Highest oxidation state of Fluoride and oxide of Mn are respectively

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$M n F_{4}, M n_{2} O_{7}$

$M n F_{7}, M n_{2} O_{7}$

$M n F_{5}, M n_{2} O_{7}$

$M n F_{6}, M n_{2} O_{7}$

answer is C.

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Detailed Solution

In covalent compounds fluorine can form single bond while oxygen forms double bond
The ability of oxygen to stabilise the higher oxidation state is more than that of fluorine. Also, the ability of oxygen to form multiple bonds with metal favours Mn₂O₇. Therefore, the highest fluoride of Mn is MnF₄ whereas highest oxide is Mn₂O₇. In Mn₂O₇, each Mn is tetrahedrally surrounded by Os including an Mn−O−Mn bridge.

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