In the ions `P^(3-),S^(2-)` and `Cl`, the increasing order of size is:

Options

A. `Cl^(-), S^(2-), P^(3-)`
B. `P^(3-), S^(2-), Cl^(-)`
C. `S^(2-), Cl^(-), P^(3-)`
D. `S^(2-), Cl^(3-), Cl^(-)`

Explanation

To determine the increasing order of size for the ions \( P^{3-}, S^{2-}, \) and \( Cl^- \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Ions**: We have three anions: \( P^{3-} \), \( S^{2-} \), and \( Cl^- \). All these ions have gained electrons, which increases their size compared to their neutral atoms.

2. **Understand Anion Size**: The size of anions increases with the addition of electrons. When an atom gains electrons to form an anion, the increased electron-electron repulsion causes the electron cloud to expand, resulting in a larger ionic radius.

3. **Compare the Number of Electrons**:
- \( P^{3-} \) has gained 3 electrons.
- \( S^{2-} \) has gained 2 electrons.
- \( Cl^- \) has gained 1 electron.

4. **Analyze the Effect of Electron Gain**:
- The more electrons an atom gains, the larger the anion becomes due to increased repulsion among the electrons.
- Therefore, \( P^{3-} \) will be the largest because it has the most electrons, followed by \( S^{2-} \), and then \( Cl^- \) which has the least number of added electrons.

5. **Establish the Order**: Based on the above analysis, we can establish the increasing order of size as:
\[ Cl^- < S^{2-} < P^{3-} \]

### Final Answer:
The increasing order of size is:
\[ Cl^- < S^{2-} < P^{3-} \]