The correct order of ionic radii for the ions `S^(2-) , Cl^(-) , P^(3-) , Ca^(2+)` is

Options

A. `Ca^(2+) gt Cl^(-) gt S^(2-) gt P^(3-)`
B. `S^(2-) gt P^(3-) gt Cl^(-) gt Ca^(2+)`
C. `P^(3-) gt S^(2-) lt Cl^(-) gt Ca^(2+)`
D. `Ca^(2+) lt Cl^(-) lt S^(2-) lt P^(3-)`

Explanation

To determine the correct order of ionic radii for the ions \( S^{2-}, Cl^{-}, P^{3-}, \) and \( Ca^{2+} \), we need to consider the following factors:

1. **Charge of the Ions**: The ionic radius is influenced by the charge of the ions. Anions (negatively charged ions) are generally larger than their neutral atoms because they have gained electrons, which increases electron-electron repulsion. Cations (positively charged ions) are smaller than their neutral atoms because they have lost electrons, which reduces electron-electron repulsion and allows the nucleus to pull the remaining electrons closer.

2. **Comparison of Anions**: Among the anions \( S^{2-}, Cl^{-}, \) and \( P^{3-} \):
- \( S^{2-} \) has gained 2 electrons.
- \( Cl^{-} \) has gained 1 electron.
- \( P^{3-} \) has gained 3 electrons.
The more electrons gained, the larger the ionic radius due to increased electron-electron repulsion. Therefore, the order of size for these anions will be \( P^{3-} > S^{2-} > Cl^{-} \).

3. **Cation Size**: The cation \( Ca^{2+} \) has lost 2 electrons, making it smaller than its neutral atom. Since cations are smaller than anions, \( Ca^{2+} \) will be smaller than all the anions.

4. **Final Order**: Combining all the information:
- The largest ion is \( P^{3-} \).
- Next is \( S^{2-} \).
- Then \( Cl^{-} \).
- The smallest is \( Ca^{2+} \).

Thus, the correct order of ionic radii from largest to smallest is:
\[ P^{3-} > S^{2-} > Cl^{-} > Ca^{2+} \]

### Final Answer:
The correct order of ionic radii is:
\[ P^{3-} > S^{2-} > Cl^{-} > Ca^{2+} \]