In the scheme given below, X and Y, respectively, are

The sequence of reactions involved in the scheme (shown in the paper) starts from $$MnO_2$$. In strongly alkaline medium and in the presence of an oxidising agent such as $$O_2$$ (or $$K_2S_2O_8$$), $$MnO_2$$ is first converted to green manganate ion:

$$MnO_2 + 4\,OH^- + O_2 \;\longrightarrow\; MnO_4^{2-} + 2\,H_2O$$

The green manganate ion $$MnO_4^{2-}$$ is unstable in acidic medium. On acidification it undergoes disproportionation (simultaneous oxidation and reduction of the same species) to give purple permanganate ion and $$Mn^{4+}$$ (as $$MnO_2$$):

$$3\,MnO_4^{2-} + 4\,H^+ \;\longrightarrow\; 2\,MnO_4^- + MnO_2 + 2\,H_2O$$

Thus the species produced after acidification, labelled X in the scheme, is the purple permanganate ion $$MnO_4^-$$.

Purple permanganate in acidic medium is a very strong oxidising agent. It can oxidise chloride ions to chlorine gas according to

$$2\,MnO_4^- + 10\,Cl^- + 16\,H^+ \;\longrightarrow\; 2\,Mn^{2+} + 5\,Cl_2 + 8\,H_2O$$

Hence the gaseous product Y obtained in the last step of the scheme is molecular chlorine $$Cl_2$$.

Therefore, X = $$MnO_4^-$$ and Y = $$Cl_2$$, which corresponds to Option C.

Option C which is: $$MnO_4^{-}$$ and $$Cl_2$$