(d) Treating bismuth with fluorine gas forms BiF5. Use the
elect...

Question

(d) Treating bismuth with fluorine gas forms BiF5. Use the 
electron configuration of Bi to explain the formation of a compound 
with this formulation.

Accepted Answer

Hello. Everyone in this video. We have this compound here because phosphorus and chlorine are reacting to make this compound and we want to write out the electric configuration for a phosphorus atom to describe why we can be able to make this compound. So let's go ahead and do that. So every look at this period table I kind of color coded of my blocks. So in yellow. So let's write that on the bottom. Yo that's going to be our s orbital or S block. I mean a block in green here, we have our D block in blue, we have our P block and in pink we have our F block. Alright, let's go ahead and also label the road numbers. So 12, 3, 4, 5, 6 and seven. So whenever doing electron configuration, I cannot think of it as we're reading a book. So we know that a phosphorous element is going to be right out here. So we're simply reading pure and table from left to right until we get to the desired element, which is right over here. In our case. So mhm let's see here for phosphorus, let's go and actually kind of do a shorthand weight which is going to be the noble gas configuration. So we want to find the noble gas that is closest but before it hits this element. So in our case we're not going forward, it's not going here, even though this is our closer no more guests, we're gonna go backwards and get to neon. So neon will be the noble gas that we're starting off at instead of starting off at this hydrogen, it'll be a lot more work. So shorthand this and put neon, we can put this exactly. Just by putting our Nobel gas in a bracket and that indicates our starting point. So we're starting right over here and continuing on, we get to the 3rd row. So we said that we're going to be in yellow, it's going to be our S block. So first spreading the row number, which is three, then stating which block we're on, which is s we have red one and two spike spaces in this block. So we'll put a little too here and moving on across the road will get to over here, same bro, so put that down. So now we're at three. The color blue means that we're in the p block. So go ahead and put heat. And then how many little boxes do we need to get to or read until we get to our phosphorus elements. So that's 12 and three. So put a little three here and that's going to be the shorthand or noble gas configuration or phosphorus adam. So as you can see here, these are going to be our valence electrons and we have a total of let's see, we have two from our S block And three From Our P. Block. So to post-3, then that is five with these five valence electrons will go ahead or go ahead to reach out to our chlorine atoms, which we have five of to make a single bond. And because we have five of those single bonds, then you'll go ahead to make this law possible. And that is going to be our final answer for this problem.

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