Editing Answers To Questions On Periodic Trends From Coluombic Force Thru Ionization Energy

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Answers to [[Questions on Periodic Trends from Coulombic Force thru Ionization Energy]]. 
 
The phrase Shielding effect is not acceptable part of your answer. You will need to explain via Coulombic Force (CF). 
 
1. Decrease. As you go down a Group, the valence electrons are in higher Energy level (n), causing the electron to be farther from the nucleus so the CF decreases (less pull of nucleus to the valence electrons). 
 
2. Increase. As you go across a Period, the valence electrons are being located in the same Energy level (so it is NOT distance that is causing CF). The nuclear charge (or number of protons in the nucleus) is increasing causing a higher CF (nucleus has a greater pull to the valence electron). 
 
3a. Top right. 
 
3b. Bottom left 
 
4. Atomic size increases as you go down a Group. 
 
5. Atomic size decreases as you go across a Period. Since the nuclear charge is increasing causing a greater pull to valence electrons (greater CF) the valence electron are closer (since they are on the outside of atom, they determine size of atom). 
 
6. Aligned from smallest to largest Atomic radii, Ne, P, Cr, Ca, Cs 
 
7a. The amount of energy needed to remove the first valence electron from an atom (atom has to be in the gas phase first). 
 
7b. Directly proportional. As CF goes up so does the Ionization Energy (IE) since it the CF that is keeping the valence electron near the nucleus (so IE can be thought of as pulling in the opposite direction than the CF). 
 
8. IE decrease as you go down the Group because since the CF is less on the valence electron due the electron are in higher Energy level, there is less pull to nucleus thereby it takes less energy (we call Ionization Energy) to remove the valence electron from the atom. 
 
9. Increase 
 
10. Ne ( or Neon) 
 
11. Be and B. It is an exception since the IE generally increase as you go across Period but between these two elements it decreases. 
 
12. N to O. The IE from N to O decrease instead of the general trend of increasing IE as you go across Period. For O, there is a 4th valence electron that goes into one of the three p orbitals. Since the two electrons in this orbital are close together they will repel and one of them will move farther away from the nucleus causing less CF and therefore less IE (we call this electron-electron repulsion in an orbital to explain this part of exception). 
 
13. F , C , Al , K (align by decreasing IE) 
 
14. Bottom left is corner of Periodic Table whose elements have the lowest IE.

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