Assign oxidation numbers to each element in the following ions. (f) V2O7 4-

Video transcript

Hello. Everyone in this video, we're trying to determine the oxidation state of each element in this ion right here. So oxygen usually has an oxidation state of negative two. With exceptions, if it's reacting with a peroxide or super oxide. But here in this case we don't have that so we don't need to worry about it. And it was just a negative two. So we can see right off the bat that the overall charge of this ion is negative two. Let's go ahead and write this out. So the overall charge Is equal to -2. This is important for our calculations. We're gonna go ahead and let x equal to the oxidation state of our unknown, which is elemental symbol W. And that's just tungsten. Again, I'll just rewrite our ion here. So W. 042 minus. So we said earlier that our oxygen provides a negative two oxidation state and then our W. Is just unknown which is X. And the overall charge is equal to negative two. So all of the oxygen oxidation states should equal to negative two. So for our tungsten we have it as an unknown and we just have one of that. So just be one X. And then we're just adding up the charges. So we have four oxygen atoms. Each provides a negative two charge. So the simple math here, four times negative two is equal to negative eight will bring the negative eight to the opposite side combining with the other negative number. So we added to the other side, we get that equal to a positive six equaling two X. So then the oxidation state Of Tungsten is equal to positive six, and our oxygen is equal to -2. This is going to be my final answer for this problem.