Transition Metals & Reactions of Aqueous Ions | A‑Level Chemistry Exam Question Walkthrough

Added: 2026-06-04

[00:00:00][music] >> Hello and welcome to this A-level chemistry exam question walkthrough. The question is available through a link in the description, and this means that if you want to, you can download the question and have a go at it yourself before you watch this video. My recommendation is that you aim to give yourself a minute per mark. During this video, I'll be modeling my thinking out loud and annotating the question, mostly in red and blue, and then the answers that will get you the marks will be shown in purple. If you find parts of this question difficult, you could always check out my explanation video about this topic by using the master links document available in the description.

[00:00:49]This question is about aqueous ions of the metal ion. When an aqueous iron 3+ complex with water ligands reacts with ethanedioate ions, an iron 3 complex ion X is formed. The only ligands in X are ethanedioate ions. Draw the structure of X, include the charge. Since we're told that the only ligands in X are ethanedioate ions, this means that all six of the water molecules are replaced in a ligand substitution reaction. So, we start with the hexaaqua iron 3 complex ion that I'm showing here. We've got the transition metal in the middle, six coordinate bonds going outwards to six water ligands. Those water ligands are neutral monodentate ligands, and so overall, we have got a 3+ charge complex. And then these ethanedioate ions are going to replace the water ligands. Ethanedioate ions have this displayed formula that I'm showing here. They are bidentate ligands, so they form two coordinate bonds from these lone pairs that I'm showing here to the central metal ion.

[00:02:09]And so, three of these ions bond to the metal to form an octahedral complex. And the equation for the reaction will look like this. We have our iron complex with six water ligands at the beginning.

[00:02:24]Three ethanedioate ions react with it, and we produce our iron complex which has got three bidentate ethanedioate ions in it. And we need to put brackets around the ethanedioate ion, so we know that there are three of what is inside the brackets. And we would have six water ligands being replaced, and so we need six water molecules as our product.

[00:02:50]And X then is this complex ion that I'm showing here. We've got iron, which is a 3+ ion, and we have three of these 2- ethanedioate ions, and so this complex will have a 3- charge overall. And so, our octahedral water complex will need the water replacing by ethanedioate ions. Each ion replaces two monodentate water ligands to give us this complex that I'm showing here. And this structure will get us our first mark.

[00:03:24]And then the second mark is for these brackets, and then the 3- charge in the top right corner.

[00:03:32]The formation of X is an example of the chelate effect. Explain the meaning of the chelate effect. Now, to an extent, this is something that you need to just simply recall and then learn the justification for the effect. We see this effect when bidentate or multidentate ligands replace monodentate ligands. And this is because the complex that is produced is more stable than the complex that we started with. And we can see this when we refer to our equation.

[00:04:08]We start with 4 moles of reactant and we produce 7 moles of products. And so that means there is a large increase in entropy. Or we could say that there is a positive entropy change or an increase in disorder. Or we could cite the number of particles that are being formed and say 4 moles of reactant are producing 7 moles of product. Whichever one of these that you say as your follow-up gets us the second mark. And just to go a little bit deeper into that explanation, the positive entropy change means that delta G will be negative. And this comes from the equation delta G equals delta H minus T delta S. Since we are breaking six coordinate bonds and making six coordinate bonds, the enthalpy change will be approximately zero. And so the fact that we have a positive value for our entropy change means delta G is negative. This won't be necessary for a two-mark question, but you should add this if the question is three or more marks or if the question specifically prompts you to use the delta G equation.

[00:05:20]Outline how iron 2+ ions catalyze the reaction between S2O82- ions, which are known as peroxodisulfate ions, and I- ions in aqueous solution.

[00:05:35]In your answer, you should include a sketch graph to show how the concentration of peroxodisulfate ions changes over time, an explanation of how iron 2+ ions catalyze the reaction, including equations, and an overall equation for the reaction. Before we start, it's worth noting that these instructions are not equally weighted.

