[Music] hello and welcome to this a-level chemistry explanation video where we're going to be taking a look at the first part of the thermodynamics topic looking at some enthalpy change definitions eny of solution latis enthalpy and born Harbor Cycles entropy and Gibs free energy will be a separate video that you can check out after this there are lots of different enthalpy change values that you need to be able to Define and to understand and apply in the thermodynamics topic and therefore a quick recap of what enthalpy changes are would be really useful so first of all enthalpy changes generally are the heat energy transferred in a chemical reaction at constant pressure the units for this enthalpy change are usually reported in kog per mole and to save time we normally write Delta H for the change in enthalpy we also normally write a little circle with a horizontal line through and this is to denote that the reaction has been taken place under standard conditions and so this is often referred to as the standard enthalpy change standard conditions are for things such as temperature which is 298 Kelvin pressure which is 100 kilopascals which is just under one atmosphere of pressure and for Solutions we're talking about Solutions which have a concentration of one mole per decim cubed the values for enthalpy changes can be exothermic or endothermic in an exothermic reaction the enthalpy change has a negative value and that's because chemical energy from the reactants has been converted into thermal energy from the surroundings and so from the point of view of the reactants the chemicals they have lost energy that chemical energy is converted into thermal energy and so the temperature goes up endothermic changes are the opposite they have a positive enthalpy change value and that's because the chemicals have gained energy where does it come from it comes from thermal energy of the surroundings and so thermal energy is converted into chemical energy and so that means there is less thermal energy than there was before and therefore the temperature will decrease the most important enthalpy change in the whole of the thermodynamic topic is something called ltis enthalpy ltis enthalpy is to do with ionic compounds and this makes sense because ionic compounds exist as a three-dimensional lce you can see from the diagram that I've got here that we've got positive ions next to negative ions in an alternating pattern I'm showing nine ions here as one plane this would of course Go off into the distance in these two directions and then off into the distance in the the third dimension as well these ions are held together by very strong electrostatic attractions in all directions and this is what ionic bonding is and you can see this positive ion in the middle here has got four oppositely charged ions that it will be attracted to that we can see there will be another one going into the screen and a sixth coming out of the screen why does ltis enthalpy matter well it influences a number of things as we shall see just to give you a few examples we have got boiling point that is heavily influenced by the latis enthalpy of an ionic compound as is the solubility of that compound and to an extent the reactivity of that compound is also influenced by the latis enthalpy as well there are actually two different types of ltis enthalpy there is firstly the ltis formation enthalpy and that is defined as the enthalpy change when one mole of an ionic compound is formed from its Gus ions and if we assume that these red and blue circles are the positive and negative ions you can see that if we turn these into a solid what we're doing is we're bringing them back into the more regular ltis Arrangements that we had on the previous slide and you can see that they were randomly organized and now they are in that ltis structure and so that is the ltis formation enthalpy and you can write this as an equation so for instance if we take sodium chloride as probably the most common example at the beginning we had Gus sodium ions and gasius chloride ions they were singly positive and singly negative and then once they've come together to make their ionic compound we have got NAC solid since we are making new attractions this is the ionic bond this will be an exothermic reaction because energy is always released when new attractions are being formed the other type of ltis enthalpy is the ltis dissociation enthalpy and this is actually the precise opposite of latis formation enthalpy so it is defined as being the enthalpy change when one mole of solid ionic compound is completely dissociated into its gaseous ions and so if I just take my solid ionic lce as I've got it here and I dissociate this I'm pulling the ions away now to do this this will require energy because these ions are held together by those strong electrostatic attractions and so this is always going to be endothermic because I'm overcoming the force of attraction between the positive and the negative ion and that will require energy to go in to make that happen and to break apart these ions into the gas form just like latis formation enthalpy we can also represent this as a chemical equation it is in fact the exact opposite of the previous example we're starting with our ionic solid nacs and we're making our positive and negative gasius ions so the sodium plus and the chloride minus and so latis dissociation enthalpy is the precise opposite of ltis formation enthalpy and so that means that the ltis formation enthalpy is negative in this example with NAC it's - 787 and the latis dissociation enthalpy is going to be + 787 so they are equal in magnitude but opposite in sign because one of them released es heat energy to the surroundings whereas the other one takes heat energy in