ALL THE TRANSITION ELEMENT REACTIONS (OCR A)

Added: 2026-05-12

[00:00:00]Hey all, I thought it would be helpful if I made a video of all the different reactions you need to know for the transition elements topic for the OCR specification.

[00:00:11]So I'm splitting the video into two parts. So part one is just the general reactions of the transition element. So lian substitution and precipitation.

[00:00:21]And then I'm going to move on to all the redux reactions that you need to know.

[00:00:26]Hope you like the video. Hope you found it helpful. And if you haven't already subscribed to the channel, obviously I'd love you to do so, but we'll get into it now. So, we'll start with lian substitution. That's where one type of liant in a complex ion is replaced by another. In many lian substitution reactions, you can see a color change.

[00:00:47]If the lians are similar in size, so for example, water and ammonia, then the coordination number, that's a number of coordinate bonds going to the central transition metal ion that doesn't change. But if the lians are different in size, so for example, water and chloride ions, then the coordination number does change.

[00:01:06]So we'll start by looking at some lian substitution reactions of copper complexes. So the first one is the hexaqua copper 2 plus ion with excess aquous ammonia.

[00:01:18]So there's the equation and you can see that four of the water lians have been substituted with ammonia lians. So this is what it looks like. So you start out with that pale blue solution and it goes to that deep blue solution.

[00:01:33]Next one is the hexaqua 2 ion with excess concentrated hydrochloric acid which is a source of chloride ions. So there's the equation.

[00:01:43]So you can see that all six of the water lians have been substituted by four chloride ions. You can see it's change in coordination number there because the chloride ions are bigger than water molecules. So what does that look like?

[00:01:58]So there's the before and there's the after photo. Now it's worth pointing out that the test tube often appears green and that's because that reaction is reversible.

[00:02:10]So the second reaction we'll look at is for a chromium complex. So it's the hexa aqua chromium 3 ion with an excess of aquous ammonia.

[00:02:20]So there's the equation and you can see this time all six of the water lians have been substituted by ammonia lians.

[00:02:29]We just look at the photos. So there's the first photo of the hexaquromium 3 ions. that's classed as being violet in color and the hexaamine chromium 3 is purple. So not much of a color change.

[00:02:44]So we'll finish this lian substitution section by looking at hemoglobin. So there's a a simplified sketch of hemoglobin. You can see it's a complex ion of iron 2 Fe2 plus in the middle there. So we've got a protein at the top here which is called globin that's attaching via a coordinate bond from a nitrogen. Then we've got this hem um quadrant liant around the middle that's got four nitrogen atoms in which can all form coordinate bonds. So five coordinate bonds being made by nitrogen atoms. The sixth coordinate bond is with an oxygen molecule. So that's when the blood's carrying oxygen. called oxyhemoglobin.

[00:03:29]So the oxygen is carried to the cells and deposited. Now we also need to know the carbon monoxide can bind very strongly with the Fe2+ and that gives us something called caroxyhemoglobin.

[00:03:41]So as that starts to happen the ability of the blood to carry oxygen is reduced and if the concentrations of caroxyheemoglobin gets too high that can actually lead to death. So we move on to precipitation reactions. So that's when you've got two aquous solutions containing ions. They react together to form an insoluble ionic solid.

[00:04:03]So aquous transition metal ions can react with aquous sodium hydroxide to form transition metal hydroxide precipitates and aquous ammonia brings about the same reaction initially.

[00:04:18]Some precipitates dissolve in excess sodium hydroxide or ammonia to form aquous complexes. So we'll start by looking at copper 2. So there's the precipitation reaction. So we've got hexaqua copper 2 reacting with two hydroxide ions to form the copper 2 hydroxide precipitate.

[00:04:39]Aquis ammonia initially brings about the same product. Now there is a simpler way of writing that equation. So you can express it like that.

[00:04:49]We need to know that the precipitate is insoluble in excess sodium hydroxide.

[00:04:55]So what does that all look like? So you're starting with a blue solution and you're getting a blue precipitate.

[00:05:04]Now if you add excess ammonia to the precipitate, it actually dissolves and it under goes lian substitution. So there's the reaction for that.

[00:05:13]So you can see in terms of color, you're going from a pale blue precipitate to a deep blue solution.

