Transcript for:
Chemistry of Transition Elements

[Music] hello my name is chris harris and i'm from alley chemistry and welcome to this topic on chemistry of transition elements so this is topic 28 for the cie specifications this is the cambridge internationals so if you are studying the cie specification then this is you well you've come to the right place basically um so obviously this is looking just at topic 28 in particular obviously with cie there's loads of different topics spanning from year one to year two and the full range of videos for year one and year two are available on alley chemistry youtube channel um hit the subscribe button just to show your support for this project that'll be absolutely brilliant plus any of this um any of this information here so these are just slides they are available to purchase which are great for your revision so you can put them in your file and write notes all of them if you wish um they are available to purchase if you click on the link in the description box below you'll be able to get a hold of them there like i said they're great value for money and i've kind of bundled all of these up into kind of bigger um kind of bundles of topics so you can you can get the full year one and year two specification if you wish so this is quite a big topic this is probably one of the biggest ones um in cie um so there's a lot of information here basically it's kind of the the main the main and only topic that looks into transition elements so some of this might be fairly new but i'm gonna bridge quite a lot of this with um some previous topics as well but it is important that you have a good understanding of orbitals um and um electrons obviously moving into different orbitals as well and obviously that you would have learned in year one chemistry so i'm going to be bringing some of this stuff back in for this here so the best place to start with this topic is d block elements so um obviously we need to know where these d block elements are in the periodic table and we can see we've got the periodic table that's up in front of us there and the d block elements actually sit in the middle of the periodic table so that's this kind of block here and to be honest a lot of the chemistry you've been looking at to be honest is mainly being kind of to the left and right of this block i don't really do a lot with it but it actually has its own topic where we just look at these chemicals here some of these elements in the middle so some of these transition elements uh some of these elements sorry are transition metals not all of them okay so the ones that we're really gonna look look at here are the ones in the top row we're not really gonna look too much at the ones below that so this is um scandium titanium vanadium chromium manganese iron cobalt nickel copper and zinc so that's these ones at the on the top row there so let's have a look at some electron configurations so a transition element is a d block element that can form at least one stable ion with a partially filled or incomplete d sub shell okay don't get the mixed up there's difference between d element d block elements and transition elements okay so transition element is a type of d block element so the d sub shell as you'll probably remember from year one chemistry can hold up to 10 electrons in total and for period four d block elements only eight of these are transition elements so scandium and zinc which are at the extreme ends of that kind of top row of the transition elements of the d block element sorry are not transition elements and this is because they don't form stab lines um with a partially filled d sub shell so we'll kind of have a look at that a little bit later on so let's have a look at the electron configurations of the eight transition elements in period four um and what you'll see here and this is the again you would have seen this in year one but we fill up these orbitals singly first and then we double up after that okay remember these electrons don't want to be um sitting in the same orbital as each other they want to kind of be as far apart as they can be and they will only occupy doubly they will only doubly occupy if they have to there is no other option for them so let's have a look so um all of these have the electric configuration of argon okay so that's what that means they're all an ar with a square bracket you might see that sometimes with transition metals instead of writing s two two p six and writing all that stuff before we just say right the electric configuration of argon plus what we're gonna show down here okay so there we are okay so there we are that's the electric configuration of argon so titanium is the first one so it has um 4s2 and it has 2 in the 3d um orbitals so this is partially filled 3d orbital so it's definitely a transition element vanadium has three chromium has five now what happens here is one of the electrons from here jumps into this one to occupy a full half shell um this is more stable than keeping two paired up in here remember these electrons want an opportunity if they can to occupy an orbital singly as long as the overall energy is is favorable and then with manganese the extra electron then goes in here okay into the 4s it's more energetically favorable to double up in this orbital before we fill up here iron we then have to double up we don't have a choice cobalt and nickel they obviously double up there and then again for the last one here copper we have a full d sub shell but we have a partially filled s subshell so the electron from here then jumps into here this is more stable to have a full d sub shell here um and this one is obviously left unoccupied here okay so chromium and copper so these behave slightly differently as you can see so an electron from the 4s orbital moves into the 3d orbital to create a more stable half full or full 3d subshell respectively so that's what we're talking about there so the electron here went into there and then this electron here jumped into this orbital here okay so so that's what happened so basically just remember these two in particular chromium and copper okay which is um which is down here okay so just remember these two in particular they're orbitals when you're filling the electrons up um they the atom or the element should i say um has this full or half 3d subshell and a full 3d subshell for copper so just make sure you wear that do not write that otherwise you are going to get that incorrect you can get that wrong okay so scandium and zinc as we kind of isolated these before these are not transition elements and we need to understand why okay so scandium only forms one stable ion okay and that's scandium three plus sc3 plus now that scandium three plus has an empty d sub shell and because it's empty it's not partially filled it's not a transition element so there's scandium as an element okay now remember you're looking at thinking about it's got a partially filled d sub shell remember what makes something a transition element is the stable ion that it forms it must form a stable line at least one stable iron that has a partially filled d subshell this is not an iron this is an element and the only sterile line that does form is scandium three plus and you can see here there is no electrons at all in the 3d subshell now it's the same with zinc zinc only forms one stable line which is zinc two plus and this has a full d sub shell and so because it's not partially filled it is not a transition element so here's zinc as a metal and here's zinc as an iron now when we remove electrons as you'll see in a moment we actually remove from the 4s orbital first then we start dipping into the 3d so um this obviously lost its two electrons in the s orbital it's got a full 3d subshell um and so therefore there is um no um well it's not a transition element you know this is not partially filled so that's the reasons why okay so let's kind of carry on looking at electron configuration of transition metals so transistor metals they lose electrons in a very specific way as i kind of just hinted at just before so let's look at the electron configuration for fe3 plus first so what it actually does is it loses three electrons two from the 4s and one from the 3d okay now all becomes quite apparent and we can see this here so you can see there's the electron configuration there okay which has got 4 s2 and then what's going to happen is we're going to remove the electrons to form fe3 plus so there we go that goes and then if you see it up here that should disappear it should do no not yet it will do in a moment so um you'll see here this is the electro configuration so the four s electrons have been removed then we remove the third electron from the d orbital and obviously that should give 23 which matches and there it is so we lose from the 4s first and then the 3d there we are and then the 3d there so that's what we should have is this this kind of setup here okay so always when you're removing electrons from a transition metal you remove from the 4s first then you start dipping into the 3d and you remove them singly first so you try and leave singly occupied orbitals if you can okay right so let's have a look at some properties of transition metals first okay so um transition metals have very specific properties um they include variable oxidation states they form colored iron solutions um they're good catalysts as well and they can actually and form complexes too okay so you've got various different properties here now what we're going to do a lot of this topic is exploring these properties and looking at the different reactions and and what they can do because actually of all the areas of chemistry although it's not kind of the biggest organic chemistry is probably the biggest area of chemistry in total but um although inorganic chemistry which this kind of sits into is not a big area um it's probably the nicest one because it's colorful there's loads of colors there um you know the the metals that we see there are used widely um in terms of jewelry in terms of catalysts in terms of you know the paints that we see the colors that we see for example all of these are absolutely vital and obviously we have transition metals in our blood iron for example um and because of iron's