`{:(,"Compounds",,"O.N."),((A),KMn^(**)O_(4),(1),+4),((B),Ni^(**)(CO)_(4),(2),+7),((C),[Pt^(**)(NH_(3))Cl_(2)]Cl_(2),(3),0),((D),Na_(2)O_(2)^(**),(4),-1):}`
The correct code for the O.N. of asterisked atom would be:

Options

A. `{:(A,B,C,D),(1,2,3,4):}`
B. `{:(A,B,C,D),(4,3,2,1):}`
C. `{:(A,B,C,D),(2,3,1,4):}`
D. `{:(A,B,C,D),(4,1,2,3):}`

Answer

To find the oxidation numbers of the specified atoms in the given compounds, we will analyze each compound step by step.

### Step 1: Determine the oxidation number of Mn in KMnO4
1. **Identify the oxidation states of other elements:**
- Potassium (K) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2. Since there are 4 oxygen atoms, the total contribution from oxygen is \(4 \times (-2) = -8\).

2. **Set up the equation:**
\[ \text{Oxidation state of K} + \text{Oxidation state of Mn} + \text{Total oxidation state of O} = 0 \]
\[ +1 + x - 8 = 0 \]
where \(x\) is the oxidation state of Mn.

3. **Solve for x:**
\[ +1 + x - 8 = 0 \implies x - 7 = 0 \implies x = +7 \]

### Step 2: Determine the oxidation number of Ni in Ni(CO)4
1. **Identify the oxidation states of other elements:**
- Carbonyl (CO) is a neutral ligand, thus has an oxidation state of 0.

2. **Set up the equation:**
\[ \text{Oxidation state of Ni} + 4 \times \text{Oxidation state of CO} = 0 \]
\[ x + 4 \times 0 = 0 \]
where \(x\) is the oxidation state of Ni.

3. **Solve for x:**
\[ x = 0 \]

### Step 3: Determine the oxidation number of Pt in [Pt(NH3)Cl2]Cl
1. **Identify the oxidation states of other elements:**
- Ammonia (NH3) is a neutral ligand, thus has an oxidation state of 0.
- Each chloride (Cl) has an oxidation state of -1. Since there are 3 chloride ions, the total contribution from Cl is \(3 \times (-1) = -3\).

2. **Set up the equation:**
\[ \text{Oxidation state of Pt} + 0 + 2 \times (-1) = 0 \]
\[ x - 2 = 0 \]
where \(x\) is the oxidation state of Pt.

3. **Solve for x:**
\[ x = +2 \]

### Step 4: Determine the oxidation number of O in Na2O2
1. **Identify the oxidation states of other elements:**
- Sodium (Na) has an oxidation state of +1. Since there are 2 sodium atoms, the total contribution from Na is \(2 \times (+1) = +2\).

2. **Set up the equation:**
\[ 2 \times \text{Oxidation state of Na} + 2 \times \text{Oxidation state of O} = 0 \]
\[ 2 + 2y = 0 \]
where \(y\) is the oxidation state of O.

3. **Solve for y:**
\[ 2y = -2 \implies y = -1 \]

### Summary of Oxidation Numbers:
- A: Mn in KMnO4 = +7
- B: Ni in Ni(CO)4 = 0
- C: Pt in [Pt(NH3)Cl2]Cl = +2
- D: O in Na2O2 = -1

### Final Answer:
The correct code for the oxidation numbers of the asterisked atoms is:
- A: 1 (Mn = +7)
- B: 2 (Ni = 0)
- C: 3 (Pt = +2)
- D: 4 (O = -1)

Thus, the final answer is **2 3 1 4**.