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![[Post New]](/templates/default/images/icon_minipost_new.gif) 6 Jan 2007 23:20:48 IST
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| if i redifine the quantum no's as n=1,2,3..............infinity L= 0,1,2,3.........n (but taking only odd values) m= -L,-(L-1)............0,............L {taking only even values (moduluswise)} S= -1/2,1/2 what should be the electronic configuration of Zn(30) | | |
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7 Jan 2007 00:06:49 IST
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ITS A VERY EASY QUESTION
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10 Jan 2007 09:47:12 IST
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Well, the Quantum numbers help in identifying the particular electron in the electronic configuration. The given quantum numbers in your question have no relation , as in my understanding, with the electronic configuration of Zinc!!!
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Lecturar, Organic Chemistry |
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10 Jan 2007 20:01:20 IST
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the electronic configuration of zinc would remain the same as in ordinary situation.if there is any correction please let me know .my email address is mipatel@gnfc.in
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14 Jan 2007 17:22:58 IST
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Electronic configuration of Zn is 1s2 2s2 2p6 3s2 3p6 3d10 4s2
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16 Jan 2007 13:01:51 IST
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to smrity mathur mam plz check the basics of quantum no's so that u will be able to understand the ans
1s2 2s2 3s2 4s2 3p6 5s2 4p6 6s2 5p6
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4 Feb 2008 22:01:47 IST
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1s2 will remain the same..
no no since s orbital is=0 means i cant use it..
so
2p6, 3p6, 3d10, 4p6, 4d 2 I think...
is it??
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Chemistry is in my blood.. |
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4 Feb 2008 22:30:54 IST
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hey lonelyguy i have not understood the ques
cud please make ur quest more clearer to interpret
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4 Feb 2008 23:03:25 IST
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n = 4
L = 1, 3 (odd values between 0 to 4)
m = -1, 0, 1 or -3, -2, -1, 0, 1, 2, 3
but since only even numbers are allowed we can only take 0.
so m = 0
S = -1/2, 1/2
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---------------------------------------------------------------
- Gaurav Ragtah (spideyunlimited)
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5 Feb 2008 18:51:21 IST
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1-s 3-p 5-d 7-d 1 orbit will not exist becoz l<1 is not there 2nd orbit will have only s and will have 2.22=8.electron...it will have 4 box so that paulie's exclusion should obey.. 3rd orbit will have only s where only 8 electron will be 2n2 violet since absence if p orbital 4 th orbit will have s & p orbital 1,3<4 will have 2.42=32 out of 32 8 will fill up into s and rest to p sequences 2s,3s,4s,5s [n+l rule],6s,4p,7s,5p...........so on
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