(Solved) Which of the following elements has the lowest ionization energy?

Another way to think of it (for clarification), is this:
Say you're playing black jack. You need your cards to add to 21 (but it doesn't matter how). Now, also say you can discard cards to equal 21 if you like. So, if your cards add to 19 or 20, you REALLY want to get at least one or two more, because then you win, but you're definitely not going to GIVE AWAY some of the points you already have. Similarly, if you have 22 or 23, you REALLY want to lose 1 or 2, but you certainly aren't going to take any MORE. That would be insane.
Atoms work the same way, only the golden number is 8. So if you're closer to 8 in your outer electron shell than you are to zero, you're going to work really hard to make eight, but you're not going to lose any. If you only have to lose one or two to get to eight (or the noble gases) you're going to work really hard to lose a couple and really resist taking on any more. This is "closest to right." It has a HIGH ionization energy, because ionization energy describes how badly you want to LOSE electrons, and if you are really close to winning the hand and only need one or two more, you aren't giving them away for anything close to free. So farthest to the right = high ionization energy.
The final piece of the puzzle comes into place if you just remember smaller is sneakier. So the smallest (closest to the top of the table) is the best at gaining/losing a couple electrons (depending on what they need to do to make eight).

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Which of the following elements has the lowest ionization energy?

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