transition metal it allows us to survive so they are obviously any area of chemistry is vital of course we're going to say that but this is it's it's quite a and quite a specialized area of chemistry i would say so transition metals so transition metals have variable oxidation states that's one of the properties of them and throughout this video we're going to look at all of them but variable oxidation states and this is because the electrons sit in that 4s and 3d energy levels and as we've seen the diagram just before um these are very close they sit very close to each other and because their energy levels are quite close the electrons can kind of jump from 4s to 3d quite readily as we've just seen before when we're removing the electrons so as a result um electrons are gained and lost with a pretty much similar amount of energy and when they form these ions and obviously we can see here we've got all of the elements kind of listed in this table here and the ions that they form and you can see obviously scandium only forms one iron and which you can see here so obviously we've seen scandium that isn't a transition element remember i've just put it on there just for reference really um but some of them have quite a few so vanadium chromium manganese iron cobalt nickel copper and titanium now you can see here these are the oxidation states along the left hand side here you would have seen that in the electrochemistry topic in year one when you looked at oxidation states if you're not familiar with that please go back and have a look at that it's really important but transitional metals can have varying oxidation states and that makes them quite quite versatile so what we can do is we can predict the maximum oxidation number for a transition element um or we can predict the maximum oxidation number for transition element it's basically the number of 4s electrons plus the number of unpaired 3d electrons and this pattern is only broken by by copper so that's a way in which we can try and predict what they are um and obviously they also form colored irons in solution and we'll look at some of these colors as well there's loads of different colors here um that that you may have seen so you might recognize some of these like say from practical work so um vanadium two plus is a lovely violet color vanadium three plus is green vanadium um some vanadium four or v or two plus is blue v or two plus is yellow so that's panadium five plus chromium is um is green or violet color so it's violet when it's surrounded by six water ligands and we'll look at ligands later on um but they're normally substituted so it looks they look actually green when you do it dichromate is orange again you might have seen these in the um alcohols topics the hydroxy group topic uh manganese mn2 plus mno42 minus is green and mno4 minus is purple ion 2 plus is pale green and three plus is yellow so you might have seen this obviously within um you know when you've done some reactions cobalt two plus is a lovely pink color nickel is green copper is um is blue so copper two plus is a blue color titanium three plus it's purple and titanium two pluses virus so you might have seen quite a few of these colors dotted around um there's some specific ones that really you should be probably more aware of than others and the chromium ones obviously you should you would have seen through like i said looking at oxidation and you need to be aware of um the cobalt um see cobalt ones that's obviously going to be quite important nickel is going to be quite important as well um obviously being aware of that um copper um is going to be quite important um so these are some of the colors that you're probably going to see a little bit more a little bit more often um so these are the ones here some of these you might not see as much like your vanadiums you might not see as much um and iron you might see a little bit of so just be aware of some of these colors here and just try and remember them we will go through some um some of the equations um here anyway that um you'll probably see uh quite a quite a bit more okay so we need to be able to compare the properties of s block element calcium with transition elements okay in particular so if we're looking at the properties we can see that the melting and boiling points of transition elements are similar to each other however calcium doesn't um doesn't really fit this trend i suppose so we can see obviously that the ionic radius for calcium two plus um let's have a look there we are okay so there's your calcium two plus there um is higher so the radius is much higher than it is for other transition elements with the same charge okay so if we kind of scan scan across the side there and this gives rise to obviously some of these properties and some of these properties as well so just make sure that obviously if you're looking at calcium look at the melting point of calcium it's actually a bit lower when you get into transition metals obviously these can form metallic bonding and all the way across here so form some quite big complexes the melting point is there and they're a little bit more dense as well so they've got a higher density they can pack together quite um a bit more so it's just really looking at um you know the properties and just being aware that obviously the radius of say group two so your calcium for example the ionic radius is going to be a good bit bigger than it is for um transition metals which uh generally tend to pull the electrons a bit closer towards them mainly because you've got an increased number of positive charge so the number of protons in the nucleus increases and so therefore that brings that radius in so it's just a brief uh brief look at that comparing transition metals with an s block element okay so titrations are really important you would have probably done quite a few of these practically um now transition metals do appear quite often when you're doing a titration in fact you might have done them you might have done manganese for example um you might have done dichromate reactions so titrations and transition metals really go together quite quite nicely so they can be used to work out the concentration of a reducing or oxidizing agent and in this example we're going to look at trying to find the concentration of mno4 minus these are manganate ions by titrating against a reducing agent a little bit like fe2 plus so here we're just going to look at find the concentration of a reducing agent obviously if we're going to work out for an oxidizing agent we just reverse the you know the chemicals what's in the buret and what's in the conical flask so in this example we have a reducing agent this for example is fe2 plus um we have that in an unknown concentration but a known volume so we don't know what the strength is of it and but we don't we do know how much we're going to put in here so for example it could be 50 centimeters cubed or 25 centimeters cubed something like that will go into the conical flask but we don't know how much is in there we don't know the strength of it so we can add obviously um excess dilute sulfuric acid into this as well and this is to ensure you have a sufficient amount of h plus signs to allow this reaction to happen and you'll see when we put our equation up um in a moment um it will be on the next slide where um you know why that's important so we have an oxidizing agent in the buret with this has a known concentration so we know how strong this is and this could be for example manganate ion so mno4 minus which is a good oxidizing agent so then what we're going to do is we're going to add the manganate ions into the buret and or in the buret we're going to add it into the conical flask and we're going to keep adding it until the faint color of manganate appears and this is known as the endpoint so in other words when you add this in let's bring this mouse back in so when you add this in this will be colorless and this will be a a nice um it's like a purpley color deep purple color so um when you add this in this will go colorless you shake this and i'll go colorless as soon as that as soon as you get a pale faint color starting to appear and it's permanent even if you shake it there's a permanent kind of color that exists there then you stop titrating and you measure the amount of manganate that you've added to this conical flask and this is the end point now practically when you do this you add it very slowly as you're getting close to it drop by drop you don't want to go over so um there's a sharp color change okay so your manganate ions um so the you've got a sharp color change basically so your manganate ions from the aqueous potassium permanganate solution are purple so lovely lovely kind of deep purple color they'll immediately react with the reducing agent until all the reducing agent is used up so you have to be really slowly adding it drop by drop um otherwise it will basically you'll kind of add too much into it so one drop at the end point can turn that solution purple okay and that's what you don't want if it goes and if it goes the same color basically if this flask is the same color as what's in here you've probably gone too far okay so it should be a nice pale color so what we could use is a colorless oxidizing agent and a colored reducing agent uh and obviously what we're doing is just looking for that color to disappear in here so it could be the other way around it's just what we're looking for with titrations is a color change mainly so we're going to read how much oxidizing agent was added remember we're going to read the bottom of the meniscus so the meniscus is like the little bubble at the bottom of the buret we've always got to read at the bottom of that and obviously always read eye level um now we're going to record our results in two decimal places okay always and repeat until you get two results that are concordant with each other okay so when we say concordant it means they kind of they're within point one centimeters cubed of each other as soon as you've got at least two results that are within point one centimeters cubed of each other then we can stop um and essentially we have um you know we can be pretty confident that the value that we've got there um it can be used to help work