What is the unabbreviated electron configuration of gold?

1 Answer
Nikka C.
Oct 31, 2015

Predicted: #1s^2# #2s^2# #2p^6# #3s^2# #3p^6# #4s^2# #3d^"10"# #4p^6# #5s^2# #4d^"10"# #5p^6# #color (red) (6s^2)# #4f^"14"# #color (red) (5d^"9")#

Actual: #1s^2# #2s^2# #2p^6# #3s^2# #3p^6# #4s^2# #3d^"10"# #4p^6# #5s^2# #4d^"10"# #5p^6# #color (red) (6s^1)# #4f^"14"# #color (red) (5d^"10")#

Explanation:

Whew. The atomic number of gold (#Au#) is 79. Therefore, its ground state electron configuration should have a total number of 79 electrons (add all the superscripts).

I have inputted two answers simply because Au is one of those elements who does not follow the predicted order of the electron configuration (the other elements are #Cu#, #Ag#, #Pd#, #Cr#, and #Mo#). This happens simply because completing the d orbital brings a lower energy (lower risk of electron repulsion) compared to filling up the s orbital first.

Now you do not have to fret if the question asks you to write the longhand electron configuration of any element. You do not even have to memorize the order of the shells just to write the electron configuration properly. Just draw this on your scratch paper and you are good to go.

http://rajpalmalra.blog.com/?page_id=157

All you need to do is follow the arrows. But you do need to memorize the anomalies above (there are only six elements so that wouldn't be so hard to do).

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