The total number of $$Mn = O$$ bonds in $$Mn_2O_7$$ is

We need to find the total number of $$Mn=O$$ bonds in $$Mn_2O_7$$ (dimanganese heptoxide). To determine the structure of $$Mn_2O_7$$, note that $$Mn_2O_7$$ is the anhydride of permanganic acid ($$HMnO_4$$). In its structure:

- Two $$MnO_3$$ units are linked by a bridging oxygen atom (Mn-O-Mn).

- Each Mn atom is in the +7 oxidation state.

- Each Mn is tetrahedrally coordinated with 4 oxygen atoms.

Identifying the types of Mn-O bonds, for each Mn atom:

- One oxygen is the bridging oxygen (Mn-O-Mn) — a single bond (Mn-O) shared between both Mn atoms.

- Three oxygen atoms are terminal — these form $$Mn=O$$ double bonds.

Counting the total $$Mn=O$$ bonds, each Mn has three terminal $$Mn=O$$ bonds. With two Mn atoms, the total $$Mn=O$$ bonds = $$3 \times 2 = 6$$.

The structure can be represented as:

$$ O=Mn(=O)(=O)-O-Mn(=O)(=O)=O $$

The correct answer is Option C: 6.