out um the concentration of iron two plus that's in your conical flask because essentially that's what we want to work out isn't it one know how much we've got in there okay so like i said redox titrations we can use them to work out the concentration of a reagent so let's put some figures to this then okay so we had we we ran the titration we've got some concordant results and we can say that 18.3 centimeters cubes there we are of 0.0250 mole per dam cubed potassium manganate that's in your buret that reacted with centimeters cubed of acidified iron ii sulfate that's in the conical flask here and we need to calculate the concentration of iron 2 plus ions so here is the equation so we've got kmno4 in here this is the lovely purple color and we've got iron sulfate in here which is like very pale green color so iron two sulfates like almost colorless to be honest so here's the the first thing we need to do is write out an equation and balance it so here we've got mno4 minus obviously going to mn2 plus and we've got fe2 plus going to fe3 plus um and obviously we then have to combine these and balance them with water and protons etcetera now you need to be familiar with this bit here and writing out half equations this you would have seen in um when we looked at redox reactions so electrochemistry i think in year one the electrochemistry topic and here one when we look at how you create half equations combine them to form a full ionic equation so you're not expected to remember this whole equation here we are expected to combine obviously the manganese half equation with the iron half equation and then obviously balance it with water and protons to form this so that's the first step though is to write up your balanced equation now they might already give you it but you need to do that so then we need to calculate the number of moles of manganate so mno4 minus so moles remember from year one concentration times volume okay so the number of moles obviously put your numbers in there so it's 0.0250 this is the concentration of manganate ions okay because that's what we've got there multiplied by the volume we added 18.3 centimeters cubed remember okay we've got to convert that to decimeters cubed first okay so we divide that number by a thousand what i've done i've put times ten to the minus three um and that means exactly the same as divided by thousands and that gives us four point five eight times by ten to the minus four moles okay so that's the number of moles of manganate that was used that was added in here then what we're going to do is use the equation to find out the molar ratio in order to work out the number of moles of fe2 plus so we know the number of moles here okay and we can see we've got a one to five ratio there it is uh we've got one to five ratio of mno4 minus signs to fe two plus so therefore the moles of fe two plus is going to be the moles of that which is your magnet multiplied by five and that means we have two point two nine times by ten to the minus three moles of fe two plus signs and this is what we had in here then we need to calculate the concentration of fe2 plus ions and again from year one concentration is moles divided by volume so the number of moles is two point two nine times ten to the minus three that's the number of moles of iron divided by the volume which is 25 centimeters cubed that's how much we had in there times by ten to the minus three gotta convert that to decimeters cubed and that gets us 0.092 moles per test meters cubed so this is basically using the titration to work out the strength of this solution here we can use that by reacting it with something else using molar ratios and obviously work out the moles there so they can also be used to work out the percentage of iron in iron tablets for example so let's have a look at an example here so we've got calculate the percentage of iron in the tablet with a mass of two and a half grams which was dissolved in dilute sulfuric acid to give a 250 centimeters cubed solution 25 centimeters cubed portion of that solution reacted with 12 and a half centimeters cubed of 0.025 molar potassium magnet solution okay so let's kind of take this strip this back a little bit okay so we want to work out how much iron is in the tablet so iron tablets don't just contain pure iron obviously um it's an iron compound and so there's other substances in that tablet um but the tablet itself so the whole thing with two and a half grams we basically dissolved that in sulfuric acid to allow it to because it just dissolves easier and um and obviously we need the h plus signs as well and we made a massive vast of this stuff okay so we've got a massive valley of acid with the dissolved tablet 250 centimeters cubed we then poured some of this big vat of solution out into a 25 or used a pure attribute took a 25 centimeter portion of this bigger solution put that in my conical flask and then um this reacted we obviously ran a burette around a titration and we needed 12 and a half centimeters cubed of your manganate which is in your buret of this strength to neutralize it basically or to react with all the ions that's in there so the first thing we need to do is write up the equation and balance it okay so in this case there's your equations you've got manganate with the fe2 plus okay the next thing as we've just seen before we calculate the number of moles of manganate ions so moles is the concentration times by volume so the moles is going to work out at 3.125 times by 10 to the minus 4 moles okay it's the total number of moles so remember this is the strength that's what we've been given there's the concentration times by the volume that we needed that's 12 and a half but we've got to convert that remember to um uh decimeters cubed okay so we'd be giving it centimeters cubed we've got to divide it by a thousand so then we're going to use the equation to find out um and the molar ratio to work out how many iron two plus ions we had in our 25 centimeter cube portion so that's in your flask remember below so we know we have a one to five ratio okay so then we can then obviously multiply the number of moles that worked out here by five and that means we have one point five six two or one point five six two five times a ten to the minus three moles in 25 centimeters cubed okay now we've just what we've worked out here is just a portion of the bigger solution remember we wanted to work out how much iron was in the tablet and that tablet was added to a massive fat of solution so we can't just use this to work out how much iron we've got in that tablet because you've only got a small percentage of that so what we need to do is we need to work out the number of moles of iron two plus in the um you know in the tablet and by for the whole iron tablet and we need to know how many moles we had in the original solution that we took from to titrate again it's obviously not going to titrate against a quarter liter of liquid you need the biggest buret in the world so that's why we're just taking a small sample but we do the maths and we scale that up to the original amount that we had there so naturally we're going to multiply that by 10 so to work out how much we have in 250 centimeters cubed which is our initial solution so we're going to take the number of balls multiply it by 10 and we get 1.5625 times by 10 to the minus 2 moles so that's how much iron was in the 250 centimeter solution now what we can do is we can calculate the mass of iron in the tablet that was added to this big solution so we do mass equals moles times by the relative atomic mass so here we know the mass of fe in the tablet or the mass of fe in tablet sorry is one mole of iron weighs 55.8 grams so you use your periodic table to find that out so one tablet contains 1.5625 times 10 to the minus 2 moles okay times by the mr and that's going to be 0.871 grams of iron okay so that's how much hinds in there and then what we have to do obviously we don't stop there because it's asked for a percentage of iron in the tablet now what we could do is calculate the percentage of iron in this tablet so the overall mass of the tablet was two and a half grams 0.871 grams of that contained iron which obviously in the iron tablet so all we do is we take the amount of iron divided by the total mass times by 100 and the actual tablet contains 34 34.8 percent iron in that whole tablet quite a lot there isn't it but hopefully you can see the logic that's what i'm trying to get at is you look at it's methodical you can you know he's trying to justify each step why are we working that out what's the purpose of that and hopefully by talking through it and justifying each of the different parts it becomes a bit clearer okay so we're going to look at a different titration reaction now we can apply this same kind of methodology or the calculation to some other ions as well um now this is actually just another form of catalysis which we've seen um in the um i think it was topic uh 26 um where why we looked at catalysis and we looked a bit in year one as well um so actually this is just an example of a a very weird reaction that actually catalyzes itself what called autocatalysis but anyway so aside from that calculation um you need to know the reaction okay so mn2 plus here um is the catalyst in a reaction and this is going to be between c2042 minus and um so this is your ethereum dioit iron that's what it's called and a manganate iron which is on obviously on that side there so remember catalyst is always reformed it's never used up obviously we've seen this obviously in the previous topic as well so mn2 plus is a product and a catalyst and this means that as the reaction proceeds the amount of product increases and so does the rate of reaction so it gets faster and faster and faster and faster it's a bit of a strange reaction isn't it so the reaction um uncatalyzed is really slow um and the reason why is we're actually trying to react two negative ions together so you've got the h4c ethan diewit ion which is two minus with the manganate ion which has a negative energy at negative energy negative charge so this means um it requires loads of energy to actually force these together they're just not interested you know the two negative charges they've already been near