In which of the following option all are isoelectronic?

Options

A. `N^(3-),O^(2-),F^(-),Na^(+)`
B. `Na^(+),N^(-3),F^(-),Li^(+)`
C. `Li^(+),N^(3-),F^(-),O^(2-)`
D. `Li^(+),Na^(+),O^(2-),F^(-)`

Answer

To determine which of the given options contains all isoelectronic species, we need to find the electronic configurations of the species provided and count the number of electrons in each.

### Step-by-step Solution:

1. **Identify the Species**: Let's consider the species mentioned: N³⁻, O²⁻, F⁻, and Na⁺.

2. **Determine the Electronic Configuration**:
- **For N³⁻**:
- Nitrogen (N) has an atomic number of 7, so it has 7 electrons.
- N³⁻ means it has gained 3 electrons:
\[ \text{Total electrons} = 7 + 3 = 10 \]
- Electronic configuration: \(1s^2 2s^2 2p^6\)

- **For O²⁻**:
- Oxygen (O) has an atomic number of 8, so it has 8 electrons.
- O²⁻ means it has gained 2 electrons:
\[ \text{Total electrons} = 8 + 2 = 10 \]
- Electronic configuration: \(1s^2 2s^2 2p^6\)

- **For F⁻**:
- Fluorine (F) has an atomic number of 9, so it has 9 electrons.
- F⁻ means it has gained 1 electron:
\[ \text{Total electrons} = 9 + 1 = 10 \]
- Electronic configuration: \(1s^2 2s^2 2p^6\)

- **For Na⁺**:
- Sodium (Na) has an atomic number of 11, so it has 11 electrons.
- Na⁺ means it has lost 1 electron:
\[ \text{Total electrons} = 11 - 1 = 10 \]
- Electronic configuration: \(1s^2 2s^2 2p^6\)

3. **Count the Electrons**:
- All species (N³⁻, O²⁻, F⁻, Na⁺) have 10 electrons.

4. **Conclusion**: Since all the species have the same number of electrons (10), they are isoelectronic.

### Final Answer:
The correct option where all species are isoelectronic is **Option A**.