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Old 04-30-2012, 04:48 PM   #1
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Default Chem Ionization / Shell


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Hi all,
So the rule is that you ionize from outer shells first (n = 4 then n = 3 etc.)

What happens when you ionize Fe which is [Ar] 4s23d6? Does it remove it first from 3d (because it would lead to half filled stability) or 4s?

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Old 04-30-2012, 04:53 PM   #2
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4s.
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Old 05-01-2012, 07:17 AM   #3
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4s.
4s? But, as the OP said, wouldn't taking it out of 3d make a more stable situation? Or does it just happen when we have ions?
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Old 05-01-2012, 02:13 PM   #4
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4s? But, as the OP said, wouldn't taking it out of 3d make a more stable situation? Or does it just happen when we have ions?

Electrons are pulled from the valence shell. As the quantum number increases, the distance of that shell from the nucleus increases. Therefore, electrons are pulled from 4s before 3d which is held tighter to the nucleus. There are exceptions to this rule.

Fe is +2 or +3; where +2 means an empty 4s shell and +3 means an empty s shell and one removed from the d shell giving half filled stability. No matter the charge, electrons in this case are removed from the 4s first.
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Old 05-01-2012, 02:33 PM   #5
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Electrons are pulled from the valence shell. As the quantum number increases, the distance of that shell from the nucleus increases. Therefore, electrons are pulled from 4s before 3d which is held tighter to the nucleus. There are exceptions to this rule.

Fe is +2 or +3; where +2 means an empty 4s shell and +3 means an empty s shell and one removed from the d shell giving half filled stability. No matter the charge, electrons in this case are removed from the 4s first.
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Old 05-02-2012, 07:19 AM   #6
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thanks!
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