each other so there's a really high activation energy so we need a catalyst to help kind of help this reaction along and so the first step in this reaction is the mn2 plus okay this catalyzes the reaction and it converts mno4 minus first into mn3 plus so the mno4 here reacts with the mn2 plus okay which is the catalyst which is there okay um so that reacts with that and what it does is it reduces the mno4 okay which is this over here to mn3 plus okay so you go yeah mn3 plus there and obviously you produce some water and then secondly the mn3 plus formed which is this year then it reacts with your ethane dioit iron which is we'll look at this a little bit more detail you'll see it as a ligand later um but this reacts with the mn2 plus that we've um that obviously we have in the in the solution um and sorry the mn2 plus signs are reformed sorry so this mn3 plus reacts with the ethernet weight mn2 plus is made back again that's how we know it's a catalyst because we start with it there and it's reformed and we produce carbon dioxide so this reaction has actually produced um mn3 plus and there we are no attached produced mn2 plus there we are there's the m2 plus there and it's produced your carbon dioxide and it's produced your water so this reaction these this reaction here has actually worked because the m and two pluses effectively kind of split them up into two different kind of steps here okay so we need to know another type of reaction and there's loads of different actions here isn't there so this is a copper percentage titration okay so this uses iodide ions as we've as we've seen before anyway in previous topics but copper ii ions oxidize iodides to iodine okay now this is particularly useful because it allows us to find the percentage of copper in an alloy such as brass so if we've got some you know like a piece of brass for example we want to work out how much actual copper is in there we can run a titration and and try and work out what we have so this one's a multi-step reaction so step one is we use an oxidate oxidizing agent to oxidize as much of the iodine as possible so the first thing is we need to dissolve a known mass of the alloy in some concentrated nitric acid so we dissolve it pour it into a 250 centimeters cube conical flask so it's quite a big flask make it up to 250 degrees cubed with de-ionized water okay and then we're going to take 25 centimeters cubed of the solution from that flask and we're going to put it into a conical flask we're going to add some sodium carbonate solution to neutralize any of the nitric acid that we added initially and we're going to keep adding until we just start to see a precipitate form okay we're then going to add several drops of ethanoic acid and to to remove this precipitate so we're trying to remove that acid um from the solution there so we're just trying to clean it up a little bit then we're going to add excess acidified potassium iodide solution which will react with the copper iron so that's going to react with the copper ions that's in there and here's the reaction so you've got your copper two plus ions now this was actually extracted from the brass as you can see from here so this is the step here react it with um your k ice potassium iodide that's going to provide your iodine and you're going to form copper iodide and iodine okay so the i minor signs are oxidized to iodine and your copper two plus is reduced to copper one plus and we get this white precipitate of copper iodide that's actually formed okay so that's the first step so we're using an oxidizing agent and we're going to try and oxidize as much of the iodide as possible okay so we're going to get rid of as much of that as we can okay so the next step so step two is we then need to work out the number of moles of iodine that is actually being produced okay because this remember the number of moles of iodine will help us to work out how much copper we had initially so we're going to take the product mixture from step one okay and we're going to titrate this against sodium thiosulfate solution and that's going to use to find the number of moles of iodine present so remember the number of moles of iodine was of the the amount of iodine is heavily dependent well it is it's dependent on the amount of copper that we had initially because remember the iodine was only formed through um the oxidation of iodides which in themselves reacted with the copper ions okay i hope you're following this so far so there's quite a few steps here so then the step three is we're going to calculate the concentration of the oxidizing agent okay so what we're going to do is work out the number of moles of copper in both the 25 centimeters cubed so that was in the bit that we've actually titrated and the original massive vat of 250 centimeters cubed of solution okay so remember to take into account the molar ratio um two moles of copper ions and one mole of iodine okay so remember that okay now we know the concentration in 250 centimeters cubed and we can then um oh now we know the concentration 250 seconds cubed we can calculate the mass of copper in the original alloy sample and then the last step is to calculate the percentage copper in the alloy completely okay so let's have a look okay oops there we are so what we're trying to do here is look at a methodology a way in which we can work this out so remember from this reaction we're trying to work out the amount of iodine that was produced we then titrate that with the sodium thiosulfate and then from that we can work out the number of moles of iodine that we had present which we can then work out the amount of copper and then scale that back to the 250 centimeter cubed solution so it's a very similar reaction in fact the methodology is very similar to the i calculate the amount of iron in the iron tablet where you're working out 25 centimeters cubed use the molar ratio to work out well it's 25 centimeters cubed times it by 10 to get 250 centimeters cubed and then work it out from there this is a very similar type of reaction okay so this with copper okay so let's have a look at some of these errors um now the average obviously with this reaction is we've got to ensure that the um the starch indicates so we're adding obviously starch to this um to basically help identify the iodine and is added at the correct point and this is when most of the iodine has reacted now normally we'll see this pale yellow colour that's formed if this didn't happen then the blue color would take a long time to um to disappear so you just want to when it just gets now you're all the iodine so you've added your thiosulfate in there you keep adding it and adding it and you get this kind of and pale yellow color that's been formed we add the starch in there okay and then we add it drop by drop until that starch disappears completely that kind of dark blue color there and now we're going to make that starch solution ready because obviously this is the starch is going to be as an indicator so you're going to use it when you're ready to use it now copper iodide the cu the cu cui molecule um is really difficult to see the color of the solution it's really tough to see that so um we've got to keep the solution as cool as possible okay so the iodine produced in that initial reaction that first step um can evaporate readily at room temperature okay so we've got to really care for that and that can leave leave lead to obviously a false tighter which is not very good um and this happens obviously the final copper percentage would be too low um and obviously we don't want that so we're going to try and keep it as cool as we can when we're adding our start indicator to to detect the end point we've got to add it when most of the iodine is actually reacted otherwise it'll just stay blue for ages like you say because iodine reacts with the starch and that's what gives that deep blue color okay so we're going to kind of move on now um so there's a lot of them with titrations and certain reactions you need to be aware of these particular reactions that's what the exam board is wanting you to want you to know so here we're going to kind of look at transition metals and redox potentials so it's quite a small kind of kind of a small area anyway so redox potentials tells us how easily an ion is reduced which is the same as electrode potential so we looked at kind of e naught values so the least basically you don't really need to know a lot about this to be honest but so there's there's not much to kind of really explain about but the least stable lines have the largest redox potential and therefore are more likely to be reduced so you can see here here's your kind of redox potentials when we're reacting these together so here we've got copper 2 plus it has an e naught value of plus 0.34 volts um and is less stable than zinc which is here which has a lower e naught value of zinc two plus sorry which is here and which has a lower um e naught value here so we're looking at the stability of these of these substances here so remember these are all um these e naught values have to follow a standard set of conditions that's what the little naught bit means or the underground symbol here says theta um temperature is 298 kelvin pressure 100 kilopascals and the concentration of the ions is one moles per decimeters cubed so there may be a difference in the redox potentials to the standard values seen in the data book so when you run the practical you might may see a slight difference obviously it depends on the environment the ions are in so you might not be at standard conditions the room might be warmer or colder than we think and the pressure might be slightly out from the standard potential so just bear in mind that there might be a slight difference there as well okay so your redox potentials are just looking at how stable something is and and whether that's likely to happen okay so not much to say on that really okay so we're going to look at d orbitals okay so we need to be aware of the shapes of some of these orbitals here um and there's five d orbitals so remember when we look back in year one there was one s orbital which is spherical we then had three p orbitals which are like figures of eight well for d orbitals we have five of them um now you've seen