GC Difficult Periodicity Problems

[Ryltar]

1. Which pair of elements is listed in order of decreasing first ionization energy?

A. Na, Mg
B. Mg, Al
C. Al, Si
D. Si, P

2. When species of F-, Na+, and Ne are arranged in order of increasing energy for the removal of an electron, what is the correct order?


Please provide an explanation to solutions, thanks!

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[NeoAkira]

1. Which pair of elements is listed in order of decreasing first ionization energy?

A. Na, Mg
B. Mg, Al
C. Al, Si
D. Si, P

2. When species of F-, Na+, and Ne are arranged in order of increasing energy for the removal of an electron, what is the correct order?


Please provide an explanation to solutions, thanks!

I'm not so sure about the second question but I think i know the answer for the first question. My answer would be B. My reasons:

A. If Na loses an electron then it reaches a noble gas configuration. Mg would have to lose 2 electrons to do that. Therefore, Na wants to give up it's first electron more than Mg, meaning it has a lower first ionization energy than Mg. Or if you want, IE increases left to right. Therefore this answer is arranged in increasing first ionization energy.

B. Normally IE increases left to right, but there are some exceptions. In this case, since ionization of Al will cause it to go from a full s and 1-electron-filled p orbital to a full s and 0-electron-filled p orbital (more stable) it wants to give up its first electron (Low IE). When Mg gives up its first electron it goes from a full-filled s orbital (stable) to a half filled s orbital (less stable) and therefore it doesn't want to give up its first electron (High IE). Assuming the above to be true, the atoms in this answer are arranged in decreasing first ionization energy.

C. IE increases left to right, so Si's IE will be higher than Al's. Therefore these atoms are arranged in increasing first ionization energy.

D. IE increases left to right, so P's IE will be higher than Si's. Therefore these atoms are arranged in increasing first ionization energy.

So I'm not sure if my rationalizations are right, but what's the real answer?

 

[Ryltar]

The answer to the first one is B, thanks. Anyone know how to do problem 2?

 

[Ryltar]

I think I understand the 2nd one. All 3 would have noble gas Neon configuration. So ionization energy would increase with decreasing size. F- would be bigger than Ne, and Na+ would be smaller than Ne. Thus, the order of increasing first ionization energy would be F-. Ne, Na+.

 

[tdkyun]

I think I understand the 2nd one. All 3 would have noble gas Neon configuration. So ionization energy would increase with decreasing size. F- would be bigger than Ne, and Na+ would be smaller than Ne. Thus, the order of increasing first ionization energy would be F-. Ne, Na+.

Shouldn't it be the other way? A stronger energy will be needed to remove an electron from F- than Na+. So lowest Na+ < Ne < F- highest. Correct me if I'm wrong.

I see it now =)

 

[Sublimation]

Shouldn't it be the other way? A stronger energy will be needed to remove an electron from F- than Na+. So lowest Na+ < Ne < F- highest. Correct me if I'm wrong.

I would suspect the difference to be negligible. They are all basically the same in regards to size. they all have the same number of electrons......if i where to guess i would go with Lowest (F-, because then it goes back to its natural state) then Neon, then Na+. However....i believe it would be too close to call. I might be wrong my Gen chem is really rusty.

 

[peyman2002]

ok fellas, its actually Na+ Ne F- in order of increasing.

 

[Ryltar]

According to ACS the increasing order is F-, Ne, Na+. I think they might have been trying to get the point across that cations are smaller and anions are larger.

 

[Ryltar]

If you directly compare them Na+ would have 10 e-, 11 p; Ne 10e-, 10p; F- 10e-, 9p. Thus the order of increasing size would be Na+, Ne, F-. And thus, increasing ionization energy would be F-, Ne, Na+.

 

[cyrano]

Na+ is a cation with a full octet. Removing another electron is difficult because there is already more positive charge on each electron than a neutral or anionic species.

Ne is a noble gas and is inert. It would not be expected to lose an electron because then it will become ionized which is only favorable if you're making an octet or decreasing overall NRG of your electrons.

F- is just going back to it's original state as F. So it's not a huge deal to lose that electron it already gained because technically it has less Z-effective (positive charge) pulling in each individual electron.

 

[chemtopper]

order of increasing first ionization energy would be F-. Ne, Na+.
yes it is the correct order because it is easy to remove electron from a negative charged atom but very difficult from a positive charge atom.
one more way to view it
All these atoms have same electronic configuration but no of protons are
increasing in order from F- ,Ne and Na+
So one with highest positive charge in the nucleus will have highest ionization energy.