1. Identify the elements and their charges:
* Strontium (Sr) is in group 2, so it loses 2 electrons to form Sr²⁺.
* Fluorine (F) is in group 17, so it gains 1 electron to form F⁻.
2. Determine the electron configurations:
* Strontium (Sr): [Kr] 5s²
* Sr²⁺ loses its 2 valence electrons: [Kr]
* Fluorine (F): [He] 2s² 2p⁵
* F⁻ gains 1 electron to fill its p orbital: [He] 2s² 2p⁶ = [Ne]
3. Combine the configurations:
Since we have two fluorine ions (F⁻) for each strontium ion (Sr²⁺), the final noble gas electron configuration for strontium fluoride is:
[Kr] + 2 [Ne]
Therefore, the noble gas electron configuration of strontium fluoride is [Kr]2[Ne].
