Overview of Group 7 Halogens

Hello and welcome to this video. This is the AQA Group 7 Halogens video. This is for revision purposes so it's just a quick overview of the Group 7 topic. My name is Chris Harris and I'm from alouetutors.com and basically like I say we're all going to have a quick look at this. This is very specific to AQA. The PowerPoint that I'm going to use here you can use and you can actually purchase from by clicking the link in the description box for this video. So if you just click on that, you'll be able to get a hold of it, and you can use it for revision, and you can use it for whatever you want, print them off or whatever. So basically, if you like what you see here, then you can get them for yourself. Okay. um also like i say these are tailored specifically to the aqa specification and so obviously the spec points are reflected therefore for this topic okay so um we're going to start with the halogens obviously because this is this is all about group seven um and obviously the halogens are made up of non-metals they're on the right hand side of the periodic table um you On the top of the halogens is fluorine, which is a pale yellow gas, and it has that electron configuration, 1s2, 2s2, 2p5. Make sure you can give the electron configurations of these. You should be able to do it for the elements such as fluorine and chlorine in particular. Chlorine gas is pale green. It's Cl2, and there's the electron configuration for it. Bromine is a browny-orange liquid, and iodine is a grey solid. And you can see the electrical configuration there is much, much longer for iodine. But it's just to kind of give you an idea. Now generally, as we go down the group, starting from fluorine down to iodine, the boiling points increase as we go down the group. It's because the molecules and the size of the atoms and the molecules here, well, these are molecules, and the relative masses of them are getting larger. So therefore, we have larger van der Waals forces. And so this means that you need a higher amount of energy to turn these into a gas. Now you can see some of these are obviously in different states. The top two are gases. Bromine is a liquid. Niadine is a solid. So that basically just explains that the van der Waals forces are increasing. And at room temperature, as we go down the group, they become more solid. The electronegativity decreases as well. When we bond these halogens to, say, like a carbon, for example, or another element in a covalent bond, then the electronegativity decreases. Now, the reason why is because the ability for one of these atoms to attract electrons towards itself weakens. the atoms get larger and the distance between the positive nucleus and the bonding electrons increases and basically we have more shielding so we have more shells between the nucleus and the outer electron so that means it's easier to the well the electronegativity is a lot less for that reason. Okay, right, so let's have a look at some displacement reactions. Now these are pretty important. You do see a lot of these for halogens and they pretty much explain the reactivity that we've described before. And the fundamental rule is that a more reactive or the most reactive, should I say, more reactive halogens will displace a less reactive halide ion. So basically, like we said before, the reactivity in halogens, they decrease as we go down group seven. OK, it's for a reaction to occur. Basically, an electron is gained and we form that halide ion. Atoms with a smaller radius, they attract electrons better than the larger ones. So that's quite important. Basically, halogens are less oxidizing as we go down the group. So remember, oxidation is the loss of electrons. So. So they become less oxidized as we go down the group. And we can show this by reacting halogens with the halide ions. Okay, so halogens will displace a halide from a solution if the halide is lower in the periodic table. So we're going to look at these two here. Okay. We're going to look at all of these, in fact, and we're going to look at the reactivity of the halogen and the reactivity of the halide ion. So remember, as we go down the group, the halogens, the reactivity of them decreases. OK, so let's look at the first one. Now, because these are two chlorines, chlorine water, Cl2, and chlorine, KCl chloride, sorry, which is KCl, potassium chloride, because they're just as reactive, we get no reaction. This one though, we do. Chlorine is more reactive than a bromide ion. So we do get a displacement reaction. You can see here, there's the chlorine, there's the bromide ion. These react and we get a displacement. So we get chloride ions and bromine is made. So we get an orange solution being formed here and that orange solution tells us that actually we're making a bromine solution here. Okay, looking at the next one. Again, chlorine is more reactive than iodine. or than iodide should we say so we do get a displacement reaction now we get a brown solution iodine when it's in solution goes like a brownie color um so this brown solution is made of iodine chlorine again reacts with the iodide ions there it is there we get displacement chloride ions are made an iodine solution is made an iodine solution is brown remember iodine is a solid is is gray and but your iodine solution is brown okay so look at bromine water no reaction here Bromine is less reactive than chloride ions, so we get no reaction. And no reaction for this one as well, because that's bromine with bromide, so they're both just as likely to react. But the final one, again, we form our brown solution. Again, it's the same reason. Bromine is more reactive than iodide ions, so we get a displacement reaction. We form 2Br-, and we get iodine produced as well. And obviously this is in solution, so it makes it turn brown. And the last one, it's no reaction here. Again, iodine is less reactive than chloride, so we get no reaction. Iodine is also less reactive than bromide. And then obviously iodine and iodide are just as equally reactive, so they won't displace either. So yeah, make sure you know your half equations and you know the colours and you know the reasons why. It's all about this displacement. Right, bleach. Okay, so there wouldn't be halogens without talking about bleach, really, let's be honest. So... This is made by using a disproportionation reaction. Okay, try to say that one quick. It's basically a reaction where we're getting oxidation, an element is simultaneously oxidized and reduced, as we'll see in a minute, I'll show you the reaction. But if we mix chlorine and sodium hydroxide together, we'll form sodium chlorate 1 solution, and this is known as bleach. Now the reaction, like I say, As you can see here, it's the sodium hydroxide plus chlorine. Literally, that's what it is. Sodium chlorate 1 is this stuff here, N-A-C-L-O. So that's that one. Plus you form your salt and you form water as well. So really important, this is your bleach. That's bleach there. So let's