In which of the following pair, both the species are isoelectronic but the first one is large in size than the second?

Options

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

Answer

To solve the problem, we need to identify which pair of species are isoelectronic (having the same number of electrons) and where the first species is larger in size than the second. Here’s a step-by-step solution:

### Step 1: Understand Isoelectronic Species
Isoelectronic species are atoms or ions that have the same number of electrons. For example, if one species has 10 electrons and another also has 10 electrons, they are isoelectronic.

**Hint:** Remember that the number of electrons can be calculated from the atomic number and the charge of the species.

### Step 2: Analyze Each Pair
We will analyze each pair given in the options to determine their electron counts.

1. **Option 1: S²⁻ and O²⁻**
- Sulfur (S) has an atomic number of 16. For S²⁻, it has 16 + 2 = 18 electrons.
- Oxygen (O) has an atomic number of 8. For O²⁻, it has 8 + 2 = 10 electrons.
- **Conclusion:** Not isoelectronic.

2. **Option 2: Cl²⁻ and S²⁻**
- Chlorine (Cl) has an atomic number of 17. For Cl²⁻, it has 17 + 2 = 19 electrons.
- Sulfur (S) has 16 + 2 = 18 electrons.
- **Conclusion:** Not isoelectronic.

3. **Option 3: F⁻ and Na⁺**
- Fluorine (F) has an atomic number of 9. For F⁻, it has 9 + 1 = 10 electrons.
- Sodium (Na) has an atomic number of 11. For Na⁺, it has 11 - 1 = 10 electrons.
- **Conclusion:** Both species are isoelectronic with 10 electrons.

4. **Option 4: Analyze Size**
- F⁻ has a negative charge, which means it has gained an electron and will be larger in size due to increased electron-electron repulsion.
- Na⁺ has a positive charge, which means it has lost an electron and will be smaller in size compared to F⁻.
- **Conclusion:** F⁻ is larger than Na⁺.

### Final Conclusion
The correct answer is **Option 3: F⁻ and Na⁺**, where both species are isoelectronic, and F⁻ is larger than Na⁺.