um sp s and p orbitals already like i say in year one now just like any orbital any other orbital they can hold a maximum of two electrons each and because there's five orbitals in total we can have ten electrons in the overall md subshell so here's the diagram of them now obviously you're not expected to be able to write them you know draw the kind of diagrams of these but they have these all sorts of these exotic shapes um now the top three these show orbitals at 90 degrees to each other as you can see so these if you kind of look at these three in the top they're 90 degrees to each other and the lores of each orbital are between each of the different axes there and and obviously the axes um here tells you where they are so this is d y z d x z and dxy okay so these are the the different axes the other two orbitals these ones at the top here these show the lobes sitting in line with the axes um and so these are given the names obviously dx squared minus y squared and dz squared so these are obviously sitting within a certain axes here so obviously all together um so these orbitals here um these orbitals here are obviously quite important in terms of how we you know how we're you know looking at the orbitals um it i know it says you expect to sketch them out um it's unlikely that you'll be asked to do that to be honest you just need to be um aware of where these orbitals are and the fact that there's five orbitals um and you need to be aware that um obviously each orbital can have um two electrons each so you have ten electrons in total in the d um the d subshell okay so let's start and kind of look into um one of the other properties of transition metals and and that's their ability to form complex ions as we mentioned before so um we need to know what a complex iron is first now the word complex sounds horrendous and it sounds as though it's really kind of um uh you know complicated and it's not too bad i think something there's a lot of colors here a lot of reactions you have to remember um but the actual kind of design of a complex is not too bad so a complex sign is basically just a big iron that's all it is so a complex sign is where you have a central transition metal ion and that's in the middle and it's surrounded by a load of ligands represented by l and you have loads of dative covalent bonds or coordinate bonds kind of bonding into the middle of the the iron there so the centromere lines there and the ligands these are about like i say represented with l they have at least a lone pair of electrons okay so they must have that and these basically donate that lone pair of electrons into the transition metal line obviously form this coordinate bond this date of covalent bond now you'll know from year one chemistry that these are quite weak as a bond you know they're not as strong as a covalent bond so obviously these can be kind of pulled away quite easily and that gives rise to quite a lot of different reactions with transition metals actually and kind of keeps them versatile so a square bracket as you can see surrounds the ligands and basically you'll have an overall charge or you may do you may not it depends on what you've got and but the overall charge if it does sits in the outside of the square bracket now complex signs they can come in loads of different shapes as well we're going to look at these later obviously this is what we call an octahedral shape but we'll look at some of the other shapes later as well okay so let's kind of kind of zoom in on that bit of a complex and we're going to look at ligands now um ligands are a ligand is an iron and it can be an atom or it can be a molecule but it must have a lone pair of electrons at least one okay now you can have mono dentite ligand you can have bident ligands or you can have polydentate ligands you know loads of different types so um ligands which only have one lone pair are called monodentate ligands sometimes they're known as unidentite so let's have a look at some examples so you've got water is a monodentate ligand so is ammonia chloride ions and hydroxyl ions so these are some of the common kind of monodentic ligands you'll see ligands which have two lone pairs of electrons are called bidante ligands and specific ones you need to know which you've seen just earlier in this video ethane dioit iron which has two lone pairs um also got ethane one two diamine which is this bit here so basically it's an amine with a lone pair of electrons but it's diamine um and then ligands which have more than one coordinate bond are called multi-dentate ligands so a classic example is edta four minus um it's a very big comp complicated ligand you don't need to draw it out um but edta four minus is an example of a multi downtick ligand it can actually form six coordinate bonds it's a bit of a you know a bit of a beast of a um of a ligand to be honest it's massive um six coordinate bonds and that can bond with the central metal line as well okay so obviously we know what some of these ligands are so we've kind of had an introduction we will come back to the ligands as well and you'll see them all the way through we're just kind of introducing different concepts here i suppose um complex shape that can change as well and actually that's dependent upon the size of the ligands that we attach to the central metal ion and the coordination number okay so the coordination number is the number of coordinate bonds okay it's not the number of ligands so for example edta is one ligand but it has six coordinate bonds in the middle so it's the coordination number for that complex would be six we've got one ligand okay so remember coordination numbers the number of coordinate bonds so some ligands are small and actually you can fit six of them around a central metal ion so you've got water you've got ammonia and you've got hydroxyl iron so i kind of imagine this a bit like and pigs around the trough so the trough is like the metalline okay it's in a fixed position and it has a certain size and if you've got little pigs you can fit loads of little pigs around that trough and they can all get to the food quite easily but if you've got much larger pigs so this could be like cl minor signs for example they're a bit fatter you can't fit as many of these pigs around the trough there just isn't enough space so larger ligands like cl minus ions you can only fit four of them around the central metal ion there just isn't enough space to put six of them around like water ammonia and hydroxide ions ethane dioit and ethane one two diamine are even bigger than chlorolines and normally we can only fit three of these around the central metalline so obviously these are a lot bigger and obviously they bond twice each anyway okay and let's have a look at some of the names of some of these then so complexes with a coordination number of six so they've got six coordinate ponds these form octahedral shapes so you can see here we've got an example this is cobalt and surrounded by six water ligands all the way around and this has got an overall charge of plus two it's also written as this so it's cobalt with a square bracket water obviously six waters and two plus charge so cobalt with your ammonia you've got obviously six ammonia ligands going around it cobalt in the middle again two plus charge and this can also be written as cobalt six nh3 so there we are with two plus and aqueous so that's how we how we can write them as well so all the bond angles in this and anything with six which can fit six ligands around so these are smaller ones these are all 90 degrees so again this is going back to the bonding topic in year one where the the obviously bond angles are um determined on the the shape of it so this is a an octahedral shape no difference to what you've seen in year one okay so complexes with the coordination number four these form tetrahedral and square planar shapes so here's one here so this is copper chloride so many chloride ligands were much bigger you can't fit as many of them around so copper chloride cucl4 your four ligands around here and these obviously form a bond there also written as cucl 40 minus um the bond angles and tetrahedral if you remember from year one 109.5 degrees and here's another example this is an example of a square planar so this is platinum with ammonia and two chlorines and this is also written as this so basically we have the square brackets we have two ammonia ligands and two chlorine ligands in that complex there and um a specific example of a square planar is an anti-cancer drug called cisplatin which you do need to know that as a specific example of a square planar molecule there's very specific ones really here and the bond angles obviously in a square planar complex are 90 degrees okay so complexes with a coordination number of two so these form linear shapes and a specific example are some silver complexes so here's a classic example here um so this is um silver with two ammonia ligands um surrounding it so ag nh32 again this is just a specific example you do need to be aware of okay bond angles and linear compounds of 180 degrees obviously that's between them points there and this is known as tollens reagents and you've seen this before so ag and h32 okay so as you've seen before another feature of right now complexes is they have an overall charge um and this is basically the same as its total oxidation state so the total oxidation state of the metal okay equals the total oxidation state of the whole complex minus the total oxidation states of the ligands as well so let's look at this example here we've got copper with four chlorine ligands or chloride ligands around it overall charge of minus two okay so that's the total charge of the whole complex now each cl minus ligand has a charge of minus one so there's four of them in total so that means we have a total oxidation state for the ligands of minus four so that must mean the total complex oxidation state is minus two because obviously that's what we've looked at there and so therefore the oxidation state of copper in the middle has got to be plus two so that's plus two there and you've got four minuses around there so that gives you the overall charge of minus two and it's as simple as that that's how we work out the oxidation state of copper