Which of the following is not isoelectronic ?

Options

A. `Na^(+)`
B. `Mg^(2+)`
C. `O^(2-)`
D. `Cl^(-)`

Explanation

To determine which of the following options is not isoelectronic, we need to count the number of electrons in each species. Isoelectronic species are those that have the same number of electrons.

### Step-by-Step Solution:

1. **Identify the Species**: We need to analyze the given species: Na⁺ (Sodium ion), Mg²⁺ (Magnesium ion), O²⁻ (Oxide ion), and Cl⁻ (Chloride ion).

2. **Count Electrons in Na⁺**:
- Sodium (Na) has an atomic number of 11, meaning it has 11 electrons.
- As a cation (Na⁺), it loses 1 electron.
- Therefore, Na⁺ has:
\[ 11 - 1 = 10 \text{ electrons} \]

3. **Count Electrons in Mg²⁺**:
- Magnesium (Mg) has an atomic number of 12, meaning it has 12 electrons.
- As a cation (Mg²⁺), it loses 2 electrons.
- Therefore, Mg²⁺ has:
\[ 12 - 2 = 10 \text{ electrons} \]

4. **Count Electrons in O²⁻**:
- Oxygen (O) has an atomic number of 8, meaning it has 8 electrons.
- As an anion (O²⁻), it gains 2 electrons.
- Therefore, O²⁻ has:
\[ 8 + 2 = 10 \text{ electrons} \]

5. **Count Electrons in Cl⁻**:
- Chlorine (Cl) has an atomic number of 17, meaning it has 17 electrons.
- As an anion (Cl⁻), it gains 1 electron.
- Therefore, Cl⁻ has:
\[ 17 + 1 = 18 \text{ electrons} \]

6. **Compare the Number of Electrons**:
- Na⁺: 10 electrons
- Mg²⁺: 10 electrons
- O²⁻: 10 electrons
- Cl⁻: 18 electrons

7. **Conclusion**:
- The first three species (Na⁺, Mg²⁺, and O²⁻) all have 10 electrons and are isoelectronic.
- Cl⁻ has 18 electrons and is not isoelectronic with the others.

Thus, the species that is not isoelectronic is **Cl⁻**.