look at this disproportionation thing again. Chlorine is an oxidation state of 0. It's an element. Chlorine here is plus 1 because oxygen is minus 2. Sodium is plus 1. So for the whole thing to be neutral, chlorine must be plus one. And chlorine here has an oxidation state of minus one because sodium is plus one. So chlorine has been simultaneously reduced and oxidized, like I say. And we call this a disproportionation reaction. And we can use bleach. Bleach is pretty useful, obviously. We can use it to treat water. We can bleach paper and fabrics, so we can make paper whiter or fabrics whiter. And obviously we can use it as a cleaning reagent. Obviously you put bleach, use it in bathrooms, cleaning toilets, etc. So yeah, it's got a pretty important use. Okay, another important use is obviously water sterilization. Now without this, we would have had outbreaks of cholera. And actually in some countries today, where they don't have efficient water sterilization techniques, they do get, unfortunately, do get spreads of cholera, which is a waterborne disease. But the whole point of adding chlorine to water is to kill the bacteria in the water, including diseases, including microbes, sorry, that cause cholera. So, yeah, adding water to chlorine will produce chlorate ions, ClO-, a little bit like the bleach that we've seen just before. These kill bacteria and they're useful in drinking waters and swimming pools. So let's have a look at these again. Now, this is slightly, slightly different. Again, we've got water and chlorine, no sodium hydroxide. We're using water instead. And what we form is H plus ions, Cl minus ions, and ClO minus. Now you might also see this as HCl and HClO. So these H pluses can be attached to this. But because they're soluble and it dissociates pretty strongly, then we normally write them as separate ions. But again, if we look at our oxidation state, chlorine has an oxidation state of zero, obviously minus one here and plus one here. So again, this is a disproportionation reaction. Look. The chlorine at the start has been simultaneously reduced and oxidised. Okay, now this is quite important, especially in things like swimming pools, because sunlight can decompose chlorinated water too. And the problem is, is that we get no ClO minus that is made. Remember, it's the ClO minus that actually kills the bacteria. So if we look at these two here, let's have a look. We've got water and we've got chlorine. Now, if we use, if we have sunlight that's present, actually what we produce is 4H plus, Cl minus and O2. Obviously this can pose a problem in swimming pools because if you get sunlight bleaching into the side of the building, maybe it's where the swimming pool is, the sunlight can actually destroy the bleach that's used in swimming pools, the ClO-. And what we form is Cl-O2. Now obviously because we don't have any ClO- because the sunlight has broken it up, then obviously we don't have any active ingredient to kill the bacteria. So... This is why the chemical reagents or the chemicals that we add to swimming pools have to be replaced quite regularly because sunlight breaks them down into these compounds here, which are obviously different to that bleach up there. Okay, drinking water. Okay, so advantages and disadvantages. Obviously, it's in the UK and we do chlorinate water, but there are like with anything, adding any chemicals or anything, there's advantages doing it and disadvantages and you need to know what they are. So the advantages, they destroy microorganisms that cause disease. So that's a pretty good thing. They're long lasting. So they reduce bacteria buildup further down the supply, which is pretty useful. And reduces the growth of algae that discolours water and give it a bad smell and taste. Last thing you want coming out of your tap is brownish water that doesn't smell too good. It would put you off pretty much. So yeah, basically the chlorine helps to reduce that. But the disadvantages. Chlorine is toxic. We can't hide that. and it can irritate the respiratory system so people who add this chemical into the water have to take obviously precautions when adding this in liquid chlorine causes severe chemical burns to the skin so if there was high levels of chlorine in drinking water that could obviously cause problems but there's systems and safeguards from doing that but i suppose the biggest one here is the fact that there is this belief by some people that the um or some scientists that the chlorine could react with organic compounds that present in the water and they make chloroalkanes and these have been linked with causing cancer but obviously the risk of not chlorinating the water and could lead to like like we said before cholera epidemic and cholera is lethal as well so the basically what we're seeing here is the the advantages outweigh the risks or the benefits outweigh the risks in this case The levels of chlorine that are added to water are very, very, very small. And scientists believe that the risk of cancer from this is obviously very low, incredibly low, because of the low levels that are added anyway. They only add just enough just to make sure that the bacteria is killed in the water. Okay, so, right, let's look at these halides. Now, these are quite difficult, actually, these things. So... And we'll actually look to halogens so far. We'll look to a little bit of halide ions with the displacement reactions. But halide ions already have that electron already. Now, these lose that extra electron they've got. And so this makes them really good reducing agents. Okay, so remember, reducing agents are oxidized themselves. Okay, and oxidation is the loss of electrons. So halide ions lose electrons because they've got that extra one. And so they make some good reducing agents. Okay. So basically, as we go down the group, the ionic radius increases. And I've just drawn that diagram there. Basically, the distance between the nucleus and the outer electrons becomes larger. There's more shielding. Use all these words in your exam. And the attractive force gets weaker. Okay, so it's not as strong. And the outer electron is lost a lot more readily. And this is the reason why I minus is more powerful reducing agent than F minus. So I minus loses that. that electron much more readily than f minus okay now there's two tests you need to know okay and to prove this to prove this trend so one's with sulfuric acid which we'll look at next and the other one is with silver nitrate solution and you've got to know the results of both of them and i think the most difficult one is the one with sulfuric acid because you get a lot of equations here so let's just talk it through make sure you know what's going on okay um So some halide ions can reduce concentrated sulfuric acid. So that must be concentrated. That's very important. So what I'm going to show you here is basically a little kind of, it looks a bit like a train map, I like to call it. I think it's probably easier maybe seeing these