Consider the ioselectronic series , `K^(o+), S^(2-), Cl^(ɵ), Ca^(2+)`, the radii of the ions decrease as

Options

A. `Ca^(2+) gt K^(o+) gt Cl^(ɵ) gt S^(2-)`
B. `Cl^(ɵ) gt S^(2-) gt K^(o+) gt Ca^(2+)`
C. `S^(2-) gt Cl^(ɵ) gt K^(o+) gt Ca^(2+)`
D. `K^(o+) gt Ca^(2+) gt S^(2-) gt Cl^(ɵ)`

Explanation

To solve the problem regarding the ionic radii of the isoelectronic series \( K^{+}, S^{2-}, Cl^{-}, Ca^{2+} \), we need to analyze the charges of these ions and how they affect their sizes. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Ions and Their Charges
We have the following ions in the isoelectronic series:
- \( K^{+} \) (Potassium ion with a +1 charge)
- \( S^{2-} \) (Sulfide ion with a -2 charge)
- \( Cl^{-} \) (Chloride ion with a -1 charge)
- \( Ca^{2+} \) (Calcium ion with a +2 charge)

### Step 2: Understand the Concept of Isoelectronic Species
Isoelectronic species are ions that have the same number of electrons. In this case, all the ions have the same electron configuration, which is similar to that of Argon (18 electrons).

### Step 3: Analyze the Effect of Charge on Ionic Radius
- **Cations (positively charged ions)**: The more positive the charge, the smaller the ionic radius. This is because the increased positive charge results in a greater effective nuclear charge (Z effective), pulling the electrons closer to the nucleus.
- For example, \( Ca^{2+} \) has a +2 charge, so it will be smaller than \( K^{+} \) which has a +1 charge.

- **Anions (negatively charged ions)**: The more negative the charge, the larger the ionic radius. This is because the additional electrons increase electron-electron repulsion and decrease the effective nuclear charge per electron.
- For example, \( S^{2-} \) has a -2 charge, making it larger than \( Cl^{-} \) which has a -1 charge.

### Step 4: Order the Ions by Ionic Radius
Based on the analysis:
1. The largest ion will be \( S^{2-} \) (most negative charge).
2. Next will be \( Cl^{-} \) (less negative charge).
3. Then comes \( K^{+} \) (least positive charge).
4. Finally, the smallest ion will be \( Ca^{2+} \) (most positive charge).

### Final Order of Ionic Radii
Thus, the order of ionic radii from largest to smallest is:
\[ S^{2-} > Cl^{-} > K^{+} > Ca^{2+} \]

### Conclusion
The correct answer is that the radii of the ions decrease as:
\[ S^{2-} > Cl^{-} > K^{+} > Ca^{2+} \]