to zero. This means that X which represents the oxidation state of nickel is equal to positive too. So my name to answer part one here is as follows die, I mean dye bromo nickel with two in Roman numerals. And in parentheses, it should be Amin with two NS by the way. And also notice how ligands are arranged in alphabetical order before the transition metal with no spaces in between. All right. So from here, let's talk about part two to identify if the square planar molecule has a geometric isomer. So recall here that square planar isomers or square planar molecules can exhibit two geometric isomers that cis and trans. Now the cis isomer is what's being shown in the model provided of the square planar molecule. This is because we have identical ligands beside each other or next to each other. So once again, this is going to be an example of the cis isomer because we have the two promo ligands and the two ami ligands on the same side of each other, which therefore means that there is going to be a trans isomer in which ligands of the same type are on opposite sides. This means that the answer to part two is going to be yes, because the square plater molecule must have a trans isomer. Now, by contrast, let's talk about part three, which is to identify if the tetrahedral molecule has a geometric isomer. The answer to this part is simply going to be no because tetrahedral complexes do not show geometrical isomerism. So there are no isomers in the case of the tetrahedral molecule. This leads us now to part four which is to classify the tetrahedral molecule as diamagnetic or paramagnetic to understand the classification. That's more appropriate here, we have to first understand the electron configuration of our central metal which is nickel. Let me just scroll down here for the sake of space. But recall that one nickel is neutral. The electron configuration is as follows. We have an argon noble gas core followed by 3d 8 and four S two. Now this is the neutral state. However, for an oxidation state of positive two, that means that two electrons had to have been removed from the highest energy orbital. In this case, the four S and so the oxidation state of nickel with an oxidation state of two is that same argon core 3d 8 because both electrons from four S two have been removed. And so from here, we have eight D electrons to accommodate. So let's go ahead and draw out these orbitals for both the tetrahedral and the square planar complexes. If I scroll down here, you can see that I went ahead and pre drew them. So I'm just going to move them to make them more visible. All right. So first, let's start with our tetrahedral diagram here. In this case, we have three higher energy orbitals that's DXYDYZ and DXZ and two lower energy orbitals. That's DX squared subtracted by Y squared and DZ squared. First, I'm going to start by placing electrons in the lower energy orbitals. First, I'm going to place one unpaired electron in each before I pair them. Now that the lower energy level is full, I'm going to place electrons into the higher energy orbitals. One unpaired electron must go in each degenerate orbital or each orbital of the same energy level before I pair. And that gives me my eight total electrons. And so based on this diagram, we can see that the tetrahedral diagram here has a total of two unpaired electrons in the higher energy orbitals. And so recall that when a substance has at least one unpaired electron, it's considered paramagnetic though the tetrahedral molecule is paramagnetic. So let's compare this with our square planar complex here, I'm going to serve the same way at the lowest energy level, placing one unpaired electron in each degenerate orbital before subsequently adding appeared electron. From here, I proceed to the second lowest level, fill that one completely before going up one more energy level. Now I have all eight D electrons and all of them happen to be paired. So in the case of the square planar complex, all electrons are paired, which if you recall means that it is diamagnetic. And so to go ahead and answer part four, which is classifying the tetrahedral molecule as diamagnetic or paramagnetic. The answer to that is paramagnetic and there you have it here are the answers for parts one through four. For part one, the complex name was die. I mean di bromo nickel too. Part two was yes. The square planar molecule has a geometric isomer. Part three is no, the tetrahedral molecule does not have a geometric isomer. And for part four, the tetrahedral molecule is paramagnetic. And so with that being said, if you watch this video all the way through, thank you so very much. And I hope you found this helpful.