Higher oxidation state of manganese in fluoride is `+4 (MnF_(4))` but highest oxidation state in oxides is `+7(Mn_(2)O_(7))` because

Options

A. fluorine is more electronegative than oxygen
B. fluorine does not posses d orbitals
C. fluorine stabilises lower oxidation state
D. in covalent compounds, fluorine can form single bond only while oxygen forms double bond

Explanation

To answer the question regarding the oxidation states of manganese in fluoride (MnF4) and oxides (Mn2O7), we can break down the explanation into clear steps:

### Step-by-Step Solution:

1. **Understanding Oxidation States**:
- The oxidation state of an element in a compound indicates the degree of oxidation (loss of electrons) of that element. In MnF4, manganese has an oxidation state of +4, while in Mn2O7, it reaches +7.

2. **Bonding Characteristics of Fluorine and Oxygen**:
- Fluorine typically forms single bonds due to its electronic configuration. Each fluorine atom can only form one bond, which limits the number of fluorine atoms that can bond with manganese.
- Oxygen, on the other hand, can form both single and double bonds. This ability allows it to bond more effectively with manganese and contribute to higher oxidation states.

3. **Analyzing MnF4**:
- In MnF4, manganese is bonded to four fluorine atoms. The oxidation state of manganese here is +4 because each fluorine contributes a -1 charge (4 x -1 = -4), resulting in Mn being +4 to balance the charge.

4. **Analyzing Mn2O7**:
- In Mn2O7, there are two manganese atoms bonded to oxygen. Here, three of the oxygen atoms form double bonds with manganese, while the remaining oxygen forms a single bond.
- The oxidation state of manganese in Mn2O7 can be calculated as follows:
- Each double-bonded oxygen contributes -2 (3 x -2 = -6) and the single-bonded oxygen contributes -2 (1 x -2 = -2).
- Therefore, the total negative charge from oxygen is -8. To balance this, the total positive charge from manganese must be +14 (2 manganese atoms, so each must be +7).

5. **Stability of Higher Oxidation States**:
- The formation of MnF7 would require manganese to achieve a +7 oxidation state with seven fluorine atoms. However, this structure is unstable due to crowding. The seven fluorine atoms would repel each other because of their lone pairs, making it difficult for such a configuration to exist.

6. **Conclusion**:
- The reason why manganese can achieve a +7 oxidation state in oxides but only a +4 state in fluorides is due to the bonding nature of oxygen (which allows for double bonds) compared to fluorine (which only allows for single bonds). This leads to the conclusion that oxides can stabilize higher oxidation states than fluorides.