destroyer numberr 84 chemistry

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arpitpatel86

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i agree with the answer but in the explanations it says Cu will fill its 3d subshell first, then the 4s.........however kaplan says 4s will fill first than the 3d..........and kaplan makes sense using the (n +l) rule....4s=(4 +0)=4 and 3d=(3+2)=5 since 4 is lower than 5 it should fill first......so is destroyer wrong or am i missing some concept?
 
there's a few exceptions for some of the atoms in the d subshell. Cu and Cr are the only ones i can remember off the top of my head. I think there might be others, not entirely positive...
 
Cu and Cr are the only 2 for the first 40 elements. Things get crazy from there on out in the transition metals.
 
The elements in Group I will have 1s1 , 2s1, 3s1....( 1 unpaired electron)
The elements in Group II will have 1s2 , 2s2 , 3 s2 ( no unpaired electron)
The elements in Group
III , IV, V, VI, VII , VIII
2p1 2p2 2p3 2p4 2p5 2p6
3p1 3p2 3p3 .................
4p1 ............................
The elements in group VIII has 6 electrons for 3 orbitals --> no unpaired electron ( noble gas)
The elements in group VII has 5 electrons for 3 orbitals --> 1 unpaired electron
The elements in group VI has 4 valence electrons for 3 orbitals --> 2 unpaired electron
The elements in group V has 3 valence electrons for 3 orbitals--> 3 unpaired electron

The transition metals will fall to d shell , and has 5 orbitals , which can fill up to 10 electron .
If you look at the periodictable , Cr, Fe , Cu are in the same row. Cr is in front of Fe and Cu . That means it need most electrong to fill up its shell . Or Cr has 4 unpaired electron and need 6 electron to fill up its shelf . Fe needs 2 more electrons to fill up its shell ; therefore , Fe has 2 unpaired electron . Cu needs 1 more electron to fill up its shell , therefore Cu has 1 unpaired electron.


So, for s shell the maximum unpaired electron is 1 .
For p shell the maximum unpaired electron is 3
For d shell the maximum unpaired electron is 5
For f shell the maximum unpaired electron is 7
 
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