The number of halogens that form `XO_3^-` ion?

Answer

To determine the number of halogens that can form the ion \( XO_3^- \), we need to analyze the electronic configurations and the ability of each halogen to expand its octet.

### Step-by-Step Solution:

1. **Identify the Halogens**: The halogens in the periodic table are Fluorine (F), Chlorine (Cl), Bromine (Br), and Iodine (I).

2. **Examine the Electronic Configuration**:
- Fluorine: \( 1s^2 2s^2 2p^5 \) (Total of 7 valence electrons)
- Chlorine: \( [Ne] 3s^2 3p^5 \) (Total of 7 valence electrons)
- Bromine: \( [Ar] 4s^2 4p^5 \) (Total of 7 valence electrons)
- Iodine: \( [Kr] 5s^2 5p^5 \) (Total of 7 valence electrons)

3. **Determine the Ability to Expand Octet**:
- **Fluorine**: Lacks vacant d-orbitals (only has 2s and 2p orbitals). It cannot expand its octet and thus cannot form \( XO_3^- \).
- **Chlorine**: Has vacant 3d orbitals, allowing it to expand its octet. It can form \( ClO_3^- \).
- **Bromine**: Has vacant 4d orbitals, allowing it to expand its octet. It can form \( BrO_3^- \).
- **Iodine**: Has vacant 5d orbitals, allowing it to expand its octet. It can form \( IO_3^- \).

4. **Conclusion**: The halogens that can form the \( XO_3^- \) ion are Chlorine, Bromine, and Iodine. Therefore, the total number of halogens that can form \( XO_3^- \) is 3.

### Final Answer:
The number of halogens that form the \( XO_3^- \) ion is **3** (Chlorine, Bromine, and Iodine).