things in this way rather than in other ways that I've seen in some notes and sheets and books, etc. I think this probably is an easier way of doing it. But you can use it however you want. Basically what I've done is I've written my reduction products of sulfuric acid on the top row. The only one which isn't a reduction product is this one, sodium hydrogen sulfate. But these ones are reduction products and I'll come on to these in a minute. And then what I've done is I've looked at sulfur. Okay, because sulfur is the key element here. And I've written the oxidation state of sulfur underneath. So we've got plus six here, plus four, zero and minus two. And what I've done is I've written our halide ions down here. These are the halide ions that we're going to be using. We're going to react it with concentrated sulfuric acid. So we're going to react the chloride ion with concentrated sulfuric acid to see what products we produce. We're going to react to bromide ion with the same stuff, see what products we produce. We're going to react to iodide ion, see what products we produce. So let's start with the first one. Here's our underground train line thing. So we're going to look at A. Now the good thing is that chlorides, bromides and iodides, they all produce this compound here. Sodium hydrogen sulfate. and we need to know the reagents of these as well the reactions of these so this is point a now point a this is not a redox reaction okay this is just a standard chemical reaction between a salt and an acid so here we are this is where sodium chloride sulfuric acid react with sodium chloride all we get is one of the hydrogens is given up um to form hcl you get white misty fumes here and you form sodium hydrogen sulfate You also get the same with sodium bromide and sodium iodide as well. However, as you will see, that these reagents here, these sodium hydrogen sulfates, will then go on to react further. So actually, you have very little of these anyway in the overall reaction. But certainly with the chloride, the only product you get with sodium chloride is that. And that's it. You can see the arrow stops there. No more product produced. Okay, so now we're going to bring our attention on to bromide. Okay, so bromide is... let's have a look okay so these are going to be looking at reaction b so bromides can actually reduce further the sulfur further they can reduce it to produce sulfur dioxide and iodides can do the same as well okay so they could produce sulfur dioxide so let's have a look at their reactions so this is point b okay you can see um in addition to a as well okay so we've got obviously all the reactions as well um but with sodium bromide because that's what we're going to look at mainly and we get the same reaction with sodium iodide as well what we do is we write down our half equations first okay and the bromide is being converted into bromine i've written the half equation there so the bromide ions are being oxidized br minus going to br2 and obviously we balance it with two electrons okay this has been oxidized and i must stress as well if you don't know how to work out your half equations you really do need to know um how to do this i have put video on that describes how to balance half equations you've got to know that as well I'm just going to go through it there's an overview here assuming that you know how to do it but and this is the bromide ion and then crucially this is this bit here we start with sulfuric acid and we're going to produce sulfur dioxide because that's one of our products what we do is we balance with water first then protons then electrons okay so we just balance this half equation this is the sulfur being reduced okay so this is a redox reaction because we've got oxidation reduction cancel out all electrons and combine the equations and we should get an overall anic equation which is this here so you can see the sulfuric acid has been reduced to sulfur dioxide we should see in this reaction an orange vapor of bromine being produced and that tells us that actually this reaction has occurred okay um so bromine tells us that we have produced sulfur we produce bromine and we've also produced sulfur dioxide as well which is a choking gas okay Let's look go further. So bromides are not as powerful reducing agents so they can't reduce that any further. But iodides can. Okay so they keep going. So iodides can reduce sulfur even further to produce sulfur which is this sulfur solid. Again we're going to do the same thing. Look balance it out. 6 I minus I2. So we're going to put I minus go into I2 and then we basically try to balance that out. We do the same here. Start with sulfuric acid forming sulfur. And then we balance with water, protons and electrons. When we balance this, we effectively worked out that we have to multiply the top one by three, so we can balance out the electrons, the six electrons. Remember, we've got to try and balance these out, cancel them out, rewrite out the equation. The overall equation here is obviously I minus going to I2, and sulfuric acid going to sulfur, so producing solid sulfur. And iodine, iodide ions are so powerful at reducing, they reduce the sulfur even further, and they produce H2S, okay? I should point out with this one as well. you produce sulfur solid, which is yellow. So this one, the final step is step D. Again, same thing. Look, we're just going from I minus to I2, just bouncing out. Sulfuric acid this time, we're forming H2S, hydrogen sulfide, we call it, which is this one here. And the overall ionic equation, obviously, just like before. Again, we're going to multiply this top one by four to make sure we get eight electrons on both sides. Classic one here, rotten egg smell. h2s is a nerve agent it's really toxic you shouldn't be breathing this in and it smells of rotten egg it's got a horrible smell it's also known as sewer gas um because it's found commonly in sewers so you probably get an idea of what uh what that what they may smell like okay so that's your reaction with hyaluronides with sulfuric acid make sure you can balance your half equations really redox is crucial here being able to balance them okay The other one, which is probably a little bit safer to use, let's be honest, is reacting it with silver nitrate. So we can use these halides, add silver nitrate, and we can confirm further with ammonia, which we'll look at later on. So this is how we do it. We test for chlorides, bromides, and iodides. All we do is we add nitric acid first, dilute nitric acid. Then we add silver nitrate solution, AgNO3. The colour of the precipitate will help you identify the halide ion. So I'll come on to the nitric acid a little bit later on as to why we add that. But you can see here we've got the test tubes. We've got different coloured test tubes. Now this one here is the white test tube. White precipitate is formed. This is chloride ions. So we add silver nitrate and if we get a white precipitate that's chloride ion. This is the reaction