The electron configuration for silver (Ag) is based upon the location meant of silver in the 5th row of the routine table in the 11th pillar of the periodic table or the ninth column of the transition metal or d block. Thus th electron configuration for silver must end as #4d^9#,

#1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^9#

This notation have the right to be written in core notation or noble gas notation by replacing the #1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6# v the noble gas .

# 5s^2 4d^9#

For several of the change metals they will certainly actually move an s electron to finish the d orbital, do silver,

# 5s^1 4d^10#

I hope this to be helpful.SMARTERTEACHER




You are watching: What is the abbreviated electron configuration for silver

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Stefan V.
Oct 30, 2015

#"Ag: " <"Kr"> 4d^10 5s^1#




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Explanation:

Silver, #"Ag"#, is located in period 5, team 11 of the regular table, and has an atomic number same to #47#.

This tells you that a neutral silver- atom will have actually a total of #47# electrons neighboring its nucleus.

Now, you have to be a small careful through silver because it is a transition metal, which means that the populated d-orbitals are actually lower in energy 보다 the s-orbitals that belong come the highest power level.

So, here"s just how silver"s electron configuration would certainly look if it complied with 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, because that the energy level #n#, the d-orbitals the belong to the #(n-1)# energy level space lower in energy 보다 the s and also p orbitals the belong come the #n# energy level.

This way that you will have to switch the 3d orbitals top top one hand, and also the 4s and 4p orbitals top top the other.

This will acquire 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 perform the same for the 4d and also 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 point to remember here is that in silver"s case, the 4d orbitals will be completely filled. That means that you won"t have two electrons in the 5s orbital, since one will be kept in the reduced 4d orbitals.

This means that the electron construction of silver will certainly 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 noble gas shorthand notation 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#