[00:05:58]Some are much more significant than others. The final bullet point, for instance, just asks for the overall equation, so they won't carry equal marks, but we will still focus on them systematically. Firstly, the sketch graph. We've been told it's a graph of concentration of peroxodisulfate against time. And so, peroxodisulfate concentration should go on the Y axis and time on the X axis. In terms of the graph's shape, we should consider that we've been told that peroxodisulfate is reacting with and so the peroxodisulfate concentration will decrease over time because that's what happens to reactants.

[00:06:41]And then, because this concentration is decreasing, the frequency of collisions will decrease as well. And that means that the rate of reaction will decrease, and so we should draw a curved line of decreasing gradient. And so, the key content for this first bullet point will be labeled axes and a line showing the peroxodisulfate concentration decreasing, and then the second key point is that it should be a curve of reducing steepness. For the second bullet point about catalysis, it's worth considering first why a catalyst is necessary at all. And this is because the two reactants are both negatively charged. And so, this means that the peroxodisulfate ions and the iodide ions will repel each other. And this means that the activation energy will be high. And then, the general reason why iron two plus is able to act as a catalyst is that it has got the ability to exist in variable oxidation states, which is one of the key features of transition metals. Specifically, iron two plus can act as a reducing agent or electron donor in a reversible reaction that produces iron three plus, which has the ability to act as an oxidizing agent or electron acceptor, and turn back into the iron two plus. So, Fe2+ can act as a catalyst and reduce one reactant by donating electrons to it and producing Fe3+ and then this Fe3+ accepts electrons from the other reactant, turning back into the Fe2+.

[00:08:32]And this means that the two negative ions never need to meet. And in fact, the positive Fe2+ ion is the opposite charge to the peroxodisulfate ion it's reacting with, and so these ions will attract each other, and this will lower the activation energy for the reaction.

[00:08:52]To be clear about why it's the peroxodisulfate ion that the catalyst reacts with, we need to look at some equations.

[00:09:00]I always think it's much easier to remember parts of the half equations, so what the reactants turn into and deduce the rest. So, iodide ions turn into iodine, which means we must have two negative iodide ions in the reactants, and to balance the charge, we need two electrons in the products. And the sulfate ions. And so, we need to produce two sulfate ions to balance the atoms, and the reactants need two electrons in to balance the charge. And so, now we have half equations. First of all, we've got the iron 2 plus donating electrons and turning to iron 3 plus and we're showing that as reversible. And then we have the oxidation half equation for the iodide ion and we have the reduction half equation for the peroxodisulfate ion.

[00:09:54]And so from here, before we combine them together, we need to make sure we have the common multiple for the electrons.

[00:10:02]The two half equations at the bottom show the loss and gain of two electrons.

[00:10:06]And so that means that we need to double the half equation for the iron and then now we have a common multiple of two for electrons. And so now we can see that the iron 2 plus must be donating electrons to the peroxodisulfate.

[00:10:21]So we need to combine the iron equation in the forwards direction with the peroxodisulfate equation. And when we do that, we have the peroxodisulfate ion plus two iron 2 plus turns into two sulfate ion plus two iron 3 plus. And then the iodide half equation combines with the iron 3 plus half equation, but this time the iron 3 plus is going backwards into iron 2 plus. This means then that we'll have two iron 3 plus reacting with two iodide ions and producing two iron 2 plus plus two iodine. When we consider these two equations, we can see that the iron 2 plus is used in the first equation and remade in the second equation, therefore proving that iron 2 plus is acting as a catalyst. And also in the first equation, we produce iron 3 plus and then we use it immediately in the second equation, which shows that iron 3 plus is an intermediate chemical. And so to deduce the overall equation, we need to add these two equations together and remove what's present on both sides of the arrow. And so, that means we remove the iron 2 plus from both sides and the iron 3 plus from both sides. And so, we have the overall equation of peroxodisulfate ions plus two iodide ions turns into two sulfate ions and iodine. This question was out of six marks and to get five or six out of six, you would need to cover all three of the stages and your descriptions would need to be generally correct and virtually complete. And by virtually complete, that means dropping a maximum of one bullet point from each stage. And to guarantee getting six out of six, you would say probably dropping a maximum of two of these bullet points, perhaps only as many as one to just be really sure that you got all six of these marks.

