00:01
Hi there.
00:02
So for this problem, for part a of this problem, we are asked to use the periodic table that is shown in here to determine the ground state configuration for the atoms of manisium, aluminum, and silicon.
00:21
So as you can see from the periodic table, you can see that this corresponds to the row 3 .8.
00:36
These elements.
00:38
You can see it in here.
00:39
This is magnesium, aluminum, and silicon.
00:45
So the columns reveal that the last shell is being filled.
00:53
It's being filled.
00:55
So the row, the numbers of the electrons in that shell.
00:58
So therefore, we will have that for 12 magnetion.
01:05
Its configuration is just simply 1 s square s 2 2 s 2 2 p 6 6 and 3 s 2 for aluminum we will have that its configuration is 1 s 2 2 2 2 2 p 6 3 s 2 and then we will have 1 in the p 6 shell.
01:44
Now for silicon, 14 silicon, we're going to have that this is 1s2, 2 s2, 2p6 and 3s2 and 3p2.
02:04
Now for part b of this problem, we are asked about to then predict the ls -caupling quantum numbers from the ground state of each atom.
02:18
So for 12 magnesium, we will have that.
02:29
The configuration represents a field shell.
02:33
And thus, field, shell, so the angular momentum, angular momenta, so all the angular momento, all angular momentum are zero, leading to then 1 s sub -zero.
03:02
Now, for 13 alumina, in this case, we know that there is a single balance electron that is s is equal to s prime, that is equal to 1 divided by 2, does 2 times s prime plus 1 should be equal to 2 and l prime should be equal to 1 given a p state and that p state it has a value g prime equals to 3 divided by 2 or 1 divided by 2 with the smaller j prime line lower and leading this then to 2 p 1 divided by two state...