Highest oxidation state of Manganese in fluorides is \(+4\) \((MnF_4)\) but highest oxidation state is oxides is \(+7\) \((Mn_2O_7)\) because ________.

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

The correct answer is option 4. In covalent compounds, fluorine can form single bond only while oxygen form double bond.

In fluorides, manganese can exhibit its highest oxidation state of +4 in compounds like \(MnF_4\). Fluorine is a highly electronegative element, and in these compounds, it acts as the electron acceptor.

On the other hand, in oxides, manganese can exhibit higher oxidation states, including +7 in compounds like \(Mn_2O_7\). Oxygen, being less electronegative than fluorine, can participate in the formation of double bonds, allowing the manganese to achieve a higher oxidation state.

The reason for the higher oxidation state in oxides compared to fluorides is linked to the bonding capabilities of oxygen and fluorine. Oxygen can form double bonds in covalent compounds, allowing it to share more electrons and accommodate higher oxidation states. Fluorine, on the other hand, typically forms only single bonds in covalent compounds.

Therefore, option (4) "In covalent compounds, fluorine can form a single bond only, while oxygen can form a double bond" explains the difference in oxidation states between fluorides and oxides of manganese.