In which of the following pairs the radii of second species is greater than of first ?

Options

A. K, Ca
B. H, He
C. `Mg^(+), Mg^(2+) `
D. `O^(2-), O^(-)`

Answer

To determine which of the following pairs has the second species with a greater radius than the first, we will analyze each pair based on their electronic configurations and the principles of atomic size.

### Step-by-Step Solution:

1. **Identify the Pairs**: The question provides several pairs of elements. We need to analyze each pair to see if the radius of the second species is greater than that of the first.

2. **Analyze the First Pair: Potassium (K) and Calcium (Ca)**:
- **Potassium (K)**: Atomic number = 19, Electronic configuration = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹
- **Calcium (Ca)**: Atomic number = 20, Electronic configuration = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
- **Comparison**: Both have the same number of electron shells (4). However, Calcium has more protons (20 vs. 19), which increases the effective nuclear charge, pulling the electrons closer and decreasing the size. Therefore, K > Ca. This pair does not satisfy the condition.

3. **Analyze the Second Pair: Hydrogen (H) and Helium (He)**:
- **Hydrogen (H)**: Atomic number = 1, Electronic configuration = 1s¹
- **Helium (He)**: Atomic number = 2, Electronic configuration = 1s²
- **Comparison**: Helium is a noble gas and has a larger atomic radius compared to Hydrogen due to its filled outer shell and inert nature. Therefore, He > H. This pair satisfies the condition.

4. **Analyze the Third Pair: Magnesium Ion (Mg⁺) and Magnesium Ion (Mg²⁺)**:
- **Magnesium (Mg)**: Atomic number = 12, Electronic configuration = 1s² 2s² 2p⁶ 3s²
- **Mg⁺**: Loses one electron → 1s² 2s² 2p⁶ 3s¹
- **Mg²⁺**: Loses two electrons → 1s² 2s² 2p⁶
- **Comparison**: Cations are smaller than their parent atoms. Mg²⁺ is smaller than Mg⁺ due to the increased effective nuclear charge on the remaining electrons. Therefore, Mg⁺ > Mg²⁺. This pair does not satisfy the condition.

5. **Analyze the Fourth Pair: Oxide Ion (O²⁻) and Oxide Ion (O⁻)**:
- **Oxygen (O)**: Atomic number = 8, Electronic configuration = 1s² 2s² 2p⁴
- **O²⁻**: Gains two electrons → 1s² 2s² 2p⁶
- **O⁻**: Gains one electron → 1s² 2s² 2p⁵
- **Comparison**: Anions are larger than their parent atoms. O²⁻ has more electrons than O⁻, leading to increased electron-electron repulsion and a larger size. Therefore, O²⁻ > O⁻. This pair satisfies the condition.

### Conclusion:
The pairs where the radius of the second species is greater than that of the first are:
- **Hydrogen (H) and Helium (He)**
- **Oxide Ion (O²⁻) and Oxide Ion (O⁻)**

### Final Answer:
The correct pairs are:
- **Option 2: Hydrogen and Helium**
- **Option 4: O²⁻ and O⁻**