00:01
All right, so this one could be a little bit tricky because it says compounds or ions.
00:05
So we're kind of going to have to take a guess at what the charge is.
00:08
Compounds would be neutral and ions could be positive or negative.
00:13
So what we're going to see here is whenever we have a bromine with an odd number of fluorine atoms, we'll have an even number of electrons and we won't need to introduce any sort of charges to it to make a molecule that makes sense.
00:27
Whenever we have a bromine with an even number of foreign atoms, that'll be two and four.
00:33
We'll need to put a charge in there, or we'll have an electron that's all by himself.
00:38
So let's jump right in with the first one, brf1.
00:45
So bromine and fouring each like to form one bond.
00:49
So that's really easy to do because they have seven valence electrons.
00:54
It looks just like this.
00:55
It looks sort of like br2 or f2, but just one or the other.
00:59
So this is going to be sp3 hybridization.
01:05
And it's also going to form a linear compound.
01:09
For brf2, we're going to encounter this odd number of electron issue first.
01:16
So if we fill out forin's electrons like this, now bromine has seven valence electrons.
01:24
Two of them are tied up in bonds.
01:26
So we have five left over.
01:30
If i put this fifth one in here, we see that we have an odd one.
01:33
Number and we've also violated the octet rule.
01:36
So we're going to instead remove that and just go with brf2 plus as our molecule.
01:44
Since bromine can expand its octet, it could be a reasonable guess to also try brf2 minus.
01:50
You'll get a different geometry than what i do, but it's still a valid structure.
01:56
So over here, again, this is going to be sp3 hybrid, and it's going to form a bent structure.
02:05
This is analogous to water here, where you have an oxygen with two electrons and two bonds.
02:14
For brf3, what we're going to see is our first time that we have to expand our octet, because bromine, again, has seven valence electrons...