1. The options suggest that the reaction scheme involves a manganese species being converted under oxidation conditions, with a halogen/hypohalite involved.

2. Among common manganese oxyanions:

   • $\mathrm{MnO_4^-}$ is permanganate, where Mn is in oxidation state $+7$.
   • $\mathrm{MnO_4^{2-}}$ is manganate, where Mn is in oxidation state $+6$.
   • $\mathrm{CrO_4^{2-}}$ is chromate, not a manganese species.

3. A very standard preparation in inorganic chemistry is oxidation of manganate to permanganate using chlorine: $2\,\mathrm{MnO_4^{2-}} + \mathrm{Cl_2} \rightarrow 2\,\mathrm{MnO_4^-} + 2\,\mathrm{Cl^-}$

   Let us verify oxidation states:

   • Mn goes from $+6$ in $\mathrm{MnO_4^{2-}}$ to $+7$ in $\mathrm{MnO_4^-}$, so it is oxidized.
   • $\mathrm{Cl_2}$ goes from $0$ to $-1$ in $\mathrm{Cl^-}$, so chlorine is reduced.

   This reaction is perfectly consistent with a scheme where:

   • $X = \mathrm{MnO_4^-}$ (product formed)
   • $Y = \mathrm{Cl_2}$ (oxidizing/reagent species in the scheme)

4. Now check options systematically:

   A: $\mathrm{CrO_4^{2-}}$ and $\mathrm{Br_2}$

   • Chromate is unrelated to the usual manganese conversion here.
   • Not consistent.

   B: $\mathrm{MnO_4^{2-}}$ and $\mathrm{Cl_2}$

   • This would assign $X$ as manganate, but the standard oxidation by chlorine gives permanganate as product, not manganate.
   • So this is not the best fit.

   C: $\mathrm{MnO_4^-}$ and $\mathrm{Cl_2}$

   • Matches the known reaction above.
   • Correct.

   D: $\mathrm{MnSO_4}$ and $\mathrm{HOCl}$

   • $\mathrm{MnSO_4}$ contains Mn in $+2$ state and does not fit the typical oxyanion interconversion suggested by the options.
   • Not correct.

5. Therefore, $X = \mathrm{MnO_4^-}, \qquad Y = \mathrm{Cl_2}$

Hence, the correct option is C.