Determine the oxidation number for the indicated element in each of the following substances: (e) Pt in PtCl4

Video transcript

Hello. Everyone in this video we're trying to calculate the oxidation number of R. E. And R. E. C. L. Six. So how we do this is, well first there's of course the set of rules that a professor of the textbook says about calculating the oxidation number. And also there are some things that we will just know from maybe periodic trends. So first things first, what I've noticed is that we have a chlorine and that always has an oxidation number of negative one unless it's bonded to oxygen or flooring. But of course the chlorine is only bonded to R. R. E. So it applies. So basically what I'm trying to say is that cl their number is going to be negative one. And the second thing that I notice is that we have a neutral compound. So what that means or how that is significant to us in calculating the oxidation number is that whenever we have a neutral compound, all the oxidation numbers in the atoms will equal to zero. So neutral compound parentheses total equals to zero. Okay, now that we have those in our kind of like our box of tools, Let's go ahead and rewrite the compound. So we have our e. c. l. six and from what we have said here Because the cl equals to negative one and we have six of those, we'll just do six times negative one. And of course we don't know what R. E. Is. I'll denote that as an X. And because we have a neutral compound I said l sum up to zero. So doing that mathematically we have our negative six plus X equals to zero. And of course we're just isolating that X, meaning we'll have to add six to both sides. If we add six to both sides, we'll get X equals to positive six. And we have said that we're having the X represent R r E and therefore the oxidation number of our E. In our ECL six r compound Is going to be positive six. So let's go ahead and highlight that answer as well. Alright, so plus six is our final answer for this problem.