Determine the oxidation number for the indicated element in each of the following substances: (c) P in AgPF6

Video transcript

Hello everyone. So in this video we're going to calculate for the oxidation number of S. E. And M. G. S. E. 202. Some things that I noticed. Well 1st we have a neutral compound in that case it means that all the oxidation numbers will equal to zero. So we have a neutral compound and therefore the sum so sigma It's going to equal to zero. Another thing that I noticed is that we have this MG atom that's in group that's going to have an oxidation number of plus two. So M. G. Group to a plus two charge. Another thing that knows is that we have an oxygen And oxygen oxygen will always have a negative two oxidation number. So O. is -2. I'm gonna go ahead and rewrite our compound. So we have our M. G. S. E. Two 02. So are MG. That's going to be a plus two charge R. S. C. Something that we don't know And we have two atoms of that. So we'll put that as two X. two because we have two atoms and X. Because we just don't know what that is now for oxygen we have two atoms of oxygen and each has a negative two oxidation number. So two times -2 is -4. Writing discount mathematically we have two plus two X minus four. And of course we have that equaling to a sum of zero. Now simplifying this will have two X minus two equal to zero adding two on both sides. We get two X equals two to isolate our X. We can divide each side by two, giving us X equals to one. And because of this calculation, we can see that the oxidation number of r S E adam In the compound is going to be positive one and that is going to be our final answer for this problem. Thank you all so much for watching.