Answer: Option (A) is the correct answer.
Explanation:
Fluorine is a halogen atom and it belongs to group 7. Atomic number of fluorine is 9.
Therefore, electronic configuration of fluorine is [tex]1s^{2}2s^{2}2p^{5}[/tex].
Atomic number of neon is 10 and its electronic configuration is [tex]1s^{2}2s^{2}2p^{6}[/tex]. Hence, we can see that electronic configuration of fluorine and neon are not the same.
Atomic number of potassium is 19 and its electronic configuration is [tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}[/tex] which is not equal to electronic configuration of fluorine.
Thus, we can conclude that out of the given options, electron configuration of a fluoride ion,F-,is 1S2 2S2 2P5.
