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Topic: Question regarding the electron configuration for gold  (Read 3575 times)

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Offline Alyssa

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Question regarding the electron configuration for gold
« on: August 06, 2012, 03:06:15 AM »
I attempted to work out the electron config. for Au (gold) and arrived at [Xe] 6s2 4f15 5d9, as there are 15 atoms in the "f-block" on the periodic table. However when I googled it to check my answer the answer that came up for the electron config. of gold was [Xe] 6s2 4f14 5d9. Why is the f group only written as 14 atoms?
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Offline Hunter2

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Re: Question regarding the electron configuration for gold
« Reply #1 on: August 06, 2012, 03:53:14 AM »
There are one s-Orbital, 3 p-orbitals, 5 d-Orbitals and 7-f-Orbitals. Each can carry maximum 2 electrons, so we have 14 elements.

Electrons on shell : e = 2 *n²   n = 1,2,3,4, etc. Shell number

s = 2, p=6, d =10, f =14

Maximum electrons: 2 + 0 = 2, 2 + 6 = 8, 8 + 10 =18, 18 +14 = 32, etc.  
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Offline AWK

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Re: Question regarding the electron configuration for gold
« Reply #2 on: August 06, 2012, 03:57:16 AM »
[Xe] 6s1 4f15 5d10
Gols is a one fom above dozen exclusion from Madelung rule.

http://en.wikipedia.org/wiki/Electron_configurations_of_the_elements_(data_page)
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Offline Hunter2

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Re: Question regarding the electron configuration for gold
« Reply #3 on: August 06, 2012, 04:02:25 AM »
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