What is the electron configuration of Ag?

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What is the electron configuration of Ag?

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Answer 1 · 2014-06-18T05:05:13

The electron configuration for silver (Ag) is based upon the place meant of silver in the fifth row of the periodic table in the 11th column of the periodic table or the 9th column of the transition metal or d block. Therefore th electron configuration for silver must end as $4 d^{9}$ $4d^9$ ,

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{2} 4 d^{9}$ $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^9$

This notation can be written in core notation or noble gas notation by replacing the $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6}$ $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6$ with the noble gas [Kr].

$[K r] 5 s^{2} 4 d^{9}$ $[Kr] 5s^2 4d^9$

For some of the transition metals they will actually transfer an s electron to complete the d orbital, making silver,

$[K r] 5 s^{1} 4 d^{10}$ $[Kr] 5s^1 4d^10$

I hope this was helpful.
SMARTERTEACHER

Answer 2 · 2015-10-30T19:09:02

$Ag: [Kr] 4 d^{10} 5 s^{1}$ $"Ag: " ["Kr"] 4d^10 5s^1$

Explanation:

Silver, $Ag$ $"Ag"$ , is located in period 5, group 11 of the periodic table, and has an atomic number equal to $47$ $47$ .

This tells you that a neutral silver atom will have a total of $47$ $47$ 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 electron configuration would look if it followed the Aufbau principle to the letter

$Ag: 1 s^{2} 2 s^{2} 2 s p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 4 p^{6} 3 d^{10} 5 s^{2} 4 d^{9}$ $"Ag: " 1s^2 2s^2 2sp^6 3s^2 3p^6 color(blue)(4s^2 4p^6) color(red)(3d^10) 5s^2 4d^9$

Now, for the energy level $n$ $n$ , the d-orbitals that belong to the $(n - 1)$ $(n-1)$ energy level are lower in energy than the s and p orbitals that belong to the $n$ $n$ 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

$Ag: 1 s^{2} 2 s^{2} 2 s p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 5 s^{2} 4 d^{9}$ $"Ag: " 1s^2 2s^2 2sp^6 3s^2 3p^6 color(red)(3d^10) color(blue)(4s^2 4p^6) 5s^2 4d^9$

Now do the same for the 4d and 5s orbitals

$Ag: 1 s^{2} 2 s^{2} 2 s p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{9} 5 s^{2}$ $"Ag: " 1s^2 2s^2 2sp^6 3s^2 3p^6 color(red)(3d^10) color(blue)(4s^2 4p^6) 4d^9 5s^2$

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

$Ag: 1 s^{2} 2 s^{2} 2 s p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{10} 5 s^{1}$ $"Ag: " 1s^2 2s^2 2sp^6 3s^2 3p^6 color(red)(3d^10) color(blue)(4s^2 4p^6) 4d^10 5s^1$

Using the noble gas shorthand notation will get you

$Ag: [Kr]      1 s^{2} 2 s^{2} 2 s p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{10} 5 s^{1}$ $"Ag: " overbrace(1s^2 2s^2 2sp^6 3s^2 3p^6 color(red)(3d^10) color(blue)(4s^2 4p^6))^(color(green)(["Kr"])) 4d^10 5s^1$

$Ag: [Kr] 4 d^{10} 5 s^{1}$ $"Ag: " ["Kr"] 4d^10 5s^1$

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What is the electron configuration of Ag?

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