Consider the following species:
`N^(3-)`,`O^(2-)`,`F^(ө)`,`Na^(o+)`,`Mg^(2+)` and `Al^(3+)`
a. What is common in them?
b. Arrange them in the order of increasing ionic radii.

Options

A. All have same ionic radii.
B. All are isoelectronic species having 10 electrons.
C. All of them belong to the third period.
D. The nature of oxides of all the ions is basic.

Answer

To solve the question, we will break it down into two parts: identifying the common feature among the given species and arranging them in order of increasing ionic radii.

### Part a: Common Feature

1. **Identify the Species**: The given species are:
- \(N^{3-}\) (Nitride ion)
- \(O^{2-}\) (Oxide ion)
- \(F^{1-}\) (Fluoride ion)
- \(Na^{1+}\) (Sodium ion)
- \(Mg^{2+}\) (Magnesium ion)
- \(Al^{3+}\) (Aluminum ion)

2. **Determine the Number of Electrons**:
- **Nitrogen (N)**: Atomic number = 7. Gaining 3 electrons gives \(7 + 3 = 10\) electrons.
- **Oxygen (O)**: Atomic number = 8. Gaining 2 electrons gives \(8 + 2 = 10\) electrons.
- **Fluorine (F)**: Atomic number = 9. Gaining 1 electron gives \(9 + 1 = 10\) electrons.
- **Sodium (Na)**: Atomic number = 11. Losing 1 electron gives \(11 - 1 = 10\) electrons.
- **Magnesium (Mg)**: Atomic number = 12. Losing 2 electrons gives \(12 - 2 = 10\) electrons.
- **Aluminum (Al)**: Atomic number = 13. Losing 3 electrons gives \(13 - 3 = 10\) electrons.

3. **Conclusion**: All the species have the same number of electrons, which is 10. Therefore, they are **isoelectronic species**.

### Part b: Order of Increasing Ionic Radii

1. **Understanding Ionic Radii**:
- Cations (positively charged ions) are smaller than their neutral atoms because they lose electrons, resulting in a higher nuclear charge relative to the number of electrons.
- Anions (negatively charged ions) are larger than their neutral atoms because they gain electrons, resulting in a lower nuclear charge relative to the number of electrons.

2. **Ranking the Ions**:
- For cations, the size decreases with increasing positive charge:
- \(Al^{3+}\) < \(Mg^{2+}\) < \(Na^{1+}\)
- For anions, the size increases with increasing negative charge:
- \(N^{3-}\) > \(O^{2-}\) > \(F^{1-}\)

3. **Combining the Sizes**:
- The order from smallest to largest ionic radii is:
- \(Al^{3+}\) < \(Mg^{2+}\) < \(Na^{1+}\) < \(F^{1-}\) < \(O^{2-}\) < \(N^{3-}\)

### Final Answer
- **Common Feature**: All species are isoelectronic with 10 electrons.
- **Order of Increasing Ionic Radii**:
\[ Al^{3+} < Mg^{2+} < Na^{1+} < F^{1-} < O^{2-} < N^{3-} \]