Question

Highest oxidation state of manganese in fluoride is \[\ce{+4 (MnF4)}\] but highest oxidation state in oxides is \[\ce{+7 (Mn2O7)}\] because ______.

विकल्प

• fluorine is more electronegative than oxygen.

• fluorine does not possess d-orbitals.

• fluorine stabilises lower oxidation state.

• in covalent compounds fluorine can form single bond only while oxygen forms double bond.

Answer 1

Highest oxidation state of manganese in fluoride is \[\ce{+4 (MnF4)}\] but highest oxidation state in oxides is \[\ce{+7 (Mn2O7)}\] because in covalent compounds fluorine can form single bond only while oxygen forms double bond.

Explanation:

Oxygen has the capacity to form multiple bonds which enables it to form a variety of covalent compounds. 

In \[\ce{(Mn2O7)}\] also, 6 oxygen are doubly bonded to two manganese atoms and one oxygen is forming bridge between two. 

While in \[\ce{(MnF4)}\], four fluorine atoms are singly bonded to manganese atom giving it a +4 oxidation state.

Therefore, due to capability of oxygen to have multiple bonds in covalent compounds, manganese is having higher oxidation state of +7 in \[\ce{(Mn2O7)}\].