00:01
In this problem, we're being asked to write the electron configuration for a variety of different ions.
00:05
The first one is going to be an o2 minus, so an oxygen ion.
00:12
And so when we're looking at our configuration, we want to figure out where oxygen is first of all, which is right here.
00:20
So from there, we can go ahead and start writing our orbitals.
00:24
So if it was just oxygen, it would be 1s2, and then 2.
00:31
2 because it has, it fills the s orbital there.
00:36
And then it's going to be 2p.
00:39
Now, typically, it'll be 1, 2, 3, 4, 2p4.
00:42
However, because we've gained two more electrons, it's actually going to be 2p6.
00:49
So we could also write it just as neon.
00:54
It has the same configuration as neon.
00:57
For our next one, the next one is bromine 1 minus, and bromine is located right here.
01:11
So again, we can go ahead and start doing our configuration.
01:17
So we would put, oops, wrong one, there we go.
01:21
We would put argon, so we don't have to write out all of those initial s &ps.
01:27
This is period 4, so it would be 4s2, 3d10, because it fills the entire d -shell, and then it would be 1, 2, 3, 4 ,5, 4p5, but because again we've gained one electron, it's going to be 4p6, which is another way of saying it has the same configuration as krypton.
01:58
Next, we have strontium 2 plus.
02:05
So find strontium is right here...