Highest oxidation state of Manganese in fluorides is \(+4\) \((MnF_4)\) but highest oxidation state in oxides is \(+7\) \((Mn_2O_7)\) because ________.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → In covalent compounds, fluorine can form single bond only while oxygen form double bond
A high oxidation state is stabilised by the small, highly electronegative elements O and F. The highest state reached depends on how many bonds the element can form with the metal. Fluorine can form only single bonds (M–F), so a high oxidation state needs as many F atoms as the oxidation number, and seven F atoms cannot be crowded around one Mn atom; the highest manganese fluoride is \( \ce{MnF4} \) (+4). Oxygen can form multiple bonds (M=O), so fewer oxygen atoms are needed for the same oxidation state.
In \( \ce{Mn2O7} \), 2x + 7(−2) = 0, so x = +7: with Mn=O double bonds, four oxygen atoms around each Mn are enough for the +7 state, whereas +7 in a fluoride would need seven F atoms on one Mn.
Why the other options are incorrect
Option (1) Fluorine is more electronegative than oxygen — Incorrect. True, but higher electronegativity would favour the fluoride, not the oxide, reaching the higher state; it does not explain the difference.
Option (2) Fluorine does not possess d-orbitals — Incorrect. Oxygen has no d-orbitals either, so this cannot explain why oxygen does better.
Option (3) Fluorine stabilises lower oxidation state — Incorrect. Fluorine stabilises high oxidation states (as in \( \ce{VF5} \) and \( \ce{CrF6} \)); it is iodine and bromine that favour lower states.
NCERT (2026-27 edition):
Trends in Stability of Higher Oxidation States – …
Note: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| – | – | – | 25-05 S3 | 18-08 S1 |