Step 1: Role of each species:
$Mn^{2+}$ is the one being oxidised, so the other reactant, $S_2O_8^{2-}$, acts as the oxidiser. An oxidiser gets reduced, so sulphur ends in a lower oxidation state than the peroxy form.
Step 2: Half reactions:
Reduction: $S_2O_8^{2-} + 2e^{-} \to 2SO_4^{2-}$.
Oxidation: $Mn^{2+} + 4H_2O \to MnO_4^{-} + 8H^{+} + 5e^{-}$.
Step 3: Combine:
Multiply the reduction half by 5 and the oxidation half by 2 so both involve 10 electrons. Adding gives $2MnO_4^{-}$ and $10SO_4^{2-}$.
Step 4: Pick the option:
The pair $MnO_4^{-}$ and $SO_4^{2-}$ is option 2. Manganate ($MnO_4^{2-}$) appears when permanganate is reduced in alkali, not here.
Final Answer:
\[ \boxed{MnO_4^{-} \text{ and } SO_4^{2-}} \]