Write the complete ground-state electron configuration of Ag.
The Correct Answer and Explanation is :
The ground-state electron configuration of silver (Ag) is:
[ \text{Ag}: [Kr] \, 4d^{10} \, 5s^1 ]
Explanation:
- Atomic Number of Silver (Ag):
Silver has an atomic number of 47, meaning it has 47 electrons in a neutral atom. - Filling of Orbitals:
The electrons fill the orbitals in order of increasing energy levels, following the Aufbau principle and the Pauli exclusion principle, which state that each electron occupies the lowest energy orbital available, and no two electrons in the same atom can have the same set of quantum numbers. The general order of filling is based on the sequence of orbitals:
[
1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^6 \, 5s^2 \, 4d^{10} \, 5p^6 \dots
] - Noble Gas Notation:
Instead of writing out all the orbitals, we use the noble gas configuration to simplify the electron configuration. The previous noble gas to silver is krypton (Kr), which has the electron configuration:
[
\text{Kr}: [Ar] \, 3d^{10} \, 4s^2 \, 4p^6
]
So, for silver, we start with the core configuration of krypton. - Electron Configuration for Silver (Ag):
After krypton, the 5s and 4d orbitals need to be filled. In most cases, you would expect the electron configuration to be ([Kr] 4d^9 \, 5s^2), but silver is an exception. To achieve a more stable and symmetric configuration, one electron from the 5s orbital moves into the 4d orbital. This leads to a fully filled 4d orbital ((4d^{10})) and a single electron in the 5s orbital ((5s^1)), resulting in:
[
[Kr] \, 4d^{10} \, 5s^1
]
This configuration is more stable because a fully filled (d)-subshell provides additional stability.