A solution of `KMnO_4` is reduced to various products depending upon its pH. At pH lt 7 it is reduced to a colourless solution (A), at pH = 7 it forms a brown precipitate (B) and at pH gt 7 it gives a green solution ( C), (A),(B) and( C) are

Options

A. `{:("(A)","(B)","(C)"),(Mn^(2+),MnO_2,MnO_4^(2-)):}`
B. `{:("(A)","(B)","(C)"),(MnO_2,mn^(2+),MnO_4^(2-)):}`
C. `{:("(A)","(B)","(C)"),(Mn^(2+),MnO_4^(2-),MnO_2):}`
D. `{:("(A)","(B)","(C)"),(MnO_4^(2-),Mn^(2+),MnO_2):}`

Explanation

To solve the question regarding the reduction products of KMnO4 at different pH levels, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the pH conditions**: The question states that the reduction products of KMnO4 vary with pH:
- pH < 7 (acidic)
- pH = 7 (neutral)
- pH > 7 (basic)

2. **Reduction in Acidic Medium (pH < 7)**:
- In acidic conditions, KMnO4 (MnO4^-) is reduced to Mn^2+.
- The balanced equation for this reaction is:
\[ \text{MnO}_4^- + 8 \text{H}^+ + 5 \text{e}^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]
- The product formed (A) in this case is **Mn^2+**, which is colorless.

3. **Reduction in Neutral Medium (pH = 7)**:
- In neutral conditions, KMnO4 is reduced to MnO2.
- The balanced equation for this reaction is:
\[ \text{MnO}_4^- + 2 \text{H}_2\text{O} + 3 \text{e}^- \rightarrow \text{MnO}_2 + 4 \text{OH}^- \]
- The product formed (B) in this case is **MnO2**, which appears as a brown precipitate.

4. **Reduction in Basic Medium (pH > 7)**:
- In basic conditions, KMnO4 is reduced to MnO4^2- (manganate ion).
- The balanced equation for this reaction is:
\[ \text{MnO}_4^- + \text{e}^- \rightarrow \text{MnO}_4^{2-} \]
- The product formed (C) in this case is **MnO4^2-**, which is green in color.

5. **Summarizing the Products**:
- A (pH < 7): **Mn^2+** (colorless)
- B (pH = 7): **MnO2** (brown precipitate)
- C (pH > 7): **MnO4^2-** (green solution)

6. **Final Answer**:
- The products A, B, and C are:
- A = Mn^2+
- B = MnO2
- C = MnO4^2-