Remember the stability of different oxidation states of manganese. Mn⁺⁷ (in MnO₄⁻) and Mn⁺⁴ (in MnO₂) are relatively stable products. In strongly acidic conditions, Mn⁺² is the most stable state, but for the disproportionation of MnO₄²⁻, MnO₂ is the typical reduction product.
Step 1: Understanding the Question:
The question asks for the products of the disproportionation reaction of the manganate ion (MnO₄²⁻) in an acidic medium. A disproportionation reaction is one where a single species is simultaneously oxidized and reduced. Step 2: Key Formula or Approach:
1. Determine the oxidation state of Manganese (Mn) in the reactant, MnO₄²⁻.
2. Identify the possible higher and lower oxidation states that Mn can achieve from this state.
3. Write the balanced chemical equation for the reaction in an acidic medium. Step 3: Detailed Explanation: 1. Oxidation state of Mn in MnO₄²⁻:
Let the oxidation state of Mn be \(x\). The oxidation state of oxygen is -2.
\[ x + 4(-2) = -2 \]
\[ x - 8 = -2 \]
\[ x = +6 \]
So, Mn is in the +6 oxidation state in the manganate ion. 2. Disproportionation:
In a disproportionation reaction, Mn(+6) will be oxidized to a higher oxidation state and reduced to a lower oxidation state.
\begin{itemize}
\item Oxidation: A common stable higher oxidation state for manganese is +7, found in the permanganate ion (MnO₄⁻).
\item Reduction: A common stable lower oxidation state for manganese in an acidic or neutral medium is +4, found in manganese dioxide (MnO₂). Another possibility is +2 (Mn²⁺).
\end{itemize}
3. Balancing the Reaction:
Let's write the reaction with the identified products, permanganate (MnO₄⁻, Mn=+7) and manganese dioxide (MnO₂, Mn=+4).
The unbalanced reaction is:
\[ \text{MnO}_4^{2-} \rightarrow \text{MnO}_4^- + \text{MnO}_2 \]
Here, Mn(+6) → Mn(+7) is oxidation (loss of 1 electron), and Mn(+6) → Mn(+4) is reduction (gain of 2 electrons).
To balance the electrons, we need 2 atoms of Mn(+6) to be oxidized for every 1 atom of Mn(+6) that is reduced. This means we start with 3 atoms of Mn(+6).
\[ 3\text{MnO}_4^{2-} \rightarrow 2\text{MnO}_4^- + \text{MnO}_2 \]
Now, we balance the charge and atoms in an acidic medium by adding H⁺ and H₂O.
\begin{itemize}
\item Reactant side charge: \(3 \times (-2) = -6\)
\item Product side charge: \(2 \times (-1) = -2\)
\end{itemize}
To balance the charge, we add 4 H⁺ ions to the reactant side:
\[ 3\text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^- + \text{MnO}_2 \]
Now, balance the hydrogen and oxygen atoms.
\begin{itemize}
\item Reactant side: 12 O, 4 H
\item Product side: (2x4) + 2 = 10 O
\end{itemize}
To balance O and H, we add 2 H₂O molecules to the product side:
\[ 3\text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O} \]
The equation is now fully balanced. The products are MnO₄⁻ and MnO₂. Step 4: Final Answer:
The disproportionation of manganate ion (MnO₄²⁻) in an acidic medium yields permanganate ion (MnO₄⁻) and manganese dioxide (MnO₂). This matches option (D).
