Write the formula for each ionic compound. c. silver nitrate

Video transcript

Welcome back everyone. We need to determine the formula of the given compound. Gold one sulfide. Let's begin by distinguishing the atoms that make up this compound, we have gold with a roman numeral one. Recall that gold is in our transition metal D. Block on our periodic table. And the roman numeral one tells us that we form the gold plus one. Carry on. So next we have sulfide recall that we're going to focus on the Adams sulfur which is located in Group six A. On the periodic table where we have the following symbol, us with a exponents of minus two. This is our sulfide an ion which is a non metal. Again because sulfur is a non metal and it's an ion. So in this case we're going to now do the criss cross method. And this is because since we have the combination of a metal and a nonmetal, we can understand that this compound is an ionic compound, meaning that we have the transfer of electrons between these two atoms. And so via the criss cross method we would cross the two charges. So beginning with our metallic atom. First gold, we have a subscript of two because two electrons are transferred two gold from sulfur and next with our nonmetal adam, sulfur or sulfide. In this case we're going to have a subscript of one which is implied because self sulfide is going to have one electron withdrawn based on on the plus one charge of gold. And so for our final answer we've successfully determined the formula of gold one sulfide as a U. Sub two S. This will correspond to choice C. In the multiple choice as the correct answer. I hope this made sense and let us know if you have any questions.