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Topic: Which electron affinity process would liberate the most energy?  (Read 11557 times)

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Offline Sis290025

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Which electron affinity process would liberate the most energy?

  a. [He] 2s2 + e- --> [He] 2s2 2p1    
  b. [He] 2s2 2p3 + e-  --> [He] 2s2 2p4    
  c. [He] 2s2 2p2 + e- -->  [He] 2s2 2p3    
  d. [He] 2s2 2p6 + e- --> [He] 2s2 2p6 3s1  

This is my reasoning so far:

Rewritten equations
[He] 2s2 + e- --> [He] 2s2 2p1    =  Be(1s2 2s2) + e- --> Be(1s2 2s2 2p1)

[He] 2s2 2p3 + e-  --> [He] 2s2 2p4   = N(1s2 2s2 2p3) + e- --> N(1s2 2s2 2p4)


[He] 2s2 2p2 + e- -->  [He] 2s2 2p3 = C(1s2 2s2 2p2) + e- --> C(1s2 2s2 2p3)


[He] 2s2 2p6 + e- --> [He] 2s2 2p6 3s1 = Ne(1s2 2s2 2p6) + e- --> Ne(1s2 2s2 2p6 3s1)

For a release of the most E, E_ea must be the most - number. This rules out Ne and Be because they have 0 or + E_ea and leaves N and C.

How do I know which has a higher -E_ea from N and C? (Is it C and, thus, c.?)
 
« Last Edit: November 02, 2005, 11:31:27 PM by Mitch »

Offline jdurg

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Re: Which electron affinity process would liberate the most energy?
« Reply #1 on: November 03, 2005, 08:52:23 AM »
First off, what makes an element chemically stable?  A full shell of electrons, right?  This is why chlorine takes an electron to form Argon, and why sodium gives one up to become neon.  The presence of that unpaired electron causes instability and instability does not release energy.  (However when that unstable situation then becomes stable, a great deal of energy is released, so the formation of stability liberates more energy and the formation of instability requires energy).

So let's take a look at each situation:

A):  You have Be gaining an electron and putting a lone e- in its 2p orbital.  This leaves it with an unpaired electron AND a negative charge.  That's not too stable.

B):  You have Nitrogen gaining an electron and putting it in its 2p orbital leaving you with two unpaired electrons.  All of the 2p shells, however, are of equal energy.  So if the 2px, 2py, and 2pz subshells each have the same number of electrons in there, a bit of stability exists as all of the electrons in the 2p shell are feeling the same effects and are essentially equal.  In the question, you're adding another electron which breaks up this stability by filling up one of the 2p shubshells.  As a result, instability exists.

C):  You're taking Carbon with 2 unpaired 2p electrons and adding a third electron.  The carbon atom now has three electrons in its 2p shell and gains the same electron configuration as the Nitrogen started with in part B.  You now have three, essentially equal electrons in the 2p shell.  This provides a bit of stability to the atom.  Stability releases energy.

D):  You're taking Neon and adding an electron to it.  To do so, you have to start filling up a higher energy 3s shell.  This gives it the same electrical configuration as sodium, and we all know how reactive sodium metal is.  To put another electron on Neon requires you to input a lot of energy in order to get it into that 3s subshell.  This is therefore not stable.

So looking back, yes.  You are correct.  Option C would liberate the most energy because it forms the most stable species.
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Offline Sis290025

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Re: Which electron affinity process would liberate the most energy?
« Reply #2 on: November 03, 2005, 03:47:13 PM »
Thank you.  :s-sault:

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