Write formulas for the following binary compounds: (a) Vanadium(III) chloride (b) Manganese(IV) oxide (c) Copper(II) sulfide (d) Aluminum oxide

Video transcript

Hi everyone for this problem. It reads provide the chemical formula for each of the given binary compounds below. So let's just remember that in a chemical formula element symbols and numerical subscript show the type and number of each atom present in the smallest unit of the substance. So let's go ahead and start with answer choice A. So for this one we have palladium for sulfide. So let's identify what we have here. So we know we have palladium specifically it's palladium four. So we see that we have four which tells us that this is a positive for charge. Okay. And with that positive for charge that becomes palladium for plus. And this is our cat ion for an ion we have sulfide and this sulfide comes from sulfur. And if we look at our periodic table we'll see that sulfur is in group six a. And Group six A. has a charge of -2. Okay so that means that our sulfur is as two minus. And this is our an ion. So we have an ionic compound here. And when we have ionic compounds, the charge without the sign of one ion becomes the subscript of the other. So what we're going to do here is swap and drop, okay so when we rewrite this we're going to write palladium two, Sulfur four. Okay because we're swapping and dropping the charges but we need to simplify this, we're simplifying this because we're going to show the number of each atom present and the smallest unit possible. So that means for this we can divide by two. So we're gonna go ahead and divide by two. And then when we get our final answer it is p. d. S. two. This is our chemical formula for a. Alright so we'll go ahead and do the same thing for the rest. Okay so for B. We have rubidium. Selena died. And let's start off with rubidium. Okay so rubidium. We see from our periodic table that it's in Group one A. Okay so group one A. Has a plus one charge. Okay so that means our rubidium is just going to have a plus charge and this is our cat ion for Selena died. This comes from selenium and selenium on our periodic table is in group six A. And Group six A. has a negative two charge. So we know that our an ion is going to be this. Alright so we're gonna go ahead and do our swap and drop and when we do our swap and drop we get R. B. Let me do it in black. So we get our B. To S. E. This is going to be our chemical formula for B. Okay. For C. Let's go ahead and continue for C. We have for C. We have Millennium six. I died. Okay We have seen um six. I died. So let's start off with our molybdenum six. Okay so we identify this as group six With this plus six charge. Excuse me not. Group six. The it's molybdenum six. So we have a plus six charge. And so this gives us the sim the chemical symbols M. O. And we have a six plus charge. And that is Arkan ion for an ion. We have iodide and iodide comes from iodine and iodine is in group seven a. Of the periodic table. So group seven A. Has a negative one charge. And so our chemical symbol is I minus. So we're going to swap and drop our charges. So our formula becomes M. O. I. six. Okay, so this is our chemical formula for C. And lastly we have D. Okay so for D. We have are Niobium five sulfide. So let's go ahead and start off with our niobium OK. Nyo Beom five. So we recognize that we have the five and this is a plus five charge. Which gives us the chemical symbol is N. B. And the charge is five plus. And this is our cat ion for an ion we have sulfide and sulfide comes from sulfur. And from our periodic table Sulfur is in group six a. And Group six A. Has a minus to charge. So our chemical symbol for sulfide is S. And we know we have a two minus. So we need to swap and drop our charges. So our formula becomes N B. Two S. five. And that's our chemical formula for D. So these are all of our answer choices here. So let's go ahead and write them up above. So for a. We have P. D. To S. Four. For B, we have R B two, E four, C. We have M o. I six and four D. We have N B two S five. So those are our chemical formulas for each of the given binary compounds below. That's it for this problem, I hope this was helpful.