Electron Configuration of a Cation (hw problem 3.21 part d)

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Steven Garcia 1H
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Joined: Fri Sep 28, 2018 12:16 am

Electron Configuration of a Cation (hw problem 3.21 part d)

Postby Steven Garcia 1H » Thu Nov 01, 2018 5:46 pm

Hi!

Why is the electron configuration for Ag+
[Kr] 4d^10

I originally thought it would be 4d^8 5s^2. I thought adding the s orbital electrons to the d orbitals (in attempt to fill the d- orbitals) only applies for certain cases such as Cu and Cr? Would this example be considered an exception as well since it's using its s-orbital electrons to fill up the d orbitals ?

-Steven

Chris Freking 2G
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Joined: Fri Sep 28, 2018 12:29 am

Re: Electron Configuration of a Cation (hw problem 3.21 part d)

Postby Chris Freking 2G » Thu Nov 01, 2018 5:56 pm

Ag has an electron configuration of [Kr] 4d^10 5s^1 because the element is much more stable when the electrons are configured in the element to have a full d shell and a half-full s shell.

When Ag is ionized (Ag+), the electron is removed from the outermost shell 5s. Therefore, the electron configuration for Ag+ is [Kr] 4d^10.

Kate Chow 4H
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Joined: Fri Sep 28, 2018 12:28 am

Re: Electron Configuration of a Cation (hw problem 3.21 part d)

Postby Kate Chow 4H » Thu Nov 01, 2018 6:02 pm

Additionally, since 5s is a higher energy level than 4d, electrons would be removed from 5s before they're removed from 4d.

Steven Garcia 1H
Posts: 79
Joined: Fri Sep 28, 2018 12:16 am

Re: Electron Configuration of a Cation (hw problem 3.21 part d)

Postby Steven Garcia 1H » Thu Nov 01, 2018 6:05 pm

Oh thank you, I understand now!

As a follow up, was the initial configuration of Ag 4d^10 5s^1 because it used the other s orbital electron to fill up the d orbital ?

Anna O 2C
Posts: 98
Joined: Fri Sep 28, 2018 12:19 am

Re: Electron Configuration of a Cation (hw problem 3.21 part d)

Postby Anna O 2C » Thu Nov 01, 2018 6:38 pm

Yep! Ag sits one electron away from having a full d orbital so the second electron from the 5s orbital instead occupies the d orbital to bring it to a full and therefore more stable state. Therefore the 4d becomes 4d10 and the 5s becomes 5s1.


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