A series of anions is shown below:

The anion on the far right is called 'BARF' by chemists, as its common abbreviation sounds similar to this word. (d) Tetrabutylammonium, (CH3CH2CH2CH2)4N + is a bulky cation. Which anion, when paired with the tetrabutylammonium cation, would lead to a salt that will be most soluble in nonpolar solvents?

Video transcript

welcome back everyone in this example, we have a list of an ions and Canadians shown below. We need to choose the pair of Catalan and an ion that would form an ionic compound that is more likely to be soluble in a non polar solvent. So we need to first recall the trend that the larger the ions corresponds to greater dispersion forces. And because we have these greater dispersion forces, we would therefore have greater Celje bility in non polar solvents. So we got we need to recall the trend on our periodic table with regards to ionic radius. We want to recall that ionic radius is going to be increasing as we go towards the bottom left of our periodic table. So looking at our options, it looks like options one through four represent our Catalans because we can see our central atoms here have a positive charge, where we should recognize that between our central atom being nitrogen as the Catalan versus phosphorus as the caddy on. We would agree that since phosphorus is located in group five a just like nitrogen is but is further down in period three of our periodic table, we would agree that phosphorus has a larger ionic radius and so therefore these two ions of phosphorus are larger cat ions than our nitrogen atoms here. So we're going to rule out choices one and two and now to pick between these two molecules of our phosphorous cat ions, we would recognize that we have these very large aromatic groups surrounding our phosphorus catalon in molecule four and we want to recall that at each corner of these aromatic groups are carbon atoms where these carbon atoms have bonds, single bonds to hydrogen atoms. Now they're all implied, which is why we don't have to really draw these hydrogen atoms atoms in. So we can see that versus compound three. We have a less amount of carbon atoms and hydrogen atoms and molecule four. Sorry, we have more carbon atoms in each of these aromatic rings. And because we have more carbon and hydrogen atoms in each of these aromatic groups, we have an increase in polarity and non polarity due to the fact that we have more of these carbon hydrogen atoms in our aromatic groups, creating more dispersion forces. So we'll make note of the fact that we have aromatic groups meaning we have more carbon and hydrogen atoms then molecule three. And so because we have these more carbon hydrogen atoms, we have an increase and none polarity of the molecule and an increase in dispersion forces. Which is why we can now rule out molecule three, meaning that molecule four would be the best caddy on that we could choose out of our set of cat ions. And now we want to figure out the best an eye on that we can that we can choose to pair this caddy on up with. So looking down below, we're comparing our an ion of boron to oxygen to phosphorus, to nitrogen. And recalling upon our trend of ionic radius Again, we said that as we move towards the left, towards the bottom left specifically of our periodic table are ionic radius increases and out of all of the ann ions that we're comparing to phosphorus is the lowest being in period three. So it will have the greatest ionic radius, meaning we can rule out choices 65 and then eight leaving us with choice seven as the best an ion to pair up Arkady on with again, because it has the largest ionic radius based on all of our other or compared to all of our other an ions. And this would make sense because we will have greater dispersion forces and therefore a greater scalability in a non polar solvent. And we know that like dissolves like so we want to choose our pair that are both the greatest in their character of being non polar. So choices four and seven are our final answers. To complete this example. I hope everything that I reviewed was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video