Fluorine, being highly electronegative and lacking vacant \( d \)-orbitals, always adopts an oxidation state of \( -1 \), unlike other halogens that show variable oxidation states.
Because fluorine is the most electronegative element and lacks accessible \( d \)-orbitals, it cannot expand its octet. Consequently, fluorine invariably exhibits an oxidation state of \( -1 \) in its compounds. In contrast, other halogens like bromine, iodine, and chlorine can display variable oxidation states because their atoms possess vacant \( d \)-orbitals, permitting octet expansion. Final Answer:\[\boxed{\text{Fluorine}}\]