Question

Given below are two statements:
Statement-I : Heating NaCl with concentrated $\text{H}_2\text{SO}_4$ and $\text{MnO}_2$ results in oxidation of Mn.
Statement-II : Heating NaI with concentrated $\text{H}_2\text{SO}_4$ and $\text{MnO}_2$ results in reduction of Mn.
In light of the above statements, choose the most appropriate answer from the options given below:

• A. Statement-I is incorrect but Statement-II is correct
• B. Both Statement-I and Statement-II are correct
• C. Both Statement-I and Statement-II are incorrect
• D. Statement-I is correct but Statement-II is incorrect

Hint:

To avoid confusion with redox terminology: - An oxidizing agent (like $\text{MnO}_2$) oxidizes other species while always undergoing reduction itself. - Consequently, in any successful oxidation reaction driven by $\text{MnO}_2$, the manganese ion will always be reduced from $\text{Mn}^{+4}$ to $\text{Mn}^{+2}$.

Answer 1

The Correct Option is A

Concept: Manganese dioxide ($\text{MnO}_2$) is a robust and well-known oxidizing agent in acidic media. When a metal halide salt (NaX) is heated in the presence of concentrated sulfuric acid ($\text{H}_2\text{SO}_4$) and $\text{MnO}_2$, the concentrated acid first reacts with the halide salt to generate the corresponding hydrogen halide gas (HX). Because $\text{MnO}_2$ is a powerful electron acceptor, it oxidizes the halide ions ($\text{X}^-$) into their elemental halogen form ($\text{X}_2$). Concurrently, the manganese atom inside $\text{MnO}_2$ undergoes reduction, changing its oxidation state from $+4$ down to a $+2$ state, forming manganese(II) sulfate ($\text{MnSO}_4$).

Step 1: Analyzing Statement-I (Reaction with NaCl).
When sodium chloride (NaCl) is heated alongside concentrated $\text{H}_2\text{SO}_4$ and $\text{MnO}_2$, chlorine gas is evolved. Let us write down the balanced redox chemical equation for this process: \[ 2\text{NaCl} + \text{MnO}_2 + 3\text{H}_2\text{SO}_4 \longrightarrow 2\text{NaHSO}_4 + \text{MnSO}_4 + \text{Cl}_2\uparrow + 2\text{H}_2\text{O} \] Let us trace the oxidation states of the elements to understand the electron transfer:

• In Reactants: Inside $\text{MnO}_2$, oxygen has an oxidation state of $-2$, meaning manganese exists in the $+4$ oxidation state ($\text{Mn}^{+4}$). The chloride ion inside NaCl has an oxidation state of $-1$.

• In Products: Inside $\text{MnSO}_4$, the sulfate ion carries a $-2$ charge, meaning manganese has shifted down to a $+2$ oxidation state ($\text{Mn}^{+2}$). The evolved elemental $\text{Cl}_2$ gas has an oxidation state of $0$.
Because the oxidation state of Manganese drops from $+4$ to $+2$, Manganese undergoes reduction, while the chloride ions are oxidized to chlorine gas. Statement-I claims that this reaction "results in oxidation of Mn", which is false. Thus, Statement-I is incorrect.

Step 2: Analyzing Statement-II (Reaction with NaI).
When sodium iodide (NaI) is heated under identical conditions with concentrated $\text{H}_2\text{SO}_4$ and $\text{MnO}_2$, dark violet iodine vapor is evolved. The corresponding chemical equation is: \[ 2\text{NaI} + \text{MnO}_2 + 3\text{H}_2\text{SO}_4 \longrightarrow 2\text{NaHSO}_4 + \text{MnSO}_4 + \text{I}_2\uparrow + 2\text{H}_2\text{O} \] Let us trace the oxidation numbers for this system:

• In Reactants: Manganese inside $\text{MnO}_2$ is at a $+4$ oxidation state. The iodide ion ($\text{I}^-$) has an oxidation state of $-1$.

• In Products: Manganese inside $\text{MnSO}_4$ is at a $+2$ oxidation state. Elemental iodine ($\text{I}_2$) is at $0$.
Here, the oxidation number of manganese decreases from $+4$ to $+2$, meaning Manganese undergoes reduction, while iodide is oxidized to iodine. Statement-II claims that this reaction "results in reduction of Mn", which is accurate. Therefore, Statement-II is correct.

Step 3: Final Selection.
Since Statement-I is incorrect and Statement-II is correct, the correct choice is Option (1).