Chemical Forums
Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: Mitch on February 10, 2006, 04:18:07 AM
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Why does Zinc have a lower 2nd ionization energy than Cupper? From simple Zeff considerations you would expect it to be higher.
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Cu
1st: remove one 4s e-
2nd: remove one 3d e-
Zn
1st: remove one 4s e-
2nd: remove one 4s e-
For Cu 2nd ionization, you need to remove one e- from full-filled 3d-orbitals.
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Is there any good reason why the second electron removed from Zinc is a d-electron?
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s-electron is more penetrating to the nucleus than d-electron?
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I was always under the impression the 3d spends more time near the nucleus than the 4s electron does.
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Is it true that a 3d-electron is removed in 2nd ionization of zinc.
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Isn't that what you said?
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I said the 3d-orbitals of Zinc will remain full-filled after 2nd ionization originally, but you want a reason for removing 3d e- rather than 4s e-.
I am confused now!
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Cu
1st: remove one 4s e-
2nd: remove one 3d e-
You said it would remove a 3de- first?
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Because this is copper.
Cu: [Ar]3d104s1
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So wait, what's the answer? :P
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Cu: [Ar]3d104s1
1st I.E. is relatively small (one 4s e- is removed)
2nd I.E. is relatively large (one 3d e- is removed from full-filled 3d-orbitals, extra E is required)
Zn: [Ar]3d104s2
1st I.E. is relatively large (one 4s e- is removed from full-filled 4s-orbital, Zeff also increases from Cu to Zn)
2nd I.E. is relatively small (another 4s e- is removed to achieve all full-filled orbitals, relatively stable)