that's in the middle so relatively straightforward okay so kind of looking at a um so we're going to look at the kind of orbital level a little bit further and look at looking at the orbitals with the shapes what we've just seen there and we need to understand about d orbital splitting okay now this can be a little bit tricky this bit here but the d sub shell is split into two when ligands bond with the central metal ion okay it's a bit weird this okay so let's have a look at this in a bit more detail so we've got the 3d electron configuration for cobalt two plus ions the co2 plus when no ligands are attached to it so it's just the iron imagine it has no ligands around it at all that's the um the d orbital when we at when we attach ligands to this ion though some of the orbitals gain energy and we get an energy gap that's created and this is what we call delta e so this is um for example cobalt with six waters around it so this is hydrated and now the d orbital splits into two parts so you've got um some electrons in here and some electrons at the top and we create an energy gap here which is also known as delta e so what happens is when electrons absorb light energy so in other words you've got a bk you put it on the window sill and obviously it'll absorb the light energy from obviously the window and then um some of the electrons move from the lowest energy state which is called the ground state okay to an excited level called or higher energy level called an excited state so there's your light energy there because your light energy has gone in and it's moved one of the electrons into the higher state there so you can see there's the electron there that's now flipped and gone into the upper orbital here and that's been absorbed by light energy so this is the excited state this bit here because we've got an electron there now in order for this to happen the energy from the light that's obviously been absorbed must equal the energy gap that's delta e so the size of delta e the size of this gap here is dependent on the central metal line um and its oxidation state in other words what we have it also depends on what ligands we have bonded to it in this case we've got water but you might get different different energy gaps use chloride ions for example and the coordination number hence the shape so this is obviously an octahedral deorbital splitting and this is where we get three orbitals at the bottom and two at the top now when we form a tetrahedral complex we get exactly the same we still get d orbital splitting but they split slightly differently and actually we have three orbitals at a higher energy and two that are lower down okay so let's have a look this is the cobalt one exactly the same example as what i mentioned before but actually when we react it with say and when we have a tetrahedral structure so we have four coordination number four so we have c or cl4 instead then actually the splitting kind of flips the way around so with the octahedral structure we had three at the bottom two at the top here we have three at the top and two at the bottom so we've got a slightly different set up here okay so we still have an energy gap that's created so the principle is the same and obviously when electrons absorb that light energy some will move from the lower energy state here and we'll move to a higher energy excited state so you can see here we had four electrons in the bottom we've now got one of the electrons is migrated up into the top here and obviously light energy has been absorbed at this point so you can see um the this kind of splitting of the d orbital this kind of unique property of d orbitals um in terms of this or transition elements in particular um is that actually they can split depending on what they're bonded with the type of ligands and the type of metal that you've got and you can now probably if you're kind of one step ahead here you're probably getting to see why we have so many different colors with transition elements because when you absorb light energy okay and then some of these electrons will drop back down to their original state they'll release some light energy and depending on obviously the size of this energy gap will depend on the color that you get and obviously this is dependent on loads of different factors such as the type of ligand the shape the type of metal that you've got and all of a sudden you've got a whole library of potential different colors that could be produced from this and i hope it kind of comes back to it as well that actually this is why you need um partially filled d sub shells to get these colors because if all of these were full like for example with zinc then you won't be able to get any of this migration of electrons and so therefore you won't actually get any fancy colours okay so let's have a look and kind of delve a little bit further into coloured complexes so some frequencies of visible light these are absorbed by transition metal complexes and the frequency absorbs obviously like we said depends on the size of delta e so you can see we've got our kind of spectrum of color here so you've got red orange yellow green blue indigo and violet all the way across here and this is actually what makes a white light now the larger the energy gap delta e that we've seen before the higher the frequency of light that can be absorbed so this is kind of at the top end here now this is what's actually absorbed any frequencies which are not absorbed okay are reflected either bounce back or they're transmitted in other words they go straight through their kind of test tube so the combination of all of these frequencies okay these ones here they create a complementary color and that's what we observe that's exactly the color that we see so when you see these um these reactions you hold up the test tube you get all these lovely different colors we're seeing a mixture of the colors that have not been absorbed by the um by the solution so for example um copper and six so hexa aqua copper complex i'd say that one quickly and that absorbs frequencies that produce red light okay so that absorbs that and the complementary color this is your color wheel here is light blue or cyan and actually when we look at uh hexa aqua copper two plus we do see a pale blue solution so it's absorbing this red but um obviously the mixture of all the other colors will produce this kind of cyan color here which is this pale blue so that's the complementary color and so for those complexes where we have a full or empty 3d subshell this is what i'm saying before you have no electrons can migrate to these higher energy levels and this means we just see these as white and colorless so this is like scandium and zinc for example okay so like we say the size of delta e is dependent on the types of ligands and we're going to explore some copper complexes with water ammonia hydroxide and some chloride ligands so we're going to look at various different types here so the order of ligands which cause the most amount of d orbital splitting is shown below so as we go down the list the splitting of the d orbital increases so we've got ammonia um is right at the bottom that increases the kind of gap and chloride lines don't increase that gap as much so as the splitting increases okay so as we go down the group the light absorbs will shift more towards the violet color this is what the complex absorbs okay so it needs more energy remember so c ucl minus c sorry c of cl4 here two minus absorbs frequencies that produce blue light okay so remember that the splitting isn't as great here so it'll absorb kind of the yellow color so a little bit lower down okay um so sorry it absorbs the frequencies that produce blue light so it's kind of up here and the comp the complementary color is yellow so that's the color that we observe with that one if we look at um hydroxide ions some hydroxide and water this absorbs frequencies that produces the red or magenta colour so the complementary colour is um green so we get this greeny color here and then for um the copper with just the water so just the pure water this is the one down here this absorbs frequencies that produce red light okay so the complementary color is light blue or cyan so that's what we see there and then finally with the ammonia ligand in there as well this absorbs frequencies that produce yellow light and the complementary color is dark blue and that'll see that's the color we observe so what we're trying to say here is you get loads of different colors that are produced and the colors that are produced is dependent on the ligands that we have the type of ligands and obviously um and the obviously the metal line is staying the same but we're just changing the ligands and with the chlorine when we're changing the shape as well so you can see this is why i get this variety of different color loads of different colors here okay so let's have a look and move on to isomerism because um transition metals can form isomerism as well in particular optical isomerism now you've seen this in year one so we're just kind of dragging some of the information from year one and applying it to transition metals so complex ions they show optical isomerism and this is a type of stereo isomerism so remember um complexes are optical um when they are not superimposable so if you'd seen the previous video for year one i had my hands out and i was overlapping them and trying to turn them over each other um anything that is um non-superimposable isomers which are non-superimposable and will be optical isomers so they're mirror images of each other so you can see with this complex here um we put a mirror line down the middle of this and the they are definitely optical isomers because they're mirror images of each other and so therefore octahedral complexes with right with three bident ligands um these short optical isomers like we can see the one on the left so you just need to be aware really in this example here is that um transition metal complexes can form optical isomers and just be prepared to be able to draw the mirror images of these isomers as well and obviously they have the same behaviors as them so um optical anything that's optically active will rotate plain polarized light as we've as we've seen earlier okay and