The decreasing size of `K^(+),Ca^(2+),Cl^(-) & S^(2-)` follows the order:

Options

A. `K^(+)gtCa^(+2)gtS^(-2)gtCl^(-)`
B. `K^(+)gtCa^(+2)gtCl^(-)gtS^(-2)`
C. `Ca^(+2)gtK^(+)gtCl^(-)gtS^(-2)`
D. `S^(-2)gtCl^(-)gtK^(+)gtCa^(+2)`

Answer

To determine the decreasing size of the ions \( K^+, Ca^{2+}, Cl^-, \) and \( S^{2-} \), we need to analyze their properties as isoelectronic species and how their ionic charges affect their sizes.

### Step-by-Step Solution:

1. **Identify the number of electrons in each ion**:
- \( K^+ \): Potassium (K) has an atomic number of 19. When it loses one electron to form \( K^+ \), it has 18 electrons.
- \( Ca^{2+} \): Calcium (Ca) has an atomic number of 20. When it loses two electrons to form \( Ca^{2+} \), it also has 18 electrons.
- \( Cl^- \): Chlorine (Cl) has an atomic number of 17. When it gains one electron to form \( Cl^- \), it has 18 electrons.
- \( S^{2-} \): Sulfur (S) has an atomic number of 16. When it gains two electrons to form \( S^{2-} \), it has 18 electrons.

All these ions are isoelectronic, meaning they have the same number of electrons (18 electrons).

2. **Consider the effect of ionic charge on size**:
- In isoelectronic species, the size of the ions is influenced by their nuclear charge (the number of protons in the nucleus).
- A higher positive charge (more protons) means a stronger attraction between the nucleus and the electrons, resulting in a smaller ionic size.
- Conversely, a higher negative charge means a weaker attraction, leading to a larger ionic size.

3. **Analyze the charges of the ions**:
- \( K^+ \) has a +1 charge (19 protons).
- \( Ca^{2+} \) has a +2 charge (20 protons).
- \( Cl^- \) has a -1 charge (17 protons).
- \( S^{2-} \) has a -2 charge (16 protons).

4. **Rank the ions based on size**:
- \( Ca^{2+} \) (smallest) < \( K^+ \) < \( Cl^- \) < \( S^{2-} \) (largest)
- This is because \( Ca^{2+} \) has the highest positive charge, pulling the electrons closer and making it smaller. \( S^{2-} \) has the highest negative charge, leading to a larger size.

5. **Final order of decreasing size**:
- The order of decreasing size is \( S^{2-} > Cl^- > K^+ > Ca^{2+} \).

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