Fluorine exhibits only a -1 oxidation state because it is the most electronegative element.
It readily gains one electron to achieve a stable electron configuration, resulting in a -1 charge in compounds.
The reason fluorine always exhibits this oxidation state is that it has no d orbitals available for bonding. As a result, it cannot expand its valence shell to accommodate positive oxidation states, which is possible for elements with available d orbitals.
Thus, Option D: It has no d orbitals is the correct explanation.
Fluorine exhibits only the -1 oxidation state because it is highly electronegative and lacks d orbitals, preventing it from adopting positive oxidation states. Its strong pull on electrons keeps it stable in the -1 state.
List I | List II |
---|---|
(A) Cr2O72− | (I) +3 |
(B) MnO4− | (II) +5 |
(C) VO3− | (III) +7 |
(D) FeF63− | (IV) +6 |
List-I | List-II | ||
(A) | Oxidation state of Mn in MnO2 is | (I) | +2 |
(B) | Most stable oxidation state of Mn is | (II) | +3 |
(C) | Most stable oxidation state of Mn in \(MnO_{4}^{-}\) is | (III) | +4 |
(D) | Characteristic oxidation state of lanthanoids is | (IV) | +7 |