2A problem 5

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EvaLi_3J
Posts: 53
Joined: Wed Oct 02, 2019 12:16 am

2A problem 5

Postby EvaLi_3J » Sat Nov 02, 2019 8:33 pm

On the textbook, this problem asks us to write the electronic configuration of Cu+, Bi3+, Ga3+, and Tl3+. Except for Bi3+, the other three ions seem to be exceptions to the energy level rule. Can anyone explain why that happened? For example, for Cu+, I wrote [Ar]3d33p54s2, but the answer is [Ar]3d10. Are there any other exceptions? When do they occur??

Trinity Vu 1D
Posts: 53
Joined: Fri Aug 30, 2019 12:15 am

Re: 2A problem 5

Postby Trinity Vu 1D » Sat Nov 02, 2019 8:55 pm

The answer is [Ar]3d10 because even though the 4s block on the periodic table is seen before the 3d block, 3d is still a shell below 4s and therefor the 3d shell would fill up before you could add any electrons to the 4s orbital. Because Cu+ has 28 valence electrons and you know that it comes after Ar, you put Ar first as a short hand to represent the configuration of the first 18 electrons. After this, you have 10 electrons left and since 3d comes before 4s, you'd fill 3d first. Since 3d "fits" all remaining 10 electrons you don't move onto 4s. I'm not sure where you got the 3p5 from though because Ar is the last for the 3p block so regardless of 4s or 3d you wouldn't write 3p since that's included in the shorthand [Ar]

EvaLi_3J
Posts: 53
Joined: Wed Oct 02, 2019 12:16 am

Re: 2A problem 5

Postby EvaLi_3J » Sat Nov 02, 2019 9:02 pm

Trinity Vu 1D wrote:The answer is [Ar]3d10 because even though the 4s block on the periodic table is seen before the 3d block, 3d is still a shell below 4s and therefor the 3d shell would fill up before you could add any electrons to the 4s orbital. Because Cu+ has 28 valence electrons and you know that it comes after Ar, you put Ar first as a short hand to represent the configuration of the first 18 electrons. After this, you have 10 electrons left and since 3d comes before 4s, you'd fill 3d first. Since 3d "fits" all remaining 10 electrons you don't move onto 4s. I'm not sure where you got the 3p5 from though because Ar is the last for the 3p block so regardless of 4s or 3d you wouldn't write 3p since that's included in the shorthand [Ar]


Thank you! Yes I was wrong about 3p . But I'm still kind of confused. So in which case would electrons fill 4s before 3d? Is it when the valence electrons would not fill up the entire 3d orbital? For example, if we have 9 outer electrons instead of 10, would we have [Ar]3d9?


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