Oxidation States Help. !

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  • #1
Oxidation States Help. URGENT!

Hey, I've got this question to answer, any help would be greatly appreciated.

In a compound, aluminum (Z=13) exists in only the +3 oxidation state but silicon (Z=14) can exist in either the +2 or the +4 oxidation state. Why?
 

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  • #2
Monique
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According to my chemistry book Al only has oxidation state +3 and Si only +4..

There are many other elements with lots of oxidation states, especially in group 7A of the periodic table:

Cl [Ne]3s23p5
+7
+5
+3
+1
-1

My theory: it has to do with the fillingof the orbital shells. Either Cl can loose all of its 7 electrons and become [Ne]+7, it can loose all its electrons in the p subshell (+5), it can fill its p subshell (-1). The +3 and +1 don't make much sense to me.. I would have expected +2.
 
  • #3
Silicon has oxidation states of +2, +4, and -4. My main confusion is, if Si can lose 2 electrons to fill the s orbital shell, why can't Aluminum do the same by losing 1 electron?
 
  • #4
Monique
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Good question.

What are the differences of the ionization energies between the two situations? Since Al has less mass, the electrons are more loosly bound. Apparently the loss of one electron doesn't stabilize the atom sufficiently and it is relatively easy to take the two additional s electrons so it will.

My speculation..
 
  • #5
Ok, thank you very much for your help Monique. I really appreciate it. Peace

JC
 
  • #6
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Elements in the right section of the periodic table (p orbital elements) only exist in oxidation states of Z, Z-2, Z-4...basically subtracting even numbers of electrons. We can do this for sillicon until 2+...subtracting two electrons. However for aluminum subtracting another 2 electrons will leave it in the 1+ oxidation state. Originally aluminum had three unpaired electrons. 1+ aluminum only has one unpaired electron drastically decreasing its paramagnetic character. This jump is unlikely. Both silicon 4+ and 2+ has two unpaired electrons (although 4+ is more stable since it has a paired electron orbital). Although 2+ is more unstable than 4+ it is possible in a relative sense to aluminum that 2+ exists although more rare than 4+.

You should have someone else check on this though...

This is only my reasoning.
 
  • #7
Monique
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? why is it even numbers of electrons that are taken ?

Can anyone explain the Cl example? Cl is [Ne]3s23p5 I would expect a Cl2+ state, since that would mean [Ne]3s23p3. I think it was Hunds rule saying that filled AND half filled orbitals are especially stable, then why not have Cl2+?
 
  • #8
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Cl has the strongest electron affinity values for which it may be difficult to remove 2 electrons stere orientation of orbitals might also restrict +2 O.S.
 
  • #9
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The Cl example:

It would be highly unlikely to get any loss of electrons from the Cl ion simply because of the size of the atom. Cl is quite small and if you remove an electron, it becomes substantially smaller causing the gigantic increase in ionization energy. It would wish to go to a +2 if forced, but its high electronegativity creates high electron affinity.

As you move into the transition elements and larger elements, it is important to not think of the electron shells as being an equal distance apart. Once the d and f shells are added to the orbital mix, two things happen:

1. The outer most electrons are "shielded" from the nucleus by the inner most filled electron shells causing little hold on them by the positive nucleus.

2. The orbitals overlap and some hybridize because there is very little energy difference between one shell and another.

Anyway, this is what I remember from Inorganic, but then again, it has been awhile.

Peace,

Galaphile
 
  • #10
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thanks for explaining facts
 

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