The reaction of NaCl with concentrated $H_{2}SO_{4}$ and $MnO_{2}$ is the laboratory preparation of chlorine gas: $$MnO_{2} + 2NaCl + 2H_{2}SO_{4} \rightarrow MnSO_{4} + Na_{2}SO_{4} + 2H_{2}O + Cl_{2}$$ In this reaction, the oxidation state of Mn changes from +4 (in $MnO_{2}$) to +2 (in $MnSO_{4}$). This is a reduction process, not an oxidation process. Thus, Statement-I is incorrect.
When NaI is heated with concentrated $H_{2}SO_{4}$ and $MnO_{2}$, the iodide ion ($I^{-}$) is a strong reducing agent. It reduces $MnO_{2}$ to $Mn^{2+}$ and is itself oxidized to $I_{2}$. $$MnO_{2} + 2NaI + 2H_{2}SO_{4} \rightarrow MnSO_{4} + Na_{2}SO_{4} + 2H_{2}O + I_{2}$$ Similar to the first case, the oxidation state of Mn changes from +4 to +2, which is a reduction. Therefore, Statement-II is correct as it states that Mn undergoes reduction.
Final Answer: (A) Statement-I is incorrect but Statement-II is correct
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