[00:05:22]So if we move on to iron now, we'll start with iron 2. So precipitation reactions look like that. There's the simplified reaction.

[00:05:31]And we need to know that the green precipitate starts to go brown when you stand it in. And I'll show you that in a photo. And also we need to know that the precipitate is insoluble in both sodium hydroxide and aquous ammonia. So we look at the photos now. So we're starting with a pale green solution and we go into a green solid. And after a while the green solid starts to look like that. So you can see at the top there we've got that sort of brownie color.

[00:06:00]That's due to the oxidation of the ion 2 up to in3. So if we move on to in3 now precipitation reactions look like that there's the simplified equation the precipitate again is insoluble in both excess sodium hydroxide ammonia. So in terms of what that looks like we start with a yellow solution and we end up with you can see the brown orangey brown solid at the top there. So that's the in3 hydroxide precipitate.

[00:06:30]So if we move on to manganese 2 now here's the precipitation equations simplified equation and we need to know that this light brown precipitate of manganese 2 hydroxide starts to darken as you stand in air.

[00:06:48]However it's insoluble in excess sodium hydroxide and aquous ammonia. So again what does that look like? very very pale pink solution going to a light brown solid precipitate.

[00:07:02]And finally, we've got chromium 3. So, precipitation reactions first. There's the abbreviated version.

[00:07:11]So, what does that look like? We're starting with this violet solution and it's going to a green or gray green precipitate. We need to know that the precipitate is soluble in excess sodium hydroxide. It's actually also soluble in excess ammonia as well. So we get a dark green complex with sodium hydroxide. So there's the equation for that.

[00:07:35]And that's what it looks like. So we're going from that green precipitate to a green solution.

[00:07:42]And finally, the precipitate is also soluble in excess ammonia and we get a purple complex this time. So there's the equation and there's the photos.

[00:07:56]In this video, I'm going to look at some reactions of the transition elements and we'll be focusing on the reaction equations and any associated color changes. So in the video I'm going to be looking at the intercon conversions between iron 2+ and ion 3+ inter conversions between chromium 3+ and the d chromate 6 ion reduction of copper 2 plus to copper 1 plus and the disproportionation of copper 1 plus. Now, just to put your minds at rest, it does say in the specification, students are not expected to remember equations, but construct them from relevant supplied information. For example, electro potentials, half equations, or oxidation numbers. So, the way I'm going to do the video is supply you with those different bits of information and we'll come up with the equations from that. So the first one I'm going to look at is the intercon conversion between in2+ and in3+ and specifically the oxidation of Fe2+ by acidified manganate 7 ions MO4 minus ions. So we've been supplied with some information and we've got to use oxidation numbers to construct the equation for the reaction. We don't need to worry about state symbols. So the first thing I'm doing is summarizing that information. So basically it's just saying what we've been told in the information. Um in 2 plus has reacted with um M4 minus ions. I'm missing out any reference to H+ ions. That's going to come up a little bit later um the way I'm going to show you how to work out this equation. So the Fe2+ are converted to Fe3+ ions and the MO4 minus ions converted to MN2 plus ions. So because we've got to use the oxidation number method, we're looking at the oxidation numbers of iron and manganese. And hopefully you can see that the iron oxidation number has increased by one.

[00:09:58]Ion's gone from 2 plus to 3+. The manganese has decreased by five. that's gone from + 7 to +2. So a rule with redux reactions is you can't have um total oxidation number changes being different because that means different numbers of electrons have been involved in the two processes. So basically we've got to get both of those oxidation number changes to total five. So the way we do that is we put fives in front of the ion species. So now the total oxidation number changes match each other. Next thing we're going to do is balance the charge left and right. And because it's acidified, we're going to use H+ ions to do this. So at the moment we've got a an overall charge on the left hand side of 9 plus. So 5 * 2 plus is 10 plus. And then factor in the 1 minus from the M& minus, we'll get 9 plus. The total charge or the overall charge on the right at the moment is 17 plus 5 * 3 + 15 plus plus the 2+.

[00:11:07]So obviously that needs to balance and we're going to use H+ ions to do that.

[00:11:12]So to get the overall charge on the left hand side to match the right hand side we need to put eight H+ ions in.

