Question

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

• 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

Hint:

When comparing the ability of oxygen and fluorine to stabilize high oxidation states of transition metals, remember that oxygen's ability to form multiple (double) bonds often outweighs fluorine's higher electronegativity. This is a common trend for many transition metals.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept.
The question addresses why oxygen can stabilize a higher oxidation state of manganese (+7) compared to fluorine (+4), despite fluorine being more electronegative.
Step 2: Analyzing the Options.
(A) Fluorine is more electronegative than oxygen: This is true, but it doesn't explain the observation. Higher electronegativity would suggest fluorine should be better at stabilizing high positive oxidation states, which contradicts the given data.
(B) Fluorine possesses d-orbitals: This is false. Fluorine is in the second period and does not have accessible d-orbitals.
(C) Fluorine stabilises lower oxidation state: This is a restatement of the observation, not an explanation for it. The question is asking *why* this is the case.
(D) In covalent compounds, fluorine can form single bond only, while oxygen forms double bond: This is the key reason.

• Fluorine, being monovalent, can only form a single covalent bond (Mn-F). To achieve a +7 oxidation state, manganese would need to bond with seven fluorine atoms, leading to significant steric hindrance and inter-electronic repulsion, making MnF\(_7\) highly unstable. The highest known stable fluoride is MnF\(_4\).
• Oxygen is divalent and has the ability to form multiple bonds (double bonds, Mn=O). This allows manganese to achieve a high oxidation state without needing to coordinate with a large number of atoms. In Mn\(_2\)O\(_7\), each manganese atom is bonded to three oxygen atoms via double bonds and shares one oxygen atom with the other manganese, allowing it to reach the +7 state.

The ability of oxygen to form \(\pi\)-bonds with d-orbitals of the transition metal is crucial for stabilizing high oxidation states.
Step 3: Final Answer.
The ability of oxygen to form multiple bonds, unlike fluorine which can only form single bonds, allows it to stabilize higher oxidation states in metals like manganese.