Q63 / 180 · NEET 2026 (Re-NEET) · Chemistry

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

  • Correct answer
    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.

Solution

The reaction of $\text{NaCl}$ with concentrated $\text{H}_2\text{SO}_4$ and $\text{MnO}_2$ is given by:

$2\text{NaCl} + \text{MnO}_2 + 2\text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{MnSO}_4 + 2\text{H}_2\text{O} + \text{Cl}_2$

In this reaction, the oxidation state of $\text{Mn}$ changes from $+4$ in $\text{MnO}_2$ to $+2$ in $\text{MnSO}_4$. Since the oxidation state decreases, $\text{Mn}$ undergoes reduction, not oxidation. Thus, Statement-I is incorrect.

The reaction of $\text{NaI}$ with concentrated $\text{H}_2\text{SO}_4$ and $\text{MnO}_2$ is given by:

$2\text{NaI} + \text{MnO}_2 + 2\text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{MnSO}_4 + 2\text{H}_2\text{O} + \text{I}_2$

Here, the oxidation state of $\text{Mn}$ also changes from $+4$ to $+2$, which means $\text{Mn}$ undergoes reduction. Thus, Statement-II is correct.

Therefore, Statement-I is incorrect but Statement-II is correct.

Answer: Statement-I is incorrect but Statement-II is correct.

Hence, the correct answer is option: 'A'.

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