Question

In the following reaction:
\[ \text{MnO}_4^{-} \xrightarrow{\text{H}^+} \text{MnO}_2 + \text{MnO}_4^{2-} \] Mangante ion undergoes disproportionation in acidic medium to form:

• A. \(\text{MnO}_2, \text{MnO}_4^{2-}\)
• B. \(\text{MnO}_4, \text{MnO}_2\)
• C. \(\text{MnO}_2, \text{Mn}_2\text{O}_3\)
• D. \(\text{MnO}_4, \text{MnO}\)

Hint:

In acidic medium, the manganate ion (\(\text{MnO}_4^{-}\)) disproportionates into manganese dioxide (\(\text{MnO}_2\)) and manganese tetraoxide (\(\text{MnO}_4^{2-}\)).

Answer 1

The Correct Option is A

The given question involves the disproportionation reaction of the manganate ion \(\text{MnO}_4^-\) in an acidic medium. Let's analyze this step-by-step:

Disproportionation reactions are chemical reactions where one substance is simultaneously oxidized and reduced to form two different products.

In this reaction:

\[ \text{MnO}_4^- \xrightarrow{\text{H}^+} \text{MnO}_2 + \text{MnO}_4^{2-} \]

The manganate ion \(\text{MnO}_4^-\) is both oxidized and reduced:

1. Reduction half-reaction: 
   \(\text{MnO}_4^- + 4\text{H}^+ + 3\text{e}^- \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O}\)
2. Oxidation half-reaction: 
   \(\text{MnO}_4^- + \text{e}^- \rightarrow \text{MnO}_4^{2-}\)

Both half-reactions show that \(\text{MnO}_4^-\) ion is undergoing disproportionation to form two different manganate species, \(\text{MnO}_2\) (manganese dioxide) and \(\text{MnO}_4^{2-}\) (manganate ion).

Therefore, in acidic conditions, the correct option of the products formed is \(\text{MnO}_2\) and \(\text{MnO}_4^{2-}\).

In conclusion, the correct answer is:

\(\text{MnO}_2, \text{MnO}_4^{2-}\)