Among the ions `Cl^(ɵ), S^(2-)` and `Na^(o+)`, the largest ion is `"______"` .

Explanation

To determine which ion is the largest among `Cl^(-)`, `S^(2-)`, and `Na^(+)`, we can follow these steps:

### Step 1: Understand the Nature of the Ions
- We have three ions: `Cl^(-)`, `S^(2-)`, and `Na^(+)`.
- `Cl^(-)` is a chloride ion with a negative charge, indicating it has gained an electron.
- `S^(2-)` is a sulfide ion with a double negative charge, indicating it has gained two electrons.
- `Na^(+)` is a sodium ion with a positive charge, indicating it has lost an electron.

### Step 2: Compare the Charges
- The charge on an ion affects its size. Generally, cations (positively charged ions) are smaller than their parent atoms because they have lost electrons, leading to a decrease in electron-electron repulsion and a stronger attraction between the nucleus and the remaining electrons.
- Conversely, anions (negatively charged ions) are larger than their parent atoms because they have gained electrons, which increases electron-electron repulsion.

### Step 3: Analyze the Size Trends
- Among the three ions:
- `Na^(+)` has lost one electron, making it the smallest.
- `Cl^(-)` has gained one electron, making it larger than `Na^(+)`.
- `S^(2-)` has gained two electrons, making it larger than both `Na^(+)` and `Cl^(-)`.

### Step 4: Establish the Size Order
- Based on the analysis:
- The size order from smallest to largest is: `Na^(+) < Cl^(-) < S^(2-)`.

### Conclusion
- Therefore, the largest ion among `Cl^(-)`, `S^(2-)`, and `Na^(+)` is `S^(2-)`.

### Final Answer
The largest ion is `S^(2-)`.

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