00:01
So here we have the problem 23 .64.
00:04
We're going to draw the crystal field energy level diagram and also show the placement of electrons for the following complexes.
00:12
So part a, we have valadium cl6, 3 minus.
00:20
So first we want to find the oxidation number of the metal valadium in this case.
00:26
So we assume it's x, and we know that the oxygen.
00:30
Oxidation number for chloride is negative 1.
00:34
That is equal to the total oxidation state, which is negative 3.
00:39
And we solve for the x equal to positive 3.
00:45
So the oxidation number for valadium is positive 3.
00:52
So v3 plus will have an a electron configuration as ar 3d2.
01:01
So now we draw the energy level diagram.
01:18
1, 2, this is the 3d orbital.
01:24
Alright, next one is part b, we have f, e, f6, 3 minus.
01:39
In this case, we assume the iron has oxidation state of x, and we know the fluoride has oxidation number of negative 1 that is equal to negative 3.
01:54
So we solve for x that is equal to positive 3.
01:59
So the oxidation number of iron in the complex is 3 plus.
02:04
So for iron 3 plus, the electron configuration would be ar 3d5.
02:16
So now we draw the energy level diagram for this.
02:23
We have, this is the d5 orbital.
02:33
There are five electrons, three, four, and five.
02:41
All right, so that is part b.
02:43
Part c, we have the complex rubidium, i, pirating, three, three plus.
02:55
And this is a low spin complex.
03:03
Alright, so now we first look for the oxidation number on the rubidium.
03:08
So we know that the bipypyridine has a zero oxidation state.
03:13
So the total oxidation state of the complex is 3 plus, so we know the rubidium has a 3 plus oxidation state.
03:22
So for a rhubidium 3 plus, the electron configuration would be 4.
03:29
So there are 5 electrons in the 4d orbital...