00:01
So in this question, write the loose structure for the following species indicate whether each is an odd electron electric deficient or significant valence.
00:10
So for the first one, we have boron and three iodines.
00:15
So we're going to just, first of all, calculate the number of valence electrons.
00:22
So in this case, boron is going to contribute three.
00:31
So we have 24 electrons total.
00:34
Therefore, you know that we have three groups of eight.
00:40
Can form up to three bonds because it is one of the elements that can result in electron deficient molecules.
00:51
So we're going to put three iodines around this boron, and then we end up getting this as a final looose structure.
01:11
And then for the next one, we have six halogens, so we do six times seven, and that's 42.
01:21
So iodine is going to be in the center because it's the these electron negative.
01:32
So then we're going to put the fluorines around it.
01:41
So you can see that we have five groups of eight.
01:44
So five times eight is 40...