1. FROM FRIDAY….. Ionization Energy:
Checklist:
___ Trend is stated
___ Shielding
___ Nuclear charge
___ Highest filled orbital
___ Coulomb’s Law (optional)
Ionization Energy:
Using your “arsenal” of definitions from yesterday, come up with what you think the trend would be across a row and down a column.
Work with your table and create a blank periodic table with arrows showing the general trend. Also write an explanation for each trend that covers the items listed on the checklist.

2. GOOD

3. GOOD

4. TOO LITTLE

5. TOO LITTLE

6. Review:
Provide an explanation for the following observations using principles of atomic structure.
1. There is an extremely large increase between the 2nd and 3rd ionization energies of Beryllium (1,757 kJ/mol up to 14,800 kJ/mol). Use an electron configuration to support your answer.
b. Cl has a 1st IE of 1251 kJ/mol while Br has a 1st IE of 1140 kJ/mol

7. Electron Affinity:
What is electron affinity? Ionization energy in reverse …
-IE is the energy required to remove an electron
-EA is the energy released when an e- is acquired
Atom + e- 
Values for EA will be negative since energy is released (exothermic)
A more negative EA means a higher affinity for e-

8. Electron Affinity: The Trend  WHY? Relate it to ionization E!
Atoms that are hard to ionize (high IE) will also have a higher affinity for gaining electrons (very negative EA)

9. Ionic Size: Cations are smaller than their atoms WHY?
After losing electrons, the atom now fills to a lower energy level, which is smaller
Ionic Size:
Cations are smaller than their atoms
WHY? Li
152 pm
Li +
78 pm
Anions are larger.
WHY?
F -1
133 pm F
71 pm
Adding electrons into the same orbital causes increased e- repulsions

10. Ionic Size:
The periodic trend is the same as that of atomic size

11. Practice:
a. Place in order of increasing atomic radius and justify your response.
O  C  Si
Si is larger than both O and C because it fills to the n=3 energy level, which is larger.
C is larger than O because it has a lower nuclear charge, so it’s electrons are less attracted to the nucleus.

12. Practice: b. Which has the largest ionization energy and why?
O has the highest IE
Si will have the lowest since it’s outer electrons are in the n=3 shell and experience the most shielding.
O is higher than C because it’s nuclear charge is greater, so the electrons are more strongly attracted to the nucleus and harder to remove.

13. Practice:
c. Which has the more negative electron affinity, O or C? Which atom will release more energy upon gaining an electron? Which atom has the greater “desire” to gain electrons?
Oxygen

14. Practice:
d. Oxygen commonly forms the oxide ion, O-2. Predict and explain how the size will change as oxygen forms this ion.
O-2 will be larger
For O to turn into O-2, 2 electrons must be added. This increases electron repulsions and increases the radius.

15. Practice:
e. If you were comparing the successive ionization energies of a carbon atom (1st, 2nd, 3rd, etc), between which values would expect to see the largest increase and why?
Carbon: 1s2 2s2 2p2
The first 4 electron removals will require less energy because they all come from the n=2 energy level and are somewhat shielded from the nucleus.
The 5th IE will be much higher because it will come from n=1 and has almost no shielding from the nucleus.

16. Practice:
f. Which particle will have the larger radius: the oxide ion, O-2 or a neon atom, Ne
Oxide ion: 1s2 2s2 2p6
Neon atom: 1s2 2s2 2p6
Both particles have same shielding and highest filled E level, only difference is nuclear charge. Neon’s is higher, so it is smaller.