and converts it into chemical energy it's a common command in a-level exam questions to compare the latis enthalpies for two different ionic compounds and the value for a latis enthalpy depends on the strength of the attraction between the positive and the negative ions and these electrostatic attractions depend on the size of the ions and the magnitude of the charge on the ions and these are often rolled into one term which is referred to as the charge density and that is quite literal it is how concentrated the charge of a particular ion is if we look at a couple of examples and say compare NAC to KCl in an exam situation they usually do try to keep one of the ions being the same because that allows you to ignore that from our consideration so I'm just drawing a single Circle for the CL minus that will be the same in both examples sodium is higher up group one than potassium and so it will be a smaller ion or it will have a smaller ionic radius and therefore its charge is more concentrated because both of them are singly positive ions and so you could say that the sodium has a higher charge density as an ion than the potassium as a result the electrostatic attractions will be stronger in NAC than it is in KCl and so the latis enthalpy will be more exothermic and that's really important that Clarity of language that we should avoid saying larger or smaller because ltis formation enthalpies are all negative values I would encourage you to say more negative or more exothermic or if it's a dissociation enthalpy you could say more positive or more endothermic if we look at adding magnesium chloride into the consideration magnesium is doubly positive and so that 2+ ion is likely to have a higher charge density than the na+ and the k+ and additionally magnesium is significantly smaller than sodium even which was the smaller the two singly positive ions so magnesium chloride would definitely have a more exothermic or a more negative latis formation enthalpy one final pairing if we compared mgcl2 to mg this time the positive ions are both the same so any differences will not be down to that ion it will be down to do with the the size of the CL minus ions charge density and the O2 minus charge density and so you'd expect the O2 minus to both have a smaller radius and also to be more highly charged and so therefore it will have a higher charge density and a more exothermic latis formation enthalpy I'm now going to take a look at how the strength of an ionic lattice can affect that compound solubility in order to dissolve an ionic compound energy needs to go in to break apart that lattice and this this is the latis dissociation enthalpy as I've already said dissociation is an endothermic process because we need to overcome the attractions between this positive and negative ion in all of those six different directions where does this energy actually come from well it comes from the water the solvent that is going to do the dissolving this works because the water solvent forms new attractions to the ions in the lattice and this creates an aquous solution in this aquous solution we say that the ions have been dissolved until we have got infinite dilution and what that means is that there is no electrostatic attraction between the positive and the negative ion anymore those attractions are purely between the ions and the water solvent and this works because water is a polar Sol solvent some regions of the water are Delta minus specifically the oxygen atom and some are Delta positive and that will be the hydrogen atoms and so what we get is we get the Delta positive hydrogen atoms in the water molecule are attracted to the negative ion from whatever the ionic compound was and similarly the oxygen atom from the water molecule will be attracted to the positive ion from the ionic compound and these form new attractions and this process of forming new attractions between the Gus ions and the water solvent to make the aquous ions is called the enthalpy change of hydration or the hydration enthalpy since we're making new attractions these processes are always going to be exothermic irrespective of what the charge of the ion is we are always making new attractions between one region of the water and the ion itself in an exam situation you might be asked to comment on the exothermicity of the enthalpy of hydration which is not as complicated as it sounds the charge density of the positive ion for instance will affect the strength of its attraction to the Delta minus oxygen in the water so for instance mg2+ or na1 plus the mg2+ will have a stronger attraction to the Delta minus oxygen because it has a higher charge density and this works exactly the same for the negative ion except the attraction is to the Delta positive hydrogen and this can have such a huge difference to the enthalpy of hydration but not only that the number of water molecules that can actually cram in a around the positive or the negative ion to the point where some highly charged small ions such as aluminium 3+ for instance can have a huge number of water molecules forming ring upon ring of water molecules around it you don't need to memorize any of these specific examples but you do need to be able to comment on charge density and speculate as to whether that will lead to a more exothermic enthalpy of hydration or less the precise value for the enthalpy of solution and so whether it is an exothermic process or endothermic process depends on the balance between this latis dissociation enthalpy and the hydration enthalpies if the magnitude of the hydration enthalpies is greater than the latis dissociation enthalpy it will be an exothermic enthalpy of solution if not then it will be endothermic we can use a h and hess's law to work out a value for the