00:01
Okay, so here we want to draw the resonance structures for the period 8 ion.
00:05
So that is i -04 minus.
00:10
So if we want to draw the resonance structure for this, we can first determine how many valence electrons we have to work with.
00:15
So oxygen has six valence, and iodine has seven.
00:21
And we have four oxygen, so multiply that by four and one iodine.
00:26
So that is 24 and 7 electrons.
00:30
And then we also have one extra electron from this negative.
00:33
Charge in the ion.
00:34
So overall that's 32 electrons.
00:40
So first what we want to do is we want to put iodine in the middle because we normally put the least electronegative atom in the middle, which is iodine in this case, and then we'll go ahead and orient the oxygens around it.
00:55
So we're putting in these four bonds, we've used up eight electrons.
00:59
So we'll go ahead and subtract that, which means we have 24 electrons left to work with.
01:05
We normally know that oxygen typically makes two bonds.
01:09
So we can go ahead and add.
01:11
In double bonds to each of these oxygens which takes up another eight electrons leaving us with 16.
01:27
And then now to satisfy the octet of all the oxygens will need to put in a lone pair for each of these, or sorry, two lone pairs on each of the oxygens so that each oxygen has a satisfied octet, which uses up the rest of our electrons.
01:53
So here we don't really see it a charge, which means that, well first we need to go ahead and calculate our formal charge because each of these has a satisfied octet.
02:02
So to calculate formal charge, formal charge is going to be the number of valence which if we do this for oxygen would be six minus the number of bonds.
02:11
Each oxygen here has two bonds and then minus the number of lone electrons which each oxygen has four electrons on it.
02:23
So none of the electrons have a formal charge.
02:26
If we look at the iodine, iodine has seven valence minus its number of bonds, which is 8 minus no lone electrons on it.
02:35
Which means that in this case, iodine has our ion.
02:42
So if we want to go ahead and draw the resonance structures for this, that's going to involve shifting around that negative ion.
02:51
And so if we shift that negative ion around, one way that we can do this is we can remove a bond from iodine and put it onto one of the oxygens.
03:03
And so that transfers the negative charge to one of the oxygens instead.
03:18
And so we end up with this resonant structure.
03:21
Now we could keep drawing this resonant structure and show the negative charge on each of these, but that would be a little bit repetitive because it's essentially the same structure.
03:31
But if we think about which one of these is going to be more stable, which one of these stabilizes the negative charge better.
03:37
So that's going to be either the oxygen or the iodine, and in this case it's the oxygen, because oxygen is a more electronegative atom.
03:45
So this is the better resonant structure.
03:48
Next, we want to draw two resonant structures for c4h8.
03:54
So carbon has four valence electrons times four of them, and hydrogen has one valence electron times six, which means we end up with 22 electrons to work with.
04:14
So if we think about how to orient this, we can typically put the carbon, in a straight line like a heart hydrocarbon and then we can go ahead and add in hydrogens...
