Which solid in each pair has the higher melting point and why? d. NaCl(s) or CaO(s)

Video transcript

Hey everyone in this example, we need to determine which of our given molecules has a lower melting point value and we need to explain why. So looking at our first molecule, we have potassium bromide, which is a solid compound, we want to recognize that potassium which is in group one a. Will form a plus one carry on. And bromine, which we recognize is in Group seven A. Of our periodic table will form a minus one an ion. And so we have the K. Plus and B. R minus an ions present within this molecule that transfer electrons between the two atoms making up potassium bromide. We want to compare this to our second molecule, given calcium sulfide, which is also a solid. We have a metallic calcium atom bonded to the non metal sulfur. So this is going to be another ionic compound. So these are both ionic compounds which involved the transfer of electrons where calcium we recognize is in group two A. So we would form the two plus ion and then sulfur we recognize is in group six A. On our periodic tables where we formed the S two minus an eye on. So we have the electrons between these two ions transferred between one another to form our calcium sulfide compound. And what we should recognize is we have doubly charged ions for calcium and sulfur within calcium sulfide versus in our potassium bromide we have lower singly or lower single charged ion densities for our potassium and bromide an ion. And so we can say that therefore potassium bromide will have weaker ionic bonds than our calcium sulfide and a lower melting point. And so we can say to complete our example that potassium bromide has a lower melting point due to its smaller charge densities of its ions. And so this statement here, sorry, would be our final answer to complete this example. So I hope that everything I went through is clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.