To determine which of the ions \( N^{3-} \), \( O^{2-} \), and \( F^{-} \) is the largest in size, we can follow these steps:
### Step 1: Identify the number of electrons in each ion
- **For \( N^{3-} \)**: Nitrogen (N) has an atomic number of 7, which means it has 7 protons and, when it gains 3 electrons, it has a total of \( 7 + 3 = 10 \) electrons.
- **For \( O^{2-} \)**: Oxygen (O) has an atomic number of 8, so it has 8 protons. When it gains 2 electrons, it has \( 8 + 2 = 10 \) electrons.
- **For \( F^{-} \)**: Fluorine (F) has an atomic number of 9, meaning it has 9 protons. With the addition of 1 electron, it has \( 9 + 1 = 10 \) electrons.
### Step 2: Determine the isoelectronic nature of the ions
All three ions \( N^{3-} \), \( O^{2-} \), and \( F^{-} \) have the same number of electrons (10 electrons). Therefore, they are isoelectronic species.
### Step 3: Compare the number of protons in each ion
- **For \( N^{3-} \)**: 7 protons
- **For \( O^{2-} \)**: 8 protons
- **For \( F^{-} \)**: 9 protons
### Step 4: Analyze the effective nuclear charge
The effective nuclear charge (Z_eff) is the net positive charge experienced by electrons in a multi-electron atom. As the number of protons increases while the number of electrons remains constant, the effective nuclear charge increases. This leads to a stronger attraction between the nucleus and the electrons, resulting in a smaller ionic size.
- **\( N^{3-} \)** has the least number of protons (7), so it has the weakest attraction for the 10 electrons, leading to the largest size.
- **\( O^{2-} \)** has 8 protons, so it has a stronger attraction than \( N^{3-} \) but weaker than \( F^{-} \).
- **\( F^{-} \)** has the most protons (9), leading to the strongest attraction for the 10 electrons, resulting in the smallest size.
### Conclusion
Since \( N^{3-} \) has the least effective nuclear charge, it will be the largest ion among the three. Therefore, the largest ion in size is:
**Answer: \( N^{3-} \)**
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