2A.21 7th Edition Question

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Mark 1D
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Joined: Fri Sep 28, 2018 12:18 am

2A.21 7th Edition Question

Postby Mark 1D » Sun Oct 28, 2018 8:15 pm

Howdy ya'll. Can someone explain why the ground state electron configuration for Ag^+ is [Kr]4d^10 in question 2A.21? Thanks

Ethan Yi 1K
Posts: 62
Joined: Fri Sep 28, 2018 12:28 am

Re: 2A.21 7th Edition Question

Postby Ethan Yi 1K » Sun Oct 28, 2018 8:27 pm

Ag is one of the exceptions where the 4d orbital will actually take an e- from the 5s. therefore, the ground state of Ag would be [Kr]4d^10 5s^1. Because in this case Ag is an ion in the form Ag^+, we remove an electron from the 5s orbital, leaving [Kr]4d^10.

yaosamantha4F
Posts: 31
Joined: Fri Sep 28, 2018 12:29 am

Re: 2A.21 7th Edition Question

Postby yaosamantha4F » Sun Oct 28, 2018 8:39 pm

The normal ground state configuration for Ag is [Kr]4d^10 5s^1. Based off of the Aufbau principle, silver's expected configuration would be [Kr]4d^9 5s^2, but since a full or half-full subshell is more stable than a partially-filled one, the s electron is transferred to the d orbital since having the full 4d subshell and partially-filled 5s subshell is a more stable configuration than having partially-filled 4d and 5s subshells. Using the configuration [Kr]4d^10 5s^1 for silver, the 5s valence electron is lost so that the final ground state configuration for Ag+ is [Kr] 4d^10.


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