What is the formula, including the charge, for each of the following complexes?

(a) An iridium(III) complex with three ammonia and three chloride ligands

(b) A chromium(III) complex with two water and two oxalate ligands

(c) A platinum(IV) complex with two ethylenediamine and two thiocyanate ligands

Video transcript

Welcome back everyone. Our next problem says deduce the formula with the charge for the co ordination compounds below. Then we have three co-ordination compounds described verbally. And our answer choices consists of just different lists of the formulas for our complexes. So rather than read them all aloud, we'll go through each of the compounds one at a time and then match them with the correct description with the correct answer. So first of all, we'll need to determine the overall charge for our compound number. One says a nickel two complex with four ammonia ligands and two nitrate ligands. So we have a nickel two complex. So we know that our central metal atom must be N I two plus. So to calculate our overall charge, we'll need to add up the charge from the metal atom and all the ligands. So we'll have a charge of positive two from our metal. We have four ammonia ligands that would be NH three. So there's four of them, but NH three is neutral. So I'll just put plus zero and then two nitrate ligands. So that would be no three minus. So nitrate carries a charge of negative one and there are two of them. So we add two multiplied by negative one. So we'll end up with positive two minus two and our overall charge will be zero. So no charge overall for this co-ordination complex. And now let's think about what our formula will look like. So we have N I and then we have the four ammonia ligands. So in parentheses, NH three and then subscript, oops I had written the wrong number, subscript four and then two nitrate ligands. So in parentheses and three, and there's two of them. So subscript two. So let's look for this correctly described among our answer choices. So if I look at choice A and look at the formula for number one, I have no charge and but it shows nickel subscript to, we only have one nickel atom. So that will not be correct there. So we can cross out choice A because right away in number one, we have the wrong formula, it also describes the wrong number of nitrate ligands. So we've just limited choice. A choice B we see correctly, no charge one nickel atom for ammonia ligands and two nitrate ligands. So the formula for number one is correct. And choice B now lets look at choice C. So answer choice C number one, no charge one nickel, but it only shows one ammonia ligand and one nitrate ligand. So choice C will not bear answer as number one does not have the correct formula. And then finally choice D, if we look at number one correctly shows no charge one nickel atam, but it shows the ligand as NH four and only one of them. So that's going to be incorrect, incorrect description of that ammonia ligand. So choice D not our answer. So we can see we've actually eliminated three of our answer choices. If I want to test, I would pick choice B and happily go on because I only have one answer choice that correctly gives a formula for number one. But we want to be thorough in this video. So we'll keep looking at our other complexes. For number two, it says a palladium two complex with two Di Ndien ligands. So our palladium two will be PD two plus. So let's look at our overall charge, we will have positive two for our metal atom. We only have one type of ligand. And then the diane ligands, diane is dile triamine. It's a trident ligament ligand. Excuse me, as you can guess by it, the fact that it has three amines or three nitrogens capable of bonding and it is neutral. So we have a zero charge from the Diane Liggins. So our charge overall for the compound will be positive too. And when we look at our formula, we'll have KD and we just have two diane ligands, we have diane and parentheses with subscript two. And then we have to put the whole thing in brackets because it does have a formal charge and add to pull us. So with that in mind, let's look over our answer choices. Again, we know which one our answer is. So we'll just double check that it's correct. And we do see that it matches what we just made PD dy N subscript two, all of that in brackets with a two plus charge. Finally, let's look at number three, a Cobalt three complex with three ethylene diamine ligands and three chloride ligands. So we have Cobalt three, so co three plus. So we'll go ahead and add a plus three in my equation for overall charge. And then we have three ethylene diamine ligands, ethylene diamine, we can guess by that diamine two nitrogens, it's bent, it's also neutral. So charge is zero from those three and then three chloride ligands, but chloride of course is CL minus. So that will be three multiplied by negative one. So that means our overall charge is positive three minus three. So it is zero and then we write our formula, we have co has three ethylene diamine. So that would be en in parentheses, subscript three and three chloride ligands. So when we look at our answer choice, put cl three, and when we look at our answer for choice, we know again, it has to be B and we have no charge overall, we have co in parentheses en subscript three cl three. So we've confirmed that indeed our correct answer here for the formulas with charge for the coordinating compounds is choice B see you in the next video.