that happens, the half equation. Ag plus, plus Cl minus forms AgCl. The cream one here is a sign of bromide ions. Cream precipitate formed. The cream precipitate is silver bromide. That's the actual precipitate. And here's the ion equation. Ag plus Br minus forms AgBr. Silver bromide, which is a solid. And the last one, the yellow one, is iodine ions. Now these form a yellow precipitate. And here's the half equation for that as well. The ion equation, sorry, of that. Ag plus plus I minus forms AgI. Silver iodide is your yellow precipitate. Okay, the nitric acid bit that we mentioned here. This is important that we add this because what this does is it reacts with any other anions and that could and any other halides for example for like we're looking at carbonates and that could precipitate out as well and when they react with the silver. Now this could give a false result because obviously we could get silver and silver ions reacting with any let's say like carbonates and we could get precipitates forming and we might think oh we've got a and chloride there but actually we don't because it's just reacted with an impurity so we just basically add the nitric acid just to mop up these kind of rogue carbonates that are floating around in the water naturally um we can also test this further as well um and because these are really difficult to see i mean obviously we can see them side by side here we've got a white screen behind here but it's really difficult to tell in isolation so we can do a further test we add ammonia to each one of these solutions so basically we add um dilute ammonia to each of these and the chloride one should dissolve the precipitate should dissolve with dilute ammonia then obviously we know that's going to be chloride so we put that to one side we should have these two left so then if we add concentrated ammonia to these two that's left the concentrated ammonia your silver bromide should dissolve in concentrated ammonia but your silver iodide shouldn't it's insoluble so bromide you get a cream precipitate dissolves in concentrated Ammonia, an iodide ions yellow precipitate is insoluble so it doesn't dissolve. So really important make sure you know these little reactions and you know the colors and what the test is for and obviously the further test with ammonia as well. Okay just the final little bit really because obviously we're looking at ions here and I thought it would just be good to kind of look at some of these other tests. So for example we can test for group 2 ions and we can use flame tests. Basically we take the solid sample you put on a nichrome wire and then you put it into the Bunsen flame we can spray it on as well. but it's a problem if your if your compounds are insoluble obviously we can't dissolve them in solution so we can't spray them but the idea is that like you say you put it in nichrome wire dip it into hydrochloric acid to just clean the wire dip it into the sample place the loop into a blue bunsen flame because it must be blue and observe the color the colors that you need to know are calcium is dark red strontium is red color just normal red color and barium is a green color okay so as long as you know these these colors here that's quite important Okay more test finds. Ammonium compounds and hydroxides. Basically we can use litmus paper for these. So testing for ammonium compounds. What we do is we add sodium hydroxide to our ammonium compound or suspected ammonium compound. Gently heat it. If ammonium is present you'll produce ammonia gas and all we do is we use a bit of damp litmus paper, red litmus and it'll turn blue if ammonia gas has been produced. Here's the overall reaction look. Ammonium ions plus ammonium. hydroxyl ions will produce ammonia plus water okay and testing for hydroxides and basically these are alkaline and they'll turn red litmus blue which is pretty straightforward but it's really difficult because a lot of things turn red litmus blue it's not necessarily a sign of a hydroxide so you're going to do further tests to effectively discount the other ones and just to confirm that is a hydroxide okay more tests for ions so carbonates and sulfates using hydrochloric acid and barium chloride so this is a little bit of mixture of group two as well um so test for carbonates uh dead easy all we do is we're going to add an acid to a carbonate so hydrochloric acid if we think it's a carbonate if we add it to a carbonate we should get carbon dioxide gas given off um and then obviously carbon dioxide when the gas given off we can test for carbon dioxide bubble it through lime water and it'll turn cloudy okay tested for sulfates We've seen this already in group 2 and we'll just kind of recap this. So basically if we need to find out if something contains sulphite ions, all we do is we add hydrochloric acid. It just removes any carbonates in there. Again, we don't want to see them because they could give a false result. And then all we do is we add barium chloride to the solution. And if there is sulphates present, we should see a white precipitate formed. And this white precipitate is barium sulphate. And obviously, as you know... as you may know barium sulfate is insoluble so that's why we see this precipitate here's the ionic reaction here barium two plus plus sulfur ions forms barium sulfate there's your solid look at your state symbols aqueous aqueous solid that means precipitate is formed okay and um basically the final thing is just make sure you test these things in a particular order just be methodical with your practical technique um and this just prevents any um misconceptions or false positives so The first thing really you should do is test for carbonates first. That'll basically tell you if you have got any carbonates in there, obviously add your acid. If there is carbon dioxide produced, brilliant, you've just identified a carbonate. If there isn't, then you need to test for sulfates. So basically take your sample, add barium sulfate to it. Do you get a white precipitate formed? If you do, fantastic, you've got a sulfate. If you don't, you go on to the final test, and this is basically testing for your halides. You add your silver nitrate solution, remember, and your nitric acid to mop up any rogue carbonate ions that may be floating around. Test for your halides. White, cream, yellow, and then add ammonia. Dilute will dissolve your chlorides. Concentrate will dissolve your bromide, salt, and iodides don't dissolve. So then test for your halides. And basically doing it this way means you can be certain that, or you try and narrow down what compound you've got. And that's it. So that's group seven. hope that was pretty useful it's just a very quick overview really of the of the topic and like I say if you want to purchase these powerpoints that could be really useful for your revision then if you just click on the link in the description box below the video and you'll be able to get them and get them there all right thanks very much bye