[00:11:22]And then the final thing we need to do is just balance any remaining atoms. So you can see we've introduced hydrogen on the left. There's none of that on the right. We're going to use water H2O to balance any remaining atoms. So we need four H2O's. And that's the equation done.

[00:11:41]So we'll just finish with the associated color change for that reaction. So an aquous solution of MO4 minus ions will be purple. So, we're going to go from a purple solution and we're creating um aquous ion 3 plus ion. So, all depends on the concentration of these. But that is a yellow or a pale yellow solution.

[00:12:02]The chances are this is going to be so dilute, so lackening concentration that you're not really going to see that yellow color. So, purple to pale yellow or you could even go purple to colorless.

[00:12:19]And the other intercon conversion involving ion2 plus and in3+ is this one here. The reduction of in3+ to in2 plus by aquous iodide ions. And the supplied information this time is going to be the half equations. So they are going on the screen now.

[00:12:39]So when you combine half equations together to come up with the overall redux reaction, you've got to have the electrons um disappearing from the overall equation. And they're not going to do that at the moment because we've only got one electron in the reduction half equation, but we've got two in the oxidation half equation. So with the way we sort that out is we multiply out the in this case the reduction half equation. We're going to multiply it out by two. That's going to give us two electrons in the equation. Obviously, it's going to put twos in front of the ion 3+ and 2+ as well. But when we add that to the oxidation half equation, the electrons will cancel cuz you got two on each side. So, the overall equation looks like that. So, hopefully you found that fairly straightforward. All we've got to do now is think about the associated color changes. Well, we've already seen the Fe3+ ions pale yellow solution and the predominant color on the right hand side, the product side is the iodine's brown color. So, we're going to go from pale yellow to brown.

[00:13:47]Moving on to some intercon conversions between chromium 3+ and the dicromate 6 ion now. So the supplied information is electro potential data and we've got to use this information to explain the reactions and color changes for each stage.

[00:14:06]So starting with stage one, we're interested in this system here because it's got the dromate 6 ion. It's also got H+ ions as well, which ties in with the acidified bit. And we're interested in this system here because that's got the zinc in it. So this is reacting with the zinc. So if we look at the electro potentials values you can see that this is plus 1.33 vol this one's -0.76 volt. So what that means is the dromate 6 ion uh system is going to move forwards more positive um and the other one's going to go backwards.

[00:14:42]So you can see that these these species here are going to react with the zinc become these species here and zinc 2+ and you can see that the electrons don't match. We got six in that one and two in that one. So we're going to need to multiply this one by three and then add it to that one. So this is the same sort of idea as the the second method, the one we've just looked at, but the information's been supplied in a in a different way. So when you do that, there's the overall equation for that stage. We'll do the um color changes at the end. So we'll move on to stage two now.

[00:15:21]So you'll notice in the reaction here, we've just generated chromium 3 plus ions, which also feature in um system two. So we're now interested in this one. Still interested in the zinc because it's in excess. So looking at these um electro potential values, you can see that this one here is more positive. The zinc one's still the most negative. So these are going to run in those directions there. And again, we just need to add these together.

[00:15:53]Electrons need to balance. So we need to double this chromium 3 + 2 + 1 uh before we add it to that one. And we get that.

[00:16:04]So moving on to color changes now. So the substances with color are the dicromate 6 ion, the chromium 3 plus ion and the chromium 2 plus ion. So the color changes there are orange to purple to blue.

[00:16:24]And the other intercon conversion involving chromium 3+ is this one here.

[00:16:30]So, we've got some supplied information and we have got to come up with the two half equations for the two processes and then put them together for the overall redux equation for the reaction. We don't need to worry about state symbols.

[00:16:45]So, starting with the chromium system, there's the change there. It's gone from CH63 minus to C42 minus. So, we're going to start with the oxidation number change in this one. So it's gone from + three, it's gone up to plus six. So effectively the chromium's lost three electrons. So we put them on the right hand side there. Now remember this is all in alkaline solution. So we're now going to balance the charges left and right, but we're going to use O minus ions, not H+ ions. So the overall charge on the left is 3 minus. Overall charge on the right 2 minus plus 3 minus 5 minus. So, we're going to put two hydroxide ions on the left hand side to give us an overall charge now of 5 minus.

