 You are looking at the lattice enthalpy which is defined as the enthalpy change when one mole of anionic compound dissociates into its ions in a gaseous state. So, to take an example, if we look at this reaction where we have sodium chloride dissociating into Na plus and Cl minus ions and both of them in gaseous state. So, the enthalpy change for this reaction is the lattice enthalpy and its value is given here and you can see how the enthalpy here is positive because to break this ionic bond in NaCl we will have to provide energy and so by a sign convention the lattice enthalpy is positive. Sometimes in textbooks lattice enthalpy is defined for the reverse of this reaction like here you have sodium and chloride ions forming NaCl and because this is the reverse of this reaction the enthalpy is negative in this case but calling both of them lattice enthalpy can be confusing. So, a better way to think about this is that the case above where NaCl is dissociating to give sodium and chloride ions is called lattice dissociation enthalpy and the other case where these ions form NaCl is called lattice formation enthalpy and you can see how the sign of the enthalpy follows from what we discussed before because energy is required to break this NaCl the enthalpy is positive and here the formation of NaCl releases energy and so the enthalpy is negative. So, in numerical problems you have to be careful about the sign of the enthalpy depending on whether the lattice is being formed or it is being dissociated. So, if we take one more example we have this MgCl2 or magnesium chloride dissociating into magnesium ions and chloride ions and the lattice enthalpy for this reaction is given here and again we see that this is also positive because you are providing energy to break the ionic bonds in this case and one more thing you will notice here is that the lattice enthalpy of MgCl2 is much higher than the lattice enthalpy of NaCl. So, one way in which we can think why this is the case is that if we look at these reactions what we are doing is we are trying to break these bonds and if you think of the ions as charges we know that there will be an electrostatic force acting on these charges and the magnitude of this force is proportional to the product of the charges and inversely to the distance. So, if we compare these two in case of magnesium the magnitude of the charge is more so the electrostatic force is stronger and therefore to break this stronger bond a higher energy is required. So, following this reasoning we can write that the lattice enthalpy is proportional to the charge of the ion so higher the charge of the ion more is the lattice enthalpy. So, if we were to compare NaCl and MgO we want to find out which of these will have a higher lattice enthalpy and here we know something about the crystal structure of both of these since both of these have the same crystal structure the arrangement of ions in space in both these cases is similar. So, if we go back to our comparison with electrostatic force we can think of distance not playing a factor here so the electrostatic force will be stronger when the magnitude of charge is more in this case the magnitude of charge is more in the case of magnesium. So, we can say that the lattice enthalpy of magnesium oxide will be higher than that of NaCl but in this comparison because we know that both of these have the same crystal structure we have not considered the effect of arrangement of ions in space. So, let's take one more scenario now if we were to compare the lattice enthalpies of all four of these compounds we can see that the magnitude of charge on each of them is the same. So, we'll have to look at the distance factor which is affecting the electrostatic force and accordingly we can guess what will be the order of lattice enthalpy. Now, as we go towards the right we know that as we go down the column the ionic radii increases so iodide will have a much larger size compared to fluoride and because the size of iodide is more the distance between the sodium ion and iodide will be far more than the distance between sodium and fluoride and again in the case of iodide since the distance is larger the electrostatic force will be smaller which means it will be far more easier to break the sodium iodide bond than the sodium fluoride bond. So, accordingly we can say that the lattice enthalpy will be highest in this case and will keep decreasing as we go from fluoride to iodide. So, we can say that the lattice enthalpy is inversely proportional to the size of the ion. Now, let's see how the lattice enthalpy is calculated using a Born-Heber cycle. So, as we saw before we are defining the lattice dissociation enthalpy here as the enthalpy when NaCl dissociates to form sodium ions and chloride ions. So, I want to calculate the enthalpy for this reaction but I don't know where to start but I know one reaction for NaCl which is the formation reaction. So, let me just write it down. So, I know this reaction where solid sodium and chlorine gas react to give NaCl which is solid and I know that for this reaction the enthalpy will be equal to the enthalpy of formation. Now, I know something about the sodium as well. I know of a way to convert this sodium to an ion but here the ion is in gaseous state. So, I can do it in a two-step process. First, if this sodium undergoes sublimation I know that it is going to form this sodium in gaseous state and the enthalpy for this reaction is going to be this delta H sub which is the enthalpy of sublimation and similarly for chlorine I can convert this chlorine molecule into an atom of chlorine gas and the enthalpy for this reaction is given as the bond dissociation enthalpy which is the enthalpy change for breaking one mole of chlorine molecules into two atoms of chlorine but here we only have one atom of chlorine. So, the enthalpy for this reaction will be half of bond dissociation energy. Now, instead of writing this separately I can combine these enthalpies and write it as enthalpy of sublimation plus half of bond dissociation enthalpy. So, that is the total enthalpy change when we go from here to here. Now, next if we look at this sodium and chlorine which are in gaseous form to convert them into ions all we have got to do is to provide the ionization energy and the electron gain enthalpy because we want the sodium which is in gaseous form as an ion and for the chlorine it will gain one electron. So, for this step the total enthalpy will be the enthalpy of ionization plus the electron gain enthalpy for chlorine and now we can just complete this step and we already know that this is our lattice formation enthalpy. So, following our sign convention I am going to write it as minus delta HL. So, if we look at this set of reactions we can get from here to here via two parts. One is this part where the enthalpy is the enthalpy of formation and second is via a combination of these three steps. So, by Hess's law we can write a relation between these enthalpies which will look like this. So, we know that the enthalpy for the first path which is delta HF is equal to the sum of the enthalpies across the other parts which is the sum of all of these including this lattice enthalpy term. So, if the values of the other enthalpies are known we can use this relation and calculate the value of the lattice enthalpy. So, this process of applying Hess's law for an ionic compound to get the lattice enthalpy like this is called the Born Haber cycle. We use the Born Haber cycle like this to determine the lattice enthalpy because it is not directly possible to experimentally determine the lattice enthalpy which is why using Hess's law since we have this relation the other enthalpies can be experimentally calculated and so using these values we can get a value of lattice enthalpy. Now, the same idea can be represented in a different form which looks something like this. It's slightly more complicated looking but it's the same thing that we saw before. So, earlier we saw that we had two paths to get from this reaction of sodium plus chlorine gas forming NaCl. So, first was the direct path which I have marked here by a delta HF which is the enthalpy of formation and the other path was the sum of all other enthalpies and all of these enthalpies are the same that we saw before. It's just that I've written them individually. So, earlier we had club the atomization and half of the Born dissociation energy but here I've written all of them individually. So, we have the atomization enthalpy, the ionization enthalpy, half of the Born dissociation enthalpy, the electron gain enthalpy and the lattice enthalpy. By Hess's law, we get the same relation that we saw before that the enthalpy of formation is equal to the sum of all of these enthalpies which we can write like this. But then what is different in this representation? First, as I said, I've written down each of these enthalpies individually and second, if you notice the arrows, the arrows tell us the sign of the enthalpy. So, the enthalpy of formation is negative which is why this is given as a downward arrow. The electron gain enthalpy and the lattice formation enthalpy are also negative which is why they are given by downward arrows and all of these enthalpies are positive. So, these are denoted by upward arrows. So, this is just a different way of representing the same idea that you sometimes see in textbooks or other places and here also using this relation, if we know the other enthalpies which are experimentally determined, we can rearrange this to calculate the lattice enthalpy.