KCET · Chemistry · d and f Block Elements
The highest oxidation state of manganese in fluoride is \(+4\) (\(\text{MnF}_4\)), but the highest oxidation state in oxides is \(+7\) (\(\text{Mn}_2\text{O}_7\)), because
- A Fluorine is more electronegative than oxygen
- B Fluorine possesses d-orbitals
- C Fluorine stabilises lower oxidation state
- D In covalent compounds, fluorine can form single bond only, while oxygen forms double bond
Answer & Solution
Correct Answer
(D) In covalent compounds, fluorine can form single bond only, while oxygen forms double bond
Step-by-step Solution
Detailed explanation
The ability of oxygen to stabilize the highest oxidation states of transition metals (such as \(+7\) for \(\text{Mn}\) in \(\text{Mn}_2\text{O}_7\)) 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 \(\text{MnF}_7\) does not exist.
In \(\text{Mn}_2\text{O}_7\), manganese forms double bonds with oxygen atoms, requiring fewer ligands around the central metal atom and thereby reducing steric hindrance.
Answer: In covalent compounds, fluorine can form single bond only, while oxygen forms double bond
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 \(\text{MnF}_7\) does not exist.
In \(\text{Mn}_2\text{O}_7\), manganese forms double bonds with oxygen atoms, requiring fewer ligands around the central metal atom and thereby reducing steric hindrance.
Answer: In covalent compounds, fluorine can form single bond only, while oxygen forms double bond
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