How do you write the noble-gas electron configuration for Ag?

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How do you write the noble-gas electron configuration for Ag?

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Answer 1 · 2016-09-29T17:02:45

Where is your Periodic Table?

Explanation:

For $silver$ $"silver"$ , $Z = 47$ $Z=47$ . It is thus 11 protons removed from the last Noble Gas, which is $krypton$ $"krypton"$ , $Z = 36$ $Z=36$ .

And thus the electronic configuration of silver metal is: $[K r] 4 d^{10} 5 s^{1}$ $[Kr]4d^(10)5s^1$ . The $A g (+ I)$ $Ag(+I)$ oxidation state can be rationalized on this basis.

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How do you write the noble-gas electron configuration for Ag?

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