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Report (any type)/Brief report
22/09/2021
##”Ag: ” [“Kr”] 4d^10 5s^1##
Silver, ##”Ag”##, is located in period 5, group 11 of the , and has an equal to ##47##.
This tells you that a neutral silver atom will have a total of ##47## electrons surrounding its nucleus.
Now, you have to be a little careful with silver because it is a transition metal, which implies that the occupied d-orbitals are actually lower in energy than the s-orbitals that belong to the highest energy level.
So, here’s how silver’s would look if it followed the Aufbau principle to the letter
##”Ag: ” 1s^2 2s^2 2sp^6 3s^2 3p^6 color(blue)(4s^2 4p^6) color(red)(3d^10) 5s^2 4d^9##
Now, for the energy level ##n##, the d-orbitals that belong to the ##(n-1)## energy level are lower in energy than the s and p orbitals that belong to the ##n## energy level.
This means that you will have to switch the 3d orbitals on one hand, and the 4s and 4p orbitals on the other.
This will get you
##”Ag: ” 1s^2 2s^2 2sp^6 3s^2 3p^6 color(red)(3d^10) color(blue)(4s^2 4p^6) 5s^2 4d^9##
Now do the same for the 4d and 5s orbitals
##”Ag: ” 1s^2 2s^2 2sp^6 3s^2 3p^6 color(red)(3d^10) color(blue)(4s^2 4p^6) 4d^9 5s^2##
The thing to remember here is that in silver’s case, the 4d orbitals will be completely filled. That implies that you won’t have two electrons in the 5s orbital, since one will be kept in the lower 4d orbitals.
This means that the electron configuration of silver will be
##”Ag: ” 1s^2 2s^2 2sp^6 3s^2 3p^6 color(red)(3d^10) color(blue)(4s^2 4p^6) 4d^10 5s^1##
Using the will get you
##”Ag: ” overbrace(1s^2 2s^2 2sp^6 3s^2 3p^6 color(red)(3d^10) color(blue)(4s^2 4p^6))^(color(green)([“Kr”])) 4d^10 5s^1##
##”Ag: ” [“Kr”] 4d^10 5s^1##
I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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