Final answer:
The ion that satisfies the octet rule for fluorine is the fluoride ion, F⁻, as it gains one electron to achieve an electron configuration similar to the noble gas neon, thereby achieving stability with an octet of electrons in its valence shell.
Step-by-step explanation:
To determine the ion that would satisfy the octet rule for fluorine, we consider the atomic structure of fluorine. Fluorine has an atomic number of 9 and seven valence electrons in its valence shell. Following the octet rule, fluorine aims to achieve a full set of eight valence electrons for stability. The most efficient way for fluorine to satisfy the octet rule is by gaining one electron, which results in a negatively charged fluoride ion, written as F⁻.
When fluorine becomes a fluoride ion, it achieves an octet configuration, becoming more stable. This is denoted with a noble gas configuration, similar to the electron arrangement of neon. As such, a fluoride anion, with a 1- charge, satisfies the octet rule for fluorine by balancing out the number of protons and electrons. It is easier for fluorine, which has a high electronegativity, to accept an electron rather than donating seven, which would require a significant energy shift. Therefore, the correct ion formation for fluorine in order to satisfy the octet rule is F⁻, not F²⁻ as the question suggests.