1. Trends in the Periodic Table
The trends (atomic radius, ion radius, ionization energy and electronegativity) can be explained using the concepts of energy levels and Coulomb’s Law.

2. Coulomb's Law
The electrical force between 2 charged objects depends on the amount of charge on each object and the distance between them.
In 1783, Charles Coulomb used the pictured apparatus to determine how charge and distance affected the measured force….
Thank you, Mr. Plattner, for your physics notes!

3. His conclusion? The electrical force is _________ related to each of the charge, q, of the objects.~
directly
“is proportional to”
Coulomb also discovered that if he increased the distance between the charged spheres by a factor of two, the force decreased by a factor of ______!
four
His conclusion? The electrical force is _________ related to the _______ of the distance between the objects. ~
inversely
square
“is proportional to”

4. The magnitude of k indicates that electrical forces are very strong.
Coulomb summarized these results in equation form in what is now known as “Coulomb’s Law”: e
Where….
d = distance between centers of objects
q = charge on each object
k = the proportionality constant
The magnitude of k indicates that electrical forces are very strong.

5. Like charges repel; Opposite charges attract
Coulomb’s Law
Summary:
Fe between valence electrons and the nucleus increases as the (+) charge in the nucleus increases… (charge ↑, Fe ↑)
Fe decreases on the electron (−) and the nucleus (+) as the distance between them increases… (d ↑, Fe ↓)
Like charges repel; Opposite charges attract
This equation is NOT on the AP Equation Sheet, so you will have to memorize these relationships!

6. Trend 1: Relative Atomic Sizes (Radius)

7. Atomic Radius (Zeff charge ↑, Fe↑)
Atoms increase in size moving down group of elements due to the fact that electrons are being added to higher energy levels which are naturally farther from the nucleus.
Moving across a period from left to right, the size decreases. Why?
More electrons are being added to the same energy level which would not significantly increase the atom’s radius because at the same time, there is an increase in the number of protons in the nucleus of the atom.
This increases the effective nuclear charge of the atom, and the nucleus pulls more strongly on the entire electron cloud…more “Coulombic Force”…so the atom becomes smaller in size.
(Zeff charge ↑, Fe↑)

8. Effective Nuclear Charge
The effective nuclear charge (often symbolize as Zeff or Z*) is the net positive charge experienced by an electron in a multi-electron atom.
The term "effective" is used because the shielding effect of negatively charged electrons prevents higher orbital electrons from experiencing the full nuclear charge by the repelling effect of inner-layer electrons.
Core electrons are generally closer to the nucleus than valence electrons, and they are considered to “shield” the valence electrons from the full electrostatic attraction of the nucleus.

9. Effective Nuclear Charge (Example)
Consider a sodium cation, a fluorine anion, and a neutral neon atom. They are “isoelectronic”… each has 10 electrons.
The number of non-valence electrons is 2…10 total electrons − 8 valence, but the effective nuclear charge varies because each has a different atomic number Zeff = (nuclear charge – core electrons)
So, the sodium cation has the largest effective nuclear charge, and thus the smallest atomic radius since its electrons would be experiencing the greatest Coulombic force pulling the electron cloud inward. (Zeff charge ↑, Fe↑)

10. Trend 2: Atomic Radius vs. Ion Radius
Removing e−’s : Cations are smaller than the original atom.
Adding e−’s : Anions are larger than the original atom.
Why?

11. Atomic Radius vs. Ion Radius
For the positive cations, there’s less electron-electron repulsion with the fewer electrons in the cloud, so the cation radius will be smaller.
The protons in the nucleus hold the remaining electrons more strongly as the radius shrinks.
For the negatively charged anions, there’s more electron-electron repulsion with additional electrons in the cloud, so the anion radius will be larger.
The nuclear attractive force decreases so the electrons are less tightly held by the nucleus.

12. First Ionization Energies
Trend 3:
(Units: kJ/mol)

13. Trend 3: Ionization Energy (IE)
(…how IE is actually measured)
Trend 3: Ionization Energy (IE)
The ionization energy of an atom is the energy required to remove an electron from the atom in the gaseous phase.
Example: Li (g) kJ/mol  Li+(g) + 1 e−
Ionization energy decreases moving down a group.
The increased distance between electrons and the nucleus and increased shielding by a full principal energy level means that it requires less energy to remove an electron.
The effect of increased nuclear charge is balanced by the effect of increased shielding, and the number of energy levels becomes the predominant factor. (d ↑, Fe ↓)
Ionization energy increases moving across a period.
The reason for this, as is the case with periodic trends in atomic radii, is that as the nucleus becomes more positive, the effective nuclear charge increases its pull on the electrons and it becomes more difficult to remove an electron. (Zeff charge ↑, Fe ↑)

14. Although removing the first electron from an atom requires energy, the removal of each subsequent electron requires even more energy.
This means that the second IE is greater than the first, the third IE is greater than the second, and so on…Why?
The reason it becomes more difficult to remove additional electrons is that they’re closer to the nucleus and thus held more strongly by the positive charge of the protons. (d ↓, Fe↑)
When you see a huge increase in the subsequent IE, it indicates that you have decreased the energy level by 1, and the e− being removed is significantly closer to the nucleus. (d ↓, Fe↑) (It is also a noble gas configuration which is more stable than usual.)

15. Important Exceptions to the Ionization Energy Trends
For a test, I would only ask that you could explain why the exception is happening, not predict that an exception would occur.
When electron pairing first occurs within an orbital, electron-electron repulsions increase, so that removing an electron takes less energy (it’s easier); thus the IE drops at this time.
Example: Nitrogen (1400) vs. Oxygen (1310)
Less energy is required to remove an electron from oxygen’s valence in spite of an increasing Zeff because oxygen’s p4 electron is the first to pair within the orbital.
The repulsion created lowers the amount of energy required to remove either electron.
Differences in electron-electron repulsion are responsible for the differences in energy between electrons in different orbitals in the same shell…2p3 vs. 2p4
There is also a drop in ionization energy from s2 to p1—also in spite of an increasing Zeff. Example: Magnesium (736) vs. Aluminum (577)
The p electrons are less tightly held because they do not penetrate the electron cloud toward the nucleus as well as an s electron does so they experience less Fe.

16. Photoelectron Spectroscopy: PES Measuring and Interpreting Ionization Energy Data
What is PES?

17. Trend 4: Electronegativity

18. Trend 4: Electronegativity
Electronegativity is a measure of the attraction an atom has for electrons when it is involved in a chemical bond.
It is a relative scale from 0 to 4.
Noble Gases =0 Fluorine = 4.0
Elements that have high ionization energy will also have high electronegativity since their nuclei strongly attract electrons.
The reasoning behind the electronegativity trend is the same as the reasoning behind the IE trend…larger atoms are less attracted not only to their own electrons but also to other electrons in a chemical bond. (d ↑, Fe ↓)