This is very specific to AQA. The PowerPoint that I'm going to use here you can use and you can actually purchase from by clicking the link in the description box for this video. So if you just click on that, you'll be able to get a hold of it, and you can use it for revision, and you can use it for whatever you want, print them off or whatever.

So basically, if you like what you see here, then you can get them for yourself. Okay. um also like i say these are tailored specifically to the aqa specification and so obviously the spec points are reflected therefore for this topic okay so um we're going to start with the halogens obviously because this is this is all about group seven um and obviously the halogens are made up of non-metals they're on the right hand side of the periodic table um you On the top of the halogens is fluorine, which is a pale yellow gas, and it has that electron configuration, 1s2, 2s2, 2p5.

Make sure you can give the electron configurations of these. You should be able to do it for the elements such as fluorine and chlorine in particular. Chlorine gas is pale green.

It's Cl2, and there's the electron configuration for it. Bromine is a browny-orange liquid, and iodine is a grey solid. And you can see the electrical configuration there is much, much longer for iodine.

But it's just to kind of give you an idea. Now generally, as we go down the group, starting from fluorine down to iodine, the boiling points increase as we go down the group. It's because the molecules and the size of the atoms and the molecules here, well, these are molecules, and the relative masses of them are getting larger. So therefore, we have larger van der Waals forces.

And so this means that you need a higher amount of energy to turn these into a gas. Now you can see some of these are obviously in different states. The top two are gases.

Bromine is a liquid. Niadine is a solid. So that basically just explains that the van der Waals forces are increasing. And at room temperature, as we go down the group, they become more solid. The electronegativity decreases as well.

When we bond these halogens to, say, like a carbon, for example, or another element in a covalent bond, then the electronegativity decreases. Now, the reason why is because the ability for one of these atoms to attract electrons towards itself weakens. the atoms get larger and the distance between the positive nucleus and the bonding electrons increases and basically we have more shielding so we have more shells between the nucleus and the outer electron so that means it's easier to the well the electronegativity is a lot less for that reason. Okay, right, so let's have a look at some displacement reactions. Now these are pretty important.

You do see a lot of these for halogens and they pretty much explain the reactivity that we've described before. And the fundamental rule is that a more reactive or the most reactive, should I say, more reactive halogens will displace a less reactive halide ion. So basically, like we said before, the reactivity in halogens, they decrease as we go down group seven.

OK, it's for a reaction to occur. Basically, an electron is gained and we form that halide ion. Atoms with a smaller radius, they attract electrons better than the larger ones. So that's quite important.

Basically, halogens are less oxidizing as we go down the group. So remember, oxidation is the loss of electrons. So.

So they become less oxidized as we go down the group. And we can show this by reacting halogens with the halide ions. Okay, so halogens will displace a halide from a solution if the halide is lower in the periodic table. So we're going to look at these two here.

Okay. We're going to look at all of these, in fact, and we're going to look at the reactivity of the halogen and the reactivity of the halide ion. So remember, as we go down the group, the halogens, the reactivity of them decreases.

OK, so let's look at the first one. Now, because these are two chlorines, chlorine water, Cl2, and chlorine, KCl chloride, sorry, which is KCl, potassium chloride, because they're just as reactive, we get no reaction. This one though, we do.

Chlorine is more reactive than a bromide ion. So we do get a displacement reaction. You can see here, there's the chlorine, there's the bromide ion. These react and we get a displacement.

So we get chloride ions and bromine is made. So we get an orange solution being formed here and that orange solution tells us that actually we're making a bromine solution here. Okay, looking at the next one.

Again, chlorine is more reactive than iodine. or than iodide should we say so we do get a displacement reaction now we get a brown solution iodine when it's in solution goes like a brownie color um so this brown solution is made of iodine chlorine again reacts with the iodide ions there it is there we get displacement chloride ions are made an iodine solution is made an iodine solution is brown remember iodine is a solid is is gray and but your iodine solution is brown okay so look at bromine water no reaction here Bromine is less reactive than chloride ions, so we get no reaction. And no reaction for this one as well, because that's bromine with bromide, so they're both just as likely to react.

But the final one, again, we form our brown solution. Again, it's the same reason. Bromine is more reactive than iodide ions, so we get a displacement reaction.

We form 2Br-, and we get iodine produced as well. And obviously this is in solution, so it makes it turn brown. And the last one, it's no reaction here.

Again, iodine is less reactive than chloride, so we get no reaction. Iodine is also less reactive than bromide. And then obviously iodine and iodide are just as equally reactive, so they won't displace either.

So yeah, make sure you know your half equations and you know the colours and you know the reasons why. It's all about this displacement. Right, bleach. Okay, so there wouldn't be halogens without talking about bleach, really, let's be honest.

So... This is made by using a disproportionation reaction. Okay, try to say that one quick.

It's basically a reaction where we're getting oxidation, an element is simultaneously oxidized and reduced, as we'll see in a minute, I'll show you the reaction. But if we mix chlorine and sodium hydroxide together, we'll form sodium chlorate 1 solution, and this is known as bleach. Now the reaction, like I say, As you can see here, it's the sodium hydroxide plus chlorine.

Literally, that's what it is. Sodium chlorate 1 is this stuff here, N-A-C-L-O. So that's that one. Plus you form your salt and you form water as well.

So really important, this is your bleach. That's bleach there. So let's look at this disproportionation thing again.

Chlorine is an oxidation state of 0. It's an element. Chlorine here is plus 1 because oxygen is minus 2. Sodium is plus 1. So for the whole thing to be neutral, chlorine must be plus one. And chlorine here has an oxidation state of minus one because sodium is plus one. So chlorine has been simultaneously reduced and oxidized, like I say.