[00:17:34]And then all we need to do now is use water to balance any remaining atoms.

[00:17:38]And you can see straight away that we've got hydrogen on the left. We've got eight hydrogens's on the left. We've got none at all on the right. So, we're going to need four H2Os. So, that's that one sorted.

[00:17:53]Moving on to the hydrogen peroxide. Much simpler this one. If we put a two in front of the O minus ions, we balance all the atoms straight away. So that's all we need to do there. And now we just need to sort the overall charge out. So we've got no charge at the moment on the left, but we've got two minus on the right. So we need two electrons on the left hand side.

[00:18:17]Now we've got the half equations. We just need to get the electrons um to disappear when we add them together. So you can see that we're going to need to multiply the top half equation, the chromium one by two. That'll give us six electrons in that one. So everything in that chromium half equation is going to be doubled the hydrogen peroxide one. We need to treble. So everything gets multiplied by three. So there's the overall equation there. I'll just quickly explain the hydroxide ions and then I'll come on to the color. Although it's pretty obvious what the color change is going to be. So remember we doubled the chromium half equation. So this the hydroxide ions will have gone to four. We trebled the hydrogen peroxide one. So they'll have gone to six.

[00:19:02]So left over when you cancel down is two hydroxide ions on the right hand side.

[00:19:08]And finally, the the obvious color change is going to be from green to yellow because those um CR63 minus ions are green. Surprise, surprise. And the um chromate 6 ion C42 minus ion is yellow.

[00:19:26]And now moving on to some inter conversions involving copper. So we're looking at the reduction of copper 2+ to copper 1 plus in this slide. So there's our supplied information. copper 2 plus ions can be reduced to copper 1 plus ions by iodide ions and there's the unbalanced equation. All we've got to do is balance the equation and we're going to give the associated color changes or observations for that one. So this is probably the most straightforward one of the lot because all we've got to do is put twos in front of the copper species.

[00:19:59]See that's going to give us four I's on the right. So we need four in front of the I minus on the left. I will explain the oxidation number change rule. So we've got copper 2+ going to copper 1 plus. So it's been reduced by one but we've got two moles the way we've balanced the equation. So the overall decrease in the copper oxidation number is two.

[00:20:26]And then if we think about the iodines, these are all all four of these are at minus1 at the start of the reaction.

[00:20:33]We've got two still at minus1.

[00:20:37]But we've got two that have changed from minus1 to zero in I2. So the overall increase in the iodine is also two.

[00:20:49]Moving on to the color changes. So we're starting with aquous copper 2 plus. So they're going to be blue. So we've got blue solution. Now there's two observations or color changes we need to talk about here. Copper one uh compounds. So this is solid copper one iodide. that's white and that's because the copper is in the plus one oxidation state. So it's got a 3d10 configuration.

[00:21:13]So when you've got a full 3D um subshell, there's no color. So the solid will be white. Um iodine, we've already seen iodine solution is brown. So we'd have to make both of those um observations there. Blue solution to white solid and brown solution.

[00:21:33]And finally, hooray, the disproportionation of copper 1 plus. So our supplied information this time is a statement. Copper 1 plus ions are unstable and spontaneously disproportionate.

[00:21:46]And we've got some electro potential information which we're going to use to explain the statement.

[00:21:52]Okay. So looking at the electro potential values, you can see the second um system is more positive. So that'll run forwards. The top one will run backwards.

[00:22:03]electrons are the same. So, we don't need to multiply these out. We'll literally just need to add them together, which gives us that, which we can tidy up to that. And that is the disproportionation of copper 1 plus because copper has been oxidized. It's gone from + one at the start to +2 in the copper 2 plus ion, but it's also been reduced to the element in the zero oxidation state.

[00:22:31]In terms of observations, we've got a copper one plus ion. So that will be in solution at the start. So that'll be colorless. Remember, we've just said on the previous slide, copper 1 plus ions have a 3d10 configuration. So in the solid state, their compounds will be white. Um, but in the solution, they'll be colorless. We've got copper metal, which is a sort of pinky brown um solid.

[00:22:57]And we've just also said in the previous one, copper 2 plus ions are blue. So the observation is going to be a colorless solution to a blue solution and that pink brown solid.