enthalpy change of solution for an ionic compound hess's law states that the enthalpy change for a reaction is independent of the route taken for that reaction it depends only on the initial and final energies of the chemicals if we take a look at the three equations that we need for a h cycle here we've got the enthalpy of solution equation and in general this is where you start with one mole of an ionic solid and you make that into aquous ions if we go for an example sodium chloride the most common example we start with NAC solid and we end up with na+ aquous and cl minus aquous which is sometimes okay to write as NAC aquous in one form but I tend to separate those out and then if we look at the ltis dissociation enthalpy the definition there is one mole of an ionic compound in solid form turns into its gaseous ions and so nacs turns into na+ gasius and cl minus gasius and last of all this new term of the enthalpy of hydration this is where you've got the gaseous ions so na plus G turns into aquous ions so na plus aquous and the same thing for the chloride ions and you can take these enthalpies of hydration separately or you can hydrate them both simultaneously but the enthalpy change for this reaction will be the sum of the individual hydration enthalpies now if we construct these into a simple H cycle you can see that we can make a triangle H Cycles are typically triangles and they're made up of overlapping equations and what I mean by that is each vertex of the triangle Le is formed where two equations overlap and say the reactants are the same for those two equations and so if we take the enthalpy of solution we've got the ionic solid at the beginning and that's the same for the latis dissociation enthalpy so an ionic solid forms one corner of the triangle however the equations Go in different directions because one of them turns into the aquous ions and the other turns into the gasius ions and then the hydration enthalpy is when gaseous ions turn into aquous ions so this Arrow of the triangle is referred to as the sum of the hydration enthalpies and we can use this template for sodium chloride we start with NAC solid in one corner we turn it into na+ and cl minus aquous along the top and na+ and cl minus gas down at the bottom corner and then the na+ gas arrow goes up to the na+ aquous and the CL minus is all part of the same arrow and it's this arrow that you could separate out into two separate steps if you wish and then you can use this H cycle that we have constructed to work out what the enthalpy change of solution will be and so you can see that the enthalpy change of solution is along the top and that is equal to the ltis dissociation enthalpy plus the sum of the hydration enthalpies and so for sodium chloride the latis dissociation enthalpy is + 787 K per mole and the hydration enthalpy of sodium is -46 K per mole and for chloride it is - 364 so the enthalpy of hydration in total is - 770 and so we can say that our enthalpy of solution is + 787 Plus - 770 and that gives us + 17 K per mole and so that means that the enthalpy change of solution for sodium chloride is a slightly endothermic value there is a second type of enthalpy change cycle that you need to know about in the thermodynamics topic and this is called the born Harbor cycle named after two German scientists Max Bourne and Fritz Harbor and that's the same Harbor who devised the harbor process born Harbor Cycles work in exactly the same way as H Cycles which is that you can use them to work out an unknown enthropy change value but also the enthropy change is still independent of root it doesn't matter if you're using a h cycle or a born Harbor cycle there are two main differences between born Harbor cycles and H cycles and the first is one of shape born Harbor Cycles are rectangular in shape which seems a little bit odd but what this ends up meaning is that the chemical equations are written at 90° to the angle that you would expect them to be written at and so the dissolving of sodium chloride is something that we would typically see written like this but if it was in a born Harbor cycle we would write it at 90° the reactants sodium chloride solid would then point upwards to the sodium ions and chloride ions in their aquous state so that's the first difference the second difference is that we actually have a y AIS to these cycles and that y- AIS is energy and the significance of this is that the direction that your arrow points in a born Harbor cycle tells you if this reaction is exothermic or endothermic in other words do the chemicals end up with more energy at the end of the reaction endothermic or less which would be exothermic and so for my example that I've drawn here this is an endothermic process because the chemicals are gaining in energy and there will be a temperature decrease and so we can construct a full born Harbor cycle for the dissolving of sodium chloride like so the sodium chloride solid can turn into gasius positive and negative ions that's a regular equation as I've written here but since this is going to be significant iFly endothermic plus 787 as discussed previously then this is going to be a large endothermic Arrow going quite a long way up on the cycle and we've got the gaseous ions along the top which can be hydrated which means they can become aquous ions and this is going to be exothermic because new attractions are being formed once we've got this cycle we use it in the exact same way as using a h cycle and so you can see we've got plus 787 and that then is added to minus 770 to give us our enthalpy change of + 17 which is of course the exact same number we got from using the H cycle and you can use this for any of the unknown three values in a born Harbor cycle it doesn't always have to be