00:01
In answering these questions, it's important to remember that the periodic table represents the arrangement of electrons or the electron configuration in for any given atom.
00:11
Each section corresponds to a different sub -level.
00:16
And so the first part here, the first two rows are referred to as the s block.
00:22
The rows over here are the p block.
00:25
These are the d block.
00:27
Down here is the f block.
00:30
And each row is a different energy level.
00:39
Remembering that the ds are one behind, so even though this is the fourth row, these are three ds.
00:48
And the fs are two behind.
00:50
So even though these fall, number 58 should fall in the sixth row, these are the 4f, and these are the 5f.
00:59
So if you can identify the element that you're looking for the electron configuration, you can use the periodic chart to find the valence electrons.
01:09
So for example, in our first problem a, it asks to find the element who has one unpaired electron in the 5p orbital that covalently bonds with fluorine.
01:24
So 5p, we look at our rows 5, would be this section here.
01:29
And we have two options for having one electron unpaired in the p orbital.
01:39
Could have one, or we could have a total of five electrons, which also has only one unpaired electron.
01:54
So if we're looking at an element that has one p electron in the 5 p orbital, it would be i .n.
02:01
If we're looking at one that has five, it would be two, three, four, five i.
02:07
So we have to choose between those two.
02:09
Because it covalently bonds with fluorine, we're looking for the non -metal, which is iodine.
02:15
So the correct answer to a is iodine.
02:19
And the electron configuration for iodine, i'm going to use our noble gas configuration.
02:24
So we write in brackets the preceding noble gas, which is krypton, which brings us here to 5s2, then to the 4d.
02:43
And we go all the way across, so 4d10, and then 5p, 1, 2, 3, 4, 4, 5...