[00:12:24]A student adds dilute ammonia solution to a solution containing Fe(H2O)6 2+ ions. Those are the hexa aqua iron 2+ ions. Give the formula of the precipitate that forms. Well, we're starting with this complex ion with six water ligands. And whenever you add dilute ammonia to a complex like this, the ammonia is going to act as a base and remove hydrogen atoms from some of the water ligands until the complex becomes neutral. And we form this complex ion here and two ammonium ions.

[00:13:02]So, our job here is to give the formula of the precipitate formed, this iron 2 hydroxide complex. And so, it would be Fe(H2O)4(OH)2.

[00:13:14]The student adds sodium carbonate solution to a solution containing Fe(H2O)6 2+ ions. State one observation the student would make and give an equation for the reaction. This will be one mark for each. Aqueous metal two plus ions are very weakly acidic. So, when you add the weak base to it, the sodium carbonate, an acid-base reaction does not occur like it did with ammonia, and we instead produce a carbonate precipitate. And in this case, we would produce iron two carbonate, which is a green precipitate. And that would be the observation. And in terms of the equation, we start with our hexa aqua iron two plus, we add sodium carbonate to it, and we produce the iron carbonate, that's the precipitate, and that means those six water ligands need to have been removed from the iron, and the two sodium ions will just be separated away from the carbonate ion.

[00:14:13]Or if you prefer, you could write this equation without the sodium ions, which are effectively spectator ions, and the equation looks like this. I prefer this type of equation because it looks slightly simpler.

[00:14:26]And part F says, "A solution containing hexa aqua iron two plus ions changes to a yellow-brown color after several hours in contact with air. The student adds sodium carbonate to the yellow-brown solution. Give an equation for the reaction with sodium carbonate." It's important to note that this color change is due to the oxidation of the iron. We have changed from iron two plus to iron three plus, and so that's what the yellow-brown color is, Fe(H2O)6 three plus. Now, aqueous metal three plus ions are more acidic than metal two plus ions. So, what they are acidic enough to react with the weak base and produce metal hydroxides in the same way that happened with the dilute ammonia.

[00:15:17]The exact same way, except instead of two two atoms being removed, three get removed in order to make it neutral because the starting complex was three plus. And as well as the hydroxide precipitate, we're going to produce carbon dioxide. And so the equation here will have Fe(H2O)6 3 + + sodium carbonate produces our hydroxide precipitate with three hydroxide ions this time, and carbon dioxide, and water, and sodium ions. In order to balance this equation, it's important to note that it's the carbonate ions reacting with the hydrogen ions that are removed that produces carbon dioxide and water. And we would need to remove two hydrogen atoms for each carbonate. And so in order to remove three hydrogen atoms from iron, we actually need to remove six hydrogen atoms from two iron. That means six hydrogen atoms are being removed. We need three carbonate ions to do it. And so that means our reactants need a two and a three on the left-hand side. And then that must mean that we're producing two of our hydroxide precipitate, three carbon dioxide, and three water. And since we've got six sodium on the left-hand side in the reactants, we'll need six sodium in the products to balance this out. So this is a complicated equation, but it's made more straightforward if you understand what's happening in terms of the carbonate ions acting as a base. Again, if you prefer, this equation gets slightly simpler to look at if you remove the sodium spectator ions. So broadly speaking, this is the same. We just aren't making Na+ as our products, and the reactants are the carbonate ions and the hexaqua iron 3+ complex ion.