the enthalpy of solution that you're calculating and the most tricky example might be if they gave you the enthalpy of solution they gave you the latis dissociation enthalpy and they gave you one of the hydration enthalpies and you had to treat this hydration enthalpy Arrow as being equal to the enthalpy of hydration of sodium which say we didn't know plus the enthalpy of hydration of chloride which we did know as minus 364 and so we'd have this Arrow being + 17 this Arrow being plus 787 and this Arrow being X plus - 364 which we could then rearrange and solve that for obviously getting us the answer that we know it should be of - 406 we can calculate a value for the ltis enthalpy using the electric field strength equation that you might encounter in physics but it's not something that you can measure directly in order to determine a value for the eny change of ltis formation or dissociation via an experiment you have to do this indirect ly through a series of other different enthalpy change experiments and combine them together in a born Harbor cycle you will have encountered these enthalpy changes earlier in a level chemistry for instance the enthalpy of formation is the enthalpy change when one mole of a compound is formed from its elements in their standard States now in the thermodynamics topic you're going to be making an ionic compound but it doesn't have to be for the eny change of formation by definition and so for instance if you had magnesium chloride you'd have solid magnesium and then cl2 in the gas form and you'd make mgcl2 in the solid form if we had NAC this would be a little bit different because you have to by definition make one mole of an ionic compound and that means there has to be a one in front of the NAC which means you only need one mole of chlorine atoms which means you only only need half a mole of chlorine gas molecules in the atomic structure topic you first encounter the ionization energy concept and this is the energy change when one mole of Gus singly positive ions or one plus ions are formed from one mole of Gus atoms and so to stick with the elements I've already been looking at we've got mg gas turns into mg1 plus gas plus an electron it's absolutely essential to include that electron that has been lost or for sodium we've got the same thing but we've just changed the magnesium symbol for sodium we can also ionize some things more than once and this is the enthropy change when one mole of 2 plus gasius ions are formed from one mole of 1 plus gasius ions and we can write this as an equation as well in this form exactly in the same way as previously except we can see that we're starting with 1 plus ions and finishing with two plus ions and that electron ionization is always an endothermic process since we're having to overcome the electrostatic attraction between the nucleus of the atom and that outer electron that is being removed and that always requires energy and the second ionization energy is always going to be larger than the first ionization energy electron Affinity is effectively the opposite of ionization energy for a couple of reasons first of all by definition it is the enthalpy change when one mole of gasius one minus ions are made from one mole of gasius atoms this means we might be starting with one mole of single gasius oxygen atoms they're going to gain an electron and we're going to make o minus gas or we might have chlorine gas plus an electron making CL minus gas both of these are making new attractions between the atom and the electron that is being gained and therefore this is going to be an exothermic process because a new attraction is being formed the exception to this is the second electron affinity this is the enthropy change when one mole of Gus's 2 minus ions are made from one mole of Gus 1 minus ions you can see again I'm using oxygen my example this time the oxide 1 minus ion is gaining electron to become the oxide 2 minus ion this will actually be an endothermic process even though we are making a new attraction and the reason for this is we are having to overcome the repulsion between the negative electron and the negative O minus ion that is gaining it and overall energy has to go in to make this reaction happen eny change of atomization is defined for an element as the enthalpy change when one mole of Gus atoms is formed from an element in its standard State and so that might be sodium for instance na solid turns into na gas we might say it has been atomized this by the way is in this instance the same process as sublimation alternatively we might make one mole of Gus chlorine atoms and for that to be the case we would have to atomize half a mole of cl2 because in order to produce one mole of chlorine atoms we would actually need half as many chlorine molecules because they are diatomic if we broke apart a full moles worth of cl2 to make 2 moles of Cl gas this would actually be the bond dissociation enthalpy so that's the enthalpy change when all the bonds of the same type in one mole of gaseous molecules are broken so in this instance the cl to CL calent Bond and the relationship between bond dissociation enthalpy for a diatomic molecule and atomization is that bond dissociation will be precisely double the atomization and these are always endothermic because energy is going in to break those bonds and so the bond dissociation enthalpy for chlorine for instance might be 24 2 K per mole and that means that the enthalpy change of atomization of chlorine will be 121 kog per mole exactly half because what you're doing is you're breaking half a moles worth of cl to CL bonds instead of breaking a whole mole of cl to CL bonds a definition of atomization that you encounter far less frequently is if you are atomizing a compound and that