The correct order of decreasing ionic radii among the following isoelectronic species is

Options

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

Answer

To determine the correct order of decreasing ionic radii among the given isoelectronic species, we need to consider the relationship between ionic charge and ionic radius. Here’s a step-by-step solution:

### Step 1: Identify the Isoelectronic Species
First, we need to identify the isoelectronic species in the question. Isoelectronic species are atoms or ions that have the same number of electrons. For example, species like \( \text{Ca}^{2+}, \text{K}^{+}, \text{Cl}^{-}, \text{S}^{2-} \) all have the same electron configuration as argon (18 electrons).

### Step 2: Understand the Relationship Between Charge and Ionic Radius
The ionic radius is influenced by the charge of the ion:
- **Negative Charge**: As the negative charge on an ion increases (i.e., more electrons), the ionic radius increases. This is because the additional electrons repel each other, causing the ion to expand.
- **Positive Charge**: Conversely, as the positive charge on an ion increases (i.e., fewer electrons), the ionic radius decreases. This is due to the increased effective nuclear charge (Z effective), which pulls the remaining electrons closer to the nucleus.

### Step 3: Analyze the Given Ions
Now, let’s analyze the charges of the given ions:
- \( \text{Ca}^{2+} \): +2 charge (smallest radius due to high positive charge)
- \( \text{K}^{+} \): +1 charge (larger than \( \text{Ca}^{2+} \))
- \( \text{Cl}^{-} \): -1 charge (larger than \( \text{K}^{+} \))
- \( \text{S}^{2-} \): -2 charge (largest radius due to high negative charge)

### Step 4: Order the Ionic Radii
Based on the analysis:
1. \( \text{Ca}^{2+} \) has the smallest ionic radius.
2. \( \text{K}^{+} \) is next.
3. \( \text{Cl}^{-} \) follows.
4. \( \text{S}^{2-} \) has the largest ionic radius.

Thus, the correct order of decreasing ionic radii is:
\[ \text{S}^{2-} > \text{Cl}^{-} > \text{K}^{+} > \text{Ca}^{2+} \]

### Final Answer
The correct order of decreasing ionic radii among the given isoelectronic species is:
\[ \text{S}^{2-} > \text{Cl}^{-} > \text{K}^{+} > \text{Ca}^{2+} \]

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