`underline (A) overset( 4 K OH, O_(2))to 2underline (B) underset(("Green"))(+2H_(2)O)`
`3 underline(B) overset(4HCl) to 2 underline (C) underset(("Puple"))(+ MnO _(2) +) 2 H _(2)O `
`2 underline (C) overset(H_(2) O, KI) to2 underline (A) + 2KOH+ underline(D)`
In the above sequence of reaction `underline (A) and underline (D)` respectively, are:

Options

A. `KI and K MnO _(4)`
B. `MnO _(2) and KIO _(3)`
C. `KIO_(3)and MnO _(2)`
D. `KI and K_(2) MnO _(4)`

Explanation

To solve the given problem, we need to identify the compounds A and D in the sequence of reactions provided. Let's break down the reactions step by step.

### Step 1: Identify Compound B
The first reaction is:
\[ A + 4 \text{KOH} + \text{O}_2 \rightarrow 2B + 2\text{H}_2\text{O} \]
The product B is described as green in color. Since we are dealing with manganese compounds, and knowing that potassium manganate (\( \text{K}_2\text{MnO}_4 \)) is green, we can infer that:
\[ B = \text{K}_2\text{MnO}_4 \]

### Step 2: Identify Compound A
Now, we need to identify compound A. Since we have established that B is \( \text{K}_2\text{MnO}_4 \), we can deduce that A must be the precursor to this compound. The reaction indicates that A is oxidized to form B. The oxidation of manganese dioxide (\( \text{MnO}_2 \)) in the presence of potassium hydroxide and oxygen produces \( \text{K}_2\text{MnO}_4 \):
\[ A = \text{MnO}_2 \]

### Step 3: Identify Compound C
The second reaction is:
\[ 3B + 4\text{HCl} \rightarrow 2C + \text{MnO}_2 + 2\text{H}_2\text{O} \]
Since B is \( \text{K}_2\text{MnO}_4 \), we can substitute it into the equation:
\[ 3\text{K}_2\text{MnO}_4 + 4\text{HCl} \rightarrow 2C + \text{MnO}_2 + 2\text{H}_2\text{O} \]
In acidic conditions, \( \text{K}_2\text{MnO}_4 \) is reduced to potassium permanganate (\( \text{KMnO}_4 \)), which is purple in color:
\[ C = \text{KMnO}_4 \]

### Step 4: Identify Compound D
The third reaction is:
\[ 2C + \text{H}_2\text{O} + \text{KI} \rightarrow 2A + 2\text{KOH} + D \]
Substituting \( C \) into the equation gives:
\[ 2\text{KMnO}_4 + \text{H}_2\text{O} + \text{KI} \rightarrow 2\text{MnO}_2 + 2\text{KOH} + D \]
In this reaction, \( \text{KMnO}_4 \) acts as a strong oxidizing agent, oxidizing iodide ions (\( \text{I}^- \)) to iodate (\( \text{IO}_3^- \)):
\[ D = \text{KIO}_3 \]

### Final Answer
Thus, the compounds A and D are:
- \( A = \text{MnO}_2 \)
- \( D = \text{KIO}_3 \)