Electron Configuration

[Cjc555777]

What are teh quantum numbers taht could define an outer valence electron of Co3+?

a. n=3, l=2, ml=2, ms=+1/2
b. n=3, l=0, ml=0, ms=-1/2
c. n=4, l=0, ml=0, ms=+1/2
d. n=4, l=2, ml=2, ms =-1/2

The answer is A, they give the reasoning that "the electron configuration of Co is [Ar]4s2 3d7. When Co forms the Co+3 ion, its electron configuration becomes [Ar]3d6. Hence, an outer valence electron is in the 3d orbital. Then n=3 and l=2. Only Choice has these quantum numbers listed, chose (A)"

I'm just confused on why none of the electrons stay in the 4s subshell. Then the electron configuration would be [Ar]4s1 3s5, and all of the shells would be half full, which I thought was more stable and lower energy than having one subshell be full and 4 of them not... I dont know, anyone explain this better?

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[Bernoull]

What are teh quantum numbers taht could define an outer valence electron of Co3+?

a. n=3, l=2, ml=2, ms=+1/2
b. n=3, l=0, ml=0, ms=-1/2
c. n=4, l=0, ml=0, ms=+1/2
d. n=4, l=2, ml=2, ms =-1/2

The answer is A, they give the reasoning that "the electron configuration of Co is [Ar]4s2 3d7. When Co forms the Co+3 ion, its electron configuration becomes [Ar]3d6. Hence, an outer valence electron is in the 3d orbital. Then n=3 and l=2. Only Choice has these quantum numbers listed, chose (A)"

I'm just confused on why none of the electrons stay in the 4s subshell. Then the electron configuration would be [Ar]4s1 3s5, and all of the shells would be half full, which I thought was more stable and lower energy than having one subshell be full and 4 of them not... I dont know, anyone explain this better?

This has to do with degeneracy of transtion elements. You need to keep two concepts in mind.

1. For representative elements, they form ions to be isoelectronic with the nearest noble gas and to have similar electronic configurations. For instance, Cl forms Cl- ions and K forms K+ ions in order to be isoelectronic with Ar .. [Ar]

2. Transition elements are wierd. They form ions by losing electrons from the highest principal quantum number first. 4s is higher than 3d in terms of n therefore it's emptied out first before e's are lost from 3d.

Finally note that generally, some transition metals have degenerate 3d and 4s orbitals (equivalent energy level) therefore it's difficult to predict electronic configurations. MCAT can test this degeneracy concept but they won't trick u by having both possiblities as answer choices UNLESS if the passage explained how orbitals are filled...

Goodluck

 

[12345a]

^^ very true

or you could remember that L = (n-1) and only one of the answers correctly gives n in relation to L

 

[shaggybill]

These were both excellent answers. One for understanding, and the other a nice shortcut.