is when you take one mole of a compound in its standard state and you convert that into gaseous atoms for an organic molecule such as methane this will involve the breaking of four carbon to hydrogen bonds so the atomization of methane is going to be four of the bond dissociation enthalpies of carbon to hydrogen but if we had an ionic compound for instance sodium chloride the process would be turning one mole of sodium chloride solid into 1 mole of sodium gas and one mole of chlorine gas we're now going to take a look at how you can use a Bor Harbor cycle to work out one of the enthalpy changes that we've just been discussing I'm going to use sodium chloride as my example again but I'm also going to try to generalize at the same time first of all it's important to begin with the end in mind we are working with an ionic compound and so sodium chloride is our ionic compound and that will be on the bottom of our cycle we can make sodium chloride in two different ways we can make it from its elements in the enthalpy of formation and this is going to be enthalpy change number one that I'm going to write on the bottom left alternatively we can make it from its gaseous ions and that's going to be the arrow on the right hand side and this is the ltis formation enthalpy for sodium chloride both of the arrows are pointing down towards the na CL solid implying that this is going to be exothermic you're not actually expected to memorize whether or not sodium chloride would have a negative enthalpy of formation or not but you are absolutely expected to know that it would have a negative latis formation enthalpy and then what we need to think about for the rest of our cycle and the way that I recommend that you remember how to construct these is how can we get from the elements sodium and chlorine on the left hand side round to the the gaseous sodium ions and the gaseous chloride ions well the answer is sequentially one substance at a time we need to convert them and work towards them becoming ions so for instance we've got sodium solid at first we need to atomize it and so this first Arrow that's going to be pointing upwards is the atomization of sodium and it's going to be endothermic energy has to go in one of the rules for these arrows on the cycle that I'm going to use to connect my elements with my gaseous ions is that you only change one substance at a time so while the sodium has been atomized the chlorine is unchanged so we just write half a cl2 G again now we deal with the chlorine we're going to atomize the chlorine and this is going to produce CL gas and this means that the sodium is Untouched by this process and so this Arrow which I'm labeling as eny change number three is the enthalpy of atomization of chlorine the sodium is unchanged now a quick check over onto the right hand side reminds us that we've got sodium ions and so our next Arrow is going to be to ionize that gaseous sodium and turn it into a positive ion and so this is enthropy change number four that is the first ionization energy of sodium and so as a result of that change the gas sodium has turned into the sodium plus gas so the ion and as a result it has lost an electron and so we need to include that electron in the cycle and the chlorine is untouched it is still the gas and the final step to complete our cycle is to convert that chlorine which is a gas into the chloride ion and so it gains that electron that the sodium has just lost and so the sodium ion is completely unchanged but the chlorine goes from being a gas to a negative ion and that is the electron affinity of chlorine now we've constructed our born Harbor cycle I know it looks much more complicated than a h cycle but how we use it is exactly the same so suppose we wanted to calculate this enthalpy of formation so Delta H number one that means that we're starting at the base of this arrow and we're going to the head of the arrow and remember the rule of hess's law it's the it doesn't matter where you start and finish the enthalpy change will be the same so that means that this enthalpy change Arrow number one will be the equivalent of going with enthalpy change number two and enthalpy change number three and enthalpy change four and enthalpy change five and enthalpy change six so using this born Harbor cycle we can see that the enthalpy of formation is actually equivalent to the sum of all of those five other enthalpy change arrows provided the ltis enthalpy is ltis formation enthalpy equally we could calculate any of those other values we could calculate the latis enthalpy itself which really is why the born Harbor Cycles are used but in an in an exam situation it doesn't have to be and so enthropy change number six is going to be starting at the base of that arrow and going to the bottom and if we work our way around this cycle anticlockwise you can see that we are going against four of these arrows and so the sign would flip over and so enthalpy change number six is equal to enthalpy change number one minus enthalpy change 2 and three and four and five that have all been added together as a general rule then for born Harbor Cycles you have your ionic lce at the bottom you have the elements in their standard States above that and to the left with an arrow connecting the two you've got the gasius ions probably above that and to the right and then you've got one of the elements has been atomized and then the other element has been atomized so that's two atomization one after the other then we ionize the metal gas atom and then the non-metal gas gains that electron that the metal has just lost and this completes the cycle don't forget the electrons need to be included at the top of this cycle if we take a look now at a more complicated born Harbor cycle for an ionic