00:01
In this problem, we're going to take a look at carbon dioxide or co2 and carbon monoxide, or co.
00:08
To begin with the lewis structures, we know that each carbon atom has four valence electrons.
00:15
We know that each oxygen atom has a total of six valence electrons.
00:20
However, because we have a subscript of two on carbon dioxide, we need to multiply that by a value of two.
00:28
So in reality, we end up having 4 plus 12, which gives us a total of 16 valence electrons.
00:38
When we think about our carbon monoxide, now we have 4 plus 6, so now we know that we only have 10 valence electrons.
00:49
In our carbon dioxide, we are going to use the least electronegative atom in the center, followed by our more electronegative atoms on the outside.
00:58
Now, since we know that carbon has 4 valence electrons, we know that we want to end up to be able to be able to the center, have having four electrons that are donated from the oxygens in order to kind of get us this nice little octet for our carbon.
01:11
Because there are eight electrons involved with these pairs of electrons, we can subtract eight electrons out of here and end up with eight valence electrons.
01:22
Now one thing you can do since we know that each electron travels as a pair on there, we can divide this by two, and this will tell us that we know that we need to have four more pairs of electrons.
01:33
Oxygen in its neutral state likes to have two lone pairs of electrons, and it likes to have two lone pairs of electrons along with those two bonded electrons pairs.
01:43
So therefore, in each of these particular cases, we know that our oxygen has reached its octet, our carbon has reached its octet, and our other oxygen has reached its octet...