The electron configuration for silver (Ag) is based upon the place meant of silver- in the 5th row the the routine table in the 11th column of the regular table or the ninth column that the transition metal or d block. Thus th electron construction 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 deserve to be composed in main point 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 some of the change metals they will actually transfer an s electron to complete the d orbital, make silver,

# 5s^1 4d^10#

I expect this was helpful.SMARTERTEACHER

You are watching: Ground state electron configuration for ag

Answer connect

Stefan V.
Oct 30, 2015

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

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Silver, #"Ag"#, is situated in period 5, team 11 of the routine table, and also has an atom number same to #47#.

This speak you the a neutral silver- atom will have a complete of #47# electrons bordering its nucleus.

Now, you need to be a little 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 the belong come the highest energy level.

So, here"s just how silver"s electron configuration would certainly look if it followed the Aufbau principle come 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)# power level are lower in energy than the s and p orbitals the belong come the #n# energy level.

This method that friend will have to switch the 3d orbitals on one hand, and the 4s and also 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 carry out the exact 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 indicates that friend won"t have actually two electrons in the 5s orbital, due to the fact that one will certainly be maintained in the reduced 4d orbitals.

This way 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 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#