complex signs also show cis trans isomerism which is another type of stereo isomerism so this is a branch of these heads so your octa your octahedral complexes with four ligands of the same type um and two ligands of a different type these display your cis trans isomerism so you can see here that we've got two chlorines obviously in the polar regions there we are so right top and bottom with your four um water molecules kind of in the square planar arrangement whereas with this one here the two chlorines are within the square planar arrangement and then you've got your two water ligands that are in the axial positions at the top there so there we are so if the two different ligands are opposite each other see the ones on the left this is trans isomers obviously they're opposite so it's trans um cobalt complex and if they are on the same or they're adjacent to each other like we see there this is a cis isomer so just be aware of these different types of isomerism in complexes as well okay so just to kind of remind it then this is what we call a non-polar molecule okay so this one here so the chlorine ligand um is the most electronegative out with these ones and but they cancel out because they're opposite ends of the of the um complex so this is a very symmetrical complex here so this is non-polar when we're looking at obviously looking at the polarities of this whereas on the other hand this one is a polar molecule because the negative um of the cl minus ligands are all on one side of the complex rather than kind of spread equally so um this is a polar molecule so just be aware that you are expected to kind of um discuss polarity of complexes as well okay your square planers they also show some cis trans isomerism um so you can see here this is cis platinum this is your anti um oh well it's one of them is anyway it's an anti-cancer drug that's cis platinum um so let's have a look at them a little bit more detail so this one is trans platinum so if you've got two ligands that are opposite each other as you can see there so they're opposite the platinum okay and and vice versa with chlorine and then these are trans isomers and then if you see on this side um this is a cis isomer because the ligands are on the same side okay of the platinum there's chlorine and ammonia and those you can't move so this is a non-polar molecule because we've got an even distribution of charge across the molecules so they're on opposite sides whereas this one is polar because the um chlorines let's say because these are electronegative and they kind of sit on one side of this um complex here so therefore it's weighted to one side so it is a polar molecule okay so we've kind of seen a little bit of cis platinum there um now cisplatin is an anti-cancer drug um it's a square play in a complex uh it has a platinum metal iron in the middle it has two ammonia ligands and two chloride ions as we've just seen before so cisplatins are really obviously it's a really important chemical um especially used in medicine um it's used as part of chemotherapy and what's this platinum does it binds to the dna in cancer cells so these chloride ions in the complex they're easy to displace they're easy to break off and they can detach from the complex and then this can then bond to the nitrogen atoms in the dna in a in a cancer cell for example and obviously because this complex is obviously attached to the dna it stops that cell from reproducing now that cell is a cancer cell and we don't want that to reproduce and spread so we want that cell to just die and and it's under it's unable to repair itself and that's the idea with um drugs like cisplatin the issue though and obviously if you know anybody or if you have yourself in the unfortunate position to obviously suffer cancer um what we do know is actually especially with chemotherapy that it does destroy or it can destroy healthy cells um at the same time so for example it affects cells in the blood which can suppress the immune system for somebody who's um perhaps going through chemotherapy and it can risk the infection and potentially kidney damage so some people may have a side effect where they may lose their hair as well because it destroys um some of the hair cells so um you know chemotherapy has is obviously very important it can help to target cancer cells but it does target healthy cells as well so obviously scientists are putting an incredible amount of research in to try and develop drugs which are more specific so they target particular um tumors and target the cancer cells rather than having these quite nasty side effects really i mean i suppose you know they are i mean i'm fortunate never to have cancer um but you know i can imagine that there isn't really too much of a choice if you know you want to try and destroy the cancer cell so but it's quite a rough journey you know for for many many people but you know these drugs have saved millions of lives globally so they are very very useful as you can see okay so we're going to look i suppose towards the kind of back end of this this video now we're kind of moving towards the end of it anyway we're going to look at some substitution reactions and look at some of the colors that you need to be aware of and then we're going to look at stability so okay stab towards the end so um a color change can exist and we've seen a lot of these color changes and why these things are colored but we need to know about some reactions as well so a color change can exist when ligands in a complex exchange or substitute okay so these substitution reactions show ligands of a similar size being exchanged so we're swapping kind of like for like but changing the ligands so here we've got cobalt with water and this is reacting with ammonia as you can see on here nh3 and this will form a hexa ammonia cobalt complex and the six waters are kicked off so this is an example of a substitution reaction so you've got a pink uh substance here this is obviously octahedral and we form a straw coloured um and again still octahedral shape here so both the ammonia and water obviously similar in size as we've seen before so they've got the same charges they're both neutral and so the complex shape remains octahedral we've got a color change though because the ligand has changed so if we look at copper so copper here we've got six water ligands here around copper um but the amount of ammonia that we've got um is obviously not is only four and we form this complex here now it's blue to start off with and it's a dark blue on that side but you'll notice with this reaction in particular there's only a partial substitution okay we don't get a full substitution here so this reaction occurs when we've got copper um h2o6 here this reacts with excess ammonia so we've got loads of ammonia we will get the maximum substitution that we can which is here these bonds are just a little bit too strong and the water copper bonds the polar bonds a little bit too strong for ammonia to break them off so these kind of remain stubbornly on the complex so we get this partial um substitution here so just be really wary of that okay you do have to remember that okay so let's look at the same thing but the substitutions um we're going to show ligands of a different size being exchanged in these examples so remember cl minus iron is a larger ligand than water of ammonia so we can only fit four chloride iron ligands around this shape so we do have a change in shape and coordination number okay we'll look at some examples here so there's your cobalt um h2o6 you've got six water ligands around cobalt adding a chloride iron in there that'll form copper chloride um and six waters now here we're literally going from a pink to a blue color and but what we're doing whoops sorry so what we're doing is basically kicking off all of the water ligands here displacing them completely so displacement's a better word and so we're displacing these water ligands here replacing with chloride ligands and we're getting a very different complex so here's copper h2o um reacting with four chloride ions um so this is obviously copper rather than cobalt and again we're going from blue to a yellow color and we're going tetrahedral so the shape is actually changing here and here's iron um again adding chlorine we're going from yellow to yellow so the color is actually pretty much the same there's no difference in color there um but it's about comments and obviously on the shapes of these as well okay so let's look at a bit of a um like summarizing i suppose some of these so we're going to look at some complex iron solution in colors so metal ions dissolved in water okay so these only have the formula of that okay so it's metal with the h2o6 what i've done down here okay is i've simplified that to mn plus and you'll probably see it here so we're gonna add we're going to look at the kind of reaction changes when we add it in aqueous solution okay and then when we add sum or h minus ion or nh3 to aqueous solution when we add excess to it and then when we add excess ammonia as well to the to the precipitate so let's have a look so copper and when we add copper in water obviously we form cu h2o six two plus and it's a blue color iron is um pale green iron three plus is a yellow color and cobalt two plus is pale pink as you can see there so these are when we add it in water you have six water ligands surrounding it so when we add some hydroxide so just a small amount or ammonia to this solution what do we get well in this case we add a small amount to copper two plus and we get a pill blue precipitate and you can see we get four water ligands still attached and two o h m two so it's a solid this overall has a charge any complex that has a charge will always be aqueous okay so it'll always be in solution if it doesn't have a charge like here because the ligands are obviously cancelling out the overall charge here then it will form a precipitate okay so just a tip just to watch out for that okay so this doesn't have an overall charge and so therefore is a solid it's a precipitate iron um again we need to add um some hydroxide ions we've got two ohs there and we've got water ligands so just enough to form that precipitator if we add a small amount we'll get a precipitate again it's got no charge