Periodic Trends
Question 2: Exceptions to the rule (we’ll do this one in class on Monday first thing)
Ionization energy and electron affinity have several examples of exceptions. This includes exceptions within the p-block that you are responsible for knowing. Atomic radius and electronegativity do not have as many exceptions, and none are notable enough that you must know. a) Rank in increasing order: carbon, nitrogen, oxygen, according to the following trends:
Ionization energy:
Electron affinity:
Electronegativity: c) Explain why ionization energy and electron affinity have these exceptions, while atomic radius and electronegativity do not.
Key for these that we won’t get to or effectively already covered in class unofficially. Remember to ask questions on facebook/discussion and office hours if you have questions.
Question 3: Understanding the basis of the trends a) Effective Nuclear charge is the reason behind all the other trends. Draw the horizontal trend into the periodic table above. Explain.
Increasing to the right.
As you go across the periodic table you increase the number of protons and electrons while not increasing the number of energy shells (therefore neither distance from nuclei or shielding effects are increased). This is like making a stronger

Periodic Trends and stronger magnet so the electrons are “held tighter” (anthropomorphizing a bit for explanation purposes) to the nucleus. b) Draw the electron energy level diagram for lithium and sodium (don’t forget to label everything including the energy axis and names of orbitals). Use this to explain why the ionization energy and electron affinity decrease as you go down the periodic table (Hint, there are two reasons).
As you go down the periodic table there are more energy levels. This means that the outside electrons are further form the nucleus (reason 1), and also that there is more shielding caused by the core electrons (reason 2). c) Explain why atomic radius is the reverse of each of the other trends.
As effective nuclear charge increases it holds the electrons tighter to the nucleus. The tighter hold on electrons decreases the size of the electron cloud, thereby decreasing the radius. However this tighter control means that it requires more energy to remove an electron, the atom is more likely to gain electrons, and that in a bond it will be more capable of taking a greater amount of the electron density. d) Explain why, in general, negative ions have a larger radius than positive ions.
The higher the proton to electron ratio, the tighter the electron density is. This means positive ions, which have MORE protons than electrons, will hold their electrons the closest, while negative ions, with less protons than electrons will not hold their electrons as tightly.
Note: This is of course only for atoms of approximately the same atomic number. The general trend can have exceptions if vastly different number of protons/electrons are present.

Periodic Trends
Challenge Question: ******(i.e. this is not at all testable, but serves to teach you interesting things about how science works so if you are interested in any sort of research science look at it).
Energies for orbitals can be determined theoretically as we saw when discussing the Schrodinger equation. When you move onto lab, you will see how these can be determined using computer software which utilizes these theoretical calculations.
Photoemission spectroscopy be can be used to determine the value of these orbitals experimentally for comparison with the theoretical calculations. This works of the principles of the photoelectric effect, the experiment that Einstein used to show the wave particle duality of light.
Remember that in the photo electric effect, energy is provided in the form of photons. If these photons have a high enough energy, they will eject electrons, whose kinetic energy and velocity are dependent on the energy of the photons and the energy with which they are held onto the metal (work function, or binding energy).
A PES measures the velocity of electrons by passing them through a region of magnetic field which deflects their path.
By changing the filed strength the path of the electrons are altered and the velocity measurement is obtained. From here kinetic energy of the electrons and therefore the energy level of the orbital can be both be calculated.
The following peaks were observed in the PES spectra for an element. Identify the element and electron configuration.
Note: Use the internet to compare the values of the binding energy found with values of ionization energies listed.
21 MJ/mol, 2.4 MJ/mol and 0.80 MJ/mol
When comparing to things with three ionization energies, relatively close, antimony’s listed values are closest. While not an exact match, in science experiments things seldom are. There are other possibilities that you could have picked,
Bismuth for instance. Further experiments/observations would have to be made to separate between these possibilities.