The correct decreasing order of ionic size among the following species is `K^+, Cl^-, S^-2` and `Ca^(+2)`.

Options

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

Explanation

To determine the correct decreasing order of ionic size among the species \( K^+, Cl^-, S^{2-}, \) and \( Ca^{2+} \), we will follow these steps:

### Step 1: Identify the number of electrons in each ion
1. **Potassium ion (\( K^+ \))**:
- Atomic number of K = 19
- Electrons in \( K^+ \) = 19 - 1 = 18

2. **Chloride ion (\( Cl^- \))**:
- Atomic number of Cl = 17
- Electrons in \( Cl^- \) = 17 + 1 = 18

3. **Sulfide ion (\( S^{2-} \))**:
- Atomic number of S = 16
- Electrons in \( S^{2-} \) = 16 + 2 = 18

4. **Calcium ion (\( Ca^{2+} \))**:
- Atomic number of Ca = 20
- Electrons in \( Ca^{2+} \) = 20 - 2 = 18

### Step 2: Determine that all ions are isoelectronic
All four ions (\( K^+, Cl^-, S^{2-}, Ca^{2+} \)) have the same number of electrons (18). Therefore, they are isoelectronic species.

### Step 3: Apply the concept of ionic size in isoelectronic species
In isoelectronic species, the size of the ion is affected by the charge:
- More positive charge results in a smaller ionic size because the effective nuclear charge pulls the electrons closer to the nucleus.
- More negative charge results in a larger ionic size because the additional electrons repel each other and spread out more.

### Step 4: Compare the charges of the ions
- \( Ca^{2+} \) has the highest positive charge (+2), so it will be the smallest.
- \( K^+ \) has a +1 charge, so it is larger than \( Ca^{2+} \).
- \( Cl^- \) has a -1 charge, making it larger than both \( Ca^{2+} \) and \( K^+ \).
- \( S^{2-} \) has the highest negative charge (-2), so it will be the largest.

### Step 5: Write the correct decreasing order of ionic size
Based on the above analysis, the correct decreasing order of ionic size is:
\[ S^{2-} > Cl^- > K^+ > Ca^{2+} \]

### Final Answer
The correct decreasing order of ionic size among the species \( K^+, Cl^-, S^{2-}, \) and \( Ca^{2+} \) is:
\[ S^{2-} > Cl^- > K^+ > Ca^{2+} \]