with iron iii we need to add a little bit more hydroxide um ions in there or ammonia and obviously this is going to form an orange precipitate so we get oh3 h2o3 so we just need a little bit more than what we do up here but nonetheless we're still forming a precipitate and then cobalt again we form a blue precipitate here so if we take these then okay so we've got a little valid like drop by drops small amounts got lovely precipitate if we then kind of go all out and say right we're just going to add loads of hydroxide to this what do we get with these reactions so this one is insoluble okay in excess so basically nothing happens we can add loads of hydroxide and we've just got this it doesn't do anything the next one is also insoluble again no change that's all we produce the next one is also insoluble we don't actually get any difference here um and cobalt is also insoluble so basically we get no change when we add excess hydroxide to any of these precipitates so let's do the same thing so we've done that i think right well that hasn't worked so let's say we get this um pale blue precipitate and now we're going to add excess ammonia to it instead what do we get well here we do get a change okay actually instead of just um forming an oh2 here so basically the ammonia is just taking a proton off the water mark this was a water ligand that we've just taken a proton from it what actually happens is and we get a substitution okay so we get part ligand substitution it basically replaces all of the ligands it goes in pushes them all out with real most of them anyway leaves two water ligands to one side and basically boots off two waters two hydroxide ions and replaces it with ammonia ligands and we get this dark blue solution it's aqueous again because we've got a charge okay remember we've got a complex of the charge it's always in solution um it's insoluble ion2 is insoluble in excess ammonia so it's not that powerful also insoluble in iron iii um with cobalt um it does um continue further so cobalt um here's your precipitate here that's formed if we add excess ammonia it kicks off all of these ligands here and we have six ammonia ligands attached to it we've got a brown yellow solution that's formed so it's aqueous okay so let's have a look at some of these equations that's involved here so copper two plus um so copper two plus reaction with hydroxide ions will form this so this is a small amount of hydroxide ions okay this is some and with ammonia and effectively the ammonia is picking off some of the protons from two of the ligands here to form oh2 and obviously you get two ammonium ligands or two ammonium molecules should i say and then if we add excess hydroxide or excess ammonia it's insoluble for this one but um for this one obviously it forms we go from this and it forms this okay so we basically get a ligand substitution a partial ligand substitution now with iron obviously we had a small amount of hydroxide a small amount of ammonia and obviously these are the reactions here so we are um substituting um well we're taking a proton from the the uh two protons from here to form oh2 same with the ammonia and obviously if we add excess they're insoluble for both so there's no change ion3 plus again a small amount will produce a solid precipitate as we can see there so that's the small amounts and obviously they're insoluble for both of them at the other end uh cobalt two plus again adding small amounts of each of them will get this um solid precipitate so we're gonna precipitate um the precipitate will dissolve um when we add excess ammonia so um this obviously forms cobalt hexa ammonia so we get full substitution so all these ligands are removed and we get that that's produced and it goes to aqueous solution okay so kind of looking towards the back end then we need to look at some stability complex or k stab um now some complexes are more stable than others um and we can obviously quantify their relative stabilities using an equilibrium constant and obviously we've seen a lot of these have we so we know that ligands can be substituted for other ligands and we've seen that all the way through this topic um and they can they can do it one at a time or until most of them are replaced so the overall equations let's look at iron and copper will show this so you can see here we've got iron um obviously reacting this time with the cyanide ligand here and obviously it removes all of the water ligands and you have six cyanide ligands attached to it or you've got um your diamine here um reacting with copper hexa aqua complex and obviously you've got three of these ligands attached to it and six waters so we can represent the stability of this or the the k star expression for fe3 plus for example if we use this as an example um by doing this now we don't include water water's excluded so we put product which is fecn63 minus over the reactants which are these here and that's obviously the expression that we use to represent k-stab so the size of k-stab that tells us how stable a complex is and quite simply the larger the number the more stable it is um so let's have a look at some examples and see how this actually works in practice because it's obviously gonna dictate certain reactions and why we see certain reactions anyway okay so let's look at the reactions of copper two plus complexes and ligand substitution reactions that they can undergo so we're looking at the relative stabilities of complexes as our complexes formed by reacting um hexa aqua copper ion with ammonia and hydrochloric acid so we actually formed two products here so we've got one times ten to the minus one sorry one times ten to the 13 for this complex here so this is the with two ammonia and two waters and then we've got copper chloride cucl2 minus is four times ten to the five so clearly this complex here with the water and ammonia is more stable than the cucl2 minus so these cl minus signs remember these readily substitute all of the water ligands and it creates this pale blue solution that we can see here but obviously that is more is more stable now if we then go and add more water to this equilibrium so we're going to add more to this then equilibrium will shift to the left it'll move to this side because it'll try and use up the water that you've added in now what we will see is we'll actually see um we'll have more dark green um more dark green color because we're producing more of the copper hexa aqua ions that we've seen before so we will get a color change and then if we look at the reaction with ammonia instead here so this substitution with ammonia we get this partial substitution as we've seen up here um now we get this now this is more stable as i mentioned before than the copper chloride okay so if we add more water to this okay it will not shift equilibrium left it won't actually move it back to here and this is because this complex what it's formed here is so stable it's it's almost like stubborn i suppose you could call it k stubborn so um effectively this complex here is so stable even if you add more water it isn't going to break apart to form this because that's just weaker so um you know the kind of forces of stability kind of keep this equilibrium well over onto this side and we get this nice stable complex so with this one adding more water will break this apart to form this because the case stab is not very high but with this one because k stab is so high adding more water to this won't shift it that way okay so let's look at at competing equilibria so we're going to add an extra layer of complexity through this and this may well occur through like say ligand exchange reactions so let's look at the hexa aqua copper complex okay which is uh this one here okay and we're going to add concentrated hydrochloric acids to it so that's going to provide us with the four cl minus signs and we form cucl4 minor two minus and six water ligands now let's assume that we're going to add to this reaction some concentrated ammonia solution so now we have another equilibrium that's at play so we've got the copper hexa aqua ions that are already in there anywhere the ones that still exist react that with some ammonia okay concentrated ammonia and we form this complex here okay and water now as the first equation okay it's the first equation is cuh2 or 6 okay so this one here this obviously produces um there we are yeah it's the first equation so this produces c h2o6 in the reverse direction then the ammonia okay will actually react with this okay to produce more of the stable complex of this so the ammonia here is going to react with the cu h2o6 remember this this one here okay is more um is more stable this complex is more stable than this complex here so if we add ammonia ammonia is going to react with this equilibrium is going to shift backwards to form more of this to replace this and effectively overall you'll get more of this complex here because it is a lot more stable than that one up there so we've got two kind of complexes at player so as ammonia like say it reacts with the cu h2o6 equilibrium at the top shifts left to replace this and this means we'll have less of the copper chloride so it's like bullied its way through and this has been dismantled to feed this reaction to produce this which is a lot more stable so why does it happen well like we say we know that this is obviously more stable than that one and so this actually wins out because the case stop is higher so there is an awful lot in that topic like i said it's one of the bigger ones that you'll see so that's the obviously end of um this topic you'd be pleased to hear there's a lot of reactions in there make sure you kind of familiarize yourself with it and like say the full range of a level videos per year from topic one all the way through to topic 37 is available on alloy chemistry all i ask is you hit the subscribe button just to show your support for this project that'll be absolutely fantastic um also and these are available to purchase i got the full range of a level notes which you can purchase from the um ted shop if you click on the link in the description box below you'll be able to get ahold of them they're great for adding into your file and writing notes on um yeah go and have a look right that's it bye bye