The anion `O^(2-)` is iso-electronic with

Options

A. `F^(+)`
B. `N^(2-)`
C. `F^(-)`
D. `N^(3+)`

Explanation

To determine which species is iso-electronic with the anion \( O^{2-} \), we need to follow these steps:

### Step 1: Determine the number of electrons in \( O^{2-} \)
- The atomic number of oxygen (O) is 8, which means it has 8 protons and 8 electrons in its neutral state.
- Since \( O^{2-} \) has gained 2 additional electrons, we calculate the total number of electrons:
\[ \text{Total electrons in } O^{2-} = 8 + 2 = 10 \]

### Step 2: Identify species that have 10 electrons
Now we need to find other species that also have 10 electrons. We will evaluate the given options.

### Step 3: Evaluate each option
1. **Option: \( F^{+} \)**
- The atomic number of fluorine (F) is 9. In \( F^{+} \), it has lost 1 electron:
\[ \text{Total electrons in } F^{+} = 9 - 1 = 8 \]
- Not iso-electronic with \( O^{2-} \).

2. **Option: \( N^{2-} \)**
- The atomic number of nitrogen (N) is 7. In \( N^{2-} \), it has gained 2 electrons:
\[ \text{Total electrons in } N^{2-} = 7 + 2 = 9 \]
- Not iso-electronic with \( O^{2-} \).

3. **Option: \( F^{-} \)**
- The atomic number of fluorine (F) is 9. In \( F^{-} \), it has gained 1 electron:
\[ \text{Total electrons in } F^{-} = 9 + 1 = 10 \]
- This is iso-electronic with \( O^{2-} \).

4. **Option: \( N^{3+} \)**
- The atomic number of nitrogen (N) is 7. In \( N^{3+} \), it has lost 3 electrons:
\[ \text{Total electrons in } N^{3+} = 7 - 3 = 4 \]
- Not iso-electronic with \( O^{2-} \).

### Step 4: Conclusion
The only species that is iso-electronic with \( O^{2-} \) is \( F^{-} \), as both have 10 electrons.

### Final Answer:
The anion \( O^{2-} \) is iso-electronic with \( F^{-} \).

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