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

#

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

#

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#