Name five species that are isoelectronic with `F^(ɵ)` ion

Explanation

To find five species that are isoelectronic with the fluoride ion (F⁻), we need to follow these steps:

### Step 1: Determine the number of electrons in F⁻
- The fluoride ion (F⁻) is derived from the neutral fluorine atom, which has 9 electrons. The negative charge indicates that there is one additional electron.
- Therefore, the total number of electrons in F⁻ is:
\[ 9 \text{ (from F)} + 1 \text{ (extra electron)} = 10 \text{ electrons} \]

### Step 2: Identify other species with 10 electrons
- We need to find other ions or neutral atoms that also have a total of 10 electrons.

### Step 3: List the isoelectronic species
1. **O²⁻ (Oxide ion)**:
- Oxygen (O) has 8 electrons. The O²⁻ ion has gained 2 additional electrons.
- Total electrons: \(8 + 2 = 10\)

2. **N³⁻ (Nitride ion)**:
- Nitrogen (N) has 7 electrons. The N³⁻ ion has gained 3 additional electrons.
- Total electrons: \(7 + 3 = 10\)

3. **Ne (Neon)**:
- Neon is a noble gas with a complete outer shell, having 10 electrons.
- Total electrons: \(10\)

4. **Na⁺ (Sodium ion)**:
- Sodium (Na) has 11 electrons. The Na⁺ ion has lost 1 electron.
- Total electrons: \(11 - 1 = 10\)

5. **Mg²⁺ (Magnesium ion)**:
- Magnesium (Mg) has 12 electrons. The Mg²⁺ ion has lost 2 electrons.
- Total electrons: \(12 - 2 = 10\)

### Conclusion
The five species that are isoelectronic with the F⁻ ion are:
1. O²⁻
2. N³⁻
3. Ne
4. Na⁺
5. Mg²⁺