00:01
In this problem, we're looking at trying to determine what element is being described based upon four different statements.
00:11
The first statement says an element that has one unpaired 5p electrons that will form a covalent bond with fluorine.
00:21
So the first thing we have to think about is if it's 5p, what area of the periodic table are we looking at? so we're going to figure out, so this is 1p, 2p, 3p, 4p, 5p is right in here.
00:40
So our 5p options are going to be indium through xenon.
00:47
Now, right off the bat, we can get rid of xenon.
00:51
Oops, let's do it this way.
00:53
You can get rid of xenon because it's not going to have any paired electrons, it's a noble gas.
01:00
We know that they fill in order and that our p -orbital is going to have six electron pairs, potential electron pairs.
01:13
So indium is an option.
01:17
Ten is not because it would have two electrons.
01:22
Antimony is not because it would have three.
01:24
It would be going like this.
01:28
Tilarium is not because then we would start pairing off those electrons.
01:33
Iodine is also an option because, again, with the pairing, it would have that one unpaired electron.
01:43
So our options are indium or iodine.
01:46
But the statement also says that it needs to combine covalently with fluorine.
01:51
Therefore, the only option that we have for this is going to be iodine.
01:57
So iodine is the answer to a, the element with one unpaired 5p electron, that can form a covalent bond with fluorine.
02:07
Okay, and we're also asked to write the ground state configuration.
02:11
So our ground state configuration is going to be krypton because that's the nearest noble gas.
02:20
And then it will be 5s2, 4d ,000.
02:30
10 because all of its d shell will be filled and then 5 p 5 because it has 5 of the 6 in its p shell...