And we call this a disproportionation reaction. And we can use bleach. Bleach is pretty useful, obviously.

We can use it to treat water. We can bleach paper and fabrics, so we can make paper whiter or fabrics whiter. And obviously we can use it as a cleaning reagent. Obviously you put bleach, use it in bathrooms, cleaning toilets, etc. So yeah, it's got a pretty important use.

Okay, another important use is obviously water sterilization. Now without this, we would have had outbreaks of cholera. And actually in some countries today, where they don't have efficient water sterilization techniques, they do get, unfortunately, do get spreads of cholera, which is a waterborne disease.

But the whole point of adding chlorine to water is to kill the bacteria in the water, including diseases, including microbes, sorry, that cause cholera. So, yeah, adding water to chlorine will produce chlorate ions, ClO-, a little bit like the bleach that we've seen just before. These kill bacteria and they're useful in drinking waters and swimming pools. So let's have a look at these again. Now, this is slightly, slightly different.

Again, we've got water and chlorine, no sodium hydroxide. We're using water instead. And what we form is H plus ions, Cl minus ions, and ClO minus. Now you might also see this as HCl and HClO. So these H pluses can be attached to this.

But because they're soluble and it dissociates pretty strongly, then we normally write them as separate ions. But again, if we look at our oxidation state, chlorine has an oxidation state of zero, obviously minus one here and plus one here. So again, this is a disproportionation reaction.

Look. The chlorine at the start has been simultaneously reduced and oxidised. Okay, now this is quite important, especially in things like swimming pools, because sunlight can decompose chlorinated water too.

And the problem is, is that we get no ClO minus that is made. Remember, it's the ClO minus that actually kills the bacteria. So if we look at these two here, let's have a look. We've got water and we've got chlorine. Now, if we use, if we have sunlight that's present, actually what we produce is 4H plus, Cl minus and O2.

Obviously this can pose a problem in swimming pools because if you get sunlight bleaching into the side of the building, maybe it's where the swimming pool is, the sunlight can actually destroy the bleach that's used in swimming pools, the ClO-. And what we form is Cl-O2. Now obviously because we don't have any ClO- because the sunlight has broken it up, then obviously we don't have any active ingredient to kill the bacteria. So... This is why the chemical reagents or the chemicals that we add to swimming pools have to be replaced quite regularly because sunlight breaks them down into these compounds here, which are obviously different to that bleach up there.

Okay, drinking water. Okay, so advantages and disadvantages. Obviously, it's in the UK and we do chlorinate water, but there are like with anything, adding any chemicals or anything, there's advantages doing it and disadvantages and you need to know what they are.

So the advantages, they destroy microorganisms that cause disease. So that's a pretty good thing. They're long lasting.

So they reduce bacteria buildup further down the supply, which is pretty useful. And reduces the growth of algae that discolours water and give it a bad smell and taste. Last thing you want coming out of your tap is brownish water that doesn't smell too good.

It would put you off pretty much. So yeah, basically the chlorine helps to reduce that. But the disadvantages.

Chlorine is toxic. We can't hide that. and it can irritate the respiratory system so people who add this chemical into the water have to take obviously precautions when adding this in liquid chlorine causes severe chemical burns to the skin so if there was high levels of chlorine in drinking water that could obviously cause problems but there's systems and safeguards from doing that but i suppose the biggest one here is the fact that there is this belief by some people that the um or some scientists that the chlorine could react with organic compounds that present in the water and they make chloroalkanes and these have been linked with causing cancer but obviously the risk of not chlorinating the water and could lead to like like we said before cholera epidemic and cholera is lethal as well so the basically what we're seeing here is the the advantages outweigh the risks or the benefits outweigh the risks in this case The levels of chlorine that are added to water are very, very, very small. And scientists believe that the risk of cancer from this is obviously very low, incredibly low, because of the low levels that are added anyway. They only add just enough just to make sure that the bacteria is killed in the water.

Okay, so, right, let's look at these halides. Now, these are quite difficult, actually, these things. So...

And we'll actually look to halogens so far. We'll look to a little bit of halide ions with the displacement reactions. But halide ions already have that electron already.

Now, these lose that extra electron they've got. And so this makes them really good reducing agents. Okay, so remember, reducing agents are oxidized themselves.

Okay, and oxidation is the loss of electrons. So halide ions lose electrons because they've got that extra one. And so they make some good reducing agents. Okay.

So basically, as we go down the group, the ionic radius increases. And I've just drawn that diagram there. Basically, the distance between the nucleus and the outer electrons becomes larger.

There's more shielding. Use all these words in your exam. And the attractive force gets weaker.

Okay, so it's not as strong. And the outer electron is lost a lot more readily. And this is the reason why I minus is more powerful reducing agent than F minus.

So I minus loses that. that electron much more readily than f minus okay now there's two tests you need to know okay and to prove this to prove this trend so one's with sulfuric acid which we'll look at next and the other one is with silver nitrate solution and you've got to know the results of both of them and i think the most difficult one is the one with sulfuric acid because you get a lot of equations here so let's just talk it through make sure you know what's going on okay um So some halide ions can reduce concentrated sulfuric acid. So that must be concentrated. That's very important.

So what I'm going to show you here is basically a little kind of, it looks a bit like a train map, I like to call it. I think it's probably easier maybe seeing these things in this way rather than in other ways that I've seen in some notes and sheets and books, etc. I think this probably is an easier way of doing it. But you can use it however you want. Basically what I've done is I've written my reduction products of sulfuric acid on the top row.

