What is the shorthand electron configuration for a barium ion?
The Correct Answer and Explanation is :
The shorthand electron configuration for a barium ion (( \text{Ba}^{2+} )) is:
[Xe] 6s²
Explanation:
Barium (Ba) is a chemical element with an atomic number of 56. This means a neutral barium atom has 56 electrons. The electron configuration for neutral barium can be written as follows:
- Barium’s neutral electron configuration:
[
\text{Ba}: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2
]
This is the full electron configuration, showing all the electrons distributed across the various orbitals. - Shorthand notation:
To make this more concise, we use the shorthand notation by referencing the electron configuration of the nearest noble gas preceding barium in the periodic table, which is Xenon (Xe). Xenon has an atomic number of 54, so its electron configuration is: [
\text{Xe}: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6
] Therefore, the shorthand electron configuration for barium (Ba) is:
[
[Xe] 6s^2
] - For the barium ion (( \text{Ba}^{2+} )):
When barium forms a ( 2+ ) ion (i.e., ( \text{Ba}^{2+} )), it loses two electrons. These electrons are lost from the outermost shell, specifically the 6s orbital, since these are the highest-energy electrons. After the loss of two electrons, the electron configuration for ( \text{Ba}^{2+} ) becomes: [
\text{Ba}^{2+}: [Xe]
] This means the ( \text{Ba}^{2+} ) ion has the same electron configuration as Xenon, a noble gas, with a stable, filled electron shell structure.