Write a formula for the ionic compound that forms between each pair of elements. a. silver and chlorine b. sodium and sulfur c. aluminum and sulfur d. potassium and chlorine

Video transcript

Hey, everyone. And welcome back to another video. Write a formula for the ionic compound that forms between each pair of elements. A silver and chlorine B, sodium and sulfur C aluminum and sulfur and D potassium and chlorine. We have four answer choices. They are all combinations of these elements. And we want to determine which one is the correct answer for this problem. So let's begin solving this problem. We will start with a, we will write the elements that are given to us. We have silver. We noticed that silver is a transition metal on the periodic table with a symbol of A G. And also we have chlorine cl which belongs to halogens or basically group seven A. Let's recall that the charge of silver is always positive one. This is something that we want to recall possibly memorize because it's the only charge of silver that we're going to use in this course. Now, for chlorine, because it's in group seven A, it's a non metal, it only needs one electron to get an octet and all of the seven A group elements would have a negative charge in ionic compounds. Now, what we're going to notice is that if we add them together, our goal is to get a compound with a net zero charge, right? Because if we want to form a compound, the number of positives should be equal to the number of negatives. So if we have one positive and one negative, they immediately cancel each other out to form a compound. A GCL, one unit of each. And how do we know that? Well, essentially we can also use the Crisscross method which tells us that silver would have a subscript equivalent to the charge of chlorine. So we can ignore the negative sign and we can say we have one negative. So that's one as a subscript for silver and the subscript of chlorine would be one as well because we bring down the positive one charge which gives us a number of one well done. So we can continue this way and take a look at part B. We have sodium group one A metal because it belongs to group one A, it will have a positive charge, right? All of those metals as well as hydrogen. They need to lose one electron to become cations. And now sulfur because it's in group six A, it would have a two negative charge just as oxygen because sulfur needs two electrons to get on tap. So now if we have two negatives, that means we need two positives to balance out the two negative charge. We can check that once again, using the crisscross method, sulfur has a two negative charge. We bring down the two for sodium and sodium has a charge of one. So we bring this down as a subscript of one which is implicit for sulfur. And that's how we get N A two S. Now for part C aluminum, that's group three A. So that means aluminum will have a three positive charge. And our sulfur, we already know that it has a two negative charge. This one might be a bit more complex because we cannot immediately balance those charges out. We need the least common multiple between two and three, which is six. So what does that mean? Well, essentially the easiest way to write down the formula would be to use the crisscross method. We bring down the two from sulfur and this becomes the subscript of aluminum and we bring down the three from aluminum which becomes the subscript for sulfur, which makes complete sense because what we notice is that we have six positives, two multiplied by three and six negatives. Finally, we have our final compound which is a compound formed between potassium K. It's in group one A meaning it also has a positive one charge. And chlorine, the L negative, we already know that chlorine has a negative one charge. And when we combine them together, we immediately notice that positives cancel out the negatives. It's a 1 to 1 ratio because we have one positive and one negative similarly to what we had in part A. So now which one is the correct option here, starting with the first compound, we got a GCL. So we can eliminate A because it's a GCL two. Now, for B we got N A two S so we can eliminate option B, we have Nas and C, it says NAS two. So N A two S then we have a L 2 S3 and KC. So we can essentially conclude that the correct option to this problem is option D A GCL and A two SAL, 2 S3 and KCL are the compounds formed between the elements given. Thank you for watching.