Answer the question on the basis of passage given below:
In the periodic table, the d-block contains the elements of group 3 to 12. The d-orbitals are progressively filled in each of the four long periods. The elements of d-block referred as transition metals have partly filled d-orbitals and exhibit certain characteristic properties such as variety of oxidation states, formation of coloured ions, act as catalyst and show paramagnetic behaviour.
The two inner transition metals series 4f and 5f are known as Lanthanoids and Actinoids respectively. The lanthanoids resemble one another more closely as compared to ordinary transition elements in any series.
Highest oxidation state of manganese in fluoride 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) → Fluorine can form single bond only while oxygen forms double bond, in covalent compounds
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 |