For each of the following, (i) give the systematic name of the compound and specify the oxidation state of the transi-tion metal, (ii) draw a crystal field energy-level diagram and assign the d electrons to orbitals, (iii) indicate whether the complex is high-spin or low-spin (for d4 - d7 complexes), and (iv) specify the number of unpaired electrons.

(a) (NH4)[Cr(H2O)6](SO4)2

Video transcript

Hi, everyone. Let's take a look at our next problem. It says, consider the complex with the formula. And then we have a complex that is K and then within brackets, co within parentheses, NH three subscript six, that's all within brackets. And then within parentheses, co three subscript two. And we're asked to do five things relating to this complex. Number one, provide the name of the compound. Number two, calculate the oxidation number of co number three sketch its crystal field energy level diagram. Number four, classify the complex as either high spin or low spin. And number five identify the number of unpaired electrons if any. Well, this can look a little daunting with all these steps, but many of them relate to each other. For instance, when we look at number two, we'll need the oxidation number of co to be able to name the compound. So it's not quite as bad as it looks because some of the steps we have to take, take in order to get to the answer for others. So we're gonna start with our name. And as we said, that means we need to calculate the oxidation number first. So whenever we look at oxidation number, we base base it on looking at the overall charge of the complex to calculate our metals oxidation number. Now, the one tricky thing here is we've got our transition metal complex with cobalt, but we also have two counter ions, which is a little, you know, at first a little confusing because there's two of them. So we have potassium and then we have the C 32 of them. So potassium being a transition or not transition, excuse me, the potassium being a group one, a metal will pretty much always take the charge of plus one as an ion. But we also have these anions over here, they're carbonate anions and a carbonate anion has a charge of negative two. So we have two counter ions here with two with opposite charges. And then we have our transition complex and we know that our overall compound is neutral, it has no charge. So to calculate the oxidation number of Cobalt, we would use the overall charge of the entire compound, which is zero. And we say it's equal to the sum of the charges of all the bits. So we have zero equals positive one for our potassium plus, then we need the charge of our complex. So in parenthesis, I would have oxidation number or oxidation state. So o period S period of Cobalt plus, we need to account for any charges from our ligands. But our ligand is an H three, which is neutral, there's six of them, but it has no charge. So we'd say plus zero to account for a ligand. So that's in parentheses. And then we need to add the charge from the carbonate anions. So there's two of them. So we have to say two, multiplied by negative two. So we're going to say zero equals positive one plus the oxidation state of cobalt zero for the ligands and then minus four, let's get oxidation state of Cobalt isolated on one side of the equation. And that will equal zero minus one plus four, just going to equal positive three. So our oxidation state of Cobalt will be plus three. So we've solved part two of our problem here. So with that, let's go ahead and name our compound. Well, the hardest part of this name will definitely be the name of our transition metal complex. So let's look through the naming rules for that. So for step one, in this naming process, we name the ligands in alphabetical order, we've only got one type of ligand, the ammonia, whose name would be Amin as a ligand. So it's, we have to think about is our ligand, neutral or charged. If it's neutral, we use, we don't alter the name. So it will just be amine. Now, the one tricky thing to note about this name is note that this amine has two MS, we're used to amino groups whose name is amine with one M. So that's a bit confusing. The reason this one has two is basically mainly historical. Um these kind of compounds were discovered first use two MS because ammonia has because ammonia has two MS. And when they got to amino groups, this name was already taken. But just something to keep in mind when naming these compounds. So we don't, we don't have to change the name because it's neutral. All we need is to account for how many groups there are. There are six and we indicate that with a prefix. So we name this with a prefix hex for six, we have hexamine for the name of our ligands. There shouldn't be any space there. I just since I had to write it in front, it has a little gap. So now we have our ligands accounted for. Now, we need to move on to naming our transition metal. And the two things we need to account for here is, is our complex a cion or an anion. Well, when we look at our whole complex, we just calculated our charges here. We know that our, I mean, I mean, our ammonia groups, excuse me, are neutral charge of zero and our cobalt has a charge of plus three. So our complex, our metal is a cion. And if it's a cion, we just use the element's name, we don't change it. So we're going to say Cobalt as the name of our transition metal. And then we have to use a Roman numeral to indicate its oxidation number, which is three. So Cobalt and then in parentheses, Roman numeral three. Now we smush all these together to name our transition male complex with no spaces with the ligands first. So that part of the name will be hea amine Cobalt three. So all we have left are ions and they're pretty straightforward. So the cion gets its name just unaltered. So we have potassium. So put potassium in front, I'm gonna have to put it a little elevated because I ran out of space potassium hexamine, Cobalt three. And then our ion, the anion gets the ending eight to show it's an anion. So we have carbonate, which we already mentioned was its name. So our overall name here is potassium hexamine, Cobalt three carbonate. And one final note is that while the name of the transition metal complex has no spaces, the ion names do have spaces there. So potassium space hexamine, Cobalt three, space carbonate. So now we've done our number one, we have our compound named, we have its oxidation number. I'm just going to cross out those one and two so that I keep track of what I've done. And now kind of the last three parts crystal field energy level diagram. Is it high spin or low spin are their unpaired electrons? Well, these are all connected with each other. So it's kind of all one process. So what do we need for our crystal field energy level diagram. Well, we need to know is it a high spin or low spin complex? Because in order to fill in where the electrons go, we need to know is there a large delta or a small delta to fill those in? And that tells us whether it's high spin or low spin. So this depends on the ligands. So we have ammonia as a ligand and ammonia. And H three is a strong field ligand, which means it causes a large delta between those two D orbital energy levels in crystal field splitting. So when we draw our energy level diagram, we will have our two higher energy level D orbitals and our three lower and we know that this is a large delta, large delta means the lower orbitals are going to get filled completely before any electrons go to the top ones that will result in more paired electrons. And so we end up with a low spin configuration. So we know that going in even before we've filled it in so large delta equals low spin. But now we've got to figure out how many electrons are going in here. So for that, we need to think about the electron configuration of cobalt. So first we start with neutral cobalt and look over at our periodic table. So Cobalt, the closest noble gas is argon. So we put argon in brackets and then Cobalt has the arrangement 3d 74 S two, we know we're dealing with Cobalt three. So it's cobalt with a plus three charge, it's lost three electrons. So we'll have the argon and brackets. It's four s two electrons will be lost and one of its 3d. So it's left with 3d 6 as its electron number. So we have six D electrons to fill in. So now we go to our diagram, we know that we have a large delta. And so we're going to fill in all the lower level before we go up. So it's 6123 and then 456. And we have pairs of electrons in each of those three, we fill them all in, none go up higher because of that large delta. So we have our diagram with six electrons and three pairs in the lower levels and nothing in the upper levels. So we have done that we have classified it, it is low spin as you can see because with more paired electrons, there's fewer unpaired and then finally identify the number of unpaired electrons. Well, that is zero since our six electrons completely filled in the lower three orbitals. So we have done with number five. So we've gotten all our answers here. When we look at our complex, the name of it is facia hexamine, cobalt three carbonate. The oxidation number of cobalt is positive three. We've sketched this crystal field energy level diagram with all six electrons paired in the lower of those D orbitals. And the complex is low spin with zero unpaired electrons. See you in the next video.