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Video transcript

a student learns that ionic compounds have significant covalent character when a cation has a polarizing effect on a large anion so what are they talking about so if I have a cation so this is my cation and then this is my large anion my large anion right over here actually let me do it this way let me make it a little bit clearer because a cation still has electrons so you have your nucleus here of your cation you have your electron cloud you have your electron cloud of your cation right over here but you have fewer electrons than you have protons so you're going to have a positive charge and that's why we call it a cation and the anion it still has a positive nucleus it has an electron cloud I'm a drawing a large anion here a large a large a large anion right over here that's electric compact electron cloud but it has more electrons in the cloud than it has protons in the nucleus so it has a negative charge that is the anion and what they're saying is is that you start to have significant covalent character remember covalent bonds or electrons are essentially shared between two atoms when a cation has a polarizing effect on a large anion so one way to think about it is these electrons that are out here well they're gonna want to get away from all their negative friends here negative charges repel each other and they might want to spend more time closer to this cation and so they're gonna start having covalent character they're gonna be more and more shared between the two and there really is a spectrum between ionic bonds and covalent bonds and over here you might have a polarizing effect these things spend more time on the left-hand side that we have drawn it then on the right-hand side so you're gonna have more of the electrons and more of the negative charge hang out over here then on the right that's what they're talking about and as a result the student hypothesizes that Salt's composed of small cations and large anions should have relatively low melting points all right so it should be easy to take less energy to break them out of there lat they're kind of the salt lattice structure when they're in solid form select two compounds from the table and explain how the data supports the students hypothesis let's see these are all salts here and there on the if the if these were to disassociate or if you think about what the the cations would be the cations the cations that make up that make up these these compounds you have lithium lithium you have sodium and you have potassium and then the anions the anions that make up these compounds you have iodide you have iodide and you have fluoride something right this way fluoride and iodide so which of these are small in which of these are big well let's compare lithium Patek let's compare lithium sodium and potassium and we see lithium sodium and potassium they are all Group one elements and in general as we go down a group as we go down a column in the periodic table we're adding shells and so size increases so size increases as we go down like this size size increases and so of these lithium is the smallest and potassium is the largest of these three so lithium let me write that down lithium is the smallest smallest and potassium is the largest of these of these three now if you look at the anions well you have once again fluorine and iodine are in the same are in the same group and iodine is below it it has way more electrons and you've added more shells here's is actually the important part and so you can see that since iodine is larger than fluorine well iodide which is which is when they gain an electron it's going to be larger than fluoride so iodide iodide this one is larger and the fluorine is smaller so let's see lithium iodide is a case of small cation small cation and large and large anion and so that's actually the the direction that's kind of the best example of small of what the student is talking about small cation and large anion lithium iodide lithium iodide is a great example of that and as we can see it has the lowest melting point of everyone on the table now let's take the other extreme what happens if we take a large cation and a small anion well let's see they don't they don't have that combination here but let's see if we hold if we see sodium well sodium floor of sodium fluoride yes so slow sodium fluoride is interesting because we have a larger we have a larger cation so this is a larger cation and then we have a smaller anti on smaller anion and notice that has the highest that has the highest melting point and so these are two that are good to compare so part a I'm doing that blue compare compare lithium iodide which has small small cation plus large anion to two sodium fluoride which has larger cation plus smaller anion smaller smaller anion and we see and we see that their melting points their melting points melting points consistent with hypothesis consistent with hypothesis the small cation or lithium iodide so small I'll just write its melting point melting point lower lower and actually much lower than sodium chlorides sodium fluoride melting melting point so at least if you compare those two and they still seem to be good extremes we took the lowest melting point which completely typifies what the student was saying is having a low melting point and we looked at the highest melting point which had a larger cation and a smaller anion and so just looking at that it seems to be consistent with the students hypothesis alright Part B identify a compound from the table that can be dissolved in water to produce a basic solution write the net ionic equation for the reaction that occurs to cause the solution to be basic alright so let's think about how we can form a basic solution so essentially we would need to nab some hydrogen's from the water molecules in order to have some hydroxide laying around and so there's a couple of candidates here that could do that you have all of these all of these halides you have the iodide do you have the fluoride and the important thing to recognize is that hydrogen iodide and you don't see or hydrochloric acid since we're dissolving it in water now Heidrich that's that hydro core hydro is it so c-h-i hydroiodic acid strong acid strong acid while hydrofluoric acid is a weaker acid is a weaker as so a strong acid is not going to be good at nabbing hydrogen's in fact it's a strong acid it wants to give away its hydrogen's really really really badly so so this so the the iodine does not seem like a good candidate the floor the the or the iodide the fluoride does seem like an interesting candidate so let's let's take one of these let's take one of these candidates out here you could use either the lithium fluoride or the sodium fluoride let's just take I know we've already dealt with the sodium fluoride so let's just keep using that so if you take sodium fluoride sodium fluoride and so let me since we're they want us they say write the net ionic equation for the reaction that occurs to cause the solution to be basic so we're gonna focus on sodium fluoride sodium sodium fluoride is what we select we select sodium fluoride that's that's the part the first part we identified a compound that can be dissolved in water to produce a basic solution we select that and now let's draw let's do a first the ionic equation then we'll do the net ionic equation so if you dissolve this in water you're going to have sodium cations dissolve in our aqueous solution plus fluoride anions dissolved and are dissolved in our aqueous dissolved in our aqueous solution and it is going to be in equilibrium in equilibrium with I'll actually let me draw it let me draw it to this way let me draw it this way plus plus h2o because that's what's going to react with and obviously I don't have to say it's an aqueous solution it is the aqueous solution so it's making the solution aqueous this is going to be in equilibrium with and I'll go to the next row here if I was actually taking the AP test well actually let me just let me just copy and paste it here just so we don't have to deal with it on two rows so put it right over there so that is going to be in that is going to be in equilibrium we get the right tool out that is going to be in equilibrium with this this fluoride NAB being a hydrogen so hydrofluoric acid we could say h hf in our aqueous solution plus o h minus in our aqueous solution and then you still have the sodium cation in our aqueous solution and so this is you can see we're going to perform an equilibrium and so you're gonna have more hydroxide around so you're forming your basic you're forming your basic solution if this was an iodide right over here then this reaction would go strongly in that direction if this was hydrogen iodide right over here this wouldn't be some kind of nice equilibrium this would be a strong acid go strongly it would definitely want to get rid of its hydrogen's so that's why we don't want to use hydrogen iodide we want to use hydrogen fluoride and you put in water becomes hydrofluoric it's it's likely to donate it we when it's in its gaseous form or it's by itself we would call it a hydrogen fluoride but now when we put it in a solution we can consider to be hydrofluoric acid and so if you want your net ionic equation well we don't have to worry too much about these sodium's they're on both sides of this and so if you just focus on this right over here this is your net ionic net ionic equation
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