To determine which ions have a radius greater than \( F^- \) (fluoride ion), we will analyze the size of each ion in comparison to \( F^- \).
### Step-by-Step Solution:
1. **Understanding the Fluoride Ion \( F^- \)**:
- The fluoride ion has gained one electron compared to the neutral fluorine atom. Fluorine has an atomic number of 9, meaning it has 9 protons and, in the ion form, 10 electrons.
- The effective nuclear charge (the net positive charge experienced by electrons) is relatively high due to the presence of 9 protons attracting 10 electrons, which results in a smaller ionic radius.
2. **Comparing with \( H^- \) (Hydride Ion)**:
- The hydride ion has 1 proton and 2 electrons. The effective nuclear charge is low because there is only 1 proton to attract 2 electrons.
- Therefore, the radius of \( H^- \) is greater than that of \( F^- \) due to the lower nuclear charge in hydrogen.
3. **Comparing with \( Cl^- \) (Chloride Ion)**:
- Chlorine has an atomic number of 17, meaning it has 17 protons and, in the ion form, 18 electrons.
- As we move down the group in the periodic table (from fluorine to chlorine), the number of electron shells increases, leading to a larger ionic radius. Thus, \( Cl^- \) has a greater radius than \( F^- \).
4. **Comparing with \( Br^- \) (Bromide Ion)**:
- Bromine has an atomic number of 35, meaning it has 35 protons and, in the ion form, 36 electrons.
- Similar to chlorine, bromine is further down the group, which means it has more electron shells and a larger ionic radius than \( F^- \).
5. **Comparing with \( I^- \) (Iodide Ion)**:
- Iodine has an atomic number of 53, meaning it has 53 protons and, in the ion form, 54 electrons.
- Iodine is even further down the group than bromine, resulting in an even larger ionic radius compared to \( F^- \).
### Conclusion:
The ions that have a radius greater than \( F^- \) are:
- \( H^- \)
- \( Cl^- \)
- \( Br^- \)
- \( I^- \)
Thus, all the ions listed (1, 2, 3, and 4) have a radius greater than \( F^- \).
