00:01
So we're going to draw some lewish structures here.
00:05
The one thing we're told is that they all violate the octet rule in some way.
00:10
Well, this one's going to be a little straightforward because we know we can only form single bonds.
00:16
Okay.
00:17
Barillium has two valence electrons, plus two more for the hydrogen is four electrons.
00:23
So that's actually our lewish structure.
00:25
Barillium always forms two bond and has no lone pairs.
00:28
So in the second one, we're going to try co with a negative charge.
00:36
So our valance electrons are 4 plus 6 plus 1 for the charge is 11.
00:41
So since we have an odd number of electrons, we're obviously going to violate our octet rule.
00:46
So let's see, let's pretend it's got single bonds, give everybody a full octet.
00:52
And we'll see that we have 14 electrons.
00:55
So is that 2 extra, or is that 4 extra? well, it's two, but not quite four, so we're going to use the rule for two extra, which means we need a double bond.
01:08
So we'll go ahead and rewrite this with a double bond here.
01:14
Again, we'll give everybody a full octet, but we still have 12 electrons now.
01:19
So i need to remove one electron, and i'm going to remove it from the element with the smaller electron negativity, which is carbon...