compound that raises a couple of different issues the way that we need to construct the cycle is exactly the same though so you start by S putting the na2o solid down on the bottom line the elements that make that up will be 2 na solid plus half an O2 gas and we are going to be making this ionic lattice ultimately from two sodium 1 plus gas ions and 1 O2 minus gas ion and so this is three of our steps completed the cycle continues in a similar way to the previous cycle we need to atomize our solid sodium and so we're going to end up turning our two sodium solid into two sodium gas and so this means that this arrow that I've just drawn is not the enthalpy change of atomization of sodium it is 2 * the enthalpy change of atomization of sodium because I needed to put that energy in twice to atomize both the moles of sodium solid and then we're going to atomize the oxygen which is going to produce o gas one mole of it at the end and the sodium is unchanged by this second upwards Arrow if the question was being confusing it might not give us enthalpy change of atomization for oxygen remember it might give us the bond dissociation enthalpy and in that case we would have to Harve it the next step is we need to ionize our sodium and so not only are we ionizing one mole of sodium there are two moles of sodium gas that needs to be ionized so this arrow is Two Times by The ionization energy for sodium as a result we make two sodium 1 plus gas ions and we produce two electrons one from each of those sodium gas atoms that has been ionized and so our oxygen is still a single gas atom and we've now got our gasius sodium ions the oxygen needs to be gaining those electrons one after the other the first electron affinity is where one of those electrons has been gained and that will be exothermic as normal the second electron affinity is where the o1 minus gas ion gains the electron the one electron that there was left over that needs to be included still and then it becomes an O2 minus ion which is a gas and so we've now got our gasius ions on the top right remember this arrow is pointing upwards because it is an endothermic process because energy needs to be put in to overcome the repulsion between the O minus ion and the electron that is being gained so they are both the same negative charge using this born Harbor cycle is exactly the same as for sodium chloride the exception is that any enthalpy change that is to do with sodium and not the oxygen would need to be doubled because we need to atomize two sodiums and we need to ionize two sodiums and if they were being mean and they were asking you to work out from this born Harbor cycle what the enthalpy of atomization of sodium was you would do this in the exam exact same way that you would expect so enthalpy change number two for instance and so this enthalpy change of atomization arrow that I'm putting the little star next to starting from the base of that arrow and going to its head is the same as going this anticlockwise journey around the cycle and so in other words the enthalpy of formation minus the ltis enthalpy of formation minus the second electron affinity minus the first electron affinity minus the ionization Times by two minus the atom ation of the oxygen all of those are equivalent to 2 * the atomization of sodium so that means that when you calculate those numbers and you get your value you then need to divide that by two and that will get you your atomization of 1 mole of sodium the completed born Harbor cycle for calcium sulfide looks like this you can see that we've got eight arrows this time that makes it look pretty complicated but it still follows the typical born Harbor cycle format we've got the ionic compound at the bottom one mole we've got the elements on the line above and to the left and we've got the gasius ions at the top and right everything looks normal so far we've got the atomization of calcium so step one atomization of sulfur is a separate second step and then we've got the first ionization energy for calcium don't forget to include the electron that has just been lost when the calcium forms the gasius positive ions and then the second ionization energy of calcium so this time we're making two electrons we had one already the second electron has now been lost and so we've got calcium as a 2+ ion and all the while this sulfur is unchanged and so this is the new Arrow then we've got the first electron affinity of sulfur remember it's exothermic because a new attraction is being formed and then the second electron affinity for sulfur endothermic because we're overcoming the repulsion between those two negative substances and then we've got the ltis formation enthalpy on the right hand side and so the typical six arrows have been joined by the second ionization and the second electron affinity but how we use the born Harbor cycle is exactly the same let's suppose we were working out this enthalpy of formation it is still equal to when you follow all of these arrows they're all going the same direction and so the enthalpy of formation is still equal to the sum of all of the other arrows the ions in an ionic compound in that ltis are held together by electrostatic attractions which are very strong and they require a lot of energy to break you can actually calculate the strength of this electrostatic attraction using the electric field strength equation that you might know from physics and so that means that you can calculate a theoretical value for a ltis formation enthalpy when we consider ltis enthalpies like this this is referred to as the purely ionic model or the perfectly ionic model and this makes one really big assumption which is that ions are perfectly spherical which when we do the maths is a essential assumption but when we