Consider the following species: `P^(3-), S^(2-), CI^(-), K^(+), Ca^(2+)` and `Sc^(3+)`
(i) What is common in them?
(ii) Arrange them in the order of decreasing ionis radii.

Answer

### Step-by-Step Solution:

**(i) What is common in them?**

1. **Identify the Species**: The given species are \( P^{3-}, S^{2-}, Cl^{-}, K^{+}, Ca^{2+}, Sc^{3+} \).
2. **Determine the Atomic Numbers**:
- Phosphorus (P): Atomic number 15
- Sulfur (S): Atomic number 16
- Chlorine (Cl): Atomic number 17
- Potassium (K): Atomic number 19
- Calcium (Ca): Atomic number 20
- Scandium (Sc): Atomic number 21
3. **Calculate the Number of Electrons**:
- For \( P^{3-} \): 15 + 3 = 18 electrons
- For \( S^{2-} \): 16 + 2 = 18 electrons
- For \( Cl^{-} \): 17 + 1 = 18 electrons
- For \( K^{+} \): 19 - 1 = 18 electrons
- For \( Ca^{2+} \): 20 - 2 = 18 electrons
- For \( Sc^{3+} \): 21 - 3 = 18 electrons
4. **Conclusion**: All these species have the same number of electrons (18 electrons). Therefore, they are **isoelectronic species**.

**(ii) Arrange them in the order of decreasing ionic radii.**

1. **Understanding Ionic Radii**: For isoelectronic species, the ionic radius is inversely proportional to the number of protons (atomic number). More protons result in a stronger nuclear charge, pulling the electrons closer and resulting in a smaller ionic radius.
2. **List the Number of Protons**:
- \( P^{3-} \): 15 protons
- \( S^{2-} \): 16 protons
- \( Cl^{-} \): 17 protons
- \( K^{+} \): 19 protons
- \( Ca^{2+} \): 20 protons
- \( Sc^{3+} \): 21 protons
3. **Arrange by Number of Protons**: Since we want to arrange them in decreasing order of ionic radii, we start with the species with the least number of protons:
- \( P^{3-} \) (15 protons) > \( S^{2-} \) (16 protons) > \( Cl^{-} \) (17 protons) > \( K^{+} \) (19 protons) > \( Ca^{2+} \) (20 protons) > \( Sc^{3+} \) (21 protons)
4. **Final Order**: The order of decreasing ionic radii is:
\[ P^{3-} > S^{2-} > Cl^{-} > K^{+} > Ca^{2+} > Sc^{3+} \]