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:


  1. Cyrus, i like the way you used visual informaTION TO GO ALONG WITH THE QUESTION. iT REALLY HELPED, and i never realized the periodic table does that. good job

  2. Quite terrific my old bean, jolly good! I liked the way you explained this confusing concept in an organized manner. You might want to shorten it a bit, old chum, for this is quite a lot of information to process. I scarceley had time to finnish my scone and tea this evening.

  3. Thank you so much for the explanation! The pictures, graphs, and electron orbital filling diagrams for the different elements was a helpful representation to show the electron stability. It was all really helpful, but the paragraph as a little long. It would have been easier to read and review with if it were a little shorter and less detailed. It was still really helpful though. Good job!

  4. Your answer is excellent, Cyrus. To truly make this an college level answer, you need to be careful about your use of the word "want." A number of scientists frown upon the idea of using a word so full of human emotions. Atoms cannot feel and therefore cannot want something. It is more that everything in nature strives for a lower-energy state. In each case, the atom is gaining a more stable, lower energy state through an electron configuration with filled or half-filled sub-levels. Noble gases are the most stable because their highest occupied sub-levels are all full.

    Also, knowing the perfectionist in you, there is a typo in the paragraph below the second set of pictures and it should be "all too perfect trend" near the end of your first paragraph