do the experiments we find out that it is not the case and the reason for this is that the positive and the negative ions do not exist as isolated perfect spheres in reality what happens is the electron clouds around each of these ions get distorted or polarized by attraction from the other ions now the good news is for a level chemistry they allow you to isolate this as an effect on the negative ion and so we always talk about polarization of of the negative ion by the positive ion but in practice it would in fact go in both directions and it follows that the charge density of the positive ion is going to have a really big influence on how much polarization occurs of the negative ion for instance a small highly charged positive ion will be highly polarizing whereas a large singly negative ion will be highly polarizable and so as a result you can get significant differences in polarization of the electron cloud around the negative ion for instance aluminium chloride would have significantly more polarization than lithium chloride because aluminium is a 3+ ion the polarized negative ion leads to there being electron density between between these two ions so we've still got the positive ion attracting the negative ion but we've got some electron density in between them as well and that's what a CO valent bond is it's some electrons between two nuclei and the nuclei are held together by a mutual attraction for those electrons and that's what a CO valent bond is and so this ionic substance has actually got some Co valent character in addition to the electrostatic attraction that is unchanged by this polarization and this leads to the ltis being stronger than you might expect and so the ltis formation enthalpy would be more exothermic than you would expect it to be a good way to explain the purely ionic model is to compare it to the purely calent model and in fact the purely coal valent model is easier to understand because we know that in coent substances the electro negativity difference can lead to a polar bond or a non-polar Bond and the best way of saying something might be purely Cove valent is if we had a diatomic molecule that was an element so hydrogen oxygen nitrogen Florine Etc this will be purely calent because there is no electro negativity difference between the atoms that are sharing that pair of electrons in the Cove valent Bond we know that you do sometimes get an electro negativity difference though for instance in hydrogen chloride you end up with a polar calent bond because the hydrogen is electron deficient because the chlorine has pulled the electrons away towards itself and we've got this polar bond forming because the chlorine is more electr negative and so this is a CO valent substance with some polar nature for a perfectly or a purely ionic substance you need to have perfectly spherical ions with no polarization and the best way of getting this is if you've got a low charge density positive ion in other words it's not very polarizing something like potassium that is big and singly positive and a very small singly negative negative ion so it's not very polarizable so a good example might be pottassium fluoride there would be very little additional Co valent character because there would be very little polarization at all whereas as you get something such as magnesium oxide which has got a positive2 ion it would be quite polarizing and so there would be some polarization and some additional Cove valent character and so we've got a Continuum with purely ionic and purely calent at the end and not many substances actually existing at either end and you've got most substances being somewhere in the middle being either ionic or Co valent but having something of the nature of the other in an exam situation you might be asked to consider the difference between the theoretical and the experimental latis enthalpy values and this could take a number of different forms for instance which would you expect to have the largest difference between Theory and experiments sodium bromide or magnesium bromide now in both of these examples the negative ion is the same that means the bromide ion is going to be equally polarizable in each instance and so the difference comes down to the positive ion the sodium is 1+ and larger than the Magnesium which is 2+ and so therefore the charge density will be greater on the magnesium ion so there will be more polarization of the bromide ion in magnesium bromide and as a result of that there will be more of a difference between experiment and theory and there will be more additional calent character in magnesium bromide alternatively they might ask you to compare two values that you have been given so for instance sodium chloride its theoretical latis enthalpy is - 766 and its experimental value is 787 in this instance these values are really quite close together and so since the values are close together you can say that there is very little polarization of the negative ion and so therefore there is very little additional Cove valent character in sodium chloride and that's likely to be because the chloride ion is quite little and not very polarizable and the sodium ion similarly is not very polariz in or they might give you an example where there might be a big difference so for instance magnesium iodide where you've got -1944 for the theoretical value but the real latis enthalpy is much more exothermic because it's minus 2327 and so in this case you would be saying that there is significant polarization of the negative ion which leads to significant additional Cove valent character and therefore the ltis is much stronger than you would expect we've got a much more exothermic value okay that's the end of this video I hope it was useful I'll see you again soon