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| ArezList |
 Posted: Mon Jun 30, 2008 12:43 am Post subject: The |
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Forum Freshman
Joined: 22 Mar 2008
Posts: 52
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(Cu2+) is available, as copper has the electron configuration that is [Ar]4s1,3d10.......
just wonder how the conf. will be very stable? |
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| Ironmaiden |
| Posted: Mon Jun 30, 2008 1:58 am Post subject: |
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Forum Freshman
Joined: 30 Jun 2008
Posts: 9
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EC: 1s2 2s2 2p6 3s2 3p6 3d10 4s1
2 electrons are removed from 3d orbital, and thus it has remaining 8 electrons in the 3d orbital. The octet is complete, and hence stable.
This is what I think the answer is. |
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| PritishKamat |
| Posted: Mon Jun 30, 2008 9:46 am Post subject: |
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Joined: 17 Sep 2007
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Location: Mumbai, India
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Its mainly because the 3d10 shell is half filled and gives more stability than 3d9,4s2
This stability of a half filled shell is described by Pauli's exclusion principle
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Beyond Equations,
Pritish |
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| ArezList |
| Posted: Mon Jun 30, 2008 5:55 pm Post subject: |
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Joined: 22 Mar 2008
Posts: 52
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if so then there's only one elcectron on the 4s . 4s is not stable enough
Of course the 3d8 is of stability, but the 4s?
and 4s orbital has the lower energy then the 3d, why you didn't satisfied the lower sudshells but to stablize the higher ones? |
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| PritishKamat |
| Posted: Tue Jul 01, 2008 1:44 am Post subject: |
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Yes, the 4s1 is less stable than 4s2, but the stability given by the 3d10 well surpasses the difference
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Beyond Equations,
Pritish |
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