The only one which isn't a reduction product is this one, sodium hydrogen sulfate. But these ones are reduction products and I'll come on to these in a minute. And then what I've done is I've looked at sulfur.

Okay, because sulfur is the key element here. And I've written the oxidation state of sulfur underneath. So we've got plus six here, plus four, zero and minus two.

And what I've done is I've written our halide ions down here. These are the halide ions that we're going to be using. We're going to react it with concentrated sulfuric acid.

So we're going to react the chloride ion with concentrated sulfuric acid to see what products we produce. We're going to react to bromide ion with the same stuff, see what products we produce. We're going to react to iodide ion, see what products we produce.

So let's start with the first one. Here's our underground train line thing. So we're going to look at A. Now the good thing is that chlorides, bromides and iodides, they all produce this compound here.

Sodium hydrogen sulfate. and we need to know the reagents of these as well the reactions of these so this is point a now point a this is not a redox reaction okay this is just a standard chemical reaction between a salt and an acid so here we are this is where sodium chloride sulfuric acid react with sodium chloride all we get is one of the hydrogens is given up um to form hcl you get white misty fumes here and you form sodium hydrogen sulfate You also get the same with sodium bromide and sodium iodide as well. However, as you will see, that these reagents here, these sodium hydrogen sulfates, will then go on to react further. So actually, you have very little of these anyway in the overall reaction. But certainly with the chloride, the only product you get with sodium chloride is that.

And that's it. You can see the arrow stops there. No more product produced.

Okay, so now we're going to bring our attention on to bromide. Okay, so bromide is... let's have a look okay so these are going to be looking at reaction b so bromides can actually reduce further the sulfur further they can reduce it to produce sulfur dioxide and iodides can do the same as well okay so they could produce sulfur dioxide so let's have a look at their reactions so this is point b okay you can see um in addition to a as well okay so we've got obviously all the reactions as well um but with sodium bromide because that's what we're going to look at mainly and we get the same reaction with sodium iodide as well what we do is we write down our half equations first okay and the bromide is being converted into bromine i've written the half equation there so the bromide ions are being oxidized br minus going to br2 and obviously we balance it with two electrons okay this has been oxidized and i must stress as well if you don't know how to work out your half equations you really do need to know um how to do this i have put video on that describes how to balance half equations you've got to know that as well I'm just going to go through it there's an overview here assuming that you know how to do it but and this is the bromide ion and then crucially this is this bit here we start with sulfuric acid and we're going to produce sulfur dioxide because that's one of our products what we do is we balance with water first then protons then electrons okay so we just balance this half equation this is the sulfur being reduced okay so this is a redox reaction because we've got oxidation reduction cancel out all electrons and combine the equations and we should get an overall anic equation which is this here so you can see the sulfuric acid has been reduced to sulfur dioxide we should see in this reaction an orange vapor of bromine being produced and that tells us that actually this reaction has occurred okay um so bromine tells us that we have produced sulfur we produce bromine and we've also produced sulfur dioxide as well which is a choking gas okay Let's look go further. So bromides are not as powerful reducing agents so they can't reduce that any further. But iodides can.

Okay so they keep going. So iodides can reduce sulfur even further to produce sulfur which is this sulfur solid. Again we're going to do the same thing.

Look balance it out. 6 I minus I2. So we're going to put I minus go into I2 and then we basically try to balance that out.

We do the same here. Start with sulfuric acid forming sulfur. And then we balance with water, protons and electrons.

When we balance this, we effectively worked out that we have to multiply the top one by three, so we can balance out the electrons, the six electrons. Remember, we've got to try and balance these out, cancel them out, rewrite out the equation. The overall equation here is obviously I minus going to I2, and sulfuric acid going to sulfur, so producing solid sulfur. And iodine, iodide ions are so powerful at reducing, they reduce the sulfur even further, and they produce H2S, okay?

I should point out with this one as well. you produce sulfur solid, which is yellow. So this one, the final step is step D. Again, same thing. Look, we're just going from I minus to I2, just bouncing out.

Sulfuric acid this time, we're forming H2S, hydrogen sulfide, we call it, which is this one here. And the overall ionic equation, obviously, just like before. Again, we're going to multiply this top one by four to make sure we get eight electrons on both sides.

Classic one here, rotten egg smell. h2s is a nerve agent it's really toxic you shouldn't be breathing this in and it smells of rotten egg it's got a horrible smell it's also known as sewer gas um because it's found commonly in sewers so you probably get an idea of what uh what that what they may smell like okay so that's your reaction with hyaluronides with sulfuric acid make sure you can balance your half equations really redox is crucial here being able to balance them okay The other one, which is probably a little bit safer to use, let's be honest, is reacting it with silver nitrate. So we can use these halides, add silver nitrate, and we can confirm further with ammonia, which we'll look at later on. So this is how we do it. We test for chlorides, bromides, and iodides.

All we do is we add nitric acid first, dilute nitric acid. Then we add silver nitrate solution, AgNO3. The colour of the precipitate will help you identify the halide ion. So I'll come on to the nitric acid a little bit later on as to why we add that.

But you can see here we've got the test tubes. We've got different coloured test tubes. Now this one here is the white test tube. White precipitate is formed. This is chloride ions.

So we add silver nitrate and if we get a white precipitate that's chloride ion. This is the reaction that happens, the half equation. Ag plus, plus Cl minus forms AgCl.

