Identify the correct electron-dot structure for XeF 5+ (a) (b) (c)

Video transcript

Hello everyone today. We are being given the following problem, draw the lewis structure for perry repair I a date or this following formula here. So the first thing we want to do is you want to calculate the number of valence electrons. So calculate the number of valence electrons. So in this pair I a date ion we have iodine and we have oxygen. So for iodine, iodine is a group 78 element and therefore has seven valence electrons. Oxygen on the other hand is a group 68 element and has six valence electrons. However there are four of them. So we're going to take those six valence electrons and multiplied by the four oxygen that we have giving us 24 valence electrons. And then since there is a negative charge that is going to contribute to how many electrons we have. So we're just going to add one more electron. So we have seven, We have 24 and we have one electron here, Giving us a total of 32 electrons. So we have a total of 32 electrons or valence electrons that we can use. So then we have to note that iodine is special because iodine can expand its octet and so why can it expand its architect? So it can expand its architect do to it having an empty D orbital. So it has an empty D orbital that can accept electrons. And this is actually true for any element in period three of the periodic table and lower. So since we have one iodine and four oxygen's. What we can do in constructing this formula here, we can draw our one iodine here and we can surround it by our four oxygen's. We see that when we do this, this does not account for all of the electrons that we possibly have. So going back to what elements were working with iodine and oxygen oxygen forms two bonds can form up to two bonds, and iodine, iodine usually forms one bond but it can expand its octet. And so how can we represent that? We can actually go ahead and draw double bond for up to three of our oxygen I. D bonds. And so if we take note of that, we can see That we've used seven bottom lines. Each bottom line represents two electrons. So we've used 14 electrons so far. And so that means we have some that we can represent around our oxygen atoms by destroying them as lone pairs. So three of these oxygen, three of these oxygen's will have two lone pairs and one will have an additional loan pair with that negative charge that we saw in the formula. And so this is going to be our structure for per I date. And so with that we have solved the problem overall, I hope that this helped until next time