Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
KCET2026Chemistryd and f Block Elements

The highest oxidation state of manganese in fluoride is +4 ( MnF ₄ ), but the highest oxidation state in oxides is +7 ( Mn ₂ O ₇ ), because

Options

  1. AFluorine is more electronegative than oxygen
  2. BFluorine possesses d-orbitals
  3. CFluorine stabilises lower oxidation state
  4. DIn covalent compounds, fluorine can form single bond only, while oxygen forms double bond

Correct answer

D. In covalent compounds, fluorine can form single bond only, while oxygen forms double bond

Step-by-step solution

The ability of oxygen to stabilize the highest oxidation states of transition metals (such as +7 for Mn in Mn ₂ O ₇ ) is due to its ability to form multiple bonds (double bonds) with the metal atoms. Fluorine can only form single bonds. To achieve a +7 oxidation state with fluorine, a single manganese atom would have to accommodate seven fluorine atoms around it, which is sterically highly unfavorable and thus MnF ₇ does not exist. In Mn ₂ O ₇ , manganese forms double bonds with oxygen atoms, requiring fewer ligand

Practice d and f Block Elements on Quantrex Academy →

More from d and f Block Elements

Pick out the correct option Assertion(A): Mercury is not considered as a transition element Reason (R): Mercury is a liquid 2026Which of the following does not correctly represent the order of the property indicated against it? 2026The element with the highest third ionisation enthalpy is: 2026The product and its colour when MnO₂ is fused with KOH in presence of O₂ : 2026The ion that has a spin only magnetic moment of 5.9 BM is: 2026Which one of the following species will impart colour to an aqueous solution? 2026The basic character of transition metal monoxides follows the order: 2026K ₂ Cr ₂ O ₇ on heating with aqueous NaOH gives ------------------------- 2026 Full d and f Block Elements list All KCET PYQs