The cream one here is a sign of bromide ions. Cream precipitate formed. The cream precipitate is silver bromide.

That's the actual precipitate. And here's the ion equation. Ag plus Br minus forms AgBr. Silver bromide, which is a solid. And the last one, the yellow one, is iodine ions.

Now these form a yellow precipitate. And here's the half equation for that as well. The ion equation, sorry, of that. Ag plus plus I minus forms AgI.

Silver iodide is your yellow precipitate. Okay, the nitric acid bit that we mentioned here. This is important that we add this because what this does is it reacts with any other anions and that could and any other halides for example for like we're looking at carbonates and that could precipitate out as well and when they react with the silver.

Now this could give a false result because obviously we could get silver and silver ions reacting with any let's say like carbonates and we could get precipitates forming and we might think oh we've got a and chloride there but actually we don't because it's just reacted with an impurity so we just basically add the nitric acid just to mop up these kind of rogue carbonates that are floating around in the water naturally um we can also test this further as well um and because these are really difficult to see i mean obviously we can see them side by side here we've got a white screen behind here but it's really difficult to tell in isolation so we can do a further test we add ammonia to each one of these solutions so basically we add um dilute ammonia to each of these and the chloride one should dissolve the precipitate should dissolve with dilute ammonia then obviously we know that's going to be chloride so we put that to one side we should have these two left so then if we add concentrated ammonia to these two that's left the concentrated ammonia your silver bromide should dissolve in concentrated ammonia but your silver iodide shouldn't it's insoluble so bromide you get a cream precipitate dissolves in concentrated Ammonia, an iodide ions yellow precipitate is insoluble so it doesn't dissolve. So really important make sure you know these little reactions and you know the colors and what the test is for and obviously the further test with ammonia as well. Okay just the final little bit really because obviously we're looking at ions here and I thought it would just be good to kind of look at some of these other tests. So for example we can test for group 2 ions and we can use flame tests.

Basically we take the solid sample you put on a nichrome wire and then you put it into the Bunsen flame we can spray it on as well. but it's a problem if your if your compounds are insoluble obviously we can't dissolve them in solution so we can't spray them but the idea is that like you say you put it in nichrome wire dip it into hydrochloric acid to just clean the wire dip it into the sample place the loop into a blue bunsen flame because it must be blue and observe the color the colors that you need to know are calcium is dark red strontium is red color just normal red color and barium is a green color okay so as long as you know these these colors here that's quite important Okay more test finds. Ammonium compounds and hydroxides. Basically we can use litmus paper for these. So testing for ammonium compounds.

What we do is we add sodium hydroxide to our ammonium compound or suspected ammonium compound. Gently heat it. If ammonium is present you'll produce ammonia gas and all we do is we use a bit of damp litmus paper, red litmus and it'll turn blue if ammonia gas has been produced.

Here's the overall reaction look. Ammonium ions plus ammonium. hydroxyl ions will produce ammonia plus water okay and testing for hydroxides and basically these are alkaline and they'll turn red litmus blue which is pretty straightforward but it's really difficult because a lot of things turn red litmus blue it's not necessarily a sign of a hydroxide so you're going to do further tests to effectively discount the other ones and just to confirm that is a hydroxide okay more tests for ions so carbonates and sulfates using hydrochloric acid and barium chloride so this is a little bit of mixture of group two as well um so test for carbonates uh dead easy all we do is we're going to add an acid to a carbonate so hydrochloric acid if we think it's a carbonate if we add it to a carbonate we should get carbon dioxide gas given off um and then obviously carbon dioxide when the gas given off we can test for carbon dioxide bubble it through lime water and it'll turn cloudy okay tested for sulfates We've seen this already in group 2 and we'll just kind of recap this. So basically if we need to find out if something contains sulphite ions, all we do is we add hydrochloric acid.

It just removes any carbonates in there. Again, we don't want to see them because they could give a false result. And then all we do is we add barium chloride to the solution. And if there is sulphates present, we should see a white precipitate formed. And this white precipitate is barium sulphate.

And obviously, as you know... as you may know barium sulfate is insoluble so that's why we see this precipitate here's the ionic reaction here barium two plus plus sulfur ions forms barium sulfate there's your solid look at your state symbols aqueous aqueous solid that means precipitate is formed okay and um basically the final thing is just make sure you test these things in a particular order just be methodical with your practical technique um and this just prevents any um misconceptions or false positives so The first thing really you should do is test for carbonates first. That'll basically tell you if you have got any carbonates in there, obviously add your acid. If there is carbon dioxide produced, brilliant, you've just identified a carbonate. If there isn't, then you need to test for sulfates.

So basically take your sample, add barium sulfate to it. Do you get a white precipitate formed? If you do, fantastic, you've got a sulfate.

If you don't, you go on to the final test, and this is basically testing for your halides. You add your silver nitrate solution, remember, and your nitric acid to mop up any rogue carbonate ions that may be floating around. Test for your halides.

White, cream, yellow, and then add ammonia. Dilute will dissolve your chlorides. Concentrate will dissolve your bromide, salt, and iodides don't dissolve. So then test for your halides. And basically doing it this way means you can be certain that, or you try and narrow down what compound you've got.

And that's it. So that's group seven. hope that was pretty useful it's just a very quick overview really of the of the topic and like I say if you want to purchase these powerpoints that could be really useful for your revision then if you just click on the link in the description box below the video and you'll be able to get them and get them there all right thanks very much bye

Transcript for:Overview of Group 7 Halogens

Transcript for:
Overview of Group 7 Halogens