Which of the following ion is not isoelectronic with `O^(2-)` ?

Options

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

Answer

To determine which of the given ions is not isoelectronic with \( O^{2-} \), we need to find the total number of electrons in each ion and compare them to the number of electrons in \( O^{2-} \).

### Step-by-Step Solution:

1. **Identify the number of electrons in \( O^{2-} \)**:
- The atomic number of oxygen (O) is 8, which means a neutral oxygen atom has 8 electrons.
- The \( O^{2-} \) ion has gained 2 additional electrons, so:
\[ \text{Total electrons in } O^{2-} = 8 + 2 = 10 \]

2. **Calculate the number of electrons in \( Mg^{+} \)**:
- The atomic number of magnesium (Mg) is 12, so a neutral magnesium atom has 12 electrons.
- The \( Mg^{+} \) ion has lost 1 electron, so:
\[ \text{Total electrons in } Mg^{+} = 12 - 1 = 11 \]

3. **Calculate the number of electrons in \( Na^{+} \)**:
- The atomic number of sodium (Na) is 11, so a neutral sodium atom has 11 electrons.
- The \( Na^{+} \) ion has lost 1 electron, so:
\[ \text{Total electrons in } Na^{+} = 11 - 1 = 10 \]

4. **Calculate the number of electrons in \( N^{3-} \)**:
- The atomic number of nitrogen (N) is 7, so a neutral nitrogen atom has 7 electrons.
- The \( N^{3-} \) ion has gained 3 additional electrons, so:
\[ \text{Total electrons in } N^{3-} = 7 + 3 = 10 \]

5. **Calculate the number of electrons in \( F^{-} \)**:
- The atomic number of fluorine (F) is 9, so a neutral fluorine atom has 9 electrons.
- The \( F^{-} \) ion has gained 1 additional electron, so:
\[ \text{Total electrons in } F^{-} = 9 + 1 = 10 \]

6. **Compare the number of electrons**:
- \( O^{2-} \): 10 electrons
- \( Mg^{+} \): 11 electrons (not isoelectronic)
- \( Na^{+} \): 10 electrons (isoelectronic)
- \( N^{3-} \): 10 electrons (isoelectronic)
- \( F^{-} \): 10 electrons (isoelectronic)

### Conclusion:
The ion that is not isoelectronic with \( O^{2-} \) is \( Mg^{+} \).