Tuesday, November 9, 2010

Test Review: #3

3.  Looking at the trend in first ionization energies across period 3, there is a sudden unusual dip with Al and S.  Explain this.

Well, at first we look at this problem and say what, this doesn't make sense.  Ionization energy is supposed to increase across periods, not drop!  Well, there just so happens to be an anomaly in this all too perfect trend.  To better understand this question, look at the visual display of first IEs below.

Also, to take a look at another representation, click here

Now, what the problem is asking for is this: why does IE drop from Mg to Al and P to S (you can ignore Si)?

Well, the answer in a nutshell is that it is a matter of stability.  Look at the orbital filling diagrams below:

Let's look at the first pair: Mg and Al.  Mg has a completely empty 3p sublevel while Al has a lone electron floating in that same corresponding sublevel.  Mg has added stability because of its completely filled 3s sublevel and it wouldn't benefit the atom if one of those electrons were ripped away.  Taking away an electron from Mg would be ruining the added stability of having a completely filled sublevel.  Al, on the other hand, would benefit from losing an electron because then its 3p sublevel would be completely empty.  An empty sublevel is more stable than one with a single electron roaming around.  That is why, if you steal one of Al's electrons, more stability would be added.  So, because Mg avoids ruining the stability it has and Al aims to get the stability Mg has, there is a dip in IE.

The next pair, P and S, can be explained with the same reasoning.  P has a half-filled 3p sublevel and S has 4 electrons in that same sublevel.  Again its about stability.  P  has already achieved stability by only having 2 electrons in 3 orbitals but S aims to have 3 electrons in 3 orbitals.  So, for the matter of stability again, P avoids ruining its stability while S benefits from gaining P's.  Therefore, there is a dip in IE.

*The easier it is to remove an electron (in other words, the more an atom tries to lose an electron-for stability in this case), the less energy is required to remove it.


The visual display of IE:

Second display: