HW problem 3.21

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Rosamari Orduna 1D
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Joined: Fri Apr 06, 2018 11:04 am

HW problem 3.21

Postby Rosamari Orduna 1D » Fri May 11, 2018 8:54 pm

Part D.)
This question asks to find the ground state electron configuration and number of unpaired electrons for :
Ag+
The answer key says its configuration is : [Kr]4d^10, and has no unpaired electrons, but how?
I got :[Kr]4d^9 5s^1, which should be just one unpaired electron....

someone please explain, thanks!!

Nandhini_2K
Posts: 60
Joined: Fri Apr 06, 2018 11:03 am

Re: HW problem 3.21

Postby Nandhini_2K » Fri May 11, 2018 9:06 pm

The ground state electron configuration for Ag+ is [kr] 4d^10 and has no unpaired electrons.

The ground state electron configuration for just Ag is [kr] 4d^9 5s^2 but remember we cannot have 4d^9. An electron from 5s^2 will transfer to 4d^9 to become stable. The 4d^9 will become 4d^10.

The Ag+ is [kr] 4d^9 5s^2 but lost an electron from 5s^2 (because of the +) and the last electron on 5s^1 transferred to 4d^9, which gives you [kr] 4d^10 for Ag+.

Salena Chowdri 1I
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Joined: Fri Apr 06, 2018 11:02 am

Re: HW problem 3.21

Postby Salena Chowdri 1I » Sat May 12, 2018 5:52 pm

The answer key is right; the ground state electron configuration for Ag+ is [kr] 4d^10 and has no unpaired electrons.

The ground state electron configuration for just Ag is [kr] 4d^10 5s^1. Since Ag+ has a charge of +1, that means it is losing an electron to become more positive. Since 5s has a higher energy level, the first electron will be taken from this subshell causing Ag+ to have a configuration of [kr] 4d^10. It has no unpaired electrons because the 4d subshell accommodates 10 electrons, which is its limit (5 orbitals w/ 2 electrons each).

Haison Nguyen 1I
Posts: 30
Joined: Fri Apr 06, 2018 11:04 am

Re: HW problem 3.21

Postby Haison Nguyen 1I » Mon May 14, 2018 12:23 pm

The ground state configuration of AG is [Kr] 4d^10 5s^1 because of the half shell rule. Since the question is asking for the configuration of AG+, you would take away an electron leaving just [Kr] 4d^10.

Alexander Hari 1L
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Joined: Wed Feb 14, 2018 3:02 am

Re: HW problem 3.21

Postby Alexander Hari 1L » Mon May 14, 2018 6:41 pm

Remember that the d orbital likes to be full so the extra electron that would have been in the 5s block is instead placed in the 4d^9 orbital to make 4d^10. This would mean you have no unpaired electrons, so the answer key is correct


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