00:01
So now we'll talk about problem 5 from chapter 22.
00:07
In this problem we need to write electron configurations for several ions.
00:13
We wrote ground state or neutral configurations in problem 4, but here we have different ions.
00:25
So the first one is vanadium 5 plus.
00:30
So the electron configuration of neutral vanadium is argon, noble gas configuration of argon, followed by 3d5 4s2.
00:43
So there's five valence electrons and all of them are being removed.
00:47
So the electron configuration would be the same as argon.
00:56
Next is chromium 3 plus.
01:00
Chromium has six valence electrons, typically five in d and one in s.
01:10
So where do the electrons come from? so it has argon configuration still.
01:16
And then the s electron is lost first, followed by two from the d, leaving 3d3.
01:27
Next we have manganese 2.
01:32
Manganese has seven valence electrons.
01:36
So what happens here is that we lose from the s orbital.
01:45
Because when we lose two electrons from the s orbital, then leave five in the d, we have d...