00:01
This problem is great practice to build our intuition of coordination chemistry, isomerism, and crystal field splitting.
00:10
The first thing that we need to do is draw to optically active isomers.
00:16
And these are what the isomers are going to look like.
00:19
These are the exact same compound, but they're essentially mirror images of one another.
00:24
We essentially reflect them across this line to see if they are actually near images, if they're superimposable is what we say.
00:36
And again, these are what our two isomers are going to look like.
00:41
And now what we need to do is find the charge of our cobalt in each of these isomers.
00:49
What we have to do in order to find that is determine the charge of each of the ligands.
00:55
Each ammonia ligand is neutral.
00:57
That would be nh3, where we see here.
01:01
Nh3 is neutral.
01:02
H -o -minus is negative 1.
01:08
Chlorine has a negative 1 charge, which means that our cobalt has to have a plus 3 charge.
01:16
So we're dealing with cobalt 3 in this case.
01:19
And because we know that we have cobalt 3, we can find the electron configuration of cobalt.
01:26
So we have cobalt 3.
01:28
We would look at the periodic table...